Lectures on Diffusion Problems and Partial Differential Equations

Lectures on
Diffusion Problems and Partial Differential
Equations
By
S. R. S. Varadhan
Notes by
P. Muthuramalingam
Tara R. Nanda
Tata Institute of Fundamental Research, Bombay
1989
Author
S. R. S. Varadhan
Courant Institute of Mathematical Sciences
251, Mercer Street
New York. N. Y. 10012.
U.S.A.
c Tata Institute of Fundamental Research, 1989
ISBN 3-540-08773-7. Springer-Verlag, Berlin, Heidelberg. New York
ISBN 0-387-08773-7. Springer-Verlag, New York. Heidelberg. Berlin
No part of this book may be reproduced in any
form by print, microfilm or any other means without written permission from the Tata Institute of
Fundamental Research, Bombay 400 005
Printed by N. S. Ray at the Book Centre Limited
Sion East, Bombay 400 022 and published by H. Goetze
Springer-Vertal, Heidelberg, West Germany
PRINTED IN INDIA
Contents
1 The Heat Equation
1
2 Kolmogorov’s Theorem
11
3 The One Dimensional Random Walk
15
4 Construction of Wiener Measure
19
5 Generalised Brownian Motion
31
6 Markov Properties of Brownian Motion
33
7 Reflection Principle
39
8 Blumenthal’s Zero-One Law
53
9 Properties of Brownian Motion in One Dimension
57
10 Dirichlet Problem and Brownian Motion
63
11 Stochastic Integration
71
12 Change of Variable Formula
91
13 Extension to Vector-Valued Itô Processes
95
14 Brownian Motion as a Gaussian Process
101
iii
iv
Contents
15 Equivalent For of Itô Process
105
16 Itô’s Formula
117
17 Solution of Poisson’s Equations
129
18 The Feynman-Kac Formula
133
19 An Application of the Feynman-Kac Formula....
139
20 Brownian Motion with Drift
147
21 Integral Equations
155
22 Large Deviations
161
23 Stochastic Integral for a Wider Class of Functions
187
24 Explosions
195
25 Construction of a Diffusion Process
201
26 Uniqueness of Diffusion Process
211
27 On Lipschitz Square Roots
223
28 Random Time Changes
229
29 Cameron - Martin - Girsanov Formula
237
30 Behaviour of Diffusions for Large Times
245
31 Invariant Probability Distributions
253
32 Ergodic Theorem
275
33 Application of Stochastic Integral
281
Appendix
284
Contents
v
Language of Probability
284
Kolmogorovs Theorem
288
Martingales
290
Uniform Integrability
305
Up Crossings and Down Crossings
307
Bibliography
317
Contents
vi
Preface
THESE ARE NOTES based on the lectures given at the T.I.F.R.
Centre, Indian Institute of Science, Bangalore, during July and August
of 1977. Starting from Brownian Motion, the lectures quickly got into
the areas of Stochastic Differential Equations and Diffusion Theory. An
attempt was made to introduce to the students diverse aspects of the
theory. The last section on Martingales is based on some additional
lectures given by K. Ramamurthy of the Indian Institute of Science. The
author would like to express his appreciation of the efforts by Tara R.
Nanda and PL. Muthuramalingam whose dedication and perseverance
has made these notes possible.
S.R.S. Varadhan
1. The Heat Equation
LET US CONSIDER the equation
(1)
1
1
ut − ∆u = 0
2
which describes (in a suitable system of units) the temperature distribution of a certain homogeneous, isotropic body in the absence of any heat
sources within the body. Here
u ≡ u(x1 , . . . , xd , t);
ut ≡
∂u
;
∂t
∆u =
d
X
∂2 u
i=1
∂x2i
,
t represents the time ranging over [0, ∞) or [0, T ] and x ≡ (x1 . . . xd )
belongs to Rd .
We first consider the initial value problem. It consists in integrating
equation (1) subject to the initial condition
(2)
u(0, x) = f (x).
The relation (2) is to be understood in the sense that
Lt u(t, x) = f (x).
t→0
Physically (2) means that the distribution of temperature throughout
the body is known at the initial moment of time.
We assume that the solution u has continuous derivatives, in the
space coordinates upto second order inclusive and first order derivative
in time.
1
1. The Heat Equation
2
It is easily verified that
(3)
2
u(t, x) =
!
|x|2
1
exp
−
;
2t
(2πt)d/2
|x|2 =
d
X
x2i ,
i=1
satisfies (1) and
u(0, x) = Lt u(t, x) = δ(x)
(4)
t→0
Equation (4) gives us a very nice physical interpretation. The solution (3) can be interpreted as the temperature distribution within the
body due to a unit sourse of head specified at t = 0 at the space point
x = 0. The linearity of the equation (1) now tells us that (by superposition) the solution of the initial value problem may be expected in the
form
Z
(5)
u(t, x) =
f (y)p(t, x − y)dy,
Rd
where
p(t, x) =
|x|2
1
.
exp
−
2t
(2πt)d/2
Exercise 1. Let f (x) be any bounded continuous function. Verify that
p(t, x) satisfies (1) and show that
R
(a) p(t, x)dx = 1, ∀t > 0;
R
(b) Lt p(t, x) f (x)dx = f (0);
t→0+
(c) using (b) justify (4). Also show that (5) solves the initial value
problem.
(Hints: For (a) use
R∞
−∞
2
e−x dx =
√
π. For part (b) make the substitution
x
y = √ and apply Lebesgue dominated convergence theorem).
π
3
3
Since equation (1) is linear with constant coefficients it is invariant
under time as well as space translations. This means that translates of
solutions are also solutions. Further, for s ≥ 0, t > 0 and y ∈ Rd ,
(6)
u(t, x) =
|x − y|2
1
exp
−
2(t + s)
[2π(t + s)]d/2
and for t > s, y ∈ Rd ,
(7)
u(t, x) =
1
|x − y|2
exp
−
2(t − s)
[2π(t − s)]d/2
are also solutions of the heat equation (1).
The above method of solving the initial value problem is a sort of
trial method, viz. we pick out a solution and verify that it satisfies (1).
But one may ask, how does one obtain the solution? A partial clue to this
is provided by the method of Fourier transforms. We pretend as if our
solution u(t, x) is going to be very well behaved and allow all operations
performed on u to be legitimate.
Put v(t, x) = û(t, x) where bstands for the Fourier transform in the
space variables only (in this case), i.e.
Z
v(t, x) =
u(t, y)ei x−y dy.
Rd
Using equation (1), one easily verifies that
1
vt (t, x) = |x|2 v(t, x)
2
(8)
with
v(0, x) = fˆ(x).
(9)
The solution of equation (8) is given by
(10)
2
v(t, x) = fˆ(x)e−t|x| /2 .
We have used (9) in obtaining (10).
1. The Heat Equation
4
Exercise 2. Verify that
4
!
t|x|2
p̂(t, x) = exp −
.
2
Using Exercise 2, (10) can be written as
v(t, x) = û(t, x) = fˆ(x) p̂(t, x).
(11)
The right hand side above is the product of two Fourier transforms
and we know that the Fourier transform of the convolution of two funtions is given by the product of the Fourier transforms. Hence u(t, x) is
expected to be of the form (5).
Observe that if f is non-negative, then u is nonnegative and if f
is bounded by M then u is also bounded by M in view of part (a) of
Exercise 1.
The Inhomogeneous Equation. Consider the equation
vt −
∆v
= g,
2
with
v(0, x) = 0,
which describes the temperature within a homogeneous isotropic body
in the presence of heat sources, specified as a function of time and space
by g(t, x). For t > s,
u(t, x) =
1
|x − y|2
exp
−
2(t − s)
[2π(t − s)]d/2
1
is a solution of ut (t, x) − ∆u(t, x) = 0 corresponding to a unit source at
2
t = s, x = y. Consequently, a solution of the inhomogeneous problem is
obtained by superposition.
Let
v(t, x) =
Z Zt
Rd 0
5
i.e.
!
|x − y|2
1
dy ds
exp −
g(s, y)
2(t − s)
[2π(t − s)]d/2
5
v(t, x) =
Zt
w(t, x, s)ds
0
where
w(t, x, s) =
Z
Rd
!
|x − y|2
1
dy.
exp −
g(s, y)
2(t − s)
[2π(t, s)]d/2
Exercise 3. Show that v(t, x) defined above solves the inhomogeneous
heat equation and satisfies v(0, x) = 0. Assume that g is sufficiently
1
smooth and has compact support. vt − ∆v = Lt w(t, x, s) and now use
s→t
2
part (b) of Exercise (1).
Remark 1. We can assume g has compact support because in evaluating
1
vt − ∆v the contribution to the integral is mainly from a small neigh2
bourhood of the point (t, x). Outside this neighbourhood
!
1
|x − y|2
exp −
2(t − s)
[2π(t − s)]d/2
satisfies
1
ut − ∆u = 0.
2
2. If we put g(s, y) = 0 for s < 0, we recognize that v(t, x) = g ∗ p.
Taking spatial Fourier transforms this can be written as
v(t, ξ) =
Zt
1
g(s, ξ) exp − (t − s)|ξ|2 dξ,
2
0
or
!
∂v̂ ∂v
1
1
=
= g(t, ξ) + ∆v = g(t, ξ) + ∆v .
∂t
∂t
2
2
Therefore
∂v 1
− ∆v = g.
∂t 2
1. The Heat Equation
6
1
Exercise 4. Solve wt − ∆w = g on [0, ∞) × Rd with w = f on {0} × Rd 6
2
(Cauchy problem for the heat equation).
Uniqueness. The solution of the Cauchy problem is unique provided the
class of solutions is suitably restricted. The uniqueness of the solution
is a consequence of the Maximum Principle.
Maximum Principle. Let u be smooth and bounded on [0, T ] × Rd satisfying
ut −
∆u
≥0
2
in (0, T ] × Rd
u(0, x) ≥ 0, ∀x ∈ Rd .
and
Then
u(t, x) ≥ 0 ∀,
t ∈ [0, T ]
and
∀x ∈ Rd .
Proof. The idea is to find minima for u or for an auxillary function.
Step 1. Let v be any function satisfying
vt −
∆v
>0
2
in (0, T ] × Rd .
Claim . v cannot attain a minimum for t0 ∈ (0, T ]. Assume (to get a
contradiction) that v(t0 , x0 ) ≤ v(t, x) for some t0 > 0 and for all t ∈
[0, T ], ∀x ∈ Rd . At a minimum vt (t0 , x0 ) ≤ 0, (since t0 , 0) ∆v(t0 , x0 ) ≥
0. Therefore
!
∆v
vt −
(t0 , x0 ) ≤ 0.
2
Thus, if v has any minimum it should occur at t0 = 0.
Step 2. Let ǫ > 0 be arbitrary. Choose α such that
h(t, x) = |x|2 + αt
7
satisfies
ht −
∆h
= α − d > 0 (say α = 2d).
2
7
Put vǫ = u + ǫh. Then
∂vǫ 1
− ∆vǫ > 0.
∂t
2
As u is bounded, vǫ → +∞ as |x| → +∞, vǫ must attain a minimum.
This minimum occurs at t = 0 by Step 1. Therefore,
vǫ (t, x) ≥ vǫ (0, x0 ) for some
x0 ∈ Rd ,
i.e.
vǫ (t, x) ≥ u(0, x0 ) + ǫ|x0 |2 > 0,
i.e.
u(t, x) + ǫh(t, x) > 0, ∀ǫ.
This gives
u(t, x) ≥ 0.
This completes the proof.
Exercise 5. (a) Let L be a linear differential operator satisfying Lu =
g on Ω (open in Rd ) and u = f on ∂Ω. Show that u is uniquely
determined by f and g if and only if Lu = 0 on Ω and u = 0 on
∂Ω imply u = 0 on Ω.
1
(b) Let u be a bounded solution of the heat equation ut − ∆u = g
2
with u(0, x) = f (x). Use the maximum principle and part (a) to
show that u is unique in the class of all bounded functions.
(c) Let

2


e−1/t , if t > 0,
g(t) = 

0,
if t ≤ 0,
∞
X g(k) (t/2)x2k
,
u(t, x) =
(2k)!
k=0
on
R × R.
8
1. The Heat Equation
8
Then
u(0, x) = 0,
ut =
∆u
,
2
u . 0,
i.e. u satisfies
ut −
1 ∂2 u
= 0,
2 ∂x2
with u(0, x) = 0.
This example shows that the solution is not unique because, u is
not bounded. (This example is due to Tychonoff).
Lemma 1. Let p(t, x) =
1
|x|2
exp
−
for t > 0. Then
2t
(2πt)d/2
p(t, ·) ∗ p(s, ·) = p(t + s, ·).
Proof. Let f be any bounded continuous function and put
Z
u(t, x) =
f (y)p(t, x − y)dy.
Rd
Then u satisfies
1
ut − ∆u = 0,
2
u(0, x) = f (x).
Let
v(t, x) = u(t + s, x).
Then
1
vt − ∆v = 0, v(0, x) = u(s, x).
2
This has the unique solution
Z
v(t, x) =
u(s, y)p(t, x − y)dy.
Thus
Z
"
f (y)p(t + s, x − y)dy =
f (z)p(s, y − z)p(t, x − y)dz dy.
Rd
9
9
This is true for all f bounded and continuous. We conclude, therefore, that
p(t, ·) ∗ p(s, ·) = p(t + s, ·).
Exercise 6. Prove Lemma 1 directly using Fourier transforms.
It will be convenient to make a small change in notation which will
be useful later on. We shall write p(s, x, t, y) = p(t−s, y−x) for every x, y
and t > s. p(s, x, t, y) is called the transition probability, in dealing with
Brownian motion. It represents the probability density that a “Brownian
particle” located at space point x at time s moves to the space point y at
a later time t.
Note . We use the same symbol p for the transition probability; it is
function of four variables and there will not be any ambiguity in using
the same symbol p.
Exercise 7. Verify that
Z
p(s, x, t, y)p(t, y, σ, z)dy = p(s, x, σ, z), s < t < σ.
Rd
(Use Exercise 6).
Remark . The significance of this result is obvious. The probability
that the particle goes from x at time s to z at time σ is the sum total
of the probabilities, that the particle moves from x at s to y at some
intermediate time t and then to z at time σ.
10
10
1. The Heat Equation
In this section we have introduced Brownian Motion corresponding
1
to the operator ∆. Later on we shall introduce a more general diffusion
2
P
1P
∂2
∂
process which corresponds to the operator
ai j
+ bj
.
2
∂xi ∂x j
∂x j
2. Kolmogorov’s Theorem
Definition. LET (Ω, B, P) BE A probability space. A stochastic process 11
in Rd is a collection {Xt : t ∈ I} of Rd -valued random variables defined
on (Ω, B).
Note 1. I will always denote a subset of R+ = [0, ∞).
2. Xt is also denoted by X(t).
Let {Xt : t ∈ I} be a stochastic process. For any collection t1 ,
t2 , . . . , tk such that ti ∈ I and 0 ≤ t1 < t2 < . . . < tk and any Borel
set Λ, in Rd × Rd × · · · × Rd (k times),. define
Ft1 . . . tk (Λ) = P(w ∈ Ω : (Xt1 (w), . . . , Xtk (w)) ∈ Λ).
If
{t1 , . . . , tk } ⊂ {s1 , . . . , sℓ } ⊂ I,
with
l≥k
such that
(k)
(k)
(1)
(1)
(0)
s(0)
1 < . . . < sn0 < t1 < s1 . . . < sn < t2 . . . < tk < s1 . . . < snk ,
let then
π : Rd × · · · × Rd (1 times) → Rd × · · · × Rd (k times)
be the canonical projection. If Eti ⊂ Rd is any Borel set in Rd , i =
1, 2, . . . , k, then
π−1 (Et1 × · · · × Etk ) = Rd × · · · × Et1 × Rd × · · · × Et2 × · · · × Rd
11
2. Kolmogorov’s Theorem
12
(l times). The following condition always holds.
(*)
Et1 . . . tk (Et1 × · · · × Etk ) = F s1 . . . s1 (Π−1 (Et1 × · · · × Etk )).
If (∗) holds for an arbitrary collection {Ft1 . . . tk : 0 ≤ t1 < t2 . . . < tk } 12
(k = 1, 2, 3 . . .) of distributions then it is said to satisfy the consistency
condition.
Exercise 1. (a) Verify that Ft1 . . . tk is a probability measure on Rd ×
· · · × Rd (k times).
(b) Verify (∗). (If Bm denotes the Borel σ field of Rm , Bm+n = Bm ×
Bn ).
The following theorem is a converse of Exercise 1 and is often used
to identify a stochastic process with a family of distributions satisfying
the consistency condition.
Kolmogorov’s Theorem.
Let {Ft1 ,t2 ,...tk 0 ≤ t1 < t2 < . . . < tk < ∞} be a family of probability
distributions (on Rd × · · · × Rd , k times, k = 1, 2, . . .) satisfying the
consistency condition. Then there exists a measurable space (Ωk , B),
a unique probability measure P an (Ωk , B) and a stochastic process
{Xt : 0 ≤ t < ∞} such that the family of probability measures associated
with it is precisely
{Ft1 ,t2 ,...tk : 0 ≤ t1 < t2 < . . . < tk < ∞},
k = 1, 2, . . . .
A proof can be found in the APPENDIX. We mention a few points
about the proof which prove to be very useful and should be observed
carefully.
13
1. The space Ωk is the set of all Rd -valued functions defined on
[0, ∞):
Y
ΩK =
Rd
t∈[0,∞)
2. The random variable Xt is the tth -projection of ΩK onto Rdt .
13
3. B is the smallest σ-algebra with respect to which all the projections are measurable.
4. P given by
P(w : Xt1 (w) ∈ A1 , . . . Xtk (w) ∈ Ak ) = Ft1 ...tk (A1 × · · · × Ak )
where Ai is a Borel set in Rd , is a measure on the algebra generated
by {Xt1 , . . . Xtk }(k = 1, 2, 3 . . .) and extends uniquely to B.
Remark . Although the proof of Kolmogorov’s theorem is very constructive the space ΩK is too “large” and the σ-algebra B too “small”
for practical purposes. In applications one needs a “nice” collection
of Rd -valued functions (for example continuous, or differentiable functions), a “large” σ-algebra on this collection and a probability measure
concentrated on this family.
3. The One Dimensional
Random Walk
BEFORE WE TAKE up Brownian motion, we describe a one dimen- 14
sional random walk which in a certain limiting case possesses the properties of Brownian motion.
Imagine a person at the position x = 0 at time t = 0. Assume that at
equal intervals of time t = τ he takes a step h either along the positive
x axis or the negative x axis and reaches the point x(t) = x(t − τ) + h or
x(t) = x(t − τ) − h respectively. The probability that he takes a step in
either direction is assumed to be 1/2. Denote by f (x, t) the probability
that after the time t = nτ (n intervals of time τ) he reaches the position x.
If he takes m steps to the right (positive x-axis) in reaching x then there
are nCm possible ways in which he can achieve these m steps. Therefore,
1
the probability f (x, t) is nCm ( )n .
2
f (x, t) satisfies the difference equation
(1)
f (x, t + τ) =
1
1
f (x − h, t) + f (x + h, t)
2
2
and
(2)
x = h(m − (n − m)) = (2m − n)h.
To see this one need only observe that to reach (x, t + τ) there are
two ways possible, viz. (x − h, t) → (x, t + τ) or (x + h, t) → (x, t + τ) and
the probability for each one of these is 1/2. Also note that by definition
15
3. The One Dimensional Random Walk
16
of f ,
f (h, τ) =
(3)
15
1
= f (−h, τ),
2
so that
f (x, t + τ) = f (h, τ) f (x − h, t) + f (−h, τ) f (x + h, t).
(4)
The reader can identify (4) as a “discrete version” of convolution.
By our assumption,
(5)
f (0, 0) = 1,
f (x, 0) = 0 if
x , 0.
We examine equation (1) in the limit h → 0, τ → 0. To obtain
reasonable results we cannot let h and τ tend to zero arbitratily. Instead
we assume that
h
→ 1 as
τ
(6)
h→0
and
τ → 0.
The physical nature of the problem suggests that (6) should hold. To
see this we argue as follows. Since the person is equally likely to go in
either direction the average value of x will be 0. Therefore a reasonable
measure of the “progress” made by the person is either |x| or x2 . Indeed,
since x is a random variable (since m is one) one gets, using (2),
E(x) = 2E(m) − n = 0, E(x2 ) = h2 E((2m − n)2 ) = h2 n.
!n
!n
n
n
P
1
n P
n(n + 1)
1
n
n
)
Cm
= ,
=
(Use
m Cm
2
2 m=0
2
4
m=0
Thus
E
16
(
)
x2
h2 n h2
1
= ,
= E(x2 ) =
t
t
nτ
τ
and as t becomes large we expect that the average distance covered per
unit time remains constant. (This constant is chosen to be 1 for reasons
that will become apparent later). This justifies (6). In fact, a simple
h2
argument shows that if
→ 0 or +∞, x may approach +∞ in a finite
τ
time which is physically untenable.
17
(1) now gives
f (x, t + τ) − f (x, t) =
1
{ f (x − h, t) − f (x, t) + f (x, h, t) − f (x, t)}.
2
Assuming sufficient smoothness on f , we get in the limit as h, τ → 0
and in view of (6),
(7)
∂f
1 ∂2 f
=
∂t
2 ∂x2
h2
→ 1). This is the equation satisfied
(to get the factor 1/2 we choose
τ
by the probability density f . The particle in this limit performs what is
known as Brownian motion to which we now turn our attention.
References.
[1] GNEDENKO: The theory of probability, Ch. 10.
[2] The Feynman Lectures on physics, Vol. 1, Ch. 6.
4. Construction of Wiener
Measure
ONE EXAMPLE WHERE the Kolmogorov construction yields a proba- 17
bility measure concentrated on a “nice” class Ω is the Brownian motion.
Definition . A Brownian motion with starting point x is an Rd -valued
stochastic process {X(t) : 0 ≤ t < ∞} where
(i) X(0) = x = constant;
(ii) the family of distribution is specified by
Z
Ft1 . . . tk (A) =
p(0, x, t1 , x1 )p(t1 , x1 , t2 , x2 ) . . .
A
p(tk−1 , xk−1 , tk , xk )dx1 . . . dxk
for every Borel set A in Rd × · · · Rd (k times).
N.B. The stochastic process appearing in the definition above is the one
given by the Kolmogorov construction.
It may be useful to have the following picture of a Brownian motion.
The space Ωk may be thought of as representing particles performing
Brownian movement; {Xt : 0 ≤ t < ∞} then represents the trajectories
of these particles in the space Rd as functions of time and B can be considered as a representation of the observations made on these particles.
Exercise 2. (a) Show that Ft1 ...tk defined above is a probability measure on Rd × · · · × Rd (k times).
19
4. Construction of Wiener Measure
20
(b) {Ft1 ...tk : 0 ≤ t1 < t2 < . . . tk < ∞} satisfies the consistency
condition. (Use Fubini’s theorem).
18
(c) Xt1 − x, Xt2 − Xt1 , . . . , Xtk − Xtk−1 are independent random variables
and if t > s, then Xt − X s is a random variable whose distribution
density is given by
!
1
1
−1 2
p(t − s, y) =
exp − (t − s) |y| .
2
[2π(t − s)]d/2
(Hint: Try to show that Xt1 − x, Xt2 − Xt1 , . . . , Xtk − Xtk−1 have a
joint distribution given by a product measure. For this let φ be
any bounded real measurable function on Rd × · · · × Rd (k times).
Then
E(φ(Z1 , . . . , Zk ))
Xt1 −x,Xt2 −Xt1 ,...,Xtk −Xtk−1
=
E
Xt1 ,...,Xtk
(φ(Z1 − x, . . . , Zk − Zk−1 ))
where E(φ) is the expectation of φ with respect to the joint disXt1 ...Xtk
tribution of (Xt1 . . . , Xtk ). You may also require the change of variable formula).
Problem . Given a Brownian motion with starting point x our aim is to
find a probability P x on the space Ω = C([0, ∞); Rd ) of all continuous
funcitons from [0, ∞) → Rd which induces the Brownian motion. We
will thus have a continuous realisation to Brownian motion. To achieve
this goal we will work with the collection {Ft1 ,...,tk : 0 ≤ t1 < t2 < . . . <
tk } where ti ∈ D, a countable dense subset of [0, ∞).
19
Step 1. The first step is to find a probability measure on a “smaller”
space and lift it to C([0, ∞); Rd ). Let
Ω = C([0, ∞); Rd ),
D a countable dense subset of [0, ∞); Ω(D) = {F : D → Rd } where f
is uniformly continuous on [0, N] ∩ D for N = 1, 2, . . .. We equip Ω
with the topology of uniform convergence on compact sets and Ω(D)
with the topology of uniform convergence on sets of the form D ∩ K
21
where K ⊂ [0, ∞) is compact; Ω and Ω(D) are separable metric spaces
isometric to each other.
Exercise 3. Let
pn ( f, g) = sup | f (t) − g(t)|
for
0≤t≤n
f, g ∈ Ω
and
pn,D ( f, g) = sup | f (t) − g(t)|
for
0≤t≤n
t∈D
f, g ∈ Ω(D).
Define
ρ( f, g) =
ρD ( f, g) =
∞
X
1 pn ( f, g)
, ∀ f, g ∈ Ω,
2n 1 + pn ( f, g)
n=1
∞
X
1 pn,D ( f, g)
, ∀ f, g ∈ Ω(D).
2n 1 + pn,D ( f, g)
n=1
Show that
(i) { fn } ⊂ Ω converges to f if and only if fn → f uniformly on
compact subsets of [0, ∞);
(ii) { fn } ⊂ Ω(D) converges to f if and only if fn|D∩K → f|D∩K| uniformly for every compact subset K of [0, ∞);
(iii) {(P1 , . . . , Pd )} where Pi is a polynomial with rational coefficients 20
is a countable dense subset of Ω;
(iv) {(P1D , . . . , PdD )} is a countable dense subset of Ω(D);
(v) τ : Ω → Ω(D) where τ( f ) = f|D is a (ρ, ρD )-isometry of Ω onto
Ω(D);
(vi) if V( f, ǫ, n) = {g ∈ Ω : pn ( f, g) < ǫ} for f ∈ Ω, ǫ > 0 and
VD ( f, ǫ, n) = {g ∈ Ω(D) : pn,D ( f, g) < ǫ}
for
f ∈ Ω(D), ǫ > 0,
then
{V( f, ǫ, n) : f ∈ Ω, ǫ > 0, n = 1, 2 . . .}
4. Construction of Wiener Measure
22
is a base for the topology of Ω and
{VD ( f, ǫ, n) : f ∈ Ω(D), ǫ > 0, n = 1, 2, . . .}
is a base for the topology of Ω(D).
Remark. By Exercise 3(v) any Borel probability measure on Ω(D) can
be lifted to a Borel probability measure on Ω.
2nd Step. Define the modulus of continuity ∆T,δ
D ( f ) of a function f on
D in the interval [0, T ] by
∆T,δ
D ( f ) = sup{| f (t) − f (s)| : |t − s| < δt, s ∈ D ∩ [0, T ]}
As D is countable one has
N, 1
Exercise 4. (a) Show that f : ∆D j ( f ) ≤ 1k } is measurable in the σalgebra generated by the projections
πt : π{Rdt : t ∈ D} → Rdt
21
Proof. The lemma is equivalent to showing that B = σ(E ). As each of
the projection πt1 ...tk is continuous, σ(E ) ⊂ B. To show that B ⊂ σ(E ),
it is enough to show that VD ( f, ǫ, n) ∈ E because Ω(D) is separable. (Cf.
Exercise 3(iv) and 3(vi)). By definition
VD ( f, ǫ, n) = {g ∈ Ω(D) : Pn,D ( f, g) < ǫ}
)
∞ (
[
1
=
g ∈ Ω(D) : pn,D ( f, g) ≤ ǫ −
m
m=1
=
∞
[
m=1
{g ∈ Ω(D) : |g(ti ) − f (ti )| ≤ ǫ −
1
, ∀ti ∈ D ∩ [0, n]}.
m
The result follows if one observes that each πti is continuous.
Remark 1. The lemma together with Exercise 4(b) signifies that the
Kolmogorov probability P x on π{Rdt : t ∈ D} is defined on the topological Borel σ-field of Ω(D).
2. The proof of the lemma goes through if Ω(D) is replaced by Ω.
23
Step 3. We want to show that P x (Ω(D)) = 1. By Exercise 4(b) this is
N,1/ j
equivalent to showing that Lt P(∆D ( f ) ≤ 1k ) = 1 for all N and k.
j→∞
The lemmas which follow will give the desired result.
Lemma (Lévy). Let X1 , . . . Xn be independent random variables, ǫ > 0
and δ > 0 arbitrary. If
P(|Xr + Xr+1 + · · · + Xℓ | ≥ δ) ≤ ǫ
∀ r, ℓ such that 1 ≤ r ≤ ℓ ≤ n, then
P( sup |X1 + · · · + X j | ≥ 2δ) ≤ 2ǫ.
1≤ j≤n
(see Kolmogorov’s theorem) for every j = 1, 2, . . . , for every N = 22
1, 2, . . . and for every k = 1, 2, . . .. (Hint: Use the fact that the projections are continuous).
∞ T
∞ S
∞
N, 1
T
(b) Show that Ω(D) =
{∆D j ( f ) ≤ 1k } and hence Ω(D) is
N=1 k=1 j=1
measurable in
π{Rdt
: t ∈ D}.
Let πt1 ...tk : Ω(D) → Rd × Rd × · · · Rd (k times) be the projections
and let
d
d
Et1 ...tk = π−1
t1 ...tk (B(R )× · · · ×B(R )).
k times
Put
E = ∪{Et1 ...tk : 0 ≦ t1 < t2 < . . . < tk < ∞; ti ∈ D}.
Then, as
Et1 ...tk ∪ E s1 ...s1 ⊂ Eτ1 ...τm ,
where
{t1 . . . tk , s1 . . . s1 } ⊂ {τ1 . . . τm },
E is an algebra. Let σ(E ) be the σ-algebra generated by E .
Lemma . Let B be the (topological) Borel σ-field of Ω(D). Then B is
the σ-algebra generated by all the projections
{πti ...tk : 0 ≤ t1 < t2 < . . . < tk , ti ∈ D}.
4. Construction of Wiener Measure
24
Remark. By subadditivity it is clear that




P  sup |X1 + · · · + X j | ≥ 2δ ≤ nǫ.
23
1≤ j≤n
Ultimately, we shall let n → ∞ and this estimate is of no utility. The
importance of the lemma is that it gives an estimate independent of n.
Proof. Let S j = X1 + · · · + X j , E = { sup |S j | ≥ 2δ}. Put
1≤ j≤n
E1 = {|S 1 | ≥ 2δ},
E2 = {|S 1 | < 2δ, |S 2 | ≥ 2δ},
...
...
...
...
...
...
...
...
En = {|S j | < 2δ, 1 ≤ j ≤ n − 1, |S n | ≥ 2δ}.
Then
E=
n
[
j=1
E j , E j ∩ Ei = φ
if
j , i;


n

[
P{E ∩ (|S n | ≤ δ) = P  (E j ∩ (|S n | ≤ δ))
j=1
n[
o
≤P
(Ei ∩ (|S n − S j | ≥ δ))
n
X
≤
P(E j )P(|S n − S j | ≥ δ) (by independence)
j=1
≤ ǫP(E)
(by data).
= P{E ∩ (|S n | > δ)} ≤ P(|S n | > δ) ≤ ǫ
(by data).
Combining the two estimates above, we get
P(E) ≤ ǫ + ǫP(E).
24
If ǫ >
1 ǫ
1
, 2ǫ > 1. If ǫ < ,
≤ 2ǫ. In either case P(E) ≤ 2ǫ.
2
2 1−ǫ
25
Lemma . Let {X(t)} be a Brownian motion, I ⊂ [0, ∞) be a finite interval, F ⊂ I ∩ D be finite. Then
!
|I|2
P x Sup |X(t) − X(σ)| ≥ 4δ ≤ C(d) 4 ,
δ
t,σ∈F
where |I| is the length of the interval and C(d) a constant depending only
on d.
Remark. Observe that the estimate is independent of the finite set F.
Proof. Let F = {ti : 0 ≤ t1 < t2 < . . . < tk < ∞}.
Put
X1 = X(t2 ) − X(t1 ), . . . , Xk−1 = X(tk ) − X(tk−1 ).
Then X1 . . . Xk−1 are independent (Cf. Exercise 2(c)). Let
ǫ=
sup
1≤r≤1≤k−1
P x (|Xr + Xr+1 + · · · + X1 | ≥ δ).
Note that
P x (|Xr + · · · + X1 | ≥ δ) = P(|X(t′ ) − X(t′′ )| ≥ δ) for some t′ , t′′ in F
(*)
E(|X(t′ ) − X(t′′ )|4 )
(see Tchebyshey’s inequality in Appendix)
δ4
C ′ (t′′ − t′ )
(C ′′ = constant)
≤
δ4
C ′ |I|2
≤
.
δ4
≤
Therefore ǫ ≤
C ′ |I|2
. Now
δ4
P x ( sup |X(t) − X(σ)| ≥ 4δ)
t,σ∈P
P x ( sup |X(ti ) − X(t1 )| ≥ 2δ)
1≤i≤k
= P x ( sup |X1 + · · · + X j | ≥ 2) ≤ 2ǫ
(by previous lemma)
i≤ j≤k−1
2C ′ |I|2 C|I|2
= 4 .
δ4
δ
25
4. Construction of Wiener Measure
26
Exercise 5. Verify (∗).
(Hint: Use the density function obtained in Exercise 2(c) to evaluate the
expectation and go over to “popular” coordinates. (The value of C ′ is
d(2d + 1))).
Lemma .












Px 
sup
|X(t)
−
X(s)|
>
ρ
= P x (∆T,h



D > ρ)




|t−s|≤h


t,s∈[0,t]∩D
≤ φ(T, ρ, h) = C
h T +
1
.
ρ4 h
Note that φ(T, ρ, h) → 0 as h → 0 for every fixed T and ρ.
Proof. Define the intervals I1 , I2 , . . . by
Ik = [(k − 1)h, (k + 1)h] ∩ (0, T ], k = 1, 2, . . . .
Let I1 , I2 , . . . Ir be those intervals for which
I j ∩ [0, T ] , φ ( j = 1, 2, . . . , r).
26
Clearly there are [ Th ] + 1 of them. If |t − s| ≤ h then t, s ∈ I j for some
∞
S
j, 1 ≤ j ≤ r. Write D =
Fn where Fn ⊂ Fn+1 and Fn is finite. Then
n=1






Px 




sup
|t−s|≤h
t,s∈[0,T ]∩D


 






∞ 



[






 

 sup |X(t) − X(s)| > ρ
|X(t) − X(s)| > ρ
= Px 












  |t−s|≤h

n=1
t,s∈D∩Fn








= sup P x 
sup sup (|XI j (t) − X(s)| > ρ)




n
j t,s∈Fn
27
≤ sup
n
r
X
j=1
!
P x sup (|XI j (t) − X(s)| > ρ)
t,s∈Fn
T C(2h)2
+1
sup
h
(ρ/4)4
n
≤ φ(T, ρ, h).
by the last lemma
Theorem . P x (Ω(D)) = 1.
Proof. It is enough to show that
N, 1
Lt P x ∆D j ( f )
j→∞
!
1
≤
= 1 (See Exercise 4(b)).
k
But this is guaranteed by the previous lemma.
Remark.
1. It can be shown that the outer measure of Ω is 1.
Q
2. Ω is not measurable in Rdt .
t≥0
Let P̃ x be the measure on Ω induced by P x on Ω(D). We have al- 27
ready remarked that P x is defined on the (topological Borel σ field of
Ω(D). As P x is a probability measure, P̃ x is also a probability measure. It should now be verified that P̃ x is really the probability measure
consistent with the given distribution.
Theorem . P̃ x is a required probability measure for a continuous realization of the Brownian motion.
Proof. We have to show that
Ft1 ...tk = P̃ x π−1
t1 ...tk
for all
t1 , t2 . . . tk
in [0, ∞).
Step 1. Let t1 , . . . , tk ∈ D. Then
P x (π−1
t1 ...tk (A1 × · · · × Ak )) = P x (τπt1 ...tk (A1 × · · · × Ak ))
for every Ai Borel in Rd . The right side above is
P x (π−1
t1 ...tk (A1 × · · · × Ak )) = F t1 ...tk (A1 × · · · × Ak )
(by definition of P x ).
4. Construction of Wiener Measure
28
Step 2. We know that T t1 ,t2 ...tk = P̃ x πt1 ,t2 ,...,tk provided that t1 , t2 , . . . , tk ∈
D. Let us pick t1(n) , . . . , tk(n) , such that each ti(n) ∈ D and tk(n) → tk as
n → ∞. For each n and for each fixed f : Rd → R which is bounded
and continuous,
(n)
E F1
28
,...tk(n)
[ f (x1 , . . . , xk )] = E P̃x [ f (x(t1(n) , . . . , x(tk(n) )))].
Letting n → ∞ we see that
E Ft1 ,...,tk [ f (x1 , . . . , xk )] = E Px [ f (x(t1 ), . . . , x(tk ))]
for all t1 , . . . , tk . This completes the proof.
The definition of the Brownian motion given earlier has a built-in
constraint that all “trajectories” start from X(0) = x. This result is given
by
Theorem . P̃0 { f : f (0) = 0} = 1.
Proof. Obvious; because E P̃x [φ(x(0))] = φ(x).
Note. In future P̃ x will be replaced by P x and P̃0 = P0 will be denoted
by P.
Let T x : Ω → Ω be the map given by (T x f )(t) = x + f (t). T x
translates every ‘trajectory’ through the vector x.
Time
Let us conceive a real Brownian motion of a system of particles. The
operation T x means that the system is translated in space (along with
29
29
everything else that affects it) through a vector x. The symmetry of the
physicl laws governing this motion tells us that any property exhibited
by the first process should be exhibited by the second process and vice
versa. Mathematically this is expressed by
Theorem . P x = PT x−1 .
Proof. It is enough to show that
−1
P x (T x π−1
t1 ...tk (A1 × · · · × Ak )) = P(πt1 ...tk (A1 × · · · × Ak ))
for every Ai Borel in Rd . Clearly,
−1
T x π−1
t1 ...tk (A1 × · · · × Ak ) = πt1 ...tk (A1 − x × · · · × Ak − x).
Thus we have only to show that
Z
Z
Z
p(0, x, t1 , x1 ) . . . p(tk−1 , xk−1 , tk , xk )dx1 . . . dxk
...
A1 −x
Ak −x
Z
Z
p(0, 0, t1 , x1 ) . . . p(tk−1 , xk−1 , tk , xk )dx1 . . . dxk ,
...
=
A1
Ak
which is obvious.
Exercise. (a) If β(t, ·) is a Brownian motion s tarting at (0, 0) then
1
√ β(ǫt) is a Brownian motion starting at (0, 0) for every ǫ > 0.
ǫ
(b) If X is a d-dimensional Brownian motion and Y is a d′ -dimensional Brownian motion then (X, Y) is a d + d′ dimensional Brownian
motion provided that X and Y are independent.
j
(c) If Xt = (Xt1 , . . . , Xtd ) is a d-dimensional Brownian motion, then Xt
is a one-dimensional Brownian motion. ( j = 1, 2, . . . d).
τ(w) = inf{t : |Xt (w)| ≥ +1}
= inf{t : |w(t)| ≥ 1}
τ(w) is the first time the particle hits either of the horizontal lines 30
4. Construction of Wiener Measure
30
+1 or −1.
6.4
2. Let {Xt } be a d-dimensional Brownian motion, G any closed set in
Rd . Define
τ(w) = inf{t : w(t) ∈ G}.
This is a generalization of Example 1. To see that τ is a stopping
time use
∞
\
lim {w : w(θ) ∈ Gn },
{τ ≤ s} =
θ∈[0,s]
n=1 θ rational
where
)
(
1
.
Gn = x ∈ Rd : d(x, G) ≤
n
3. Let (Xt ) be a d-dimensional Brownian motion, C and D disjoint
closed sets in Rd . Define
τ(w) = inf{t; w(t) ∈ C and for some s ≤ t, w(s) ∈ D}.
τ(w) is the first time that w hits C after visiting D.
5. Generalised Brownian
Motion
LET Ω BE ANY space, F a σ-field and (Ft ) an increasing family of 31
sub σ-fields such that σ(∪Ft ) = F . Let P be a measure on (Ω, F ).
X(t, w) : [0, w) × Ω → Rd
is called a Brownian motion relative to (Ω, Ft , P) if
(i) X(t, w) is progressively measurable with respect to Ft ;
(ii) X(t, w) is a.e. continuous in t;
(iii) X(t, w) − X(s, w) for t > s is independent of Fs and is distributed
normally with mean 0 and variance t − s, i.e.
P(X(t, ·) − X(s, ·) ∈ A|Fs ) =
Note.
Z
A
1
|y|2
dy.
exp
−
2(t − s)
[2π(t − s)]d/s
1. The Brownian motion constructed previously was concentrated on Ω = C([0, ∞); Rd ), F was the Borel field of Ω, X(t, w) =
w(t) and Ft = σ{X(s) : 0 ≤ s ≤ t}. The measure P so obtained is
often called the Wiener measure.
2. The above definition is more general because
σ{X(s) : 0 ≤ s ≤ t} ⊂ Ft .
31
5. Generalised Brownian Motion
32
Exercise. (Brownian motion starting at time s). Let Ω = C([s, ∞); Rd ),
B = Borel field of Ω. Show that for each x ∈ Rd ∃ a probability measure
P sx on Ω such that
(i) P sx {w : w(s) = x} = 1;
32
(ii) P sx (Xt1 ∈ A1 , . . . , Xtk ∈ Ak )
Z Z
Z
=
...
p(s, x, t1 , x1 )p(t1 , x1 , t2 , x2 ) . . .
A1
A2
Ak
. . . p(tk−1 xk−1 , tk , xk )dx1 . . . dxk ,
∀ s < t1 < . . . < tk .
For reasons which will become clear later, we would like to shift the
measure P sx to a measure on C([0, ∞); Rd ). To do this we define
T : C([s, ∞); Rd ) → C([0, ∞); Rd )
by



w(t),
(T w)(t) = 

w(s),
if t ≥ s,
if t ≤ s.
Clearly, T is continuous. Put
P s,x = P sx T −1 .
Then
(i) P s,x is a probability measure on the Borel field of C([0, ∞); Rd );
(ii) P s,x {w : w(s) = x} = 1.
6. Markov Properties of
Brownian Motion
Notation.
1. A random variable of a stochastic process {X(t)}t∈I shall 33
be denoted by Xt or X(t). 0 ≤ t < ∞.
2. Fs will denote the σ-algebra generated by {Xt : 0 ≤ t ≤ s};
Fs+ = {Fa : a > s}; Fs− will be the σ-algebra generated by
∪{Fa : a < s}s > 0. It is clear that {Ft } is an increasing family.
3. For the Brownian motion, B = the σ-algebra generated by ∪{Ft :
t < ∞} will be denoted by F .
Theorem . Let {Xt : 0 ≤ t < ∞} be a Brownian motion. Then Xt − X s is
independent of Fs .
Proof. Let
0 ≤ t1 < t2 < t3 < . . . < tk ≤ s.
Then the σ-algebra generated by Xt1 , . . . , Xtk is the same as the σalgebra generated by
Xt1 , Xt2 − Xt1 , . . . , Xtk − Xtk−1 .
Since Xt − X s is independent of these increments, it is independent
of σ{Xt1 , . . . , Xtk }. This is true for every finite set t1 , . . . , tk and therefore
Xt − X s is independent of Fs .
Let us carry out the following calculation very formally.
P[Xt ∈ A | Fs ](w) = P[Xt − X s ∈ B | Fs ](w), B = A − X s (w),
33
6. Markov Properties of Brownian Motion
34
= P[Xt − X s ∈ B],
34
i.e.
P[Xt ∈ A | Fs ](w) =
Z
A
by independence,
1
|y − X s (w)|2
dy.
exp
−
2(t − s)
(2πt)d/2
This formal calculation leads us to
Theorem .
P[Xt ∈ A | Fs ](w) =
Z
A
1
|y − X s (w)|2
dy.
exp
−
2(t − s)
(2πt)d/2
where A is Borel in Rd , t > s.
Remark. It may be useful to note that p(s, X s (w), t, y) can be thought of
as a conditional probability density.
Proof.
(i) We show that
Z
fA (w) =
A
|y − X s (w)|2
1
exp
−
dy
2(t − s)
(2πt)d/2
is FS -measurable. Assume first that A is bounded and Borel in
Rd . If ωn → ω, then fA (ωn ) → fA (ω), i.e. fA is continuous and
hence Fs -measurable. The general case follows if we note that
any Borel set can be got as an increasing union of a countable
number of bounded Borel sets.
(ii) For any C ∈ Fs we show that
Z Z
Z X
exp −|y − X s (ω)|2
−1
dy dP(ω).
Xt (A)dP(ω) =
(*)
(2π(t − s))d/2
C
C
A
It is enough to verify (∗) for C of the form
C = ω : (Xt1 (ω), . . . , Xtk (ω)) ∈ A1 × · · · × Ak ; 0 ≤ t1 < . . . < tk ≤ s ,
35
where Ai is Borel in Rd for i = 1, 2 . . . k. The left side of (∗) is
then
Z
p(0, 0, t1, xt1 )p(t1 , xt1 , t2 , xt2 ) . . . p(tk , xtk , t, xt )dxt1 . . . dxt .
Ai ×···×Ak ×A
35
To compute the right side define
f : R(k+1)d → B
by
f (u1 , . . . , uk , u) = XA1 (u1 ) . . . XAk (uk )p(s, u, t, y).
Clearly f is Borel measurable. An application of Fubini’s theorem
to the right side of (∗) yields
Z
Z
dy XA1 (Xt1 (ω)) . . . XAk (Xtk (ω))p(s, X s (ω), t, y)dP(ω)
A
=
ZΩ
dy
A
=
Z
f (x1 . . . xk , xs )dFt1 ...tk , s
Rd ×···×Rd
(k+1) times
dy
A
=
Z
Z
p(0, 0, t1 , x1 ) . . . p(tk−1 , tk , xk )
A1 ×···×Ak ×Rd
Z
p(tk , xk , s, xs )p(s, xs , t, y)dx1 . . . dxk dxs
p(0, 0, t1 , x1 ) . . . p(tk−1 , xk−1 , tk , xk )p(tk , xk , t, y)
A1 ×···×Ak ×A
dx1 . . . , dxk dy
(by the convolution rule)
= left side.
Examples of Stopping Times.
6. Markov Properties of Brownian Motion
36
1. Let (Xt ) be a one-dimensional Brownian motion. Define τ by
{τ ≤ s} =
∞
\
{w : w(θ1 ) ∈ Dn , w(θ2 ) ∈ Cn },
lim
θ1 ,θ2
n=1 θ1 ,θ2 rational in [0,s]
where
36
(
)
(
)
1
1
d
Dn = x ∈ R : d(x, D) ≤
, Cn = x ∈ R : d(x, C) ≤
n
n
d
Exercise 1. Let τ be as in Example 1.
(a) If A = {w : X1 (w) ≤ τ} show that A < Fτ .
(Hint: A ∩ {τ ≤ 0} < F0 ). This shows that Fτ $ F0 .
(b) P0 {w : τ(w) = ∞} = 0.
(Hint: P0 {w : |w(t)| < 1} ≤
R
|y|≤t−1/2
2
e−1/2|y| dy ∀t).
Theorem . (Strong Markov Property of Brownian Motion). Let τ be any
finite stopping time, i.e. τ < ∞ a.e. Let Yt = Xτ+t − Xτ . Then
1. P[(Yt1 ∈ A1 , . . . , Ytk ∈ Ak ) ∩ A] = P(Xt1 ∈ A1 , . . . Xtk ∈ Ak ) · P(A),
∀ A ∈ Fτ and for every Ai Borel in Rd . Consequently,
2. (Yt ) is a Brownian motion.
3. (Yt ) is independent of Fτ .
The assertion is that a Brownian motion starts afresh at every stopping time.
Proof.
Step 1. Let τ take only countably many values, say s1 , s2 , s3 . . .. Put
E j = τ−1 {s j }. Then each E j is Fτ -measure and
Ω=
∞
[
j=1
E j , E j ∩ Ei = ∅ j , i.
37
37
Fix A ∈ Fτ .
P[(Yt1 ∈ A1 , . . . , Ytk ∈ Ak ) ∩ A]
∞
X
=
P[(Yt1 ∈ A1 , . . . , Ytk ∈ Ak ) ∩ A ∩ E j ]
j=1
=
=
∞
X
j=1
∞
X
j=1
P[(Xt1 +s j − X s j ) ∈ A1 , . . . , Xtk +s j − X s j ∈ Ak ) ∩ A ∩ E j ]
P[(Xt1 ∈ A1 ), . . . , (Xtk ∈ Ak )]P(A ∩ E j )
(by the Markov property)
= P(Xt1 ∈ A1 , . . . , Xtk ∈ Ak ) · P(A)
[nτ] + 1
. A simple calStep 2. Let τ be any stopping time; put τn =
n
culation shows that τn is a stopping time taking only countably many
values. As τn τ, Fτ ⊂ Fτn ∀n . Let Yt(n) = Xτn +t − Xτn .
By Step 1,
P[(Yt(n)
< x1 , . . . , Yt(n)
< xk ) ∩ A]
1
k
= P(Xt1 < x1 , . . . , Xtk < xk ) · P(A)
(where x < y means xi < yi i = 1, 2, . . . , d) for every A ∈ Fτ . As all
the Brownian paths are continuous, Yt(n) → Yt a.e. Thus, if x1 , . . . , xk is
a point of continuity of the joint distribution of Xt1 , . . . , Xtk , we have
P[(Yt1 < x1 , . . . , Ytk < xk ) ∩ A] = P(Xt1 < x1 , . . . , Xtk < xk )P(A)
∀ A ∈ Fτ . Now assertion (1) follows easily.
For (2), put A = Ω in (1), and (3) is a consequence of (1) and (2).
38
7. Reflection Principle
LET (Xt ) BE A one-dimensional Brownian motion. Then P( sup X s ≥ 39
0≤s≤t
a) = 2P(Xt ≥ a) with a > 0. This gives us the probability of a Brownian
particle hitting the line x = a some time less than or equal to t. The
intuitive idea of the proof is as follows.
Time,
Among all the paths that hit a before time t exactly half of them end
up below a at time t. This is due to the reflection symmetry. If X s = a
for some s < t, reflection about the horizontal line at a gives a one one correspondence between paths with Xt > a and paths with Xt < a.
Therefore
P max X s ≥ a, Xt > a = max X s ≥ a, Xt < a
0≤s≤t
0≤s≤t
39
7. Reflection Principle
40
Since P{Xt = a} = 0, we obtain
(
)
(
)
(
)
P sup X s ≥ a = P sup X s ≥ a, Xt > a + P sup X s ≥ a, Xt > a
0≤s≤t
0≤s≤t
0≤s≤t
= 2P{Xt ≥ a}
40
We shall now give a precise argument. We need a few elementary
results.
n
P
Lemma 1. Let Xn =
Yk where the Yk are independent random varik=1
ables such that P{Yk ∈ B} = P{−Yk ∈ B}∀ Borel set B ⊂ R (i.e. Yk are
symmetric). Then for any real number a,
P max Xi > a ≤ 2P{Xn > a}
1≤i≤n
Proof. It is easy to verify that a random variable is symmetric if and
only if its characteristic function is real. Define
Ai = {X1 ≤ a, . . . Xi−1 ≤ a, Xi > a}, i = 1, 2, . . . , n;
B = {Xn > a}
Then Ai ∩ A j = ∅ if i , j. Now,
P(Ai ∩ B) ≥ P(Ai ∩ {Xn ≥ Xi })
= P(Ai )P(Xn ≥ Xi ),
by independence.
= P(Ai )P(Yi+1 + · · · + Yn ≥ 0).
As Yi+1 , . . . , Yn are independent, the characteristic function of Yi+1 +
· · · + Yn is the product of the characteristic functions of Yi+1 + · · · + Yn ,
so that Yi+1 + · · · + Yn is symmetric. Therefore
P(Yi+1 + · · · + Yn ≥ 0) ≥
Thus P(Ai ∩ B) ≥
P(B) ≥
n
X
i=1
1
P(Ai ) and
2
1
.
2
 n 
1 [ 
1X
P(Ai ∩ B) ≥
P(Ai ) ≥ P  Ai  ,
2
2 i=1
41
41
i.e.
 n 
[ 
2P(B) ≥ P  Ai  ,
i=1
or
P max Xi > a ≤ 2P{Xn > a}
1≤i≤n
Lemma 2. Let Yi , . . . , Yn be independent random variables. Put Xn =
n
P
Yk and let τ = min{i : Xi > a}, a > 0 and τ = ∞ if there is no such i.
k=1
Then for each ǫ > 0,
(a) P{τ ≤ n − 1, Xn − Xτ ≤ −ǫ} ≤ P{τ ≤ n − 1, Xn ≤ a} +
n−1
P
P(Y j > ǫ).
j=1
(b) P{τ ≤ n−1, Xn > a+2ǫ} ≤ P{τ ≤ n−1, Xn −Xτ > ǫ}+
n−1
P
P{Y j > ǫ}
j=1
(c) P{Xn > a + 2ǫ} ≤ P{τ ≤ n − 1, Xn > a + 2ǫ} + P{Yn > 2ǫ}.
If, further, Y1 , . . . , Yn are symmetric, then
(d) P{max Xi > a, Xn ≤ a} ≥ P{Xn > a + 2ǫ} − P{Yn ≥ 2ǫ} −
2
1≤i≤n
n−1
P
P{Y j > ǫ}
j=1
(e) P{max Xi > a} ≥ 2P{Xn > a + 2ǫ} − 2
1≤i≤n
Proof.
n
P
P{Y j > ǫ}
j=1
(a) Suppose w ∈ {τ ≤ n − 1, Xn − Xτ ≤ −ǫ} and w ∈ {τ ≤
n − 1, Xn ≤ a}. Then Xn (w) > a and Xn (w) + ǫ ≤ Xτ(w) (w) or,
Xτ(w) (w) > a + ǫ.
By definition of τ(w), Xτ(w)−1 (w) ≤ a and therefore,
Yτ(w) (w) = Xτ(w) (w) − Xτ(w)−1 (w) > a + ǫ − a = ǫ
if τ(w) > 1; if τ(w) = 1, Yτ(w) (w) = Xτ(w) (w) > a + ǫ > ǫ.
Thus Y j (w) > ǫ for some j ≤ n − 1, i.e.
42
7. Reflection Principle
42
w∈
n−1
[
{Y j > ǫ}.
j=1
Therefore
{τ ≤ n − 1, Xn − Xτ ≤ −ǫ} ⊂ {τ ≤ n − 1, Xn ≤ a}
n−1
[
{Y j > ǫ}
j=1
and (q) follows.
(b) Suppose w ∈ {τ ≤ n − 1, Xn > a + 2ǫ} but w ∈ {τ ≤ n − 1, Xn − Xτ >
ǫ}. Then
Xn (w) − Xτ(w) (w) ≤ ǫ,
or, Xτ(w) (w) > a + ǫ so that Yτ(w) (w) > ǫ as in (a); hence Y j (w) > ǫ
for some j ≤ n − 1. This proves (b).
(c) If w ∈ {Xn > a + 2ǫ}, then τ(w) ≤ n; if w < {τ ≤ n − 1, Xn >
a + 2ǫ}, then τ(w) = n so that Xn−1 (w) ≤ a; therefore Yn (w) =
Xn (w) − Xn−1 (w) > 2ǫ. i.e. w ∈ {Yn > 2ǫ}. Therefore
{Xn > a + 2ǫ} ⊂ {τ ≤ n − 1, Xn > a + 2ǫ} ∪ {Yn > 2ǫ}.
This establishes (c).
(d) P{max Xi > a, Xn ≤ a} = P{τ ≤ n − 1, Xn ≤ a}
1≤i≤n
≥ P{τ ≤ n − 1, Xn − Xτ ≤ −ǫ} −
P
=
=
"
n−1
S
k=1
n−1
P
k=1
n−1
P
k=1
#
{τ = k, Xn − Xk ≤ −ǫ} −
P{τ = k, Xn − Xk ≤ −ǫ} −
n−1
P
P(Y j > ǫ),
j=1
n−1
P
P(Y j > ǫ)
j=1
n−1
P
P(Y j > ǫ)
j=1
P{τ = k}P{Xn − Xk ≤ −ǫ} −
n−1
P
P(Y j > ǫ)
j=1
(by independence)
by (a),
43
=
n
P
k=1
P{τ = k}P{Xn − Xk > ǫ} −
= P{τ ≤ n − 1, Xn − Xτ ≥ ǫ} −
≥ P{τ ≤ n − 1, Xn − Xτ > ǫ} −
n−1
P
n−1
P
(by symmetry)
P(Y j > ǫ)
j=1
n−1
P
P(Y j > ǫ)
j=1
≥ P{τ ≤ n − 1, Xn > a + 2ǫ} − 2
n−1
P
P{Y j > ǫ}
(by (b))
j=1
≥ P{Xn > a + 2ǫ} − P{Yn > 2ǫ} − 2
This proves (d).
P(Y j > ǫ)
j=1
n−1
P
P{Y j > ǫ}
(by (c))
43
j=1
(e) P{max Xi > a} = P{max Xi > a, Xn ≤ a} + P{max Xi > a, Xn > a}
1≤i≤n
1≤i≤n
1≤i≤n
= P{max Xi > a, Xn ≤ a} + P{Xn > a}
1≤i≤n
= P{Xn > a + 2ǫ} − P{Yn > 2ǫ} + P{Xn > a}
−2
n−1
P
P{Y j > ǫ}
(by (d))
j=1
Since P{Xn > a + 2ǫ} ≤ P{Xn > a} and
P{Yn > 2ǫ} ≤ P{Yn > ǫ} ≤ 2P{Yn > ǫ},
we get
P{max Xi > a} ≥ 2P{Xn > a + 2ǫ} − 2
1≤i≤n
n
X
P(Y j > ǫ)
j=1
This completes the proof.
44
Proof of the reflection principle.
By Lemma 1
(
jt )
> ≤ 2P(X(t) > a).
p = P max X
1≤ j≤n
n
7. Reflection Principle
44
By Lemma 2(e),
( !!
)
n
X
( j − 1)t
jt
p ≥ 2P(X(t) > a + 2ǫ) − 2
−X
P X
>ǫ .
n
n
j=1
!
( j − 1)t
are independent and identically distribun
n
t
ted normal random variables with mean zero and variance (in particn
ular they are symmetric),
!!
!
jt t ( j − 1)t
−X
− X(0) > ǫ
P X
>ǫ =P X
n
n
n
t
=P X
>ǫ .
n
Since X
jt Therefore
−X
t
p ≥ 2P(X(t) > a + 2ǫ) − 2n P X
>ǫ .
n
P(X(t/n) > ǫ) =
Z∞
ǫ
=
√
Z∞
2 2t
1
e−x / n dx
(2t/n)
√ √
ǫ n/ (2t)
or
√ √
2
e−x
ǫ n/ (2t)
√ dx ≤
√
π
x
−1∞
Z
2
xe−x dx
√ √
ǫ n/ (2t)
√
1 −ǫ 2 n/2t
(2t)
P(X(t/n) > ǫ) ≤ √ e
· √ .
2 π
ǫ n
Therefore
nP(X(t/n) > ǫ) ≤
√
2
n√
(2t)/ (πn)e−ǫ n/2t → 0 as
2ǫ
n → +∞.
45
By continuity,
P max X( jt/n) > a → P max X(s) > a .
1≤≤n
0≤s≤t
45
We let n tend to ∞ through values 2, 22 , 23 , . . . so that we get
2P{X(t) > a + 2ǫ} − 2n P{X(t/n) > ǫ}
(
)
≤P max X(t/n) > a ≤ 2P{X(t) > a},
1≤ j≤n
or
2P{X(t) > a} ≤ 2P{X(t) ≥ a} ≤P max X(t) > a
0≤s≤t
2P{X(t) > a},
on letting n → +∞ first and then letting ǫ → 0. Therefore,
P max X(s) > a = 2P{X(t) > a}
0≤s≤t
=2
Z∞
√
2
1/ (2πt)e−x /2t dx.
a
AN APPLICATION. Consider a one-dimensional Brownian motion. A
particle starts at 0. What can we say about the behaviour of the particle
in a small interval of time [0, ǫ)? The answer is given by the following
result.
P(A) ≡ P{w : ∀ǫ > 0, ∃ t, s in [0, ǫ) such that Xt (w) > 0 and
X s (w) < 0} = 1.
INTERPRETATION. Near zero all the particles oscillate about their 46
starting point. Let
A+ = {w : ∀ ∈> 0∃ t ∈ [0, ǫ) such that Xt (w) > 0},
A− = {w : ∀ǫ > 0∃ s ∈ [0, ǫ) such that X s (w) < 0}.
We show that P(A+ ) = P(A− ) = 1 and therefore P(A) = P(A+ ∩
A− ) = 1.

 \
∞ 
∞ [
∞ 
\






+
 sup w(t) ≥ 1/m
A ⊃

 sup w > 0
=
n=1
0≤t≤1/n
n=1 m=1 0≤t≤1/n
7. Reflection Principle
46
Therefore




P(A ) ≥ Lt sup P  sup w(t) ≥ 1/m
+
n→∞ m→∞
≥2
Lt
n→∞m→∞
0≤t≤1/n
sup P(w(1/n) ≥ 1/m)
(by the reflection principle)
≥ 1.
Similarly P(A− ) = 1.
Theorem . Let {Xt } be a one-dimensional Brownian motion, A ⊂ (−∞, a)
(a > 0) and Borel subset of R. Then
P0 {Xt ∈ A, X s < a ∀s such that 0 ≤ s ≤ t}
Z
Z
√
√
2
2
=
1/ (2πt)e−y /2t dy − 1/ (2πt)e−(2a−y) /2t dy
A
A
Proof. Let τ(w) = inf{t : w(t) ≥ a}. By the strong Markov property of
Brownian motion,
P0 {B(X(τ + s) − X(τ) ∈ A)} = P0 (B)P0 (X(s) ∈ A)
for every set B in Ft . This can be written as
E(X(X(τ+s)−X(τ)∈A) |Fτ ) = P0 (X(s) ∈ A)
47
Therefore
E(X(X(τ+ℓ(w))−X(τ)∈A) |Fτ ) = P0 (X(ℓ(w)) ∈ A)
for every function ℓ(w) which is Fτ -measurable. Therefore,
Z
P0 ((τ ≤ t) ∩ ((X(τ + ℓ(w)) − X(τ)) ∈ A) =
P0 (X(ℓ(w)) ∈ A)dP(w)
{τ≤t}
In particular, take ℓ(w) = t − τ(w), clearly ℓ(w) is Fτ -measurable.
Therefore,
Z
P0 (X(ℓ(w) ∈ A)dP(w)).
P0 ((τ ≤ t)((X(t) − X(τ)) ∈ A)) =
{τ≤t}
47
Now X(τ(w)) = a. Replace A by A − a to get
(*)
P0 ((τ ≤ t) ∩ (X(t) ∈ A)) =
Z
P0 (X(ℓ(w) ∈ A − a)dP(w))
{τ≤t}
Consider now
P2a (X(t) ∈ A) = P0 (X(t) ∈ A − 2a)
= P0 (X(t) ∈ 2a − A) (by symmetry of x)
= P0 ((τ ≤ t) ∩ (X(t) ∈ 2a − A)).
The last step follows from the face that A ⊂ (−∞, a) and the continuity of the Brownina paths. Therefore
P2a (X(t) ∈ A) =
Z
P0 (X(ℓ(w)) ∈ a − A)dP(w),
(using ∗)
{τ≤t}
= P0 ((τ ≤ t) ∩ (X(t) ∈ A)).
Now the required probability
P0 {Xt ∈ A, X s < a ∀s ∈ 0 ≤ s ≤ t} = P0 {Xt ∈ A} − P0 {(τ ≤ t) ∩ (Xt ∈ A)}
Z
Z
√
√
2
−y2 /2t
1/ (2πt)e
dy − 1/ (2πt)e−(2a−y) /2t dy.
=
A
A
The intuitive idea of the previous theorem is quite clear. To obtain 48
the paths that reach A at time t without hitting the horizontal line x = a,
we consider all paths that reach A at time t and subtract those paths that
hit the horizontal line x = a before time t and then reach A at time t. To
see exactly which paths reach A at time t after hitting x = a we consider
a typical path X(w).
7. Reflection Principle
48
49
The reflection principle (or the strong Markov property) allows us
to replace this path by the dotted path (see Fig.). The symmetry of the
Brownian motion can then be used to reflect this path about the line
x = a and obtain the path shown in dark. Thus we have the following
result:
the probability that a Brownian particle starts from x = 0 at t = 0
and reaches A at time t after it has hit x = a at some time τ ≤ t is the
same as if the particle started at time t = 0 at x = 2a and reached A at
time t. (The continuity of the path ensures that at some time τ ≤ t, this
particle has to hit x = a).
We shall use the intuitive approach in what follows, the mathematical analysis being clear, thorugh lengthy.
Theorem . Let X(t) be a one-dimensional Brownian motion, A ⊂ (−1, 1)
any Borel subset of R. Then
"
# Z
P0 sup |X(s)| < 1, X(t) ∈ A =
φ(t, y)dy,
0≤s≤t
A
where
φ(t, y) =
∞
X
n=−∞
√
2
(−1)n / (2πt)e−(y−2n) /2t .
49
Proof.
Let En be the set of those trajections which (i) start at x = 0 at time
t = 0 (ii) hit x = +1 at some time τ1 < t (iii) hit x = −1 at some later
time τ2 < t (iv) hit x = 1 again at a later time τ3 < t . . . and finally reach
A at time t. The number of τ’s should be equal to n at least, i.e.
En = {w : there exists a sequence τ1 , . . . τn of
stopping times such that 0 < τ1 < τ2 < . . . < tτn < t, X(τ j ) = (−1) j−1 ,
X(t) ∈ A}. Similarly, let
Fn = {w : there exists a sequence τ1 , . . . , τn of stopping times
0 < τ1 < τ2 < . . . < τn < t, X(τ j ) = (−1) j , X(t) ∈ A}
Note that
50
E1 ⊃ E2 ⊃ E3 ⊃ . . . ,
F1 ⊃ F2 ⊃ F3 ⊃ . . . ,
Fn ⊃ En+1 ; En ⊃ Fn+1 ,
En ∩ Fn = En+1 ∪ Fn+1 .
Let
"
#
φ(t, A) = P sup |X(s)| < 1, X(t) ∈ A .
0≤s≤t
Therefore
"
φ(t, A) = P[X(t) ∈ A] − P sup |X(s)| ≥ 1, X(t) ∈ A
0≤s≤t
#
7. Reflection Principle
50
=
Z
√
2
1/ (2πt)e−y /2t dy − P[(E1 ∪ F1 ) ∩ A0 ],
A
where
A0 = {X(t) ∈ A} =
Z
√
2
1/ (2πt)e−y /2t dy − P[(E1 ∩ A0 ) ∪ (F1 ∩ A0 )].
A
Use the fact that P[A ∪ B] = P(A) + P(B) − P(A ∩ B) to get
φ(t, A) =
Z
√
2
1/ (2πt)e−y /2t dy−P[E1 ∩A0 ]−P[F1 ∩A0 ]+P[E1 ∩F1 ∩A0 ],
A
as E1 ∩ F1 = E2 ∪ F2 . Proceeding successively we finally get
φ(t, A) =
Z
A
X
X
√
2
1/ (2πt)e−y /2t dy+ (−1)n P[En ∩A0 ]+ (−1)n P[Fn ∩A0 ]
∞
∞
n=1
n=1
We shall obtain the expression for P(E1 ∩ A0 ) and P[E2 ∩ A0 ], the
other terms can be obtained similarly.
E1 ∩ A0 consists of those trajectries that hit x = ±1 at some time
τ ≤ t and then reach A at time t. Thus P[E1 ∩ A0 ] is given by the
previous theorem by
Z
√
2
1/ (2πt)e−(y−2) /2t dy.
A
51
E2 ∩ A0 consists of those trajectories that hit x = ±1 at time τ1 , hit
x = −1 at time τ2 and reach A at time t(τ1 < τ2 < t).
According to the previous theorem we can reflect the trajectory upto
τ2 about x = −1 so that P(E2 ∩ A0 ) is the same as if the particle starts at
x = −2 at time t = 0, hits x = −3 at time τ1 and ends up in A at time t.
We can now reflect the trajectory
51
upto time τ1 (the dotted curve should be reflected) about x = −3 to
obtain the required probability as if the trajectory started at x = −4.
Thus,
Z
2 /2t/√(2πt)
e−(y+4)
P(E2 ∩ A0 ) =
dy.
A
Thus
φ(t, A) =
∞
X
n=−∞
=
Z
n
(−1)
Z
√
2
1/ 2πte−(y−2n) /2t dy
A
φ(t, y)dy.
A
The previous theorem leads to an interesting result:
"
# Z1
P sup |X(s)| < 1 =
φ(t, y)dy
Therefore
"
0≤s≤t
−1
#
"
#
P sup |X(s)| ≥ 1 = 1 − P sup |X(s)| < 1
0≤s≤t
0≤s≤t
52
7. Reflection Principle
52
= −1 −
φ(t, y) =
∞
X
Z1
φ(t, y)dy,
−1
√
2
(−1)n / (2πt)e−(y−2n) /2t
n=−∞
Case (i). t is very small.
In this case it is enough to consider the terms corresponding to n = 0,
±1 (the higher order terms are very small). As y varies from −1 to 1,
h 2
i
√
2
2
φ(t, y) ≃ 1/ (2πt) e−y /2t − e−(y−2) /2t − e−(y+2) /2t .
Therefore
Z1
√
φ(t, y)dy ≃ 4/ (2πt)e−1/2t .
−1
Case (ii). t is large. In this case we use Poisson’s summation formula
for φ(t, y):
φ(t, y) =
∞
X
2 2
e−(2k+1)
π t/8
Cos{(k + 1)/2πy},
k=0
to get
Z1
φ(t, y)dy ≃ 4/πe−π
2
t/8
−1
53
2
for large t. Thus, P(τ > t) = 4/πe−π t/8 .
This result says that for large values of t the probability of paths
which stay between −1 and +1 is very very small and the decay rate is
2
governed by the factor e−π t/8 . This is connected with the solution of a
certain differential equation as shall be seen later on.
8. Blumenthal’s Zero-One
Law
LET Xt BE A d-dimensional Brownian motion. If A ∈ F0+ =
then P(A) = 0 or P(A) = 1.
T
Ft , 54
t>0
Interpretation. If an event is observable in every interval [0, t] of time
then either it always happens or it never happens.
We shall need the following two lemmas.
Lemma 1. Let (Ω, B, P) be any probability space, Ca sub-algebra of
B. Then
(a) L2 (Ω, C , P) is a closed subspace of L2 (Ω, B, P).
(b) If π : L2 (Ω, B, P) → L2 (Ω, C , P) is the projection map then π f =
E( f |C ).
Proof. Refer appendix.
Lemma 2. Let Ω = C([0, ∞); Rd ), P0 the probability corresponding to
the Brownian motion. Then the set {φ(πt1 , . . . , tk ) ∈ φ is continuous,
bounded on Rd × · · · × Rd (k times), πt1 , . . . , tk the canonical projection)
is dense in L2 (Ω, B, P).
Proof. Functions of the form φ(x(t1 ), . . . , x(tk ) where φ runs over continuous functions is clearly dense in L2 (Ω, Ft1 ,t2 ,...,tk , P) and
[ [
L2 (Ω, Ft1 ,...,tk , P)
k t1 ,...,tk
53
8. Blumenthal’s Zero-One Law
54
is clearly dense in L2 (Ω, B, P).
55
Proof of zero-one law. Let
Ht = L2 (Ω, Ft , P), H = L2 (Ω, B, P), H0+ =
\
Ht .
t>0
Clearly H0+ = L2 (Ω, F0+ , P).
Let πt : H → Ht be the projection. Then πt f → π0+ f ∀ f in H.
To prove the law it is enough to show that H0+ contains only constants,
which is equivalent to π0+ f = constant ∀ f in H. As π0+ is continuous
and linear it is enough to show that π0+ φ = const ∀φ of the Lemma 2:
π0+ φ = Lt πt φ = Lt E(φ|t ) by Lemma 1,
t→0
t→0
= Lt E(φ(t1 , . . . , tk )|Ft ).
t→0
We can assume without loss of generality that t < t1 < t2 < . . . < tk .
E(φ(t1 , . . . , tk )|Ft ) =
Z
√
2
φ(y1 , . . . , yk )1/ (2π(t1 − t))e−|y1 −Xt (w)| /2(t1 −t) . . .
−|yk −yk−1 |2
p
. . . 1/ (2π(tk − tk−1 ))e 2(tk −tk−1 ) dy1 . . . dyk .
Since X0 (w) = 0 we get, as t → 0,
π0+ φ = constant.
This completes the proof.
APPLICATION. Let α ≥ 1 A = {w :
For, if 0 < s < 1, then
56
as
Rs
0
R1
s
R1
0
|w(t)|/tα < ∞}. Then A ∈ F0+ .
|w(t)|/tα < ∞. Therefore w ∈ A or not according
|w(t)|/tα dt converges or not. But this convergence can be asserted
55
by knowing the history of w upto time s. Hence A ∈ Fs . Blumenthal’s
law implies that
Z1
α
|w(t)|/t dt < ∞ a.e.w., or,
0
Z1
|w(t)|/tα dt = ∞ a.e.w.
0
A precise argument can be given along the following lines. If 0 <
s < 1,
A = {w :
Zs
|w(t)|/tα < ∞}
0
= {w : sup In,s (w) < ∞}
where In,s (w) is the lower Riemannian sum of |w(t)n |/tα corresponding
to the partition {0, s/n, . . . , s} and each In,s ∈ Fs .
9. Properties of Brownian
Motion in One Dimension
WE NOW PROVE the following.
57
Lemma . Let (Xt ) be a one-dimensional Brownian motion. Then
(a) P(lim Xt = ∞) = 1; consequently P(lim Xt < ∞) = 0.
(b) P(lim Xt = −∞) = 1; consequently P(lim Xt > −∞) = 0.
(c) P(lim Xt = −∞); lim Xt = ∞) = 1.
SIGNIFICANCE. By (c) almost every Brownian path assumes each
value infinitely often.
Proof.
∞
\
{lim Xt = ∞} =
(lim Xt > n)
n=1
∞
\
=
( lim Xθ > n) (by continuity of Brownian paths)
n=1
θ rational
First, note that
"
#
"
#
P0 sup X(s) ≤ n = 1 − P0 sup X(s) > n
0≤s≤t
0≤s≤t
57
9. Properties of Brownian Motion in One Dimension
58
Z∞
√
= 1 − 21/ (2πt)
√
= (2/πt)
Zn
2 /2t
e−y
dy
n
2 /2t
e−y
dy.
0
Therefore, for any x0 and t,


"
#


P sup X(s) ≥ n|X(t0 ) = x0 = P0  sup X(s) ≥ n − x0 
t0 ≤s≤t
0≤s≤t−t0
(independent increments) which tends to 1 as t → ∞. Consequently,
#
#
"
"
P0 sup X(t) ≥ n = EP sup X(t) ≥ n|X(t0 )
t≥t0
t≥t0
= E1 = 1.
58
In other words,
"
#
P0 lim supX(t) ≥ n = 1
t→∞
for every n. Thus
P(lim Xt = ∞) = 1.
(b) is clear if one notes that w → −w leaves the probability invariant.
(c) P(lim Xt = ∞, lim Xt = −∞)
= P(lim Xt = ∞) − P(lim Xt > −∞, lim Xt = ∞).
≥ 1 − P(lim Xt > −∞)
= 1.
Corollary . Let (Xt ) be a d-dimensional Brownian motion. Then
P(lim |Xt | = ∞) = 1.
59
Remark . If d ≥ 3 we shall see later that P( Lt |Xt | = ∞) = 1. i.e.
t→∞
almost every Brownian path “wanders” off to ∞.
Theorem . Almost all Brownian paths are of unbounded variation in
any interval.
Proof. Let I be any interval [a, b] with a < b. For n = 1, 2, . . . define
Vn (wQn ) =
n
X
i=1
|w(ti ) − w(ti−1 )| (ti = a + (b − a)i/n, i = 0, 1, 2, . . . n),
The variation corresponding to the partioin Qn dividing [a, b] into n 59
equal parts. Let
Un (w, Qn ) =
n
X
i=1
|(w(ti ) − w(ti−1 )|2 .
If
An (w, Qn ) sup |w(ti ) − w(ti−1 )|,
1≤i≤n
then
An (w, Qn )Vn (w, Qn ) ≥ Un (w, Qn ).
By continuity Lt An (w, Qn ) = 0.
n→∞
Claim. Lt E[(Un (w, Qn ) − (b − a))2 ] = 0.
n→∞
Proof.
E[(Un − (b − a))2 ]
2



n





X
2
−
X
)
−
(b
−
a/n)]
[(X
= E

t
t
j−1
j





 j=1
X
E[( (Z 2j − b − a/n))2 ], Z j = Xt j − Xt j−1 ,
= nE[(Z12 − b − a/n)2 ]
9. Properties of Brownian Motion in One Dimension
60
(because Z j are independent and identically distributed).
= n[E(Z14 ) − (b − a/n)2 ] = 2(b − a/n)2 → 0.
Thus a subsequence Uni → b − a almost everywhere. Since Ani → 0
it follows that Vni (w, Qn ) → ∞ almost everywhere. This completes the
proof.
60
Note . {w : w is of bounded variation on [a, b]} can be shown to be
measurable if one proves
Exercise . Let f be continuous on [a, b] and define Vn ( f, Qn ) as above.
Show that f is of bounded variation on [a, b] iff sup Vn ( f, Qn ) < ∞.
n=1,2,...
Theorem . Let t be any fixed real number in [0, ∞), Dt = {w : w is
differentiable at t}. Then P(Dt ) = 0.
Proof. The measurability of Dt follows from the following observation:
if f is continuous then f is differentiable at t if and only if
Lt
r→0
r rational
f (t + r) − f (t)
.
r
exists. Now
Dt =
∞
[
m=1
w:|
w(t + h) − w(t)
| ≤ M, for all h , 0, rational}
h
and
√
M h
Z
1
Xt+h − Xt
2/2
| ≤ M ∀h ∈ Q, h , 0 ≤ 2 inf
P w:|
√
e−|y| dy = 0
h
h
(2π)
0
Remark. A stronger result holds:


[ 

P  Dt  = 0.
t≥0
61
Hint:
[
0≤t≤1
Dt
!
!
)
(
k−1
71
k
−w
|≤
w:w
n
n
n
m=1 n−m i=1 k=i+1,i+2,i+3
[[
=1
n+2
[
and


P 
(
!
!
)
k
k−1
71
w:w
−w
|≤
n
n
n
i=1 k=i+1,...,i+3
n+2
[

√ 
const/ n
This construction is due to A. Dvoretski, P. Erdos & S. Kakutani.
10. Dirichlet Problem and
Brownian Motion
LET G BE ANY bounded open set in Rd . Define the exit time τG (w) as 61
follows:
τG (w) = {inf t : w(t) < G}.
If w(0) ∈ G, τG (w) = 0; if w(0) ∈ G, τG (w) is the first time w escapes
G or, equivalently, it is the first time that w hits the boundary ∂G of G.
Clearly τG (w) is a stopping time. By definition XτG (w) ∈ ∂G, ∀w and
XτG is a random variable. We can define a Borel probability measure on
∂G by
πG (x, Γ) = P x (XτG ∈ Γ)
= probability that w hits Γ.
If f is a bounded, real-valued measurable funciton defined on ∂G,
we define
Z
f (y)πG (x, dy)
u(x) = E x ( f (XτG )) =
∂G
where
E x = E Px .
In case G is a sphere centred around x, the exact form of πG (x, Γ) is
computable.
Theorem . Let S = S (0; r) = {y ∈ Rd : |y| < r}. Then
surface area of Γ
.
πS (0, r) =
surface area of S
63
10. Dirichlet Problem and Brownian Motion
64
62
Proof. The distributions {Ft1 ,...,tk } defining Brownian motion are invariant under rotations. Thus πS (0, ·) is a rotationally invariant probability
measure. The result follows from the fact that the only probability measure (on the surface of a sphere) that is invariant under rotations is the
normalised surface area.
Theorem . Let G be any bounded region, f a bounded measurable real
valued function defined on ∂G. Define u(x) = E x ( f (XτG )). Then
(i) u is measurable and bounded;
(ii) u has the mean value property; consequently,
(iii) u is harmonic in G.
Proof.
(i) To prove this, it is enough to show that the mapping x →
P x (A) is measurable for every Borel set A.
Let C = {A ∈ B : x → P x (A) is measurable}
d
d
It is clear that π−1
t1 ,...,tk (B) ∈ C , ∀ Borel set B in R × · · · × R . As
C is a monotone class C = B.
(ii) Let S be any sphere with centre at x, and S ⊂ G. Let τ = τS
denote the exit time through S . Clearly τ ≤ τG . By the strong
Markov property,
u(Xτ ) = E( f (XτG )|Fτ ).
Now
u(x) = E x ( f (XτG )) = E x (E( f (XτG ))|Fτ )
Z
u(y)πS (x, dy)
= E x (u(Xτ )) =
1
=
|S |
63
Z
∂S
∂S
u(y)dS ; |S | = surface area of S .
65
(iii) is a consequence of (i) and (ii). (See exercise below).
Exercise ′ . Let u be a bounded measurable function in a region G satisfying the mean value property, i.e.
Z
1
u(y)dS
u(x) =
|S |
∂S
for every sphere S G. Then
(i) u(x) =
1 R
u(y)dy.
v ∈ lS S
(ii) Using (i) show that u is continuous.
(iii) Using (i) and (ii) show that u is harmonic.
We shall now solve the boundary value problem under suitable conditions on the region G.
Theorem . Let G, f , u be as in the previous theorem. Further suppose
that
(i) f is continuous;
(ii) G satisfies the exterior cone condition at every point of ∂G, i.e.
for each y ∈ ∂G there exists a cone Ch with vertex at the point y
of height h and such that Ch − {y} ⊂ exterior of G. Then
lim u(x) = f (y), ∀y ∈ ∂G.
x→y,x∈G
64
Proof.
Step 1. Py {w : w(0) = y, w remains in G for some positive time} = 0.
Let An = {w : w(0) = y, w(s) ∈ G for 0 ≤ s ≤ 1/n},
∞
∞
S
T
Bn = Ω − An , A =
An , B =
Bn .
n=1
n=1
10. Dirichlet Problem and Brownian Motion
66
As An ’s are increasing, Bn ’s are decreasing and Bn ∈ F1/n ; so that
B ∈ F0+ . We show that P(B) > 0, so that by Bluementhal’s zero-one
law, P(B) = 1, i.e. P(A) = 0.
Py (B) = lim Py (Bn ) ≥ lim Py {w : w(0) = y, w(
n→∞
n→∞
1
) ∈ Ch − {y}}
2n
Thus
Py (b) ≥ lim
Z
√
1/ (2π/2n)d exp(−|z − y|2 /2/2n)dz
Ch −{y}
=
Z
√
2
1/ (2π)e−|y| /2 dy,
C∞
where C∞ is the cone of infinite height obtained from Ch . Thus Py (B) >
0.
Step 2. If C is closed then the mapping x → P x (C) is upper semicontinuous.
For, denote by XC the indicator function of C. As C is closed (in
a metric space) ∃ a sequence of continuous functions fn decreasing to
XC such that 0 ≤ fn ≤ 1. Thus E x ( fn ) decreases to E x (XC ) = P x (C).
Clearly x → E x (Fn ) is continuous. The result follows from the fact that
the infimum of any collection of continuous functions is upper semicontinuous.
65
Step 3. Let δ > 0,
N(y; δ) = {z ∈ ∂G : |z − y| < δ},
Bδ = {w : w(0) ∈ G, XτG (w) ∈ ∂G − N(y; δ)},
i.e. Bδ consists of trajectories which start at a point of G and escape for
the first time through ∂G at a point not in N(y; δ). If Cδ = Bδ , then
Cδ ∩ {w : w(0) = y} ⊂ A ∩ {w : w(0) = y}
where A is as in Step 1.
67
For, suppose w ∈ Cδ ∩{w : w(0) = y}. Then there exists wn ∈ Bδ such
that wn → w uniformaly on compact sets. If w < A∩{w : w(0) = y} there
exists ǫ > 0 such that w(t) ∈ G ∀t in (0, ǫ]. Let δ∗ = inf d(w(t), G −
0≤t≤ǫ
N(y, δ)). Then δ∗ > 0. If tn = τG (wn ) and tn does not converge to 0,
then there exists a subsequence, again denoted by tn , such that tn ≥ kǫ >
0 for some k ∈ (0, 1). Since wn (kǫ) ∈ G and wn (kǫ), w(kǫ) ∈ G, a
contradiction. Thus we can assume that tn converges to 0 and also that
ǫ ≥ tn ∀n, But then
|wn (tn ) − w(tn )| ≥ δ∗ .
(*)
However, as wn converges to w uniformly on [0, ǫ],
wn (tn ) − w(tn ) → w(0) − w(0) = 0
contradicting (*). Thus w ∈ A{w : w(0) = y}.
Step 4.
lim P x (Bδ ) = 0.
x→y,x∈G
For,
66
lim P x (Bδ ) ≤ lim P x (Cδ ) ≤ Py (Cδ ) (by Step 2)
x→y
x→y
= Py (Cδ ∩ {w : w(0) = y})
≤ Py (A) (by Step 3)
= 0.
Step 5.
|u(x) − f (y)| = |
≤
Z
Ω−Bδ
≤
Z
Z
Ω
f (XτG (w))dP x (w) −
Z
Ω
| f (XτG (w)) − f (y)|dP x (w) + |
f (y)dP x (w)|
Z
( f (XτG (w)) − f (y))dP x (w)|
Bδ
| f (XτG (w)) − f (y)|dP x (w) + 2|| f ||∞ P x (Bδ )
Ω−Bδ
and the right hand side converges to 0 as x → y (by Step 4 and the fact
that f is continuous). This proves the theorem.
10. Dirichlet Problem and Brownian Motion
68
Remark. The theorem is local.
Theorem . Let G = {y ∈ Rd : δ < |y| < R}, f any continuous function
on ∂G = {|y| = δ} ∩ {|y| = R}. If u is any harmonic function in G with
boundary values f , then u(x) = E x ( f (XτG )).
Proof. Clearly G has the exterior cone property. Thus, if
v(x) = E x ( f (XτG )),
then v is harmonic in G and has boundary values f (by the previous
theorem). The result follows from the uniqueness of the solution of the
Dirichlet problem for the Laplacian operator.
The function f = 0 on |y| = R and f = 1 on |y| = δ is of speR,0
cial interest. Denote by ∪δ,1
the corresponding solution of the Dirichlet
problem.
67
Exercise.
(i) If d = 2 then
R,0
Uδ,1
(x) =
(ii) If d ≥ 3 then
log R − log |x|
, ∀x ∈ G.
log R − log δ
R,0
Uδ,1
(x) =
Case (i): d = 2. Then
|x|−n+2 − R−n+2
.
δ−n+2 − R−n+2
log R − log |x|
R,0
= Uδ,1
(x).
log R − log δ
Now,
E x ( f (XτG )) =
Z
πG (x, dy) = P x (|XτG | = δ),
|y|=δ
i.e.
log R − log |x|
= P x (|XτG | = δ)
log R − log δ
69
Px
(the particle hits |y| = δ before it hits |y| = R).
Fix R and let δ → 0; then 0 = P x (the particle hits 0 before hitting
|y| = R).
Let R take values 1, 2, 3, . . ., then 0 = P x (the particle hits 0 before
hitting any of the circles |y| = N). Recalling that
P x (lim |Xt | = ∞) = 1,
we get
Proposition . A two-dimensional Brownian motion does not visit a
point.
Next, keep δ fixed and let R → ∞, then,
1 = P x (|w(t)| = δ
for some time t > 0).
Since any time t can be taken as the starting time for the Brownian 68
motion, we have
Proposition . Two-dimensional Brownian motion has the recurrence
property.
Case (ii): d ≥ 3. In this case
P x (w : w hits |y| = δ before it hits |y| = R)
= (1/|x|n−2 − 1/Rn−2 )/(1/δn−2 − 1/Rn−2 ).
Letting R → ∞ we get
P x (w : w hits |y| = δ) = (δ/|x|)n−2
which lies strictly between 0 and 1. Fixing δ and letting |x| → ∞, we
have
Proposition . If the particle start at a point for away from 0 then it has
very little chance of hitting the circle |y| = δ.
If |x| ≤ δ, then
P(w hits S δ ) = 1 where S δ = {y ∈ Rd : |y| = δ}.
10. Dirichlet Problem and Brownian Motion
70
Let
Vδ (x) = (δ/|x|)n−2 for |x| ≥ δ.
In view of the above result it is natural to extend Vδ to all space
by putting Vδ (x) = 1 for |x| ≤ δ. As Brownian motion has the Markov
property
=
69
Z
P x {w : w hits S δ after time t}
√
Vδ (y)1/ (2πt)d exp −|y|2 /2t dy → 0 as t → +∞.
Thus P(w hits S δ for arbitrarily large t) = 0. In other words, P(w :
lim |w(t)| ≥ δ) = 1. As this is true ∀δ > 0, we get the following
t→∞
important result.
Proposition . P( lim |w(t)| = ∞) = 1,
t→∞
i.e. for d ≥ 3, the Brownian particle wander away to +∞.
11. Stochastic Integration
LET {Xt : t ≥ 0} BE A one-dimensional Brownian motion. We want 70
R∞
first to define integrals of the type f (s)dX(s) for real functions f ∈
0
L1 [0, ∞). If X(s, w) is of bounded variation almost everywhere then we
R∞
can give a meaning to f (s)dX(s, w) = g(w). However, since X(s, w)
0
is not bounded variation almost everywhere, g(w) is not defined in the
usual sense.
R∞
In order to define g(w) = f (s)dX(s, w) proceed as follows.
0
Let f be a step function of the following type:
f =
n
X
i=1
ai X[ti ,ti+1 ) , 0 ≤ t1 < t2 < . . . < tn+1 .
We naturally define
g(w) =
Z∞
f (s)dX(s, w) =
n
X
ai (Xti+1 (w) − Xti (w))
n
X
ai (w(ti+1 ) − w(ti )).
i=1
0
=
i=1
g satisfies the following properties:
(i) g is a random variable;
71
11. Stochastic Integration
72
(ii) E(g) = 0; E(g2 ) =
P
a2i (ti+1 − ti ) = || f ||2 .
This follows from the facts that (a) Xti+1 − Xti is a normal random
variable with mean 0 and variance (ti+1 − ti ) and (b) Xti+1 − Xti are independent increments, i.e. we have

 ∞

 ∞

 Z

Z




f dX|2  = || f ||22 .
E  f dX  = 0, E |




0
0
71
Exercise 1. If
f =
g=
n
X
i=1
m
X
i=1
ai X[ti ,ti+1 ) , 0 ≤ t1 < . . . < tn+1 ,
bi X[si ,si+1 ) , 0 ≤ s1 < . . . < sm+1 ,
Show that
Z∞
( f + g)dX(s, w) =
0
and
Z∞
f dX(s, w) +
0
Z∞
(α f )dX(s, w) = α
0
Z∞
gdX(s, w)
0
Z∞
f dX(s, w), ∀α ∈ R.
0
Remark. The mapping f →
R∞
f dX is therefore a linear L2R -isometry of
0
the space S of all simple functions of the type
n
X
i=1
into L2 (Ω, B, P).
ai X[ti ,ti+1 ) , (0 ≤ t1 < . . . < tn+1 )
73
Exercise 2. Show that S is a dense subspace of L2 [0, ∞).
Hint: Cc [0, ∞), i.e. the set of all continuous functions with compact support, is dense in L2 [0, ∞). Show that S contains the closure of Cc [0, ∞).
Remark . The mapping f →
R∞
f dX can now be uniquely extended as
0
L2 (Ω, B, P).
an isometry of L2 [0, ∞) into
Next we define integrals fo the type
g(w) =
Zt
72
X(s, w)dX(s, w)
0
Put t = 1 (the general case can be dealt with similarly). It seems
natural to define
(*)
Z1
X(s, w)dX(s) =
0
Lt
sup |t j −t j−1 |→0
n
X
j=1
X(ξ j)(X(t j ) − X(t j−1 ))
where 0 = t0 < t1 < . . . < tn = 1 is a partion of [0, 1] with t j−1 ≤ ξ j ≤ t j .
In general the limit on the right hand side may not exist. Even if it
exists it may happen that depending on the choice of ξ j , we may obtain
different limits. To consider an example we choose ξ j = t j and then
ξ j = t j−1 and compute the right hand side of (∗). If ξ j = t j−1 ,
n
X
j=1
Xξ j (Xt j − Xt j−1 ) =
n
X
j=1
Xt j−1 − (Xt j − Xt j−1 )
1X
1X
(Xt j ) − (Xt j−1 ) −
(Xt − Xt j−1 )
=
2 j=1
2 j=1 j
n
n
1
1 2
[X (1) − X 2 (0)] − as n → ∞, and sup |t j − t j−1 | → 0,
2
2
arguing as in the proof of the result that Brownian motion is not of
bounded variation. If ξ j = t j ,
Lt
n
X
n→∞
Sup |t j −t j−1 |→0 j=1
Xt j (Xt j − Xt j−1 ) = 1/2X(1) − 1/2X(0) + 1/2.
11. Stochastic Integration
74
73
Thus we get different answers depending on the choice of ξ j and
hence one has to be very careful in defining the integral. It turns out
that the choice of ξ j = t j−1 is more appropriate in the definition of the
integral and gives better results.
Remark . The limit in (∗) should be understood in the sense of convergence probability.
Exercise 3. Let 0 ≤ a < b. Show that the “left integral” (ξ j = t j−1 ) is
given by
Zb
X 2 (b) − X 2 (a) − (b − a)
L X(s)dX(s) =
2
a
and the “right integral” (ξ j = t j ) is given by
R
Zb
X(s)dX(s) =
X 2 (b) − X 2 (a) + (b − a)
.
2
s
We now take up the general theory of stochastic integration. To
motivate the definitions which follow let us consider a d-dimensional
Brownian motion {β(t) : t ≥ 0}. We have
Z
√
2
E[β(t + s) − β(t) ∈ A|Ft ] =
1/ (2πs)e−|y| /2s dy.
A
Thus
E( f (β(t + s) − β(t))|Ft ] =
Z
√
2/2s
f (y)1/ (2πs)e−|y| dy.
Z
√
2
eiu.y 1/ (2πs)e−|y| /2s dy
In particular, if f (x) = eix.u ,
iu
E[e (β(t + s) − β(t))|Ft ] =
=e
74
Thus
−s|u|2
2
.
75
2 /2
E[eiu.β(t+s) |Ft ] = eiu.β(t) e−s|u|
,
or,
2
E[eiu.β(t+s)+(t+s)|u|
/2
2
|Ft ] = eiu.β(t)+t|u|
/2
.
Replacing iu by θ we get
2 /2
E[eθ.β(s)−|sθ|
2/2
| Ft ] = eθ.β(t)−t|θ| , s > t, ∀θ.
2
It is clear that eθ.β(t)−t|θ| /2 is Ft -measurable and a simple calculation
gives
2
E(eθ.β(t)−|θ| t/2| ) < ∞ ∀θ.
We thus have
Theorem . If {β(t) : t ≥ 0} is a d-dimensional Brownian motion then
exp[θ.β(t) − |θ|2 t/2] is a Martingale relative to Ft , the σ-algebra generated by (β(s) : s ≤ t).
Definition. Let (Ω, B, P) be a probability space (Ft )t≥0 and increasing
S
family of sub-σ-algebras of F with F = σ( Ft ).
t≥0
Let
(i) a : [0, ∞)×Ω → [0, ∞) be bounded and progressively measurable;
(ii) b : [0, ∞) × Ω → R be bounded and progressively measurable;
(iii) X : [0, ∞)×Ω → R be progressively measurable, right continuous
on [0, ∞), ∀ w ∈ Ω, and continous on [0, ∞) almost everywhere
on Ω;
θX(t,w)−θ
(iv) Zt (w) = e
Rt
0
2
b(s,w)ds− θ2
Rt
a(s,w)ds
0
be a Martingale relative to (Ft )t≥0 .
Then X(t, w) is called an Ito process corresponding to the parameters
b and a and we write Xt ∈ I[b, a].
N.B. The progressive measurability of X ⇒ Xt is Ft -measurable.
75
11. Stochastic Integration
76
Example . If {β(t) : t ≥ 0} is a Brownian motion, then X(t, w) = βt (w)
is an Ito process corresponding to parameters 0 and 1. (i) and (ii) are
obvious. (iii) follows by right continuity of βt and measurability of βt
relative to Ft and (iv) is proved in the previous theorem.
Exercise 4. Show that Zt (w) defined in (iv) is Ft -measurable and progressively measurable.
[Hint:
(i) Zt is right continuous.
(ii) Use Fubini’s theorem to prove measurability].
Remark . If we put Y(t, w) = X(t, w) −
Rt
b(s, w)ds then Y(t, w) is pro-
0
gressively measurable and Y(t, w) is an Ito process corresponding to
the parameters 0, a. Thus we need only consider integrals of the type
Rt
f (s, w)dY(s, w) and define
0
Zt
0
f (s, w)dX(s, w) =
Zt
f (s, w)dY(s, w) +
0
Zt
f (s, w)b(s, w)ds.
0
(Note that formally we have dY = dX − dbt).
76
Lemma . If Y(t, w) ∈ I[0, a], then
Y 2 (t, w) −
Y(t, w) and
Zt
a(s, w)ds
0
are Martingales relative to (Ft ).
Proof. To motivate the arguments which follow, we first give a formal
proof. Let
2
θY(t,w)− θ2
Yθ (t) = e
Rt
0
a(s,w)ds
.
77
Yθ − 1
is a Martingale, ∀θ.
Then Yθ (t) is a martingale, ∀θ. Therefore
θ
Hence (formally),
Yθ − 1
= Yθ′ |θ=0
lim
θ→0
θ
is a Martingale.
Step 1. Y(t, ·) ∈ Lk (Ω, F , P), k = 0, 1, 2, . . . and ∀t. In fact, for every
real θ, Yθ (t) is a Martingale and hence E(Yθ ) < ∞. Since a is bounded
this means that
E(eθY(t,·) ) < ∞, ∀θ.
Taking θ = 1 and −1 we conclude that E(e|Y| ) < ∞ and hence
E(|Y|k ) < ∞, ∀k = 0, 1, 2, . . .. Since Y is an Ito process we also get

k 
Rt

θ2
 Y(t,·)− 2 ads  
0
  < ∞
sup E e
 

|θ|≤α 
∀k and for every α > 0.
Step 2. Let Xθ (t) = [Y(t, ·) − θ
Define
φA (θ) =
Z
Rt
0
ads]Y(t) =
d
Yθ (t, ·).
dθ
(Xθ (t, ·) − Xθ (s, ·))dP(w)
A
where t > s, A ∈ Fs . Then
Zθ2
φA (θ)dθ =
θ1
Zθ2 Z
θ1
[Xθ (t, ·) − Xθ (S , ·)]dP(w)dθ.
A
Since a is bounded, sup E([Yθ (t, ·)]k ) < ∞, and E(|Y|k ) < ∞, ∀k; we
|θ|≤α
can use Fubini’s theorem to get
Zθ2
θ1
φA (θ)dθ =
Z Zθ2
A θ1
[Xθ (t, ·) − Xθ (s, ·)]dθ dP(w).
77
11. Stochastic Integration
78
or
Zθ2
φA (θ)dθ =
θ1
Z
Yθ2 (t, ·) − Yθ1 (t, ·)dP(w) −
A
Z
Yθ1 (s, ·) − Yθ1 (s, ·)dP(w).
A
Let A ∈ Fs and t > s; then, since Y is a Martingale,
Zθ2
φA (θ)dθ = 0.
θ1
This is true ∀θ1 < θ2 and since φA (θ) is a continuous function of θ,
we conclude that
φA (θ) = 0, ∀θ.
In particular, φA (θ) = 0 which means that
Z
Z
Y(t, ·)dP(w) =
Y(s, ·)dP(w), ∀A ∈ Fs , t > s,
A
78
A
i.e., Y(t) is a Martingale relative to (Ω, Ft , P).
To prove the second part we put
Zθ (t, ·) =
and
ψA (θ) =
Z
d2
Yθ (t)
dθ2
{Zθ (t, ·) − Zθ (s, ·)}dP(w).
A
Then, by Fubini,
Zθ2
θ1
ψA (θ)dθ =
Z Zθ2
A θ1
Zθ (t, ·) − Zθ (s, ·)dθ dP(w).
79
or,
Zθ2
ψA (θ)dθ = φA (θ2 ) − φA (θ1 )
θ1
= 0 if A ∈ Fs , t > s.
Therefore
ψA (θ) = 0, ∀θ.
In particular, ψA (θ) = 0 implies that
2
Y (t, w) −
Zt
a(s, w)ds
0
is an (Ω, Ft , P) Martingale. This completes the proof of lemma 1.
Definition. A function θ : [0, ∞) × Ω → R is called simple if there exist
reals s0 , s1 , . . . , sn , . . .
0 ≤ s0 < s1 < . . . < sn . . . < ∞,
sn increasing to +∞ and
θ(s, w) = θ j (w)
if s ∈ [s j , s j+1 ), where θ j (w) is Fs j -measurable and bounded.
Definition . Let θ : [0, ∞) × Ω → R be a simple function and Y(t, w) ∈
I[0, a]. We define the stochastic integral of θ with respect to Y, denoted
Zt
0
by
ξ(t, w) =
Zt
0
θ(s, w)dY(s, w)
θ(s, w)dY(s, w)),
79
11. Stochastic Integration
80
=
k
X
j=1
θ j−1 (w)[Y(s j , w) − Y(s j−1 , w)] + θk (w)[Y(t, w) − Y(sk , w)].
Lemma 2. Let σ : [0, ∞) × Ω → R be a simple function and Y(t, w) ∈
I[0, a]. Then
ξ(t, w) =
Zt
σ(s, w)dY(s, w) ∈ I[0, aσ2 ].
0
Proof.
(i) By definition, σ is right continuous and σ(t, w) is Ft measurable; hence it is progressively measurable. Since a is progressively measurable and bounded
aσ2 : [0, ∞) × Ω → [0, ∞)
is progressively measurable and bounded.
80
(ii) From the definition of ξ it is clear that ξ(t, ·) is right continuous,
continous almost everywhere and Ft -measurable therefore ξ is
progressively measurable.
2
[θξ(t,w)− θ2
(iii) Zt (w) = e
Rt
aσ2 ds]
0
is clearly Ft -measurable ∀θ. We show that
E(Zt ) < ∞, ∀t and E(Zt2 |Ft1 ) = Zt1 if t1 < t2 .
We can assume without loss of generality that θ = 1 (if θ , 1 we
replace σ by θσ). Therefore
Rt
[ξ(t,w)− aσ2 ds]
Zt (w) = e
0
.
81
Since a and σ are bounded uniformly on [0, t], it is enough to show that
E(eξ(t,w) ) < ∞. By definition,
ξ(t, w) =
k
X
j=1
θ j−1 (w)[Y(s j , w) − Y(s j−1 , w)] + θk (w)(Y(t, w) − Y(sk , w)).
The result E(eξ(t,w) ) < ∞ will follow from the generalised Holder’s inequality provided we show that
E(eθ(w)[Y(t,w)−Y(s,w)] ) < ∞
for every bounded function θ which is Fs -measurable. Now
E(eθ[Y(t,·)−Y(s,·)] |Fs ) =
constant for every constant θ, since Y ∈ I[0, a]. Therefore
E(eθ(w)[Y(t,·)−Y(s,·)] |Fs ) = constant
for every θ which is bounded and Fs -measurable. Thus
E(eθ(w)[Y(t,·)−Y(s,·)] ) < ∞.
This proves that E(Zt (w)) ∈ ∞.
Finally we show that
if t1 < t2 .
E(Zt2 |Ft1 ) = Zt1 (w),
Consider first the case when t1 and t2 are in the same interval
[sk , sk+1 ).
Then
ξ(t2 , w) = ξ(t1 , w) + σk (w)[Y(t2 , w) − Y(t1 , w)] (see definition),
Zt2
Zt2
Zt1
2
2
aσ (s, w)ds + aσ2 (s, w)ds.
aσ (s, w)ds =
0
0
t1
81
11. Stochastic Integration
82
Therefore
θ2
E(Zt2 (w)|Ft1 ) = Zt1 (w)E(exp[θσk (w)[Y(t2 , w) − Y(t1 , w)] −
2
Zt2
aσ2 ds)|Ft1 )
t1
as Y ∈ I[0, a].
(*)
θ2
E(exp[θ(Y(t2 , w) − T (t1 , w)) −
2
Zt2
a(s, w)ds]|Ft1 ) = 1
t1
and since σk (w) is Ft1 -measurable (∗) remains valid if θ is replaced by
θσk . Thus
E(Zt2 |Ft1 ) = Zt1 (w).
The general case follows if we use the identity
E(E(X|C1 )|C2 ) = E(X|C2 ) for
C2 ⊂ C1 .
Thus Zt is a Martingale and ξ(t, w) ∈ I[0, aσ2 ].
82
Corollary .
(i) ξ(t, w) is a martingale; E(ξ(t, w)) = 0;
(ii) ξ 2 (t, w) −
Rt
aσ2 ds
0
is a Martingale with
Zt
E(ξ (t, w)) = E( aσ2 (s, w)ds.
2
0
Proof. Follows from Lemma 1.
Lemma 3. Let σ(s, w) be progressively measurable such that for each
t,
Zt
E( σ2 (s, w)ds) < ∞.
0
83
Then there exists a sequence σn (s, w) of simple functions such that

 t

Z


lim E  |σn (s, w) − σ(s, w)|2 ds = 0.
n→∞ 

0
Proof. We may assume that σ is bounded, for if σN = σ for |σ| ≤ N
and 0 if |σ| > N, then σn → σ, ∀(s, w) ∈ [0, t] × Ω. σN is progressively
measurable and |σn − σ|2 ≤ 4|σ|2 . By hypothesis σ ∈ L([0, t] : Ω).
Rt
Therefore E( |σn −σ|ds) → 0, by dominated convergence. Further,
0
we can also assume that σ is continuous. For, if σ is bounded, define
σh (t, w) = 1/h
Zt
σ(s, w)ds.
(t−h)v0
σn is continuous in t and Ft -measurable and hence progressively measurable. Also by Lebesgue’s theorem
σh (t, w) → σ(t, w),
as
h → 0, ∀t, w.
Since σ is bounded by C, σh is also bounded by C. Thus
Zt
E( |σh (s, w) − σ(s, w)|2 ds) → 0.
0
(by dominated convergence). If σ is continuous, bounded and progressively measurable, then
!
[ns]
σn (s, w) = σ
,w
n
is progressively measurable, bounded and simple. But
Lt σn (s, w) = σ(s, w).
n→∞
83
11. Stochastic Integration
84
Thus by dominated convergence
 t

Z



E  |σn − σ|2 ds → 0


as n → ∞.
0
Theorem . Let σ(s, w) be progressively measurable, such that
Zt
E( σ2 (s, w)ds) < ∞
0
for each t > 0. Let (σn ) be simple approximations to σ as in Lemma 3.
Put
Zt
ξn (t, w) =
σn (s, w)dY(s, w)
0
where Y ∈ I[0, a]. Then
(i) Lt ξn (t, w) exists uniformly in probability, i.e. there exists an aln→∞
most surely continuous ξ(t, w) such that
!
Lt P sup |ξn (t, w) − ξ(t, w)| ≥ ǫ = 0
n→∞
0≤t≤T
for each ǫ > 0 and for each T . Moreover, ξ is independent of the
sequence (σ0 ).
84
(ii) The map σ → ξ is linear.
(iii) ξ(t, w) and ξ 2 (t, w) −
Rt
aσ2 ds are Martingales.
0
(iv) If σ is bounded, ξ ∈ I[0, aσ2 ].
85
Proof.
(i) It is easily seen that for simple functions the stochastic
integral is linear. Therefore
(ξn − ξm )(t, w) =
Zt
(σn − σm )(s, w)dY(s, w).
0
Since ξn − ξm is an almost surely continuous martingale
!
1
P sup |ξn (t, w) − ξm (t, w)| ≥ ǫ ≤ 2 E[(ξn − ξm )2 (T, w)].
ǫ
0≤t≤T
This is a consequence of Kolmogorov inequality (See Appendix).
Since
Zt
2
(ξn − ξm ) − a(σn − σm )2 ds
0
is a Martingale, and a is bounded,
(*)
 T

Z



2
2
E[(ξn − ξm ) (T, w)] = E  (σn − σm ) a ds .


0
 T

Z

1 

≤ const 2 E  (σn − σm )2 ds .

ǫ 
0
Therefore
Lt E[(ξn − ξm )2 (T, w)] = 0.
n,m→∞
Thus (ξn −ξm ) is uniformly Cauchy in probability. Therefore there
exists a progressively measurable ξ such that
!
Lt P sup |ξn (t, w) − ξ(t, w)| ≥ ǫ = 0, ∀ǫ > 0, ∀T.
n→∞
0≤t≤T
It can be shown that ξ is almost surely continuous.
11. Stochastic Integration
86
If (σn ) and (σ′n ) are two sequences of simple functions approxi- 85
mating σ, then (∗) shows that
E[(ξn − ξn′ )2 (T, w)] → 0.
Thus
Lt ξn = Lt ξn′ ,
n
n
i.e. ξ is independent of (σn ).
(ii) is obvious.
(iii) (*) shows that ξn → ξ in L and therefore ξn (t, ·) → ξ(t, ·) in L1 for
each fixed t. Since ξn (t, w) is a martingale for each n, ξ(t, w) is a
martingale.
(iv) ξn2 (t, w) −
Rt
aσ2n is a martingale for each n.
0
Since ξn (t, w) → ξ(t, w) in L2 for each fixed t and
ξn2 (t, w) → ξ 2 (t, w) in
L1
for each fixed t.
For ξn2 (t, w) − ξ 2 (t, w) = (ξn − ξ)(ξn + ξ) and using Hölder’s inequality, we get the result.
Similarly, since
σn → σ in L2 ([0, t] × Ω),
σ2n → σ2 in L1 ([0, t] × Ω),
and because a is bounded aσ2n → aσ2 in L1 ([0, t]×Ω). This shows
Rt
that ξn2 (t, w) − aσ2n ds converges to
0
2
ξ (t, w) −
Zt
0
aσ2 ds
87
for each t in L1 . Therefore
2
ξ (t, w) −
Zt
aσ2 ds
0
is a martingale.
86
(v) Let σ be bounded. To show that ξ ∈ I[0, σ2 ] it is enough to show
that
t
2
θξ(t,w)− θ2
e
R
aσ2 ds
0
is a martingale for each θ, the other conditions being trivially satisfied. Let
t
2
θξn (t,w)− θ2
Zn (t, w) = e
R
aσ2n ds
0
We can assume that |σn | ≤ C if |σ| ≤ C (see the proof of Lemma
3).


t
Zt


2 Z
(2θ)


aσ2n ds + θ2 aσ2n ds .
Zn = exp 2θξn (t, w) −


2
0
0
Thus
(**)


t
 2θξn (t,w)− (2θ)2 R aσ2n ds 
2


0
E(Zn ) ≤ const E e
 = const

since Zn is a martingale for each θ. A subsequence Zni converges
to
t
2
θξ(t,w)− θ2
e
R
aσ2 ds
0
almost everywhere (P). This together with (**) ensures uniform
integrability of (Zn ) and therefore
2
θξ(t,w)− θ2
e
Rt
0
aσ2 ds
11. Stochastic Integration
88
is a martingale. Thus ξ is an Ito process, ξ ∈ I[0, aσ2 ].
Definition . With the hypothesis as in the above theorem we define the
stochastic integral
ξ(t, w) =
Zt
σ(s, w)dY(s, w).
0
87
Exercise. Show that d(X + Y) = dX + dY.
Remark . If σ is bounded, then ξ satisfies the hypothesis of the previous theorem and so one can define the integral of ξ with respect to Y.
Further, since ξ itself is Itô, we can also define stochastic integrals with
respect to.
Examples.
1. Let {β(t) : t ≥ 0} be a Brownian motion; then β(t, w) is
progressively measurable (being continuous and Ft -measurable).
Also,
Z Zt
Zt
Zt Z
t
β2 (s)dP ds =
β2 (s)ds dP =
sds =
2
Ω
0
Hence
0
Zt
Ω
0
β(s, w)dβ(s, w)
0
is well defined.
Rt
2. Similarly β(s/2)dβ(s) is well defined.
0
3. However
Zt
β(2s)dβ(s)
0
is not well defined, the reason being that β(2s) is not progressively
measurable.
89
Exercise 5. Show that β(2s) is not progressively measurable.
88
(Hint: Try to show that β(2s) is not Fs -measurable for every s. To show
this prove that Fs , F2s ).
Exercise 6. Show that for a Brownian motion β(t), the stochastic integral
Z1
β(s, ·)dβ(s, ·)
0
is the same as the left integral
L
Z1
0
defined earlier.
β(s, ·)dβ(s, ·)
12. Change of Variable
Formula
WE SHALL PROVE the
89
Theorem . Let σ be any bounded progressively measurable function
and Y be an Ito process. If λ is any progressively measurable function
such that
 t

Z



E  λ2 ds < ∞, ∀t,


0
then
(*)
Zt
λdξ(s, w) =
0
Zt
λ(s, w)σ(s, w)dY(s, w),
Zt
σ(s, w)dY(s, w).
0
where
ξ(t, w) =
0
Proof.
Step 1. Let λ and σ be both simple, with λ bounded. By a refinement
of the partition, if necessary, we may assume that there exist reals 0 =
s0 , s1 , . . . , sn , . . . increasing to +∞ such that λ and σ are constant on
[s j , s j+1 ), say λ = λ j (w), σ = σ j (w), where λ j (w) and σ j (w) are Fs j measurable. In this case (*) is a direct consequence of the definition.
91
12. Change of Variable Formula
92
Step 2. Let λ be simple and bounded. Let (σn ) be a sequence of simple
bounded functions as in Lemma 3. Put
Zt
ξn (t, w) =
σn (s, w)dY(s, w)
0
By Step 1,
Zt
(**)
λdξn =
0
Zt
λσn dY(s, w).
0
90
Since λ is bounded, λσn converges to λσ in L2 ([0, t] × Ω). Hence,
Rt
Rt
by definition, λσn dY(s, w) converges to λσdY in probability.
0
0
Further,
Zt
λdξn (s, w) = λ(s0 , w)[ξn (s1 , w) − ξn (s0 , w)] + · · ·
0
+ · · · + λ(sk , w)[ξn (t, w) − ξn (sk−1 , w)],
where s0 < s1 < . . . . . . is a partition for λ, and ξn (t, w) converges to
ξ(t, w) in probability for every t. Therefore
Zt
λdξn (s, w)
Zt
λdξ(s, w).
0
converges in probability to
0
Taking limit as n → ∞ in (**) we get
Zt
0
λdξ(s, w) =
Zt
0
λσdY(s, w).
93
Step 3. Let λ be any progressively measurable function with
Zt
E( λ2 ds) < ∞, ∀t.
0
Let λn be a simple approximation to λ as in Lemma 3. Then, by Step
2,
(***)
Zt
λn (s, w)dξ(s, w) =
0
Zt
λn (s, w)σ(s, w)dY(s, w).
0
By definition, the left side above converges to
Zt
λ(s, w)dξ(s, w)
0
in probability. As σ is bounded λn σ converges to λσ in L2 ([0, t] × Ω). 91
Therefore


Zt
Zt




P  sup | λn σdY(s, w) − λσ dy(s, w)| ≥ ǫ 

0≤t≤T
0
0

 t

Z


||a||∞ 1/ǫ 2 E  (λn σ − λσ)2 ds


0
(see proof of the main theorem leading to the definition of the stochastic
integral). Thus
Zt
λn σdY(s, w)
0
converges to
Zt
λσ dY(s, w)
0
in probability. Let n tend to + in (***) to conclude the proof.
94
12. Change of Variable Formula
13. Extension to
Vector-Valued Itô Processes
Definition. Let (Ω, F , P) be a probability space and (Ft ) an increasing 92
family of sub σ-algebras of F . Suppose further that
(i)
a : [0, ∞) × Ω → S +d
is a probability measurable, bounded function taking values in the class
of all symmetric positive semi-definite d × d matrices, with real entries;
(ii)
b : [0, ∞) × Ω → Rd
is a progressively measurable, bounded function;
(iii)
X : [0, ∞) × Ω → Rd
is progressively measurable, right continuous for every w and continuous almost everywhere (P);
Z(t, ·) = exp[hθ, X(t, ·)i −
Zt
hθ, b(s, ·)ids
0
(iv)
−
1
2
Zt
hθ, a(s, ·)θids]
0
is a martingale for each θ ∈ Rd , where
hx, yi = x1 y1 + · · · + xd yd , x, y ∈ Rd .
95
13. Extension to Vector-Valued Itô Processes
96
Then X is called an Itô process corresponding to the parameters b
and a, and we write X ∈ I[b, a]
Note.
1. Z(t, w) is a real valued function.
2. b is progressively measurable if and only if each bi is progressively measurable.
3. a is progressively measurable if and only if each ai j is so.
93
Exercise 1. If X ∈ I[b, a], then show that
(i)
(ii)
Y=
d
X
i=1
Xi ∈ I[bi , aii ],
θi Xi ∈ I[hθ, bi, hθ, aθi],
where
θ = (θ1 , . . . , θd ).
(Hint: (ii) (i). To prove (ii) appeal to the definition).
Remark . If X has a multivariate normal distribution with mean µ and
covariance (ρi j ), then Y = hθ, Xi has also a normal distribution with
mean hθ, µi and variance hθ, ρθi. Note the analogy with the above exercise. This analogy explains why at times b is referred to as the “mean”
and a as the “covariance”.
Exercise 2. If {β(t) : t ≥ 0} is a d-dimensional Brownian motion, then
β(t, w) ∈ I[0, I] where I = d × d identity matrix.
Rt
As before one can show that Y(t, ·) = X(t, ·) − b(s, w)ds is an Itô
0
process with parameters 0 and a.
Definition . Let X be a d-dimensional Ito process. σ = (σ1 , . . . , σd ) a
d-dimensional progressively measurable function such that

 t

Z


E  hσ(s, ·), σ(s, ·)i > ds


0
97
is finite or, equivalently,
 t

Z



2
E  σi (s, ·)ds < ∞,


(i = 1, 2, . . . d).
0
94
Then by definition
Zt
hσ(s, ·), dX(s, ·)i =
0
d Z
X
t
i=1 0
σi (s, ·)dXi (s, ·).
Proposition . Let X be a d-dimensional Itô process X ∈ I[b, a] and let
σ be progressively measurable and bounded. If
ξi (t, ·) =
Zt
σi dXi (s, ·),
0
then
ξ = (ξ1 , . . . , ξd ) ∈ I[B, A],
where
B = (σ1 b1 , . . . , σd bd ) and
Proof.
Ai j = σi σ j ai j .
(i) Clearly Ai j is progressively measurable and bounded.
Since a ∈ S +d , A ∈ S +d .
(ii) Again B is progressively measurable and bounded.
(iii) Since σ is bounded, each ξi (t, ·) is an Itô process; hence ξ is progressively measurable, right continuous, continuous almost everywhere (P). It only remains to verify the martingale condition.
Step 1. Let θ = (θ1 , . . . , θd ) ∈ Rd . By hypothesis,
(*)
E(exp[(θ1 X1 + · · · +
θd Xd )|ts
···
Zt
s
(θ1 b1 + · · · + θd bd )du
13. Extension to Vector-Valued Itô Processes
98
1
−
2
Zt X
θi θ j ai j ds]|Fs ) = 1.
0
95
Assume that each σi is constant on [s, t], σi = σi (w) and Fs measurable. Then (*) remains true if θi are replaced by θi θi (w) and since
σi ’s are constant over [s, t], we get
Zt X
Zt
d
E(exp[
θi σi (s, ·)dXi (s, ·) − θi bi σi (s, ·)ds
i=1
0
−
1
2
Zt
0
0
X
θi θ j σi (s, ·)σ j (s, ·)ai j ds]| s )

 s
Zs
Zs
d

Z X


θi σi (s, ·)dXi (s, ·) − hθ, Bidu − 1 hθ, Aθidu .
exp 


0
i=1
0
0
Step 2. Let each σi be a simple function.
By considering finer partitions we may assume that each σi is a step
function,
finest partition
i.e. there exist points s0 , s1 , s2 , . . . , sn , s = s0 < s1 < . . . < sn+1 = t,
such that on [s j , s j+1 ) each σi is a constant and s j -measurable. Then (**)
holds if we use the fact that if C1 ⊃ C2 .
E(E( f |C1 )|C2 ) = E( f |C2 ).
99
96
Step 3. Let σ be bounded, |σ| ≤ C. Let (σ(n) ) be a sequence of simple functions approximating σ as in Lemma 3. (**) is true if σi is replaced by σ(n)
i for each n. A simple verification shows that the expression Zn (t, ·), in the parenthes is on the left side of (**) with σi replaced
by σi(n) , converges to
= Exp
Zt
0
Z(t, ·) =
Zt X
X
θi σi (s, ·)ds −
θi bi σi (s, ·)ds−
i
0
−
1
2
Zt
0
X
i, j
i
θi θ j σi σ j ai j ds
as n → ∞ in probability. Since Zn (t, ·) is a martingale and the functions
σi , σ j , a are all bounded,
sup E(Zn (t, ·)) < ∞.
n
This proves that Z(t, ·) is a martingale.
Corollary . With the hypothesis as in the above proposition define
Z(t) =
Zt
hσ(s, ·), dX(s, ·)i.
0
Then
Z(t, ·) ∈ I[hσ, bi, σaσ∗ ]
where σ∗ is the transpose of σ.
Proof. Z(t, ·) = ξ1 (t, ·) + · · · + ξd (t, ·).
Definition. Let σ(s, w) = (σi j (s, w)) be a n × d matrix of progressively
measurable functions with

 t

Z


E  σ2i j (s, ·)ds < ∞.


0
13. Extension to Vector-Valued Itô Processes
100
If X is a d-dimensional Itô process, we define

 t
d Zt
 X
Z


σi j (s, ·)dX j (s, ·).
 σ(s, ·)dX(s, ·) =
0
j=1 0
i
Exercise 3. Let
Z(t, w) =
Zt
σ(s, ·)dY(s, ·),
0
where Y ∈ I[0, a] is a d-dimensional Itô process and σ is as in the above
definition. Show that
Z(t, ·) ∈ I[0, σaσ∗ ]
is an n-dimensional Ito process, (assume that σ is bounded).
Exercise 4. Verify that

 t

Z


E(|Z(t)|2 ) = E  tr(σaσ∗ )ds .


0
Exercise 5. Do exercise 3 with the assumption that σaσ∗ is bounded.
Exercise 6. State and prove a change of variable formula for stochastic
integrals in the case of several dimensions.
(Hint: For the proof, use the change of variable formula in the one dimensional case and d(X + Y) = dX + dY).
97
14. Brownian Motion as a
Gaussian Process
SO FAR WE have been considering Brownian motion as a Markov pro- 98
cess. We shall now show that Brownian motion can be considered as a
Gaussian process.
Definition. Let X ≡ (X1 , . . . , XN ) be an N-dimensional random variable.
It is called an N-variate normal (or Gaussian) distribution with mean
µ ≡ (µ1 , . . . , µN ) and covariance A if the density function is
!
1
1
1
−1
∗
exp − [(X − µ)A (X − µ) ]
2
(2π)N/2 (det A)1/2
where A is an N × N positive definite symmetric matrix.
Note.
1. E(Xi ) = µi .
2. Cov(Xi , X j ) = (A)i j .
Theorem . X ≡ (X1 , . . . , XN ) is a multivariate normal distribution if and
only if for every θ ∈ RN , hθ, Xi is a one-dimensional Gaussian random
variable.
We omit the proof.
Definition. A stochastic process {Xt : t ∈ I} is called a Gaussian process
if ∀t1 , t2 , . . . , tN ∈ I, (Xt1 , . . . , XtN ) is an N-variate normal distribution.
101
14. Brownian Motion as a Gaussian Process
102
Exercise 1. Let {Xt : t ≥ 0} be a one dimensional Brownian motion.
Then show that
(a) Xt is a Gaussian process.
99
(Hint: Use the previous theorem and the fact that increments are
independent)
(b) E(Xt ) = 0, ∀t, E(X(t)X(s)) = s ∧ t.
Let ρ : [0, 1] = [0, 1] → R be defined by
ρ(s, t) = s ∧ t.
Define K : L2R [0, 1] → L2R [0, 1] by
K f (s) =
Z1
ρ(s, t) f (t)dt.
0
Theorem . K is a symmetric, compact operator. It has only a countable
number of eigenvalues and has a complete set of eigenvectors.
We omit the proof.
Exercise 2. Let λ be any eigenvalue of K and f an eigenvector belonging
to λ. Show that
(a) λ f ′′ + f = 0 with λ f (0) = 0 = λ f ′ (1).
(b) Using (a) deduce that the eigenvalues are given by λn = 4/(2n +
1)2 π2 and the corresponding eigenvectors are given by
√
fn = 2 Sin 1/2[(2n + 1)πt]n = 0, 1, 2, . . . .
Let Z0 , Z1 , . . . , Zn . . . be identically distributed, independent, normal
random variables with mean 0 and variance 1. Then we have
Proposition . Y(t, w) =
∞
P
√
Zn (w) fn (t) λn
n=0
converges in mean for every real t.
103
100
Proof. Let Ym (t, w) =
m
P
√
Zi (w) fi (t) λi . Therefore
i=0
2
E{(Yn+m (t, ·) − Yn (t, ·)) } =
E(||Yn+m (·) − Yn (·)||2 ≤
n+m
X
n+1
n+m
X
fi2 (t)λi ,
n+1
λi → 0.
Remark. As each Yn (t, ·) is a normal random variable with mean 0 and
n
P
variance
λi fi2 (t), Y(t, ·) is also a normal random variable with mean
i=0
zero and variance
∞
P
i=0
λi fi2 . To see this one need only observe that the
limit of a sequence of normal random variables is a normal random variable.
Theorem (Mercer).
ρ(s, t) =
∞
X
i=0
λi fi (t) fi (s), (s, t) ∈ [0, 1] × [0, 1].
The convergence is uniform.
We omit the proof.
Exercise 3. Using Mercer’s theorem show that {Xt : 0 ≤ t ≤ 1} is a
Brownian motion, where
X(t, w) =
∞
X
√
Zn (w) fn (t) λn .
n=0
This exercise now implies that
Z1
0
X 2 (s, w)ds = (L2 − norm of X)2
14. Brownian Motion as a Gaussian Process
104
=
X
λn Zn2 (w),
since fn (t) are orthonormal. Therefore
−λ
E(e
R1
X 2 (s,·)ds
−λ
) = E(e
0
∞
P
n=0
λn Zn2 (w)
)=
101
∞
Y
2
E(e−λλn Zn )
n=0
(by independence of Zn )
=
∞
Y
2
E(e−λλn Z0 )
n=0
as Z0 , Zn . . . are identically distributed. Therefore
−λ
E(e
R1
)=
∞
Y
√
1/ (1 + 2λλn )
=
∞
Y
√
1/ 1 +
X 2 (s,·)ds
0
n=0
n=0
8 8λ
(2n + 1)2 Π2
√
√
= 1/ (cosh) (2λ).
R1
APPLICATION. If F(a) = P( X 2 (s)ds < a), then
0
Z∞
−λa
e
0
= E(e−λ
R1
0
dF(a) =
Z∞
−∞
X 2 (s)ds
e−λa dF(a)
√
√
) = 1/ (cosh) (2λ).
!
15. Equivalent For of Itô
Process
LET (Ω, F , P) BE A probability space with (Ft )t≥0 and increasing fam- 102
ily of sub σ-algebras of F such that σ(U Ft ) = F . Let
t≥0
(i) a : [0, ∞)×Ω → S d+ be a progressively measurable, bounded function taking values in S d+ , the class of all d × d positive semidefinite
matrices with real entries;
(ii) b : [0, ∞) × Ω → Rd be a bounded, progressively measurable
function;
(iii) X : [0, ∞) × Ω → Rd be progressively measurable, right continuous and continuous a.s. ∀(s, w) ∈ [0, ∞) × Ω.
For (s, w) ∈ [0, ∞) × Ω define the operator
L s,w =
d
d
X
1X
∂2
∂
+
.
ai j (s, w)
b j (s, w)
2 i, j=1
∂xi ∂x j j=1
∂x j
For f , u, h belonging to C0∞ (Rd ), C0∞ ([0, ∞) × Rd ) and Cb1,2 ([0, ∞) ×
Rd ) respectively we define Y f (t, w), Zu (t, w), Ph (t, w) as follows:
Y f (t, w) = f (X(t, w)) −
Zt
0
105
(L s,w ( f )(X(s, w))ds,
15. Equivalent For of Itô Process
106
Zu (t, w) = u(t, X(t, w)) −
!
∂u
+ L s,w u (s, X(s, w))ds,
∂s
Zt
0
Ph (t, w) = exp[h(t, X(t, w)) −
1
−
2
103
Z
Zt
!
∂h
+ L s,w h (s, X(s, w)ds−
∂s
0
t
0
ha(s, w)∇x h(s, X(s, w)), ∇x h(s, X(s, w))ids].
Theorem . The following conditions are equivalent.
Rt
Rt
(i) Xθ (t, w) = exp[hθ, X(t, w)i − hθ, b(s, w)ids − hθ, a(s, w)θids]
0
is a martingale relative to (Ω, Ft , P), ∀θ ∈
0
d
R .
(ii) Xλ (t, w) is a martingale ∀λ in Rd . In particular Xiθ (t, w) is a martingale ∀θ ∈ Rd .
(iii) Y f (t, w) is a martingale for every f ∈ C0∞ (Rd )
(iv) Zu (t, w) is a martingale for every u ∈ C0∞ ([0, ∞) × Rd ).
(v) Ph (t, w) is a martingale for every h ∈ Cb1,2 [(0, ∞) × Rd ).
(vi) The result (v) is true for functions h ∈ C 1,2 ([0, ∞) × Rd ) with
linear growth, i.e. there exist constants A and B such that |h(x)| ≤
A|x| + B.
∂2 h
∂h ∂h
,
, and −
which occur under the integral
∂t ∂xi
∂xi ∂x j
sign in the exponent also grow linearly.
The functions
Remark. The above theorem enables one to replace the martingale condition in the definition of an Itô process by any of the six equivalent
conditions given above.
Proof. (i) (ii). Xλ (t, ·) is Ft -measurable because it is progressively measurable. That E(|Xλ (t, w)|) < ∞ is a consequence of (i) and the fact that
a is bounded.
107
Xλ (t, w)
is continuous for fixed t, s, w, (t > s).
Xλ (s, w)
Morera’s theorem shows that φ is analytic. Let A ∈ Fs . Then
Z
Xλ (t, w)
dP(w)
Xλ (s, w)
φ
The function λ −
→
A
is analytic. By hypothesis,
Z
Xλ (t, w)
dP(w) = 1, ∀λ ∈ Rd .
Xλ (s, w)
A
Thus
R Xλ (t, w)
dP(w) = 1, ∀ complex λ. Therefore
A Xλ (s, w)
E(Xλ (t, w)|Fs ) = Xλ (s, w),
proving (ii). (ii) ⇒ (iii). Let


Zt

 Zt
1


hθ, a(s, w)θids , θ ∈ Rd .
A(t, w) = exp −i hθ, b(s, w)ids +


2
0
0
By definition, A is progressively measurable and continuous. Also
dA
| (t, w)| is bounded on every compact set in R and the bound is indt
dependent of w. Therefore A(t, w) is of bounded variation on every interval [0, T ] with the variation ||A||[0,T ] bounded uniformly in w. Let
M(t, w) = Xiθ (t, w). Therefore
sup |M(t, w)| ≤ e1/2 T sup |hθ, aθi|.
0≤t≤T
0≤t≤T
By (ii) M(t, ·) is a martingale and since
!
E sup |M(t, w)| ||A||[0,T ] (w) < ∞, ∀T,
0≤t≤T
1
M(t, ·)A(t, ·) −
2
Zt
0
M(s, ·)dA(s, ·)
104
15. Equivalent For of Itô Process
108
is a martingale (for a proof see Appendix), i.e. Y f (t, w) is a martingale
when f (x) = eihθ,xi .
Let f ∈ C0∞ (Rd ). Then f ∈ F (Rd ) the Schwartz-space. Therefore
by the Fourier inversion theorem
Z
f (x) =
fˆ(θ)eihθ,xi dθ.
Rd
105
On simplification we get
Y f (t, w) =
Z
fˆ(θ)Yθ (t, w)dθ
Rd
where Yθ ≡ Ye ihθ, xi. Clearly Y f (t, ·) is progressively measurable and
hence Ft -measurable.
Using the fact that
E(|Yθ (t, w)|) ≤ 1 + t d|θ| ||b||∞ +
d2 2
|θ| ||a||∞ ,
2
the fact that F (Rd ) ⊂ L1 (Rd ) and that F (Rd ) is closed under multiplication by polynomials, we get E(|Y f (t, w)|) < ∞. An application of
Fubini’s theorem gives E(Y f (t, w)|Fs ) = Y f (s, w), if t > s. This proves
(iii).
(iii) ⇒ (iv). Let u ∈ C0 ([0, ∞) × Rd ).
Clearly Zu (t, ·) is progressively measurable. Since Zu (t, w) is bounded for every w, E(|Zu (t, w)|) < ∞. Let t > s. Then
E(Zu (t, w) − Zu (s, w)|F s ) =
= E(u(t, X(t, w) − u(s, X(s, w)|F s) − E(
Zt
(
∂u
+ Lσ,w u)(σ, X(σ, w)dσ|F s)
∂σ
s
= E(u(t, X(t, w) − u(t, X(s, w))|F s) + E(u(t, X(s, w) − u(s, X(s, w))|F s)−
Zt
∂u
− E( (
+ Lσ uw )(σ, X(σ, w))dσ|F s )
∂σ
s
109
Zt
(Lσ,w u)(t, X(σ, w))dσ|F s)+
Zt
(
∂u
(σ, X(s, w))dσ|F s )−
∂σ
Zt
(
∂u
+ Lσu , w)(σ, X(σ, w))dσ|F s ),
∂σ
= E(
s
+ E(
s
− E(
by (iii)
s
= E(
Zt
[Lσ,w u(t, X(σ, w)) − Lσ,w u(σ, X(σ, w))]dσ|F s)
s
+ E(
Zt
[
∂u
∂u
(σ, X(s, w)) −
(σ, X(σ, w))]dσ|F s)
∂σ
∂σ
s
= E(
Zt
(Lσ,w u(t, X(σ, w)) − Lσ,w u(σ, X(σ, w))]dσ|F s)
s
− E(
Zt
s
dσ
Zσ
Lρ,w
∂u
(σ, X(ρ, w))dρ|F s)
∂σ
s
106
The last step follows from (iii) (the fact that σ > s gives a minus
sign).
Zt
Zt
∂
= E( dσ
Lσ,w u(ρ, X(σ, w))dρ|Fs )
∂ρ
0
σ
Zt
Zσ
∂u
− E( dσ Lρ,w (σ, X(ρ, w))dρ|Fs )
∂σ
s
s
=0
(by Fubini). Therefore Zu (t, w) is a martingale.
Before proving (iv) ⇒ (v) we show that (iv) is true if u ∈ Cb1,2 ([0, ∞)
× Rd . Let u ∈ Cb1,2 .
15. Equivalent For of Itô Process
110
that
(*) Assume that there exists a sequence (un ) ∈ C0∞ [[0, ∞) × Rd ] such
un → u,
107
∂un
∂u ∂un
∂u ∂un
∂2 u
→ ,
→
,
→
∂t
∂t ∂xi
∂xi ∂xi ∂x j
∂xi ∂x j
uniformly on compact sets.
Then Zun → Zu pointwise and sup(|Zun (t, w)|) < ∞.
n
Therefore Zu is a martingale. Hence it is enough to justify (*).
For every u ∈ Cb1,2 ([0, ∞) × Rd ) we construct a u ∈ Cb1,2 ((−∞, ∞) ×
Rd ) ≡ Cb1,2 (R × Rd ) as follows. Put



u(t, x), if t ≥ 0,
u(t, x) = 

C1 u(−t, x) + C2 u(− t , x), if t < 0;
2
∂ũ ∂u
matching
,
at t = 0 and û(t, x) and u(t, x) at t = 0 and ũ(t, x) and
∂t ∂t
u(t, x) at t = 0 yields the desired constants C1 and C2 . In fact C1 = −3,
C2 = 4. (*) will be proved if we obtain an approximating sequence for
ũ. Let S : R be any C function such that if |x| ≤ 1,



1, if |x| ≤ 1,
S (x) = 

0, if |x| ≥ 2.
!
|x|2
Let S n (x) = S
where |x|2 = x21 + · · · + x2d+1 . Pur un = S n ũ.
n
This satisfies (*).
(iv) ⇒ (v). Let
h ∈ Cb1,2 ([0, ∞) × Rd ).
Put u = exp(h(t, x)) in (iv) to conclude that
h(t,X(t,w))
M(t, w) = e
−
Zt
0
108
is a martingale.
Put
h(s,X(s,w))
e
"
#
1
∂h
+ L s,w h + h∇x h, a∇x hids
∂s
2
111

 t

Z
1
∂h


(s, w) + L s,w − (s, w) + ha(s, w)∇ x h, ∇ x hids .
A(t, w) = exp − 

 ∂s
2
0
A((t, w)) is progressively measurable, continuous everywhere and
||A||[0,T ] (w) ≤ C1 ∈ C2 T
dA
where C1 and C2 are constants. This follows from the fact that | | is
dt
uniformly bounded in w. Also sup |M(t, w)| is uniformly bounded in
0≤t≤T
w. Therefore
E( sup |M(t, w)| ||A||[0,T ] (w)) < ∞.
0≤t≤T
Hence M(t, ·)A −
Rt
0
M(s, ·)dA(s, ·) is a martingale. Now
"
#
∂h
dA(s, w)
1
=−
(s, w) + L s,w h(s, w) + ha∇x h, ∇x hi
A(s, w)
∂s
2
Therefore
M(t, w) = eh(t,X(t,w)) +
Zt
eh(s,X(s,w))
dA(s, w)
.
A(s, w)
0
M(t, w)A(t, w) = Ph (t, w) + A(t, w)
Zt
eh(s,X(s,w))
dA(s, w)
A(s, w)
0
Zt
M(s, ·)dA(s, ·) =
0
Zt
eh(s,X(s,w)) dA(s, w)
0
+
Zt
0
dA(s, w)
Zs
eh(σ,X(σ,w))
dA(σ, w)
A(σ, w)
0
Use Fubini’s theorem to evaluate the second integral on the right
above and conclude that Ph (t, w) is a martingale.
15. Equivalent For of Itô Process
112
(vi) ⇒ (i) is clear if we take h(t, x) = hθ, xi. It only remains to prove
that (v) ⇒ (vi).
(v) ⇒ (vi). The technique used to prove this is an important one and
we shall have occasion to use it again.
109
Step 1. 0 Let h(t, x) = θ1 x1 + θ2 x2 + · · · θd xd = hθ, xi for every (t, x) ∈
[0, ∞) × Rd , θ is some fixed element of Rd . Let


Zt
Zt


1


Z(t) = exp hθ, Xt i − hθ, bids −
hθ, aθids


2
0
0
We claim that Z(t, ·) is a supermartingale.
Let f : R → R be a C ∞ function with compact support such that
f (x) = x in |x| ≤ 1/2 and | f (x)| ≤ 1, ∀x. Put fn (x) = n f (x/n). Therefore
| fn (x)| ≤ C|x| for some C independent of n and x and fn (x) converges to
x.
d
P
Let hn (x) = θi fn (xi ). Then hn (x) converges to hθ, xi and |hn (x)| ≤
i=1
C ′ |x| where C is also independent of n and x. By (v),


!
Zt
Zt


1
∂h


n
+ L s,w h ds −
ha∇x hn , ∇x hn ids
Zn (t) = exp hn (t, Xt ) −


∂s
2
0
0
is a martingale. As hn (x) converges to hθ, xi, Zn (t, ·) converges to Z(t, ·)
pointwise. Consequently
E(Z(t)) = E(lim Zn (t)) ≤ lim E(Zn (t)) = 1
and Z(t) is a supermartingale.
Step 2. E(exp B sup |X(s, w)|) < ∞ for each t and B. For, let Y(w) =
0≤s≤t
sup |X(s, w)|, Yi (w) = sup |Xi (s, w)| where X = (X1 , . . . , Xd ). Clearly
0≤s≤t
0≤s≤t
Y ≤ Y1 + · · · + Yd . Therefore
E(eBY ) ≤ E(eBY 1eBY 2 . . . eBY d).
113
110
The right hand side above is finite provided E(eBY i) < ∞ for each i
as can be seen by the generalised Holder’s inequality. Thus to prove the
assertion it is enough to show E(eBY i) < ∞ for each i = 1, 2, . . . d with
aB′ different from B; more specifically for B′ bounded.
Put θ2 = 0 = θ3 = . . . = θd in Step 1 to get
u(t) = exp[θ1 X1 (t) −
Zt
1
θ1 b1 (s, ·)ds − θ12
2
Zt
a11 (s, ·)ds]
0
0
is a supermartingale. Therefore
!
1
1
P sup u(s, ·) ≥ λ ≤ E(u(t)) = , ∀λ > 0.
λ
λ
0≤s≤t
(Refer section on Martingales). Let c be a common bound for both b1
and a11 and let θ1 > 0. Then (∗) reads
!
1 2
1
P sup exp θ1 X1 (s) ≥ λ exp(θ1 ct + θ1 ct) ≤ .
2
λ
0≤s≤t
Replacing λ by
2
eλθ1 e−ctθ1 −1/2ctθ1
we get
!
2
P sup exp θ1 X1 (s) ≥ exp λθ1 ≤ e−λθ1 +θ1 ct+1/2θ1 ct ,
0≤s≤t
i.e.
!
2
P sup X1 (s) ≥ λ ≤ e−λθ1 +θ1 ct+1/2θ1 ct , ∀θ1 > 0.
0≤s≤t
Similarly
!
2
P sup −X1 (s) ≥ λ ≤ e−λθ1 +θ1 ct+1/2θ1 tc , ∀θ1 > 0.
0≤s≤t
As
(
) (
)
{Y1 (w) ≥ λ} sup X1 (s) ≥ λ ∪ sup −X1 (s) ≥ λ ,
0≤s≤t
0≤s≤t
15. Equivalent For of Itô Process
114
we get
111
2
P{Y1 ≥ λ} ≤ 2e−λθ1 +θ1 ct+1/2θ1 ct , ∀θ1 > 0.
Now we get
1
E(exp BY1 ) =
B
Z∞
exp(Bx)P(Y1 ≥ x)dx
2
B
Z∞
1
exp(Bx − xθ1 + θ1 ct + θ12 ct)dx
2
(since Y1 ≥ 0)
0
≤
0
< ∞,
if
B < θ1
This completes the proof of step 2.
Step 3. Z(t, w) is a martingale. For
|Zn (t, w)| = Zn (t, w)


!
Zt
Zt


∂h
1


n
ha∇x hn , ∇x hn ids
+ L s,w hn dx −
= exp hn (Xt ) −


∂s
2
0
0


Zt




≤ exp hn (Xt ) − L s,w hn 


0
(since a is positive semidefinite and ∂hn /∂s = 0).
Therefore |Zn (t, w)| ≤ A exp(B sup |X(s, w)|) (use the fact that
0st
∂hn ∂2 hn
|hn (s)| ≤ C|x| and
,
are bounded by the same constant). The
∂xi ∂xi ∂x j
result now follows from the dominated convergence theorem and Step
2.
112
Remark. In Steps 1, 2 and 3 we have proved that (v) ⇒ (i). The idea of
the proof was to express Z(t, ·) as a limit of a sequence of martingales
proving first that Z(t, ·) is a supermartingale. Using the supermartingale
inequality it was then shown that (Zn ) is a uniformly integrable family
proving thereby that Z(t, ·) is a martingale.
115
Step 4. Let h(t, x) ∈ C 1,2 ([0, ∞)×Rd ) such that h(t, x),
∂h
∂h
(t, x),
(t, x),
∂s
∂xi
∂2 h
(t, x) are all dominated by α|x| + β for some suitable scalars α and
∂xi ∂x j
β. Let φn be a sequence of real valued C ∞ functions defined on Rd such
that



1 on |x| ≤ n
φn = 

0 on |x| ≥ 2n
and suppose there exists a common bound C for
φn ,
∂φn ∂2 φn
,
(∀n).
∂xi ∂xi ∂x j
Let hn (t, x) = h(t, x)φn (x). By (v) Zhn (t, w) is a martingale. The
conditions on the function h and φn ’s show that
!
|Zhn (t, w)| ≤ A exp B sup |X(s, w)|
0≤s≤t
where A and B are constants. By Step 2, (Zhn ) are uniformly integrable.
Also Zhn (t, ·) converges pointwise to Ph (t, ·) (since hn → h pointwise).
By the dominated convergence theorem Ph (t, ·) is a martingale, proving
(vi).
16. Itô’s Formula
Motivation. Let β(t) be a one-dimensional Brownian motion. We have 113
seen that the left integral

 t

 Z


(*)
L 2 β(s, ·)dβ = [β2 (t, ·) − β2 (0, ·) − t]


0
Formally (*) can be written as
dβ2 (t) = 2β(t)dβ(t) + dt.
For, on integrating we recover (*).
Newtonian calculus gives the result:
d f (β(t)) = f ′ (β(t))dβ(t) +
1 ′′
f (β(t))dβ2 (t) + · · ·
2
for reasonably smooth functions f and β. If β is of bounded variation,
only the first term contributes something if we integrate the above equaP
tion. This is because dβ2 = 0 for a function of bounded variation.
P
For the Brownian motion we have seen that dβ2 → a non zero value,
P 3
but one can prove that dβ , . . . converge to 0. We therefore expect the
following result to hold:
d f (β(t)) ≈ f ′ (β(t))dβ(t) +
1 ′′
f (β(t))d2 β(t).
2
We show that for a one-dimensional Brownian motion
X
(dβ)3 , sup(dβ)4 , . . .
117
16. Itô’s Formula
118
all vanish.

 n
X
 X
X
3
3

E[(β(ti+1 ) − β(ti )]3
E( (dβ) ) = E  [β(ti+1 ) − β(ti )]  =
i=0
i=0
=
n
X
0 = 0,
i=1
114
because β(ti+1 ) − β(ti ) is a normal random variable with mean zero and
variance ti+1 − ti . Similarly the higher odd moments vanish. Even moments of a normal random variable with mean 0 and variance σ2 are
connected by the formula
µ2k+2 = σ2 (2k + 1)µ2k , k > 1.
So
n
X
X
(dβ)4 =
(β(ti+1 ) − β(ti ))4 .
i=0
Therefore
n
X
X
E( (dβ)4 ) =
E([β(ti+1 ) − β(ti ))4 ]
i=0
n
X
=3
i=0
(ti+1 − ti )2 ;
the right hand side converges to 0 as the mesh of the partition goes to 0.
Similarly the higher order even moments vanish.
More generally, if β(t, ·) is a d-dimensional Brownian motion then
we expect
d f (β(t)) ≈ ∇ f (β(t)) · dβ(t) +
1 X ∂2 f
dβi dβ j .
2 i, j ∂xi ∂x j
119
However
P
dβi dβ j = 0 if i , j (see exercise below). Therefore
1
d f (β(t)) ≈ ∇ f (β(t)) · dβ(t) + ∆ f (β(t))dβ2 (t).
2
The appearance of ∆ on the right side above is related to the heat
equation.
Exercise 4. Check that dβi dβ j = δi j dt.
115
(Hint: For i = j, the result was proved earlier. If i , j, consider a
partition 0 = t0 < t1 < . . . < tn = t. Let ∆k βi = βi (tk ) − βi (tk−1 ). Then
 n
2
X
X
X


E  ∆k βi ∆k β j  =
E(∆2k βi ∆2k β j ) + 2
E[(∆k βi )(∆1 β j )];
k=1
k
k,l
the right side converges to 0 as n → ∞ because ∆k βi and ∆ℓ β j are independent for k , ℓ).
Before stating Itô’s formula, we prove a few preliminary results.
Lemma 1. Let X(t, ·) ∈ I[b, 0] be a one-dimensional Itô process. Then
X(t, ·) − X(0, ·) =
Zt
b(s, ·)ds
a.e.
0
Proof. exp[θX(t, ·) − θX(0, ·) − θ
Therefore
E(exp[θ(X(t, ·) − X(0, ·)) − θ
Rt
0
Zt
b(s, ·)ds] is a martingale for each θ.
b(s, ·)ds]) = constant = 1, ∀t.
0
Let
W(t, ·) = X(t, ·) − X(0, ·) −
Zt
b(s, ·)ds.
0
Then
E(exp θW(t, ·)) = Moment generating function of w = 1, ∀t.
Therefore θ(t, ·) = 0 a.e.
16. Itô’s Formula
120
116
Remark . If X(t, ·) ∈ I[0, 0] then X(t, ·) = X(0, ·) a.e.; i.e. X(t, ·) is a
trivial process.
We now state a theorem, which is a particular case of the theorem
on page 103.
Theorem . If h ∈ C 1,2 ([0, ∞) × Rd ) such that (i) |h(x)| ≤ A|x| + B,
∂h ∂h ∂2 h
∀x ∈ [0, ∞) × Rd , for constants A and B (ii)
,
,
also grow
∂t ∂xi ∂xi ∂x j
linearly, then
exp[h(t, β(t, ·) −
Zt
0
!
Zt
1
∂h 1
+ ∆h (s, β(s, ·) −
|∇h|2 (s, β(s, ·))ds]
∂s 2
2
0
is a martingale.
Ito’s Formula. Let f ∈ C01,2 ([0, ∞) × Rd ) and let β(t, ·) be a d-dimensional Brownian motion. Then
f (t, β(t)) − f (0, β(0)) =
Zt
∂f
(s, β(s, ·))ds+
∂s
0
+
Zt
1
h∇ f (s, β(s, ·)), dβ(s, ·)i +
2
0
Zt
∆ f (s, β(s, ·))ds.
0
where
∆≡
∂2
∂2
+
·
·
·
+
.
∂x21
∂x2d
Proof.
Step 1. Consider a (d + 1)-dimensional process defined by
X0 (t, ·) = f (t, β(t, ·)),
X j (t, ·) = β j (t, ·).
121
We claim that X(t, ·) ≡ (X0 , X1 , . . . , Xd ) is a (d + 1)-dimensional Itôprocess with parameters
!
#
"
∂f 1
+ ∆ f (s, β(s, ·)), 0, 0, . . . 0
d terms
b=
∂s 2
and
117


a00 a01 . . . a0d 

a

 10
a =  .

.
 .
Id×d 


ad0
where
a00 = |∇x f |2 (s, β(s, ·)),
!
∂
a0 j =
f (s, β(s, ·)),
∂x j
a j0 = a0 j .
For, put h = λ f (t, x) + hθ, xi, x = (x1 , . . . , xd ) ∈ Rd , in the previous
theorem. Then
∂f
∂h
∂f
∂h
= λ , ∆h = λ∆ f,
=λ
+ θ j.
∂s
∂s
∂x j
∂x j
Therefore we seen that
exp[λ f (t, β(t, ·)) + hθ, β(t, ·)i − λ
Zt
!
∂f 1
+ ∆x f (s, β(s, ·)ds
∂s 2
0
1
− λ2
2
Zt
1
|∇ f | (s, β(s, ·))ds − |θ|2 t − λhθ,
2
2
0
Zt
∇( f (s, β(s, ·)))dsi]
0
is a martingale.
Consider (λ, θ)a λθ . We have
" # "
#
P
a00 λ + dj=1 a0 j θ j
λ
a
=
,
θ
ρ+θ
 
a10 
 
ρ = λ  . . .  .
 
ad0
16. Itô’s Formula
122
Therefore
" #
d
d
X
X
λ
2
(λ, θ)a
= a00 λ + λ
a0 j θ j +
a0 j θ j +
θ
j=1
j=1
∂f
= λ2 |∇ f |2 + 2λ
θ j + |θ|2 .
∂x j
118
Thus (*) reads
exp[λ f (t, β(t, ·) + hθ, β(t, ·)i − λ
Zt
1
b0 (s, ·)ds −
2
0
Zt
hα, aαids]
0
is a martingale where α = (λ, θ) ∈ Rd+1 . This proves the claim made
above.
Step 2. Derine σ(s, ·) = (1, −∇x f (s, β(s, ·))) and let
Z(t, ·) =
Zt
hσ(s, ·), dX(s, ·)i where X ≈ (X0 , X1 , . . . , Xd )
0
is the (d + 1)-dimensional Itô process obtained in Step 1. Since f ∈ Cb1,2 ,
Z(t, ·) is an Itô process with parameters hσ, bi and σaσ∗ :
∂f
+
"∂s
a
aσ∗ = 00
ρ∗
hσ, bi =
1
∆ f,
2# "
#
ρ
1
I −∇x f
 
0
"
#  
0
a − hρ, ∇x f i
= 00 ∗
=  . 
ρ − ∇x f
 .. 
 
0
Therefore σaσ∗ = 0. Hence by Lemma 1,
Z(t, ·) − Z(0, ·) −
Zt
0
hσ, bids = 0(a.e.),
123
Z(t, ·) =
Zt
dX0 (s) −
0
Zt
h∇ f (s, β(s, ·)), dβ(s, ·)i
0
= f (t, β(t)) − f (0, β(0)) −
Zt
h∇ f ( f, β(s, ·)), dβ(s, ·)i
0
119
Hence Z(0) = 0. Thus
f (t, β(t)) − f (0, β(0)) −
Zt
h∇ f (s, β(s, ·))dβ(s, ·)i−
0
−
Zt
!
∂f 1
+ ∆x f (s, β(s, ·))ds = 0 a.e.
∂s 2
0
This estabilished Itô’s formula.
Exercise. (Itô’s formula for the general case). Let
φ(t, x) ∈ Cb1,2 ([0, ∞) × Rd ).
If X(t, ·) is a d-dimensional Itô process corresponding to the parameters b and a, then the following formula holds:
φ(t, X(t, w)) − φ(0, X(0, w))
Zt X
Zt
Zt
∂2
1
∂φ
(s, X(s, x))ds + h∇x φ, dXi +
ai j
ds.
=
∂s
2
∂xi ∂x j
0
0
0
This is also written as
dφ(s, X(s, w)) = φ s ds + h∇x φ, dXi +
1X
∂2 φ
dx.
ai j
2
∂xi ∂x j
To prove this formula proceed as follows.
16. Itô’s Formula
124
(i) Take h(t, x) = λφ(t, x) + hθ, xi in (vi) of the theorem on the equivalence of Itô process to conclude that
Y(t, ·) = (φ(t, X(t, ·)), X(t, ·))
is a (d + 1)-dimensional Ito process with parameters
!
∂φ
′
+ L s,w φ, b
b =
∂t
120
and
ha∇x φ, ∇x φi, a∇x φ
1×1
1 × d A = a∇x φ
a d×1
d × d
(ii) Let σ(t, x) = (1, −∇x φ(t, x)) and
Z(t, ·) =
Zt
hσ(s, X(s, ·)), dY(s, ·)i.
0
The assumptions on φ imply that Z is an Itô process corresponding
to
!
∂φ 1 X
∂2 φ
′
∗
(hσ, b i, σAσ ) ≡
+
,0 .
ai j
∂t 2
∂xi ∂x j
(iii) Use (ii) to conclude that
Z(t, ·) =
Zt
hσ, b′ ids
a.e.
0
This is Itô’s formula.
(iv) (Exercise) Verify that Itô’s formula agrees with the formula obtained for the case of Brownian motion.
Note. Observe that Itô’s formula does not depend on b.
125
Examples.
1. Let β(t) be a one-dimensional Brownian motion. Then
d(et φ(β(t)) = et dφ(β(t)) + φ(β(t))d(et )
= et φ′ (β(t))dβ(t) + φ(β(t))et dt+
1
+ φ′′ (β(t))et dt.
2
Rt
2. To evaluate d(e e V(β(s,·))ds u(t, β(t))) where V is a smooth function, 121
put
"
0
# " #
0
b
= 1
V((t, ·))
b2
"
#
1 0
a=
.
0 0
Let X1 (t, ·) = β(t, ·), X = (X1 , X2 ).
X2 (t, ·) =
Zt
V(β(s, ·))ds, b =
Then
exp[θ1 X1 (t, ·) − θ2 X2 (t, ·) − θ1
Zt
b1 (X(s, ·))ds
Zt
haθ, θids]
0
− θ2
Zt
1
b2 (X(s, ·))ds −
2
0
exp[θ1 X1 (t, ·) −
0
θ122
2
t];
the right side is a martingale. Therefore (X1 , X2 ) is a 2-dimensional Itô process with parameters b and a and one can use Itô’s
formula to write
Rt
V(β(s,·))ds
0
d e
u(t, β(t)) = d eX2 (t) u(t, β(t))
Rt
= e 0 V(β(s,·))
Rt
+e 0 V(β(s,·))dt
∂
u(t, β(t))dt+
∂t
Rt
∂
u(t, β(t))d(t) + e 0 V(β(s,·))ds u(t, β(t))dX2
∂x
16. Itô’s Formula
126
∂
1 Rt
+ e 0 V(β(s,·))ds u(t, β(t))dt
2
∂t
Rt
∂
∂
1 ∂2
u(t, β(t))dt
= e 0 V(β(s,·))dt u(t, β(t))dt + u(t, β(s))d(t) +
∂t
∂t
2 ∂x2
+u(t, β(t))V(β(t, ·)dt].
122
3. Let σi , fi , i = 1, 2, . . . k be bounded and progressively measurable
relative to a one-dimensional Brownian motion (Ω, Ft , P). Write
Xi (t, ·) =
Zt
σi (s, ·)dβ(s, ·) +
0
Zt
fi (s, ·)ds.
0
Rt
Then Xi (t, ·) is an Itô process with parameters ( fi (s, ·)ds, σ2i ) and
0
(X1 , . . . , Xk ) is an Itô process with parameters

 t
Zt

Z


fk (s, ·)ds ,
B =  f1 (s, ·)ds, . . . ,


0
0
A = (Ai j ) where Ai j = σi σ j δi j .
If φ ≡ φ(t, X1 (t) . . . , Xk (t)), then by Itô’s formula
∂φ
∂φ
∂φ
ds +
dX1 + · · · +
dXk
∂s
∂x1
∂xk
∂2 φ
1X
ds
σi σ j δi j
+
2
∂xi ∂x j
∂φ
∂φ
∂φ
=
ds +
dX1 + · · · +
dXk
∂s
∂x1
∂xk
k
1 X 2 ∂φ
+
ds
σ
2 i=1 i ∂xi
dφ =
Exercise. Take σ1 = 1, σ2 = 0, f1 = f2 = 0 above and verify that if
φ = eX2 (t,·) u(t, β(t)),
then one gets the result obtained in Example 2 above.
127
We give below a set of rules which can be used to calculate dφ in
practice, where φ is as in Example 3 above.
√
1. With each dβ associate a term (dt)
2. If φ = φ(t, X1 , . . . , Xk ), formally differentiate φ using ordinary 123
calculus retaining terms upto the second order to get
(*) dφ =
∂φ
∂φ
1 X ∂2 φ
∂φ
dt +
dX1 + · · · +
Xk +
dXi dX j
∂t
∂x1
∂xk
2
∂xi ∂x j
3. Formally write dXi = fi dt + αi dβi , dX j = σ j dβ j + f j dt.
4. Multiply dXi dX j and retain only the first order term in dt., For
dβi dβ j substitute δi j dt. Substitute in (*) to get the desired formula.
Illustration of the use of Itô Calculus. We refer the reader to the section on Dirichlet problem. There it was shown that
Z
u(x) =
u(y)π(x, dy) = E(u(X(τ)))
G
satisfies ∆u = 0 in a region G with u = u(X(τ)) on the boundry of G
(here τ is the first hitting time).
The form of the solution is given directly by Itô’s formula (without
having recourse to the mean value property). If u = u(X(t)) satisfies
∆u = 0 then by Itô’s formula
du(X(t)) = h∇u, dXi.
16. Itô’s Formula
128
Therefore
u(X(t)) = u(X(0)) +
Zt
h∇u(X(s)), dX(s)i
0
124
Assuming ∇u to be bounded, we see that u(X(t)) is a martingale. By
the optional stopping theorem
E(u(X(τ)) = u(X(0)) = u(x).
Thus Itô’s formula connects solutions of certain differential equations with the hitting probabilities.
17. Solution of Poisson’s
Equations
LET X(t, ·) BE A d-dimensional Brownian motion with (Ω, Ft , P) as 125
1
usual. Let u(x) : Rd → R be such that ∆u = f . Assume u ∈ Cb2 (Rd ).
2
Rt
1
Then L s,w u ≡ ∆u = f and we know that u(X(t, ·)) − f (X(s, ·))ds is
2
0
a (Ω, Ft , P)-martingale. Suppose now that u(x) is defined only on an
1
open subset G ⊂ Rd and ∆u = f on G. We would like to consider
2
Z(t, ·) = u(X(t, ·)) −
Zt
f (X(s, ·))ds
0
and ask whether Z(t, ·) is still a martingale relative to (Ω, Ft , P). Let
τ(w) = inf{t, X(t, w) ∈ ∂G}. Put this way, the question is not well-posed
because Z(t, ·) is defined only upto time τ(w) for u is not defined outside
G. Even if at a time t > τ(w)X(t, w) ∈ G, one needs to know the values
of f for t > τ(w) to compute the integral.
To answer the question we therefore proceed as follows. Let At =
[w : τ(w) > t]. As t increases, At have decreasing measures. We shall
give a meaning to the statement ‘Z(t, ·) is a martingale on At ’. Define
Z(t, ·) = u(X(τ ∧ t, ·)) −
Zτ∧t
0
129
f (X(s, ·))ds.
17. Solution of Poisson’s Equations
130
Therefore
126



on At
Z(t),
Z(t, ·) = 

Z(τ, ·), on (At )c .
Since Z(t, ·) is progressively measurable upto time τ, Z(t, ·) is Ft measurable.
Theorem . Z(t, ·) is a martingale.
Proof. Let Gn be a sequence of compact sets increasing to G such that
Gn ⊂ G0n+1 . Choose a C ∞ function φn such that φn = 1 on Gn and
1
support φn ⊂ G. Put un = φn u and fn = ∆un . Then
2
Zn (t, ·) = un (X(t, ·)) −
Zt
fn (X(s, ·))ds
0
is a martingale for each n. Put
τn = inf{t : X(t, ·) < Gn }
Then Zn (τn ∧ t, ·) is also a martingale (See exercise below). But
Zn (τn ∧ t) = Z(τn ∧ t).
Therefore Mn (t, ·) = Z(τn ∧ t, ·) is a martingale. Observe that τn ≤
τn+1 and since Gn G we have τn τ. Therefore Z(τn ∧t) → Z(τ∧t) (by
continuity); also |Mn (t, ·)| ≤ ||u||∞ + || f ||∞ t. Therefore Z(τ ∧ t) = Z(t, ·)
is a martingale.
Exercise. If M(t, ·) is a (Ω, Ft , P)-martingale, show that for my stopping
time τ, M(τ ∧ t, ·) is also a martingale relative to (Ft ).
[Hint: One has to show that if t2 > t1 ,
Z
Z
M(τ ∧ t2 , w)dP(w) =
M(τ ∧ t1 , w)dP(w), ∀A ∈ Ft1 .
A
A
131
The right side =
Z
M(t1 , w)dP(w) +
A∩(τ>t1 )
The left side
Z
=
127
M(τ, w)dP(w).
A∩(τ<t1 )
M(t2 , w)dP(w) +
A∩(τ>t1 )
Z
Z
M(τ, w)dP(w).
A∩(τ<t1 )
Now use optional stopping theorem].
Lemma . Let G be a bounded region and τ be as above. Then E x (τ) <
∞, ∀x ∈ G, where E x = E Px .
Proof. Without loss of generality we assume that G is a sphere of radius
1
R2 − |x|2
≥ 0 and satisfies ∆u = −1 in G. By
R. The function u(x) =
d
2
the previous theorem
Zτ∧t
u(X(τ ∧ t, ·) + ds
0
is a martingale. Therefore
E x (u(X(τ ∧ t, ·))) + E x (τ ∧ t) = u(X(0)) = u(x).
Therefore E x (τ ∧ t) ≤ u(x) (since u ≥ 0). By Fatou’s lemma, on
letting t → ∞, we obtain E x (τ) ≤ u(x) < ∞. Thus the mere existence of
1
a u satisfying ∆u = 1 helps in concluding that E x (τ) < ∞.
2
Theorem . Let u ∈ Cb2 (G) and suppose that u satisfies
(*)
1
∆u = f in G,
2
u = g on ∂G.
Rτ
Then u(x) = E x [g] − E x [ f (X(s, ·))ds] solves (*).
0
132
17. Solution of Poisson’s Equations
Remark . The first part of the solutin u(x) is the solution of the homogeneous equation, and the second part accounts for the inhomogeneous
term.
128
Proof. Define Z(t, ·) = u(X(τ ∧ t)) −
τ∧t
R
0
f (X(s, ·))ds. Then Z is a mar-
tingale. Also |Z| ≤ ||u||∞ + τ|| f ||∞ . Therefore, by the previous Lemma,
Z(t, ·) is a uniformly integrable martingale. Therefore we can equate the
expectations at time t = 0 and at time t = ∞ to get
Zτ
u(x) = E x (g) − E x [ f (X(s, ·))ds].
0
18. The Feynman-Kac
Formula
WE NOW CONSIDER the modified heat equation
∂u 1
+ ∆u + V(x)u(t, x) = 0, 0 ≤ t ≤ T,
∂t 2
where u(T, x) = f (x). The Feynman-Kac formula says that the solution
for s ≤ T is given by
(*)
(**)
RT
u(s, x) = E s,x (e
V(X(s))dS
s
f (X(T ))).
Observe that the solution at time s depends on the expectation with
respect to the process starting at time s.
Note . (**) is to be understood in the following sense. If (*) admits a
smooth solution then it must be given by (**). We shall not go into the
conditions under which the solution exists. Let
Rt
Z(t, ·) = u(t, X(t, ·))e
s
V(X(σ,·)dσ)
,
t ≥ s.
By Ito’s formula (see Example 2 of section 16), we get
Z(t, ·) = Z(s, ·) +
Zt
Rt
e
s
V(X(σ,·)dσ)
s
h∇u(λ, X(λ)), dX(λ)i,
provided that u satisfies (*). Assume tentatively that V and ∇u are
bounded and progressively measurable. Then Z(t, ·) is a martingale.
Therefore
E s,x (Z(T, ·)) = Ez,x (Z(s, ·)),
133
129
18. The Feynman-Kac Formula
134
or
R
E s,x (u(T, X(T )))e
T
s
V(X(σ,·))dσ
= u(s, x).
This proves the result.
130
We shall now remove the condition ∇u is bounded and prove the
uniqueness of the solution corresponding to (*) under the assumption
that V is bounded above and
α
|u(t, x)| ≤ eA+|x| ,
α < 2,
on
[s, T ).
In particular, the Feynman-Kac formula extends the uniqueness theorem for the heat equation to the class of unbounded functions satisfying
a growth condition of the form given above.
Let φ be a C ∞ function such that φ = 1 on |X| ≤ R, and φ = 0 outside
|x| > R + 1. Put uR (t, x) = u(t, x)φ,
Rt
ZR (t, x) = uR (t, x)e
s
V(X(σ)dσ)
.
By what we have already proved, ZR (t, ·) is a martingale. Let
τR (ω) = inf{t : t ≥ sω(t) ∈ S (0; R) = {|x| ≤ R}}.
Then ZR (t ∧ τR , ·)) is also a martingale, i.e.
R t∧τ
uR (t ∧ τR , X(t ∧ τR , ·))e
s
R
V(X(σ))dσ
is a martingale. Equating the expectations at time t = s and time t = T
and using the fact that
uR (t ∧ τR , X(t ∧ τR , ·)) = u(t ∧ τR , X(t ∧ τR , ·))),
we conclude that
Rτ
u(s, x) = E s,x [u(τR ∧ T, X(τR ∧ T, ·))e
R
= E s,x [X(τR ∧T ) f (X(T ))e
T
s
s
R ∧T
V(X(s))ds
RτR
+ E s,x [X(τR ≤T ) u(τR , X(τR ) e s
V(X(s))ds
]+
V(X(s))ds
]
]
135
131
Consider the second term on the right:
Rτ
|E s,x [X(τR ≤T ) u(τR , X(τR ))e
Rα
s
R
V(X(s))ds
]|
≤ A′ e P[R ≤ T ]
(where A′ is a constant given in terms of the constants A and T and the
bound of V)
α
= A′ eR P[ sup |X(σ)| ≥ R].
s≤σ≤T
2
P[ sup |X(σ)| ≥ R] is of the order of e−c(T )R and since α < 2, the
s≤σ≤T
second term on the right side above tends to 0 as R → ∞. Hence, on
letting R → +∞, we get, by the bounded convergence theorem,
RT
u(s, x) = E s,x [ f (X(T ))e
s
V(X(s))ds
]
Application. Let β(t, ·) be a one-dimensional Brownian motion. Recall
4
(Cf. Reflection principle) that P{ sup |β(s)| ≤ 1} is of the order of e −
π
0≤s≤t
π2 t
. The Feynman-Kac formula will be used to explain the occurance
8
π2
λ2
π2
in the exponent. First observe that
=
where λ is
of the factor
8
8
2
the first positive root of Cos λ = 0. Let
τ(w) = inf{t : |β(t)| ≥ 1}.
Then
P{ sup |β(s, ·)| ≤ 1} = P{τ ≥ t}.
0≤s≤t
Let φ(x) = E x [eλτ ], λ < 0. We claim that φ satisfies
(*)
1 ′′
φ + λφ = 0, |x| < 1,
2
φ = 1, |x| = 1.
132
18. The Feynman-Kac Formula
136
Assume φ to be sufficiently smooth. Using Itô’s formula we get
1
d(eλt φ(β(t)) = eλt φ′ (β(t))dβ(t) + [λφ(β(t)) + φ′′ (β(t))]eλt dt.
2
Therefore
λt
e φ(β(t)) − φ(β(0)) =
Zt
eλs φ′ (β(s))dβ(s)+
0
+
Zt
1
[λφ(β(s)) + φ′′ (β(s))]eλs ds,
2
0
i.e.
λt
e φ(β(t)) − φ(β(0)) −
Zt
1
[λφ(β(s)) + φ′′ (β(s))]eλs ds
2
0
is a martingale. By Doob’s optional sampling theorem we can stop this
martingale at time τ, i.e.
λ(t∧τ)
e
Zt∧τ
1
φ(β(t ∧ τ)) − φ(β(0)) − [λφ(β(s)) + φ′′ (β(s))]eλs ds
2
0
is also a martingale. But for s ≤ t ∧ τ,
1
λφ + φ′′ = 0.
2
Thus we conclude that φ(β(t ∧ τ))eλ(τ∧t) is a martingale. Since λ <
0 and φ(β(t ∧ τ)) is bounded, this martingale is uniformly integrable.
Therefore equating the expectation at t = 0 and t = ∞ we get (since
φ(β(τ)) = 1)
φ(x) = E x [eλτ ].
133
By uniqueness property this must be the solution. However (*) has
a solution given by
√
Cos( (2λx))
(x) =
√
.
Cos( (2λ))
137
Therefore
E0 [eλτ ] =
(1)
1
√
(λ < 0),
Cos( (2λ))
If F(t) = P(τ ≥ t), then
Z∞
eλt dF(t) = E0 (eλτ ).
0
A theorem on Laplace transforms now tells us that (1) is valid till
π2
1
√
. This occurs at λ = . By
we cross the first singularity of
8
Cos( (2λ))
the monotone convergence theorem
E0 [eτπ
R∞
2 /8
] = +∞
π2
π2
and diverges for λ ≥
.
8
8
0
R∞
π2
Thus
is the supremum of λ for which eλt dF(t) converges, i.e. sup
8
0
[λ : E0 (eλτ )] exists, i.e. the decay rate is connected to the existence or
the non existence of the solution of the system (*). This is a general
feature and prevails even in higher dimensions.
Hence
eλt dF(t) converges for λ <
19. An Application of the
Feynman-Kac Formula. The
Arc Sine Law.
LET β(t, ·) BE THE one-dimensional Brownian motion with β(0, ·) = 0. 134
Define
Zt
1
ξt (w) =
X[0,∞) (β(s, w))ds.
t
0
ξt (w) is a random variable and denotes the fraction of the time that a
Brownian particle stays above the x-axis during the time interval [0, t].
We shall calculate
P[w : ξt (w) ≤ a] = Ft (a)
1
Brownian Scaling. Let Xt (s) = √ β(ts). Then Xt is also a Brownian
t
motion with same distribution as that of β(s). We can write
ξt (w) =
Z1
X[0,∞) (Xt (s, w))ds.
0
The ξt (w) = time spent above the x-axis by the Brownian motion
Xt (s) in [0, 1]. Hence Ft (a) is independent of t and is therefore denoted
139
19. An Application of the Feynman-Kac Formula....
140
by F(a). Suppose we succeed in solving for F(a); if, now,
ξt∗ (w)
=
Zt
X[0,∞) (β(s))ds = tξt ,
0
then the amount of time β(s, ·) > 0 in [0, t] is t (amount of time Xt (s) > 0
in [0, 1]). Hence we can solve for P[ξt∗ (w) ≤ a] = Gt (a). Clearly the
solution of Gt is given by
Gt (a) = F(a/t).
It is clear that
135



0 if
F(a) = 

1 if
a ≤ 0,
a ≥ 1.
Hence it is enough to solve for F(a) in 0 ≤ a ≤ 1. Let
uλ (t, x) = E x [e−(λ
Rt
0
X[0,∞) (β(s,w))ds)
Then
−ξ(w))
u1 (t, 0) = E[e
]=
Z1
]
e−tx dF(x).
0
Also note that uλ (t, x) is bounded by 1, if λ ≥ 0. By the Feynman1
1
Kac formula (appropriately modified in case ∆ is replaced by − ∆)
2
2
u1 (t, x) satisfies
∂u 1 ∂2 u
− u, x > 0,
=
∂t
2 ∂x2
1 ∂2 u
,
x < 0,
=
2 ∂x2
(*)
and u(0, x) = 1. Let
φα (x) = α
Z∞
0
u(t, x)e−αt dt,
α > 0,
where
u = u1 ,
141
=
Z∞
u(t, x)(−de−αt ).
0
A simple integration by parts together with (*) gives the following
system of ordinary differential equations for φα :
1
+ (α + 1)φα = α, x > 0,
− φ′′
2 α
1
+ αφα = α,
x < 0.
− φ′′
2 α
These have a solution
√
√
α
φα (x) =
+ Aex (2(α+1)) + Be−x (2(α+1)) ,
α+1 √
√
= 1 + Cex (2α) + De−x (2α) , x < 0.
x > 0,
136
However u is bounded by 1 (see definition of uλ (t, x)). Therefore φα
is also bounded by 1. This forces A = D = 0. We demand that φα and
dφα
should match at x = 0. (Some justification for this will be given
dx
later on).
Lt φα (x) = Lt φα (x)
x→0+
x→0−
gives
α
+ B = C + 1.
1+α
√
√
dφα
. SolvSimilarly we get −B (2(α + 1)) = C (2α) by matching
dx
ing for B and C we get
√
α
B=
√
√
,
(1 + α)( α + (α + 1))
C=√
Therefore
1
√
√
.
(1 + α)( α + (α + 1))
√
α
α
+B= √
φα (0) =
,
α+1
(α + 1)
19. An Application of the Feynman-Kac Formula....
142
i.e.
Z∞
E[e−tξt αe−αt ]dt = √
0
√
α
.
(α + 1)
Using Fubini’s theorem this gives
E[
√ α
α
]= (
),
α+ξ
α+1
or
√
1
1
E[
]= (
),
1 + ξ/a
1 + (1/α)
i.e.
Z1
0
dF(x)
1
.
=√
1 + γx
(1 + γ)
137
This can be inverted to get
dF(x) =
2
dx
√
.
π (x(1 − x))
(Refer tables on transforms or check directly that
2
π
Z1
0
1
dx
1
√
=√
1 + βx (x(1 − x))
(1 + β)
by expanding the left side in powers of (β). Therefore
F(a) =
Hence Gt (a) =
2
π
2
π
arcsin
√
( a),
0 ≤ a ≤ 1.
√
arcsin ( ( at )), 0 ≤ a ≤ t, i.e.
P[ξt ≤ a] =
2
π
arcsin
√ a
( ( )),
t
0 ≤ a ≤ t.
This result goes by the name of arc sine law for obvious reasons.
We now give some justification regarding the matching conditions
used above.
143
The equation we solved was
1
αφ − φ′′ + Vφ = f
2
where φ was bounded V ≥ 0. Suppose we formally use Itô’s formula to
calculate
Rt
d φ(β(t)e− 0 (α+V)(β(s)ds) )
= e−
Rt
0
(α+V)(β(s,·))ds
[− f (β(s, ·))dt +
dφ
(β(t))dβ(t)]
dX
(see Example 2 of Itô’s formula). Therefore
−
(Z(t, ·) = φ(β(t, ·))e
Rt
(α+V)(β(s,·))ds
0
+
Z
0
t
f β(s, ·) exp(−
Zs
(α + V)dσ)ds
0
is a martingale. Since φ, f are bounded and V ≥ 0,
|Z(t, ·)| ≤ ||φ||∞ + || f ||∞
Z∞
138
e−αs ds ≤ ||φ||∞ + C|| f ||∞ .
0
Therefore Z(t, ·) is uniformly integrable. Equating the expectations
at time 0 and ∞ gives
(*)
φ(0) = E0
Z∞
R
[ f (β(s, ·))e−αs−
s
0
V(β(σ)dσ)
]ds.
0
This is exactly the form obtained by solving the differential equations. In order to use Itô’s formula one has to justify it. If we show
that Itô’s formula is valid for functions having a discontinuity in the
second derivative, (*) will be a legitimate solution and in general there
is no reason why the second derivatives (or higher derivatives) should be
dφ
only.
matched. This partially explains the need for matching φ and
dx
19. An Application of the Feynman-Kac Formula....
144
Proposition . Let β(t, ·) denote a one-dimensional Brownian motion.
Suppose φ ∈ Cb1 and satisfied
1
αφ − φ′′ + Vφ = f,
2
Then
φ(β(t)) −
Zt
f (β(s))ds
0
is a martingale.
Proof. Let (φǫ ) ∈ Cb2 such that αφǫ − 21 φ′′
ǫ + Vφǫ + Vφǫ = fǫ and such
that (i) φǫ converges to φ uniformly on compact sets, (ii) φ′ǫ converges to
′′
φ′ uniformly on compact sets, (iii) φ′′
ǫ converges pointwise to φ except
at 0. We may suppose that the convergence is bounded.
139
Claim.
Rt
fǫ (β(s))ds converges to
Rt
f (β(s))ds a.e. As fǫ (β(s)) converges
0
0
to f (β(s)) except when β(s) = 0, it is enough to prove that
(*) P[w: Lebesgue measure (s : β(s) = 0) > 0]= 0. Let X{0} denote
the indicator function of {0}. Then
E
Zt
X{0} (β(s))ds =
0
Zt
EX{0} (β(s))ds = 0.
0
Thus (*) holds and establishes the claim. Now
φǫ (β(t)) −
Zt
fǫ (β(s))ds
0
is a uniformly bounded martingale converging to
φ(β(t)) −
Zt
0
f (β(s))ds.
145
Therefore
φ(β(t)) −
Zt
0
is a martingale.
f (β(s))ds
20. Brownian Motion with
Drift
LET Ω = C[0, ∞; Rd ], F = BOREL σ-FIELD of Ω, {X(t, ·)} ≡ Brow- 140
nian motion, Ft = σ[X(s, ·) : 0 ≤ s ≤ t], P x ≡ probability measure
on Ω corresponding to the Brownian motion starting at time 0 at x.
F = σ( U Ft ). Let b : Rd → Rd be any bounded measurable funcit≥0
ton. Then the map (s, w)| → b(w(s)) is progressively measurable and
Zt
Zt
1
|b(X(s, ·))|2 ds]
Z(t, ·) = exp[ hb(X(s, ·)), dX(s, ·)i −
2
0
0
is a martingale relative to (Ω, Ft , P x ). Define Qtx on Ft by
Z
t
Q x (A) =
Z(t, ·)dP x ,
A
i.e. Z(t, ·) is the Radon-Nikodym derivative of Qtx with respect to P x on
Ft .
Proposition .
(i) Qtx is a probability measure.
(ii) {Qtx : t ≥ 0} is a consistent family on ∪ Ft , i.e. if A ∈ Ft1 and
t2 ≥ t1 then Qtx 1(A) = Qtx 2(A).
t≥0
Proof. Qtx being an indefinite integral, is a measure. Since Z(t, ·) ≥ 0,
Qtx is a positive measure. Qtx (Ω) = E x (Z(t, ·)) = E x (Z(0, ·)) = 1. This
proves (i). (ii) follows from the fact that Z(t, ·) is a martingale.
147
20. Brownian Motion with Drift
148
If A ∈ Ft , we define
Q x (A) = Qt(A)
x .
141
The above proposition shows that Q x is well defined and since (Ft )
S
is an increasing family, Q x is finitely additive on Ft .
t≥0
S
Exercise. Show that Q x is countably additive on the algebra Ft .
t≥0
S
Then Q x extends as a measure to F = σ( Ft ). Thus we get a
t≥0
family of measures {Q x : x ∈ Rd } defined on (Ω, F ).
Proposition . If s < t then
Q x (Xt ∈ A|Fs ) = QX(s) (X(t − s) ∈ A) a.e.
Definition . If a family of measures {Q x } satisfies the above property it
is called a homogeneous Markov family.
Proof. Let B ∈ Fs . Therefore B ∩ Xt−1 (A) ∈ Ft and by definition,
Z
Z(t, ·)dP x
Q x ((X(t) ∈ A) ∩ B)) =
B∩Xt−1 (A)
E Px (Z(t, ·)χ(w)
B χA (X(t, ·)))
Z(t, ·)
Z(s, ·)χ(w)
= E Px (E Px (
B χA (X(t, ·))|F s )
Z(s, ·)
Z(t, ·)
= E Px ([χB Z(s, ·))E Px (
χA (χ(t, ·))]|Fs )
Z(s, ·)
(since B ∈ Fs and Z(s, ·) is Fs -measurable)
(1)
Z(t, ·)
χA (χ(t, ·))|Fs )] . . .
Z(s, ·)
Zt
Zt
1
Qx
Px
|b|2 ]χA (X(t, ·))|Fs )]
= E [χB E (exp[ hb, dXi −
2
= E Qx [χB E Px (
s
0
149
Qx
= E [χB E
PX(s)
Zt−s
Zt−s
1
|b|2 ]XA (X(t − s))]
(exp[ hb, dXi
2
0
0
(by Markov property of Brownian motion)
Qx
= E (χB E
Qt−s
X(s)
(χA (χ(t − s)))
(since
dQt−s
X(s)
dPX(s)
= Z(t − s, ·)
= E Qx (XB E QX(s) χA (X(t − s, ·))
142
The result follows from definition.
Let b : [0, ∞] × Rd → Rd be a bounded measurable function, P s,x
the probability measure corresponding to the Brownian motion starting
at time s at the point x. Define, for t ≥ s,
Zt
Z s,t (w) = exp[ hb(σ, X(σ, w)), dX(σ, w)i
s
1
−
2
Zt
|b(σ, X(σ, w))|2 dσ]
s
(i) Z s,t is a martingale relative to (Fts , P s,x ).
R
(ii) Define Qts,x by Qts,x (A) = Z s,t dP s,x , ∀A ∈ Fts .
Exercise.
A
Show that
Qts,x
is a probability measure on Fts .
(iii) Qts,x is a consistent family.
(iv) Q s,x defined on U Fts by Q s,x |Fts = Qts,x is a finitely additive set
t≥s
function which is countably additive.
(v) The family {Q s,x : 0 ≤ s < ∞, x ∈ Rd } is an inhomogeneous
Markov family, i.e.
Q s,x (X(t, ·) ∈ A|Fσs ) = Qσ,X(σ,·) (X(t, ·) ∈ A), ∀s < σ < t, A ∈ Fts .
20. Brownian Motion with Drift
150
[Hint: Repeat the arguments of the previous section with obvious 143
modifications].
Proposition . Let τ be a stopping time, τ ≥ s. Then
Q s,x [X(t, ·) ∈ A|Fτs ] = Qτ,Xτ (·) (X(t, ·) ∈ A) on τ(w) < t,
= χA (X(t, ·)) on τ(w) ≥ t.
Proof. Let B ∈ Fτs and B ⊂ {τ < t} so that B ∈ Fts .
E Qs,x (χB χA (Xt )) = E Ps,x (Z s,t χB χA (Xt ))
= E Ps,x [E Ps,x (Zτ,t Z s,τ χB χA (Xt )|Fτs )]
(since Z satisfies the multiplicative property)
= E Ps,x (Z s,τ χB E Ps,x (Zτ,t χA (Xt )|Fτs )]
(since Z s,τ is Fτs -measurable)
= E Ps,x [Z s,τ XB E Pτ ,Xτ (Zτ,t χA (Xt ))]
(*)
(by strong Markov property).
Now
dQ s,x = Z s,t .
dP s,x Fts
144
so that the optional stopping theorem,
dQ s,x = Z s, on {τ < t}, ∀x.
dP s,x Fτs
Putting this in (*) we get
E Qs,x [XB χA (Xt )] = E Ps,x [Z s,τ χB E Qτ,Xτ (χA Xt ) ].
Observe that
χB E Qτ,Xτ (χA (Xt ))
is Fτs -measurable to conclude the first part of the proof. For part (ii)
observe that
Xt−1 (A) ∩ {τ ≥ t} ∩ {τ ≤ k}
151
is in Fks if k > s, so that
Xt−1 (A) ∩ {τ ≥ t} ∈ Fτs .
Therefore
E Qs,x (Xt ∈ A ∩ (τ ≥ t))|Fτs ) = χA (Xt )χ{τ≥t} ,
or
E Qs,x [(Xt ∈ A)|Fτs ] = χA (Xt ) if
τ ≥ t.
Proposition . Let b : [0, ∞) × Rd → Rd be a bounded measurable function, f : Rd → R any continuous bounded function. If
∂u 1
+ ∆u + hb(s, x), ∆ui = 0,
∂s 2
u(t, x) = f (x)
0 ≤ s ≤ t,
has a solution u, then
u(s, x) =
Z
f (Xt )dQ s,x .
Ω
Remark . b is called the drift. If b = 0 and s = 0 then we recover the 145
result obtained earlier. With the presence of the drift term, the result is
the same except that instead of P s,x one has to use Q s,x to evaluate the
expectation.
Proof. Let
Y(σ, ·) =
Zσ
s
1
hb(θ, X(θ, ·)), dX(θ, ·)i −
2
Zσ
|b(θ, X(θ, ·)|2 dθ.
s
Step 1. (X(σ, ·)−X(s, ·), Y(σ, ·)) is a (d+1)-dimensional Itô process with
parameters
1
(0, 0, . . . , 0, − |b(σ, X(σ, ·))|2 ) and
2
20. Brownian Motion with Drift
152
d terms
"
I
bd×1
a = d×d
∗
b
b1×d
#
Let λ = (λ1 , . . . , λd ). We have to show that
µ
exp[µY +
2
Zσ
|b(σ, X(σ, ·)|2 dσ + hλ, X(σ, ·) − X(s, ·)i−
s
−
1
2
Zσ
hλ, λi + 2µhλ, b + µ2 |b(σ, ·)|2 dσ]
s
is a martingale, i.e. that
exp[hλ, X(σ, ·) − X(s, ·)i + µ
Zσ
s
1
hb, dXi −
2
Zσ
s
|λ + bµ|2 dρ].
is a martingale; in other words that
Zσ
Zσ
1
exp[ hλ + bµ, dXi −
|λ + bµ|2 dρ]
2
s
s
146
is a martingale. But this is obvious because
Z(σ, ·) =
Zσ
s
hλ + µb, dXi
is a stochastic integral and hence an Itô process with parameters (0, |λ +
µb|2 ). (Refer to the section on vector-valued Itô process).
Step 2. Put φ(σ, X(σ, ·), Y(σ, ·)) = u(σ, X(σ, ·))eY(σ,·) . By Itô formula,
dφ = eY
∂u
1X
π2 φ
dt + eY h∇u, dXi + ueY dY +
,
ai j
∂t
2
∂zi ∂z j
where z = (x, y), or
dφ = eY [
∂u
µ
1
dt + h∇u, dXi + uhb, dXi − |b|2 dt + ∇udt + hb, ∇uidt+
∂t
2
2
153
1
+ u|b|2 dt]
2
= eY [h∇u, dXi + uhb, dXi].
Therefore φ is an Itô process and hence a martingale. Therefore
E(φ(t, ·)) = E(φ(s, ·))
u(s, x) = E
P s,x
R
[( f (X(t))e
t
hb,dXi− 21
s
Rt
0
|b|2 dθ
]
= E Qs,x [ f (X(t))],
which proves the theorem.
Alternate Proof.
147
Exercise . Let Y(σ, ·) be progressively measurable for σ ≥ s. Then
Y(σ, ·) is a martingale relative to (Q s,x , Fts ) if and only if Y(σ)Z s,σ is
a martingale relative to (P s,x , Fts ).
Now for any function θ which is progressively measurable and bounded,
Zt
Zt
1
|θ|2 dσ]
exp[ hθ, dXi −
2
s
s
is a martingale relative to (Ω, Fts , P s,x ). In particular let θ be replaced
by θ + b(σ, w(σ)). After some rearrangement one finds that Xt is an Itô
process with parameters b, I relative to Q s,x . Therefore
u(t, Xt ) −
Zt
s
!
∂u
1
+ hb, ∇ui + ∇u dσ
∂σ
2
is a martingale relative to Q s,x . But
∂u
1
+ hb, ∇ui + ∇u = 0.
∂σ
2
20. Brownian Motion with Drift
154
Therefore
E Qs,x (u(t, X(t)) = u(s, X).
We have defined Q s,x by using the notion of the Radon-Nikodym
derivative. We give one more relation between P and Q.
Theorem . Let T : C([s, ∞), Rd ) → C([s, ∞), Rd ) be given by
TX = Y
where
Y(t) = X(t) −
Zt
b(σ, X(σ))dσ.
s
(b is as before). Then
Q s,x T −1 = P s,x .
148
Proof. Define Y(t, w) = X(t, T w) where X is a Brownian motion. We
prove that Y is a Brownian motion with respect to Q s,x . Clearly Y is
progressively measurable because T is (Ft − Ft )-measurable for every
t, i.e. T −1 (Ft ) ⊂ Ft and X is progressively measurable. Clearly Y(t, w)
is continuous ∀ w. We have only to show that Y(t2 ) − Y(t1 ) is Q s,x independent of Fts1 and has distribution N(0; (t2 − t1 )I) for each t2 >
t1 ≥ s. But we have checked that
1
exp[hθ, Xt − xi − |θ|2 (t − s) −
2
Zt
hθ, bidσ]
s
is a martingale relative to Q s,x . Therefore
1
E Qs,x (exphθ, Yt2 − Yt1 i|Fts1 ) = exp( |θ|2 (t2 − t1 )),
2
showing that Yt2 − Yt1 is independent of Fts1 and has normal distribution
N(0; (t2 − t1 )I). Thus Y is a Brownian motion relative to Q s,x . Therefore
Q s,x T −1 = P s,x .
21. Integral Equations
Definition . A function b : Rd → Rd is said to be locally Lipschitz if 149
given any x0 ∈ Rd there exists an open set U0 containing x0 such that
b|U0 is Lipschitz.
Exercise 1. b is locally Lipschitz iff b|K is Lipschitz for every compact
set K i.e. iff b|K is Lipschitz for every closed sphere K.
Exercise 2. Every locally Lipschitz function is continuous.
Theorem . Let b : Rd → Rd be locally Lipschitz and X : [0, ∞) → Rd
continuous. Then
(i) the equation
Y(t) = X(t) +
Zt
b(Y(s))ds
(∗)
0
has a continuous solution near 0, i.e. there exists an ǫ > 0 and a
continuous function Y : [0, ǫ] → Rd such that the above equation
is satisfied for all t in [0, ǫ].
(ii) (Uniqueness) If Y1 , Y2 are continuous solutions of the above equation in [0, T ], then
Y1 = Y2
on
[0, T ].
Proof. (ii) (Uniqueness) Let f (t) = |Y1 (t) − Y2 (t)|. As Y1 , Y2 are continuous, there exists a k > 0 such that |Y1 (t)|, |Y2 (t)| ≤ k for all t in
155
21. Integral Equations
156
150
[0, T ]. Choose C such that |b(x) − b(y)| ≤ C|x − y| for |x|, |y| ≤ k and
Rt
n
C sup f (t). Then f (t) ≤ C and f (t) ≤ C f (s)ds so that f (t) ≤ (ct)
n! for
0≤t≤T
0
all n = 1, 2, 3, . . .. Thus Y1 (t) = Y2 (t), proving uniqueness.
(i) (Existence) We can very well assume that X(0) = 0. Let a =
inf{t : |X(t)| ≥ 1},
M > sup{|b(x)| : |x| ≤ 2},
α = inf{a,
1
},
M
C , 0, a Lipschitz constant, so that |b(x) − b(y)| ≤ C|x − y| for all |x|,
|y| ≤ 2. Define the iterations Y0 , Y1 , . . . by
Y0 (t) = X(t),
Yn+1 (t) = X(t) +
Zt
b(Yn (s))ds
0
for all t ≥ 0. By induction, each Yn is continuous. By induction again,
|Yn (t) − X(t)| ≤ Mt for all n, 0 ≤ t ≤ α. Again, by induction |Yn+1 (t) −
(Ct)n+1
Yn (t)| ≤ M
C (n+1)! for 0 ≤ t ≤ α. Again, by induciton |Yn+1 (t) − Yn (t)| ≤
M (Ct)n+1
C (n+1)!
for 0 ≤ t ≤ α. Thus Yn (t) converges uniformly on [0, α] to a
continuous function Y(t) which is seen to satisfy the integral equation.
Remark . Let X : [−δ, ∞) → Rd be continuous where δ > 0. Then a
similar proof guarantees that the equation (*) has a solution in [−ǫ, ǫ]
for some ǫ > 0.
Define B(X) = sup{t : (∗) has a solution in [0, t]}. The theorem
above implies that 0 < B(X) ≤ ∞. B(X) is called the exploding time.
Remark . If b is, in addition, either bounded or globally Lipschitz,
B(X) = ∞ for every continuous X : [0, ∞) → Rd .
151
Example. Let b(y) = y2 , X(t) = 0. The equation
Y(t) = x0 +
Zt
0
b(y(s))ds
157
with x0 > 0 has a solution
Y(t) =
1
x0
1
1
, ∀t < ;
x0
−t
the solution explodes at t = x−1
0 .
Proposition . If
B(w) < ∞,
then
Lt |y(t)| = +∞.
t↑B(w)
Proof. Suppose that lim |y| = R < ∞. Let (tn ) be a sequence increast→B(w)
ing to B(w) such that |y(tn )| ≤ R + 1, ∀n. Let
τn = inf{t ≥ tn : |y(t) − y(−n )| ≥ 1}.
Then
1 = |y(τn ) − y(tn )|
≤ w(τn ) − w(tn )| + (τn − tn ) sup |b(λ)| . . . , (1)
λ ∈ S (y(tn ), 1).
Since (tn ) is bounded, we can choose a constant M such that
|w(tn ) − w(t)| <
1
2
if
|t − tn | ≤ M.
Then using (1),
τn − tn ≥ inf{M, (2 sup |b(λ)|)−1
where
λ ∈ S (Y(tn ); 1)
Therefore
τn − tn ≥ inf(M, (2 sup |b(λ)|)−1 , λ ∈ S (0; R + 2)) = α(say)∀ n.
152
21. Integral Equations
158
Chose n such that τn > B(w) > tn . Then y is bounded in [tn , B(w)]
and hence it is bounded in [0, B(w)). From the equation
y(t) = X(t) +
Zt
b(y(s))ds
0
one then gets that
Lt y(t) exists. But this is clearly a contradiction
t→B(w)
since in such a case the solution exists in [0, B(w) + ǫ) for suitable ǫ,
contradicting the definition of B(w). Thus
lim |y(t)| = +∞
t→B(w)
and hence
lim |y(t)| = +∞.
t→B(w)
Corollary . If b is locally Lipschitz and bounded, then B(X) = ∞ for all
X in C([0, ∞), Rd ).
Proof. Left as an exercise.
Proposition . Let b : Rd → Rd be locally Lipschitz and bounded. Define
T : C([0, ∞), Rd ) → C([0, ∞), Rd ) by T X = Y where
Y(t) = X(t) +
Zt
b(Y(s))ds.
0
Then T is continuous.
153
Proof. Let X, X ∗ : [0, ∞) → Rd be continuous, K > 0 be given.
Let Y0 , Y1 , . . . , Y0∗ , Y1∗ , . . . be the iterations for X, X ∗ respectively. Then
|Yn (t) − X(t)| ≤ K||b||∞ for 0 ≤ t ≤ K, n = 0, 1, 2, 3, . . ., so that we can
find R such that |Yn (t)|, Yn∗ (t)| ≤ R for 0 ≤ t ≤ k, n = 0, 1, 2, . . ., Let
C ≥ 1 be any Lipschitz constant for the function b on |x| ≤ R. Then
|Yn (t) − Yn∗ (t)| ≤ sup |X(t) − X ∗ (t)| · (1 + Ct +
0≤t≤K
(Ct)2
+
2!
159
+ ··· +
(Ct)n
)
n!
for
0 ≤ t ≤ K,
n = 0, 1, 2, 3, . . . .
A b is bounded, Yn converges uniformly to Y on [0, K]. Letting
n → ∞, we get
sup |(T X)t − T X ∗ )t| ≤ eck sup |X(t) − X ∗ (t)|, . . . (2)
0≤t≤K
0≤t≤K
where c depends on sup |X(t)|, sup |X ∗ (t)|. The proof follows by (2).
0≤t≤K
0≤t≤K
22. Large Deviations
LET Pǫ BE THE Brownian motion starting from zero scaled to Brown- 154
∆
ian motion corresponding to the operator ǫ . More precisely, let
2
!
A
Pǫ (A) = P √
ǫ
where P is the Brownian motion starting at time 0 at the point 0.
Interpretation 1. Let {Xt : t ≥ 0} be Brownian motion with X(0) = x.
Let Y(t) = X(ǫt), ∀t ≥ 0. Then Pǫ is the measure induced by the process
Y(t). This amounts to stretching the time or scaling time.
√
Interpretation 2. Let Y(t, ·) = ǫX(t, ·). In this case also Pǫ is the
measure induced by the process Y(t, ·). This amounts to ‘looking at the
process from a distance’ or scaling the length.
Exercise. Make the interpretations given above precise.
(Hint: Calculate (i) the probability that X(ǫt) ∈ A, and (ii) the probability
√
that ǫX(t, ) ∈ A).
Problem. Let
1
I(w) =
2
Z1
|ẇ(t)|2 dt
0
if w(0) = 0, w absolutely continuous on [0, 1]. Put I(w) = ∞ otherwise.
161
22. Large Deviations
162
We would like to evaluate
Z
e
F(w)
ǫ
dPǫ (w)
Ω
155
for small values of ǫ. Here F(w) : C[0, 1] → R is assumed to the a
bounded and continuous function.
Theorem . Let C be any closed set in C[0, 1] and let G be any open set
in C[0, 1]. Then
Z
lim sup ∈ log Pǫ (C) ≤ −
I(w),
ǫ→0
w∈C
lim inf ǫ log Pǫ (G) ≥ − inf I(w).
w∈G
ǫ→0
Here Pǫ (G) = Pǫ (π−1 G) where π : C[0, ∞) → C[0, 1] is the canonical projection.
Significance of the theorem . If
1.
dPǫ = e
then
Pǫ(A) =
Z
−I(w)
ǫ
e
,
−I(w)
ǫ
dPǫ
A
is asymptotically equivalent to
1
exp[− inf I(w)].
ǫ w∈A
2. If A is any set such that
inf I(w) = inf I(w),
w∈A0
w∈A
then by the theorem
Lt log Pǫ (A) = inf I(w).
ǫ→0
w∈A
163
156
Proof of the theorem.
Lemma 1. Let w0 ∈ Ω with I(w0 ) = ℓ < ∞. If S = S (w0 ; δ) is any
sphere of radius δ with centre at w0 then lim ∈ log Pǫ (S ) ≥ −I(w0 ).
ǫ→0
Proof.
P(S ) =
=
Z
Z
χ(w)
S (w0 ,δ) dPǫ
0)
χS(w−w
(0;) dPǫ
!
w
=
χS (0;δ) (λw)dP √ , where λ(w) = w − w0 ,
ǫ
Z
√
=
χS (0;δ) (λ( ǫw))dP(w)
 1

Z
Z1
Z

√
1


=
χS (0,δ) ( ǫw) exp  hw0 , dXi −
|ẇ0 |2 dσ dP(w)


2
Z
0
0

 1
Z

Z
√


=
χS (0;δ) ( ǫw) exp  hẇ0 , dXi − I(w0 ) dP(w)


0
=
Z


Z1


1
 1

χS (0;δ) (w) exp −
hẇ0 , dXi − I(w0 ) dPǫ (w)
 ǫ

ǫ
0
=e
−I(w0 )
ǫ
1
Pǫ (S (0; δ))
Pǫ (S (0; δ))
Z
S (0;δ)


Z1


1


hẇ0 , dXi dPǫ
exp −

 ǫ

Z

−I(w0 )
1
 1
ǫ
Pǫ (S (0; δ))e  −
≥e
 ǫ Pǫ (S (0; δ))
0
Z1
S (0;δ) 0
by Jensen’s inequality,
=e
=e
−I(w0 )
ǫ
−I(w0 )
ǫ



hẇ0 , dXidPǫ 

Pǫ (S (0, δ))e0 (use Pǫ (w) = Pǫ (−w) if w ∈ S (0; δ))
Pǫ (S (0, δ)).
22. Large Deviations
164
157
Therefore
Pǫ (S (w0 ; δ)) ≥ e
−I(w0 )
ǫ
δ
P(S (0; √ ))
ǫ
or,
δ
ǫ log Pǫ (S (w0 ; δ)) ≥ −I(w0 ) + ǫ log P(S (0; √ ));
ǫ
let ǫ → 0 to get the result. Note that the Lemma is trivially satisfied if
I(w0 ) = +∞.
Proof of Part 2 of the theorem.
Let G be open, w0 ∈ G; then there exists δ > 0 with S (w0 , δ) ⊂ G.
By Lemma 1
lim ∈ log Pǫ (G) ≥ lim ∈ log Pǫ (S (w0 ; δ)) ≥ −I(w0 ).
ǫ→0
ǫ→0
Since w0 is arbitrary, we get
lim ∈ log Pǫ (G) ≥ − inf{I(w0 ) : w0 ∈ G}.
For part 1 we need some more preliminaries.
Lemma 2. Let (wn ) ∈ C[0, 1] be such that wn → w uniformly on [0, 1],
I(wn ) ≤ α < ∞. Then I(w) < α, i.e. I is lower semi-continuous.
Proof.
n
Step 1. w is absolutely continuous. Let {(x′i , x′′
i )}i=1 be a collection of
mutually disjoint intervals in [0, 1]. Then
n
X
i=1
|wm (x′i ) − wm (x′′
i )| ≤
n
X
i=1|
′′
Zxi
′ 1/2
|x′′
[ |wm |2 ]1/2
i − xi |
x′i
(by Hölder’s inequality)
165
1/2

x′′
1/2
 X
X
n
n Zi

 

′′
′
2
|wm |   |xi − xi |
≤ 


(again by Hölder)
i=1
i=1 x′
i
X
√
′ 1/2
≤ (2α)(
|x′′
.
i − xi |)
158
Letting m → ∞ we get the result.
Step 2. Observe that wm (0) = 0S 0 w(0) = 0. Therefore
wn (x + h) − wn (x) 2
1
|
| =|
h
h
≤
1
h
2
 x+h
Zx+h

 Z
1


wn dt|2 ≤ 2  |wn |dt

h 
x
x
x+h
Z
x
|wn |2 dt.
Hence
Z1−h
Z1−h Zh
wn (x + h) − wn (x) 2
1
|
| dx ≤
dx |(ẇn (x + t))|2 dt
h
h
0
0
≤
1
h
Zh
dt
0
1
≤ 2
2
Zh
0
1−h
Z
|ẇn (x + t)|2 dx
0
dt = 2α
0
letting n → ∞, we get
Z1−h
w(x + h) − w(x) 2
| dx ≤ 2α.
|
h
0
Let h → 0 to get I(w) ≤ α, completing the proof.
22. Large Deviations
166
159
Lemma 3. Let C be closed and put C δ =
S
S (w; δ); then
w∈C
lim ( inf I(w)) = inf I(w).
w∈C
δ→0 w∈C δ
Proof. If δ1 < δ2 , then C δ1 ⊂ C δ2 so that inf I(w) is decreasing. As
w∈C δ
C δ ⊃ C for each δ,
lim ( inf I(w)) ≤ inf I(w)
δ→0 w∈C δ
w∈C
Let ℓ = lim ( inf I(w)). Then there exists wδ ∈ C δ such that I(wδ ) →
δ→0 w∈C δ
ℓ, and therefore (I(wδ )) is a bounded set bounded by α (say).
Claim.
|wδ (t1 ) − wδ (t2 )| = |
Zt2
t1
Z
wδ dt| ≤ |(|t1 − t2 |)( |wδ |2 )1/2
√
√
≤ (2α|t1 − t2 |).
The family (wδ ) is therefore equicontinuous which, in view of the
fact that wδ (0) = 0, implies that it is uniformly bounded and the claim
follows from Ascoli’s theorem. Hence every subfamily of (wδ ) is equicontinuous. By Ascoli’s theorem there exists a sequence δn → 0 such
that wδn → w uniformly on [0, 1]. It is clear that w ∈ C. By lower
semicontinuity of I(w),
lim inf I(w) ≥ inf
δ→0 w∈C δ
w∈C
completing the proof.
160
Proof of Part 1 of the theorem. Let X be continuous in [0, 1]. For each
n let Xn be a piecewise linear version of X based on n equal intervals, i.e.
167
Xn is a polygonal function joining the points (0, X(0)), (1/n, X(1/n)), . . . ,
(1, X(1)).
Pǫ (||Xn − X|| ≥ δ), (|| · || = || · ||∞ )


!

[
δ
j
−
1

sup · sup |Xr (t) − Xr
| ≥ √  ,
≤ P 
n
2 d
n i≤ j≤n j−1 ≤t≤ j
n
n
where
≤ ndPǫ ( sup |X(t) − X(0)| ≥
0≤t≤1/n
X = (X1 , . . . , Xd ).
δ
√ (Markov property; here X is one2 d
dimensional).



δ 
≤ ndPǫ  sup |Xt | ≥ √  (since X(0) = 0)
2 d
0≤t≤1/n
δ
≤ 2nd Pǫ ( sup Xt ≥ √ )
2 d
0≤t≤1/n
δ
= 2dn P( sup Xt ≥ √ )
2 ǫd
0≤t≤1/n
δ
= 4dn P(X(1/n) ≥ √ ) (by the reflection principle)
2 ǫd
Z∞
2
1
√
e−ny /2 dy
= 4dn
2π/n
√ √
= 4d
δ n/2 ǫd
Z∞
√ √
δ n/2 ǫd
2
1
√ e−x /2 dx
2π
Now, for every a > 0,
a
Z∞
a
Thus
−x2 /2
e
dx ≤
Z∞
2 /2
xe−x
2 /2
dx = e−a
.
a
161
22. Large Deviations
168
Pǫ (||Xn − X|| ≥ δ) ≤
√
2
ǫ −nδ2 /(8∈d)
4dn e−nδ /(8∈d)
√
e
,
=
C
(n)
√
√
1
δ
δ n/2 ǫd
where C1 depends only on n. We have now
Pǫ (Xn ∈ C δ ) ≤ Pǫ (I(Xn ) ≥ ℓδ ) where ℓδ = inf{I(w)w ∈ C δ }.


!
n−1
j

 1 X
j
+
1
= P 
|2 ≥ ℓδ 
n|X
−X
2 j=0
n
n
!
2ℓδ
2
2
2
,
= P Y1 + Y2 + · · · + Ynd ≥
ǫ
√
where Y1 = n(X1 (1/n) − X1 (0)) etc. are independent normal random
variables with mean 0 and variance 1. Therefore,
2ℓδ
2
≥
P(Y12 + · · · + Ynd
)
ǫ
Z
2ℓδ −(y2 +···+y2 )1/2
nd
e 1
dy1 . . . dynd .
=
ǫ
y21 +···+y2nd
= C(n)
√
Z∞
e−r
2 /2
rnd−1 dr,
(2ℓδ /ǫ)
using polar coordinates, i.e.
2
P(Y12 + Y22 + · · · + Ynd
2ℓδ
≥
) = C ′ (n)
ǫ
Z∞
nd
e−s s 2 −1 ds
(ℓδ /ǫ)
(change the variable from r to s =
Z∞
r2
). An integration by parts gives
2
e−s sk ds = e−α (αk +
α
k!
αk−2 + · · · ).
(k − 1)!
Using this estimate (for n even) we get
2
P((Y12 + · · · + Ynd
)≥
ℓδ nd
2ℓδ
) ≤ C2 (n)e−ℓδ /ǫ ( ) 2 −1 ,
ǫ
ǫ
169
where C2 depends only on n. Thus,
Pǫ (C) ≤ Pǫ (||Xn − X|| ≥ δ) + Pǫ (Xn < C δ )
! nd2 −1
√ ǫ −nδ2 /(8∈d)
−ℓδ /ǫ ℓδ
e
+ C2 (n)e
≤ C1 (n)
δ
ǫ
! nd2 −1
√ ǫ −nδ2 /(8∈d)
−ℓδ /ǫ ℓδ
≤ 2 max[C1 (n)
e
, C2 (n)e
δ
ǫ
√ ǫ −nδ2 /(8∈d)
e
∈ log Pǫ (C) ≤ ǫ log 2 + ǫ max[log(C1 (n)
δ
ℓδ nd
log C2 (n)e−ℓδ /ǫ ( ) 2 −1 ]
ǫ
162
Let ǫ → 0 to get
(
)
−nδ2 −ℓδ
lim ∈ log Pǫ (C) ≤ max
,
.
8d
1
Fix δ and let n → ∞ through even values to get
lim ∈ log Pǫ (C) ≤ −ℓδ .
Now let δ → 0 and use the previous lemma to get
Z
lim ∈ log Pǫ (C) ≤ −
I(w).
ǫ→0
w∈C
Proposition . Let ℓ be finite; then {w : I(w) ≤ ℓ} is compact in Ω.
Proof. Let (wn ) be any sequence, I(wn ) ≤ ℓ. Then
√
|wn (t1 ) − wn (t2 )| ≤ (ℓ|t1 − t2 |)
and since wn (0) = 0, we conclude that {wn } is equicontinuous and uniformly bounded.
Assumptions. Let Ω be any separable metric space, F = Borel σ-field
on Ω. For every ǫ > 0 let Pǫ be a probability measure. Let I : Ω →
[0, ∞] be any function such that
22. Large Deviations
170
(i) I is lower semi-continuous.
(ii) ∀ finite ℓ, {w : I(w) ≤ ℓ} is compact.
163
(iii) For every closed set C in Ω,
lim sup ∈ log Pǫ (C) ≤ − inf I(w).
w∈C
ǫ→0
(iv) For every open set G in Ω
lim inf ∈ log Pǫ (G) ≥ − inf I(w).
w∈G
ǫ→0
Remark . Let Ω = C[0, 1], Pǫ the Brownian measure corresponding to
1 R1 2
the scaling ǫ. If I(w) =
|w| dt if w(0) = 0 and ∞ otherwise, then all
20
the above assumptions are satisfied.
Theorem . Let F : Ω → R be bounded and continuous. Under the
above assumptions the following results hold.
(i) For every closed set C in Ω
Z
F(w)
dPǫ ≤ sup(F(w) − I(w)).
lim sup ǫ log exp
ǫ→0
ǫ
w∈C
C
(ii) For every open set G in Ω
Z
F(w)
lim inf ∈ log exp
dPǫ ≥ sup(F(w) − I(w)).
ǫ→0
ǫ
w∈G
G
In particular, if G = Ω = C, then
Z
F(w)
dPǫ = sup(F(w) − I(w)).
lim ∈ log exp
ǫ→0
ǫ
w∈Ω
Ω
171
Proof. Let G be open, w0 ∈ G. Let δ → 0 be given. Then there exists a
neighbourhood N of w0 , F(w) ≥ F(w0 ) − δ, ∀w in N. Therefore
Z
Z
F(w0 )−δ
F(w)
F(w)
dPǫ ≥
exp
dPǫ ≥ e ǫ Pǫ (N).
exp
ǫ
ǫ
N
G
164
Therefore
ǫ log
Z
exp
F(w)
dPǫ ≥ F(w0 ) − δ + ǫ log Pǫ (N).
ǫ
G
Thus
lim log
Z
exp
F(w)
dPǫ ≥ F(w0 ) − δ + lim ǫ log Pǫ (N).
ǫ
G
≥ F(w0 ) − δ − inf I(w) ≥ F(w0 ) − I(w0 ) − δ.
w∈N
Since δ and w0 are arbitrary (w0 ∈ G) we get
Z
F(w)
lim ∈ log exp
dPǫ ≥ sup(F(w) − I(w)).
ǫ
w∈G
G
This proves Part (ii) of the theorem.
Proof of Part (i).
Step 1. Let C be compact; L = sup(F(w) − I(w)). If L = −∞ it follows
w∈G
easily that
lim sup ∈ log
ǫ→0
Z
eF/ǫ dPǫ ≤ −∞.
C
(Use the fact that F is bounded). Thus without any loss, we may assume
L to be finite. Let w0 ∈ C; then there exists a neighbourhood N of w0
such that F(w) ≤ F(w0 ) + δ and by lower semi-continuity of I,
I(w) ≥ I(w0 ) − δ, ∀w ∈ N(w0 ).
22. Large Deviations
172
By regularity, there exists an open set Gw0 containing w0 such that
Gw0 Gw0 N(w0 ). Therefore
165
!
Z
F(w)
F(w0 ) + δ
exp
dPǫ exp
Pǫ (Gw0 ).
ǫ
ǫ
G w0
Therefore
lim sup ∈ log
ǫ→0
Z
exp
F(w)
dPǫ ≤ F(w0 ) + δ + ǫ lim Pǫ (Gw0 )
ǫ→0
ǫ
Gw0
≤ F(w0 ) + δ − inf I(w)
w∈G w0
F(w0 ) + δ − I(w0 ) + δ
≤ L + 2δ.
Let Kℓ = {w : I(w) ≤ ℓ}. By assumption, Kℓ is compact. Therefore,
for each δ > 0, there exists an open set Gδ containing Kℓ ∩ C such that
Z
F(w)
lim sup ∈ log e ǫ dPǫ ≤ L + 2δ.
ǫ→0
Gδ
Therefore
lim sup ∈ log
ǫ→0
Z
Z
Z
e
F(w)
ǫ
dPǫ ≤ L + 2δ,
G δ ∩C
e
F(w)
ǫ
dPǫ ≤ eM/ǫ P(C ∩ Gcδ ).
e
F(w)
ǫ
dPǫ ≤ M + lim sup ǫ log Pǫ (Cδc ∩ C)
G cδ ∩C
Therefore
lim sup ∈ log
ǫ→0
G cδ ∩C
ǫ→0
≤ M − inf
w∈C∩G cδ
I(w).
173
166
Now
Gcδ ⊂ Kℓc ∩ C c .
Therefore
C ∩ Gcδ ⊂ C ∩ Kℓc
if w ∈ C ∩ Gcδ , w < Kℓ . Therefore I(w) > ℓ. Thus
Z
eF(w)/ǫ dPǫ ≤ M − ℓ ≤ L ≤ L + 2δ.
lim sup ∈ log
ǫ→0
G cδ ∩C
This proves that
lim sup ∈ log
ǫ→0
Z
exp
F(w)
dPǫ ≤ L + 2δ.
ǫ
C
Since C is compact there exists a finite number of points w1 , . . . , wn
in C such that
n
[
C⊂
Gwi
i=1
Therefore
lim ∈ log
Z
C
F(w)
dPǫ ≤ lim ǫ log
exp
ǫ
S
≤ lim(ǫ log n Max
1≤i≤
Z
Z
eF(w)/ǫ dPǫ
n
i=1 G wi
exp
F(w)
dPǫ )
ǫ
G wi
≤ L + 2δ.
Since δ is arbitrary.
Z
F(w)
dPǫ ≤ sup(F(w) − I(w)).
lim ∈ log exp
ǫ
w∈C
C
The above proof shows that given a compact set C, and δ > 0 there 167
exists an open set G containing C such that
Z
F(w)
dPǫ ≤ L + 2δ.
lim ∈ log exp
ǫ
G
22. Large Deviations
174
Step 2. Let C be any arbitrary closed set in Ω. Let
L = sup(F(w) − I(w)).
w∈C
Since F is bounded there exists an M such that |F(w)| ≤ M for all
w. Choose ℓ so large that M − ℓ ≤ L. Since δ is arbitrary
Z
F(w)
lim sup ∈ log exp
dPǫ ≤ sup(F(w) − I(w))
ǫ→0
ǫ
w∈C
C
We now prove the above theorem when Pǫ is replaced by Qǫx . Let
Pǫx be the Brownian motion starting at time t = 0 at the space point x
corresponding to the scaling ǫ. Precisely stated, if
τǫ : C([0, ∞); Rd ) → C([0, ∞); Rd )
′′
is the map given by (τǫ w)(t) = w(ǫt), then Pǫx = P x τ−1
ǫ . Note T 1 τǫ = T ǫ
def
and T ǫ is given by
T ǫ w = y where y(t) = w(ǫt) +
Zt
b(y(s))ds.
0
Hence
−1
ǫ −1
P x T ǫ−1 = P x (T 1 , τǫ )−1 = P x τ−1
ǫ T1 = Px T1 ;
either of these probability measures is denoted by Qǫx .
Theorem . Let b : Rd → Rd be bounded measurable and locally Lipschitz. Define
Z1
1
|X(t) − b(X(t))|2 dt
I(w) :
2
0
168
If w ∈ C([0, ∞); Rd ), w(0) = x and x absolutely continuous. Put
I(w) = ∞ otherwise. If C is closed in C[(0, 1]; Rd ), then
lim ∈ log Qǫx (C) ≤ − inf I(w).
ǫ→0
w∈C
175
If G is open in C([0, 1]; Rd ), then
lim ∈ log Qǫx (G) ≥ − inf I(w).
w∈G
ǫ→0
As usual Qǫx (C) = Qǫx π−1 (C) where
π : C([0, ∞); Rd ) → C([0, 1]; Rd )
is the canonical projection.
Remark. If b = 0 we have the previous case.
Proof. Let T be the map x(·) → y(·) where
y(t) = x(t) +
Zt
b(y(s))ds.
0
Then
Qǫx = Pǫx (T −1 ).
If C is closed
Qǫx (C) = Pǫx (T −1 C).
The map T is continuous. Therefore T −1 (C) is closed. Thus
lim sup ∈ log Q x (C) = lim sup ∈ log Pǫx (T −1 C)
ǫ→0
≤−
(*)
ǫ→0
inf
w∈T −1 (C)
= − inf
w∈C
1
2
1
2
Z1
Z1
|X|2 dt
(see Exercise 1 below)
0
|T −1 w|2 dt.
0
Now
T −1
y(·) −−−→ y(t) −
Zt
0
169
b(y(s))ds.
22. Large Deviations
176
Therefore
(T −1 y) = y − b(y(s)).
Therefore
lim sup ∈ log Qǫx (C) ≤ − inf I(w).
w∈C
ǫ→0
The proof when G is one is similar.
Exercise 1. Replace Pǫ by Pǫx and I by Ix where
1
Ix (w) =
2
Z1
|w|2 , w(0) = x, w absolutely continuous,
0
= ∞ otherwise.
Check that (*) holds, i.e.
lim sup ∈ log Pǫx (C) ≤ − inf I x (w), if C is closed,
w∈C
ǫ→0
and
lim inf ∈ log Pǫx (G) ≥ − inf I x (w).
w∈G
ǫ→0
set in Rn ,
Let G be a bounded open
with a smooth boundary Γ = ∂G.
d
d
∞
Let b : R → R be a smooth C function such that
(i) hb(x), n(x)i0, ∀x ∈ ∂G where n(x) is the unit inward normal.
(ii) there exists a point x0 ∈ G with b(x0 ) = 0 and |b(x)| > 0, ∀x in
G − {x0 }.
170
(iii) for any x in G the solution
ξ(t) = x +
Zt
b(ξ(s))ds,
0
of the vector field starting from x converges to x0 as t → +∞.
Remark.
(a) (iii) is usually interpreted by saying that “x0 is stable”.
177
(b) By (i) and (ii) every solution of (iii) takes all its values in G and
ultimately stays close to x0 .
Let ǫ > 0 be given; f : ∂G → R be any continuous bounded function. Consider the system
Lǫ uǫ =
1
∆uǫ + b(x) · ∆uǫ = 0 in G
2
uǫ = f on ∂G.
We want to study lim uǫ (x). Define
ǫ→0
I0T (X(t))
1
=
2
ZT
|X(t) − b(X(t))|2 dt; X : [0, T ] → Rd
0
whenever X is absolutely continuous, = ∞ otherwise.
Remark. Any solution of (iii) is called an integral curve. For any curve
X on [0, T ], I0T gives a measure of the deviation of X from being an
integral curve. Let
VT (x, y) = inf{I0T (X) : X(0) = x; X(T ) = y}
and
V(x, y) = inf{VT (x, y) : T > 0}.
V has the following properties.
(i) V(x, y) ≤ V(x, z) + V(z, y) ∀x, y, z.
(ii) Given any x, ∃δ → 0 and C > 0 such that for all y with |x − y| ≤ δ.
V(x, y) ≤ C|x − y|
Proof. Let X(t) =
Put
t(y − x)
+ x.
|y − x|
T = |y − x|, X(0) = x, X(T ) = y,
171
22. Large Deviations
178
I0T (X(t))
1
=
2
ZT 2
y − x − b(X + S (y − x)) ds.
T
T
0
Then
I0T
1
≤
2
ZT
!
|y − x|2
2
+ ||b||∞ ds,
2
T2
0
where
||b||∞ =
sup
b(λ),
|λ−x|≤|y−x|
or,
I0T ≤ (1 + ||b||2∞ )|y − x|.
As a consequence of (ii) we conclude that




2

V(x, y) ≤ 1 + sup |b(λ)|  |y − x|,
|λ−x≤|y−x|
i.e. V is locally Lipschitz.
The answer to the problem raised is given by the following.
Theorem .
lim uǫ (x) = f (y0 )
ǫ→0
where y0 is assumed to be such that y0 ∈ ∂G and
V(x0 , y0 ) < V(x, y), ∀y ∈ ∂G, y , y0 .
172
We first proceed to get an equivalent statement of the theorem. Let
be the Brownian measure corresponding to the starting point x, and
corresponding to the scaling ǫ. Then there exists a probability measure
Qǫx such that
dQǫx Ft = Z(t)
dPǫx
Pǫx
179
where
Z(t, ·) = exp
Zt
1
hb∗ (X(s)), dX(s)i −
2
Zt
b∗ (X(s))ds;
0
0
b∗ is any bounded smooth function such that b∗ = b on G. Further we
have the integral representation
Z
f (X(τ))dQǫx
uǫ (x) =
∂G
where τ is the exit time of G, i.e.
τ(w) = inf{t : w(t) < G}.
Z
|uǫ (x) − f (y0 )| = | ( f (X(τ)) − f (y0 ))dQǫx |
≤|
∂G
Z
( f (X(τ)) − f (Y0 ))dQǫx |+
N∩∂G
+|
Z
N c ∩∂G
( f (X(τ)) − f (Y0 ))dQǫx |
(N is any neighbourhood of y0 ).
≤
Qǫx (X(τ)
∈ N ∩ ∂G) sup | f (λ) − f (y0 )|+
λ∈N∩∂G
ǫ
+2|| f ||∞ Q x (X(τ) ∈ N c
∩ ∂G).
173
Since f is continuous, to prove the theorem it is sufficient to prove
the
Theorem .
lim Q x (X(τ) ∈ N c ∩ ∂G) = 0
ǫ→0
for every neighbourhood N of y0 .
22. Large Deviations
180
Let N be any neighbourhood of y0 . Let
V = V(x0 , y0 ), V ′ =
inf
y∈N c ∩∂G
V(x, y).
By definition of y0 and the fact that N c ∩∂G is compact, we conclude
that V ′ > V. Choose η = η(N) > 0 such that V ′ = V + η. For any δ > 0
let D = S (x0 ; δ) = {y : |y − x0 | < δ}, ∂D = {y : |y − x0 | = δ}.
Claim. We can choose a δ2 such that
3η
, ∀x ∈ ∂D2 , y ∈ N c ∂G.
4
η
(ii) V(x, y0 ) ≤ V + , ∀x ∈ ∂D2 .
4
(i) V(x, y) ≥ V +
Proof.
(i) V(x0 , y) ≥ V + η, ∀y ∈ N c ∂G. Therefore
V + η ≤ V(x0 , y) ≤ V(x0 , x) + V(x, y)
≤ C|x − x0 | + V(x, y).
η
Choose C such that C|x − x0 | ≤ . Thus
4
V+
174
3η
η
≤ V(x, y) if C|x − x0 | ≤ , ∀y ∈ N c ∂G.
4
4
C depends only on x0 . This proves (i).
η
(ii) |V(x0 , y0 ) − V(x, y0 )| ≤ V(x0 , x) ≤ C|x0 − x| ≤ if x is close to x0 .
4
Thus
η
η
V(x, y0 ) ≤ V(x0 , y0 ) + = V +
4
4
if x is close to x0 . This can be achieved by choosing δ2 very small.
Claim (iii). We can choose δ1 < δ2 such that for points x1 , x2 in ∂D1
there is a path X(·) joining x1 , x2 with X(·) ∈ D2 − D1 , i.e. it never
penetrates D1 ; and
η
I(X) ≤ .
8
181
Proof. Let C = sup{|b(λ)|2 : |λ − x0 | ≤ δ2 }. Choose X(·) to be any path
on [0, T ], taking values in D2 with X(0) = x1 ; X(T ) = x2 and such that
|X| = 1 (i.e. the path has unit speed). Then
I0T (X)
≤
ZT
2
(|X| + C)dt ≤ CT +
0
ZT
|X|dt
0
= (c + 1)T = (C + 1)|x2 − x1 |.
η
Choose δ1 small such that (C + 1)|x2 − x1 | ≤ .
8
Let Ωδ1 = {w : w(t) ∈ G − D1 , ∀t ≥ 0}, i.e. Ωδ1 consists of all
trajectories in G that avoid D1 .
Claim (iv).
inf
inf
I T (X(·))
T >0 X∈Ωδ1 ,X(0)∈∂D2 0
≥V+
3η
4
X(T ) ∈ N c ∩ ∂G
175
Proof. Follows from Claim (i) and (ii).
Claim (v).
inf
inf
T >0 X∈Ωδ1 ,X(0)∈∂D2
X(T )=y0
I0T (X(·)) ≤ V +
3η
.
8
η
Proof. By (ii) V(x, y0 ) ≤ V + ∀x ∈ ∂D2 , i.e.
4
η
I0T (X(·)) ≤ V + .
T >0 X(0)=x,X(T )=y0
4
inf
inf
η
Let ǫ > 0 be arbitrary. Choose T and X(·) such that I0T (X) ≤ V + +ǫ
4
with X(0) = x, X(T ) = y0 , X(·) ∈ G. If X ∈ Ωδ1 define Y = X. If X < Ωδ1
define Y as follows:
22. Large Deviations
182
Let t1 be the first time that X enters D1 and t2 the last time that it
gets out of D1 . Then 0 < t1 ≤ t2 < T . Let X ∗ be a path on [0, s] such
η
that (by Claim (iii)) I0s (X ∗ ) ≤ with X ∗ (0) = X(t1 ) and X ∗ (s) = X(t2 ).
8
Define Y on [0, T − (t2 − t1 ) + s] [T − (t2 − t1 ) + s, ∞) by
Y(t) = X(t) on [0, t1 ] = X ∗ (t − t1 ) on [t1 , t1 + s]
= X(t − t1 − s + t1 ), on [t1 + s, T − (t2 − t1 ) + s]
= X(t2 ), for t ≥ T − (t2 − t1 ) + s.
Then
I0T −t2 +t1 +s
1
=
2
Zt1
1
|X − b(X(s))| ds +
2
2
|X ∗ − b(X ∗ (s))|2 ds
0
0
+
1
2
ZT
t2
|X(s) − b(X(s))|2 ds
≤V+
176
Zs
η
η
+ǫ+
4
8
by choice of X and X ∗ . As Y ∈ Ωδ1 , we have shown that
I T (X(·))
inf
inf
T >0 X∈Ωδ1 ,X(0)∈∂D1 0
X(T )=y0
≤V+
3η
+ ǫ.
8
Since ǫ is arbitrary we have proved (v).
Lemma . I0T is lower semi-continuous for every finite T .
Proof. This is left as an exercise as it involves a repetiti on of an argument used earlier.
Lemma . Let Xn ∈ Ωδ1 . If T n → ∞ then I0T n (Xn ) → ∞.
This result says that we cannot have a trajectory which starts outside
of a deleted ball for which I remains finite for arbitrary long lengths of
time.
183
Proof. Assume the contrary. Then there exists a constant M such that
I0T n (X0 ) ≤ M, ∀n. Let T < ∞, so that MT = sup I0T (Xn ) < ∞.
n
Define XnT = Xn |[0,T ] .
Claim. {XnT }n=1 is an equicontinuous family.
177
Proof.
|XnT (x2 ) −
XnT (x1 )|2
=|
Zx2
XnT (t)dt|2
x1
Zx2
≤ |x2 − x1 |2
|XnT |2 dt
x1
2
≤ 2|x2 − x1 |
Zx2
|XnT
−
b(XnT )|2 ds
+
ZT
b(XnT )2 ds]
0
≤ 2|x2 − x1 |2 [2MT + T ||b||2∞ ].
Thus, {XnT }n is an equicontinuous family. Since G is bounded, {XnT }n
is uniformly bounded. By Arzela-Ascoli theorem and a “diagonal procedure” there exists a subsequence Xnk and a continuous function to X
uniformly on compact subsets of [0, ∞). As Xnk (·) ∈ G−D1 , X ∈ G−D1 .
Let m ≥ n. I0T n (Xm ) ≤ M. Xn → X uniformly on [0, T n ]. By lower semicontinuity I0T (X) ≤ M. Since this is true for every T we get on letting T
tend to ∞, that
Z∞
1
|X − b(X(s))|2 ds ≤ M.
2
0
Thus we can find a sequence a1 < b1 < a2 < b2 < . . . such that
Iabnn (X(·))
1
=
2
Zbn
an
|X(t) − b(X(t))|2 dt
22. Large Deviations
184
converges to zero with bn − an → ∞. Let Yn (t) = X(t + an ). Then
I0bn −an (Yn ) → 0
178
with bn − an → +∞,
Y n ∈ Ω δ1 .
Just as X was constructed from Xn , we can construct Y from Yn such
that Yn → Y uniformly on compact subsets of [0, ∞).
I0bn −an (Y) ≤ inf I0bn −an (Ym ) = 0
m≥n
(by lower semi-confirmity of I0T ). Thus I0bn −an (Y) = 0, ∀n, showing that
Z∞
Y(t) − b(Y(t))|2 dt = 0
0
Thus Y satisfies Y(·) = x +
Rt
b(Y(s))ds with Y(t) ∈ G − ∂D1 , ∀t.
0
Case (i). Y(t0 ) ∈ G for some t0 . Let Z(t) = Y(t+t0 ) so that Z is an integral
curve starting at a point of G and remaining away from D1 contradicting
the stability condition.
Case (ii). Y(t0 ) < G for any t0 , i.e. Y(t) ∈ ∂G for all t. Since Y(t) =
b(Y(t))hY(t), n(Y(t))i is strictly positive. But Y(t) ∈ ∂G and hence
hY(t), n(Y(t))i = 0 which leads to a contradiction. Thus our assumption is incorrect and hence the lemma follows.
Lemma . Let x ∈ ∂D2 and define
E = {X(t) exits from G before hitting D1 and it exits from N}
F = {X(t) exists from G before hitting D1 and it exits from N c }
Then
Qǫx (F)
1
3η
≤ exp − + 0
Qǫx (E)
8ǫ
ǫ
179
!!
→ 0 uniformly in x(x ∈ ∂D2 ).
Significance. Qǫx (E) and Qǫx (F) are both small because diffusion is small
and the drift is large. The lemma says that Qǫx (E) is relatively much
larger than Qǫx (F).
185
Proof. Qǫx (E) ≥ Qǫx {X(t) exists from G before hitting D1 , and exists in
1
N before time T }, = Q x (B) ≥ exp[− inf I0T (X(·))] where the infimum
ǫ
is taken over the interior of B,
!
!#
"
1
3η
1
+0
.
≥ exp − V +
ǫ
8
ǫ
Similarly,
Qǫx (F)
"
!
!#
1
1
3η
≤ exp − V +
+0
.
ǫ
4
ǫ
Therefore
"
!#
Qǫx (F)
1
3η
≤ exp − + 0
→ 0 as
Qǫx (E)
8ǫ
ǫ
ǫ → 0.
We now proceed to prove the main theorem. Let
τ0 = 0,
τ1 = first time ∂D1 is hit,
τ2 = next time ∂D2 is hit,
τ3 = next time ∂D1 is hit,
..................
and so on. Observe that the particle can get out of G only between the
time intervals τ2n and τ2n+1 . Let En = {between τ2n and τ2n+1 the path
exits from G for the first time and that it exits in N}, Fn = {between τ2n 180
and τ2n+1 the path exits from G for the first time and that it exists in N c }.
Qǫx (X(τ) ∈ N) + Q x (X(τ) ∈ N c ) = 1.
Also
Qǫx (X(τ) ∈ N c ) =
∞
X
Qǫx (X(τ) ∈ N) =
∞
X
Qǫx (Fn ),
n=1
n=1
Qǫx (En ),
22. Large Deviations
186
≤
X
∞
X
Qǫx (Fn ) =
n=1
Qǫx
E [χ(τ>τ2n ) sup
x∈∂D2
n=1
≤ 0(ǫ)
∞
X
∞
X
n=1
ǫ
Q x (F)]
ǫ
E Qx (Qǫx (Fn |Fτ2n ))
(by the Strong Markov property)
ǫ
E Qx [χ(τ>τ2n ) inf Qǫx (E)]
(as
x∈∂D2
n=1
≤ 0(ǫ)
∞
X
n=1
Qǫx (F)
→ 0)
Qǫx (E)
Qǫx (En ) = 0(ǫ)Q x (X(τ) ∈ N c ).
Therefore
Qǫx (χ(τ) ∈ N) → 1,
Q x (X(τ) ∈ N c ) → 0.
Exercise. Suppose b(x) = ∇u(x) for some u ∈ C 1 (G ∪ ∂G, R). Assume
that u(x0 ) = 0 and u(x) < 0 for x , x0 . Show that
V(x0 , y) = −2u(y).
[Hint: For any trajectory X with X(0) = x0 ,
X(T ) =
y, I0T (X)
1
=
2
ZT
0
181
2
|X + ∇u(X)| dt − 2
ZT
∇u(X) · X(t)dt ≥ −2u(y)
0
so that V(x0 , y) ≥ −2u(y). For the other inequality, let u be a solution of
duX(s)
0
X(t) + ∇u(X(t) = 0 on [0, ∞) with X(0) = y. Show that because
ds
for X(s) , 0 and x0 is the only zero of u, limitX(t) = x0 . Now conclude
t→∞
that V(x0 , y) ≤ −u(y)].
23. Stochastic Integral for a
Wider Class of Functions
WE SHALL NOW define the stochastic integral for a wider class of 182
functions.
Let θ : [0, ) × Ω → Rd be any progressively measurable function
such that for every t
Zt
|θ(s, w)|2 ds < ∞,
a.e.
0
Define, for every finite L ≥ 0,





θ(s, w),




θL (s, w) = 






0,
if
if
Rs
0
|θ(t, w)|2 dt < L < ∞,
0
|θ(t, w)|2 dt ≥ L.
Rs
We can write θL (s, w) = θ(s, w)χ[0,L) (φ(s, w)) where
φ(s, w) =
Zs
|θ(t, w)|2 dt
0
is progressively measurable. Hence θL (s, w) is progressively measur187
23. Stochastic Integral for a Wider Class of Functions
188
able. It is clear that
RT
0
|θL (s, w)|2 ds ≤ L, a.e. ∀T . Therefore
ZT
E( |θL (s, w)|2 ds) ≤ L.
0
Thus the stochastic integral ξL (t, w) =
Rt
0
hθL (s, w), dX(s, w)i is well de-
fined.
The proofs of the next three lemmas follow closely the treatment of
Stochastic integration given earlier.
183
Lemma 1. Let τ be a bounded, progressively measurable, continuous
function. Let τ be any finite stopping time. If θ(s, w) = 0, ∀(s, w) such
Rt
that 0 ≤ s ≤ τ(w) then hθ(s, w), dX(s, w)i = 0 for 0 ≤ t ≤ τ(w).
0
Proof. Define θn (s, w) = θ( [ns]
n , w). θn is progressively measurable and
by definition of the stochastic integral of θn ,
Zt
hθn (s, w), dX(s, w)i = 0,
∀t,
0 ≤ t ≤ τ(w).
0
Now

 t

Z


E  |θn (s, w) − θ(s, w)|2 ds


0
 t

!
Z

[ns]


= E  |θ
w − θ(s, w)|2 ds → 0 as n → ∞


n
0
and
Zt
0
hθn (s, w), dX(s, w)i →
Zt
0
hθ(s, w), dX(s, w)i
189
in probability. Therefore
Zt
hθ, dXi = 0 if 0 ≤ t ≤ τ(w).
0
Lemma 2. If θ is progressively measurable and bounded the assertion
of Lemma 1 still holds.
Proof. Let
1
θn (t, w) =
n
Zt
θ(s, w)ds.
(t−1/n)V0
Then

 T

Z


E  |θn (t, w) − θ(t, w)|2 dt → 0 (Lebesgue’s theorem).


0
θn is continuous and boundd, θn (t, w) = 0 for 0 ≤ t ≤ τ(w). By lemma 1 184
Zt
hθn (s, w), dX(s, w)i = 0
0
if 0 ≤ t ≤ τ(w). This proves the result.
Lemma 3. Let θ be progressively measurable such that, for all t,

 t

Z


E  |θ(s, w)|2 ds < ∞.


0
If θ(s, w) = 0 for 0 ≤ s ≤ τ(w), then
Zt
0
hθ(s, w), dX(s, w)i = 0 for
0 ≤ t ≤ τ(w).
23. Stochastic Integral for a Wider Class of Functions
190
Proof. Define



θ,
θn = 

0,
Then
Zc
if |θ| ≤ n,
if |θ| > n.
hθn , dXi = 0, if 0 ≤ t ≤ τ(w),
(Lemma 2) and
0
Rt
E( |θn − θ|2 ds) → 0. The result follows.
0
Lemma 4. Let θ be progressively measurable such that ∀t,
Zt
|θ(s, w)|2 ds < ∞
a.e.
0
Then Lt ξL (t, w) exists a.e.
L→∞
185
Proof. Define


Zs








2
τL (w) = inf 
s
:
|θ(σ,
w)|
dσ
≥
L
;







0
clearly τL is a stopping time. If L1 ≤ L2 , τL1 (w) ≤ τL2 (w) and by
assumptions of the lemma τL ↑ ∞ as L ↑ ∞. If
L1 ≤ L2 ,
θL1 (s, w) = θL2 (s, w) for 0 ≤ s ≤ τL1 (w).
Therefore by Lemma 3,
ξL2 (t, w) = ξL1 (t, w)
if 0 ≤ t ≤ τL1 (w). Therefore as soon as L is large enough such that
t ≤ τL (w), ξL (t, w) remains constant (as a function of L). Therefore
Lt ξL (t, w) exists a.e.
L→∞
191
Definition. The stochastic integral of θ is defined by
Zt
hθ(s, w), dX(s, w)i = Lt ξL (t, w).
L→∞
0
Exercise. Check that the definition of the stochastic integral given above
coincides with the previous definition in case

 t

Z


E  |θ(s, w)|2 ds < ∞, ∀t.


0
Lemma . Let θ be a progressively measurable function, such that
Zt
|θ(s, w)|2 ds < ∞, ∀t.
0
If ξ(t, w) denotes the stochastic integral of θ, then
186
 T

Z


 L2
P sup |ξ(t, ·)| ≥ ǫ ≤ P  |θ|2 ds ≥ L + 2 .

 ǫ
0≤t≤T
!
0
Proof. Let τL = inf{t :
Rt
0
θ(s, w). Also
|θ|2 ds ≥ L}. If T < τL (w), then θL (s, w) =
ξL (t, w) = ξ(t, w) for 0 ≤ t ≤ T.
Claim.
(
)
w : sup |ξ(t, w)| ≥ ǫ
0≤t≤T
(
)
w : sup |ξL (t, w)| ≥ ǫ ∪ {w : τL (w) ≤ T }
0≤t≤T
23. Stochastic Integral for a Wider Class of Functions
192
For, if w is not contained in the right side, then
sup |ξL (t, w)|2 < ǫ
and
0≤t≤T
|τL (w)| > T.
If τL > T , ξL (t, w) = ξ(t, w) ∀t ≤ T . Therefore
sup |ξL (t, w)| = sup |ξ(t, w)|
0≤t≤T
0≤t≤T
Therefore w < left side. Since




ZT








2
w
:
|θ|
ds
≥
L
{w : τL (w) > T } = 









0
we get
P sup |ξ(t, ·)| ≥ ǫ
0≤t≤T
 T
Z

≤ P 

0
!

!


2
|θ| ds ≥ L + P sup |ξL (t, ·)| ≥ ǫ

0≤t≤T

 T

Z
 L2

≤ P  |θ|2 ds ≥ L + 2 ,
 ǫ

0
187
by Kolmogorov’s inequality. This proves the result.
Corollary . Let θn and θ be progressively measurable functions such
that
(a)
Rt
0
|θn (s, w)|2 ds < ∞,
(b) Lt
n→∞
Rt
0
Rt
0
|θ(s, w)|2 ds < ∞, ∀t;
|θn (s, w) − θ(s, w)|2 ds = 0 in probability.
If ξn (t, w) and ξ(t, w) denote, respectively the stochastic integrals of
θn and θ, then sup |ξn (t, w) − ξ(t, w)| converges to zero in probability.
0≤t≤T
193
Proof. Let τn,L (w) = inf{t :
Rt
0
|θn |2 ds ≥ L}; replacing θ by θn − θ and ξ
by ξn − ξ in the previous lemma, we get
P sup |ξn (t, ·) − ξ(t, ·)| ≥ ǫ
0≤t≤T
!

 T

Z
L2


≤ 2 + P  |θn (s, ·) − θ(s, ·)|2 ds ≥ L .


ǫ
0
Letting n → ∞, we get
lim P sup |ξn (t, ·) − ξ(t, ·)| ≥ ǫ
n→∞
0≤t≤T
!
L2
.
ǫ2
As L is arbitrary we get the desired result.
Proposition . Let θ be progressively measurable such that
Zt
|θ(s, w)|2 ds < ∞,
∀t and ∀ w.
0
Then
188
 t

Zt
Z

1


Z(t, ·) = exp  hθ(s, ·), dX(s, ·)i −
|θ(s, ·)|2 ds


2
(*)
0
0
is a super martingale satisfying
(a) E(Z(t, ·)) ≤ 1;
(b) Lt E(Z(t, ·)) = 1.
t→0
Proof. Let (θn ) be a sequence of bounded progressively measurable
functions such that
Zt
0
|θn − θ|2 ds → 0 ∀t,
∀w.
23. Stochastic Integral for a Wider Class of Functions
194
(For example we may take θn = θ if |θ| < n, = 0 otherwise). Then
(∗) is a martingale when θ is replaced by θn . This martingale satisfies
E(Zn (t, ·)) = 1, and Zn (t, ·) → Z(t, ·) pointwise (a) now follows from
Fatou’s lemma:
!
lim E(Z(t, ·)) ≥ E lim Z(t, ·)
t→0
t→0
= E(1) = 1.
Therefore Lt E(Z(t, ·)) = 1. This proves (b).
t→0
24. Explosions
Exercise. Let R > 0 be given. Put bR = bφR where φR = 1 on |x| ≥ R, 189
φR = 0 if |x| ≥ R + 1; φR is C ∞ . Show that bR = b on |x| ≤ R, bR is
bounded on Rd and bR is globally Lipschitz.
Let ΩT = {w ∈ Ω : B(w) > T }. Let S T = ΩT → C[0, T ] be the map
Rt
S T w = y(·) where y(t) = w(t) + b(y(s))ds on [0, T ]. Unless otherwise
0
specifie b : Rd → Rd is assumed to be locally Lipschitz. Define the
measure QTx on (Ω, T ) by
QTx (A) = P x {w : S T w ∈ A, B(w) > T },
where P x is the probability measure corresponding to Brownian motion.
Theorem .
QTx (A)
=
Z
∀A ∈ FT ,
Z(T, ·)dP x ,
A
where

 T
ZT

Z
1

2 
|b(X(s, ·))| ds .
Z(T, ·) = exp  hb, dXi −


2
0
0
Remark . If b is bounded or if b satisfies a global Lipschitz condition
then B(w) = ∞, so that ΩT = Ω and QTx are probability measures.
Proof. Let 0 ≤ R < ∞. For any w in Ω, let y be given by
y(t) = w(t) +
Zt
0
195
b(y(σ))dσ.
24. Explosions
196
Define σR (w) = inf{t : |y(t)| ≥ R and let bR be as in the Exercise.
Then the equation
yR (t) = w(t) +
Zt
bR (yR (σ))dσ
0
190
has a global solution. Denote by S R : Ω → Ω the map w → yR . If QR,x
is the measure induced by S R , then
 t

Zt
Z

dQR,x 1


2 
|bR | ds .
Ft = Zr (t) = exp  hbR , dXi −

dP x
2
0
0
Let τR (w) = inf{t : |w(t)| > R}. τR is a stopping time satisfying
τR S R = σR . By the optional stopping theorem.
dQR,x (1)
FτR ∧T = ZR (τR ∧ T ) = Z(τR ∧ T ).
dP x
Claim. QR,x ((τR > T ) ∩ A) = QTx ((τR > T ) ∩ A), ∀A in FT .
Proof.
Right side = P x {w : B(w) > T, S T (w) ∈ A ∩ (τR > T )}
= P x {w : B(w) > T, y ∈ A, sup |y(t)| < R}
0≤t≤T
= P x {w : y is defined at least upto time T,
y ∈ A, sup |y(t)| > R}
0≤t≤T
= P x {w : yR ∈ A, sup |yR (t)| < R}
0≤t≤T
= P x {w : S R (w) ∈ A, τR S R (w) > T }
= QR,x {(τR > T ) ∩ A}
(by definition). As Ω is an increasing union of {τR > T } for R increasing,
QTx (A) =
lt
R→+∞
QTx ((τR > T ) ∩ A), ∀A in FT ,
197
= lt QR,x ((τR > T ) ∩ A) (by claim)
R→∞

 τR ∧T
τZ
R ∧T
Z

 Z
1


2 
hb, dXi −
|b| ds dP x
exp 
= lt


R→∞
2
=
=
Z
A
Z
(τR ∧T )∩A
0
(by (1))
0
 T

ZT
Z

1


exp  hb, dXi −
|b|2 ds dP x


2
0
0
Z(T )dP x .
A
191
Theorem . Suppose b : Rd → Rd is locally Lipschitz; let L =
∆
+ b.∇.
2
(i) If there exists a smooth function u : Rd → (0, ∞) such that u(x) →
∞ as |x| → ∞ and Lu ≤ cu for some c > 0 then P x {w : B(w) <
∞} = 0, i.e. for almost all w there is no explosion.
(ii) If there exists a smooth bounded function u : Rd → (0, ∞) such
that Lu ≥ cu for some c > 0, then P x {w : B(w) < ∞} > 0, i.e.
there is explosion.
Corollary . Suppose, in particular, b satisfies |hb(x), xi| ≤ A + B|x|2 for
some constants A and B; then P x (w : B(w) < ∞) = 0.
Proof. Take u(x) = 1 + |x|2 and use part (1) of the theorem.
∆
Proof of theorem. Let bR be as in the Exercise and let LR = + bR · ∇;
2
then LR u(x) ≤ cu(x) if |x| ≤ R.
Claim. u(X(t))e−ct is a supermartingale upto time τR relative to QRx ,
 t


Zt
Z


1



|bR |2 ds
d u(X(t))e−ct exp  hbR , dXi −



2
0
0
24. Explosions
198
Rt
Rt
−ct hbR ,dXi− 21 |bR |2 ds
e
0
0
|2
×
1
|bR
dt] + bR udt + |bR |2 udt}
2
2
Zt
Zt
1
= exp(−ct + hbR , dXi −
|bR |2 ds)
2
×{−cudt + h∇u, dXi + u(x)[hbR , dXi −
0
0
·[LR − c)u + h∇u, dXi + uhbR , dXi].
192
Therefore
−ct
Rt
e0
hbR ,dXi− 21
Rt
0
|bR |2 ds
u(X(t))e E


Zt
Zs
Zs




−
exp −cs + hbR , dXi − |bR |2 ds · (LR − c)u(X(s))ds


0
0
0
is a Brownian stochastic integral. Therefore


τR ∧t
τR ∧t
Z
Z


1


2 
|bR | ds −
hbR , dXi −
u(X(τR ∧ t)) exp −c(τR ∧ t) +


2
0
0


τR ∧t
Z
Zs
Zs


1


2 
−
exp −cs + hbR , dXi −
|bR | ds (LR − c)u(X(s))ds


2
0
0
0
is a martingale relative to P x , FτR ∧t . But bR (x) = b(x) if |x| ≤ R. Therefore


τR ∧t
τR ∧t
Z
Z


1


|b|2 ds −
hb, dXi −
u(X(τR ∧ t)) exp −c(τR ∧ t) +


2
0
0


τR ∧t
Zs
Zs
Z




−
exp −cs + hb, dXi − |b|2 ds (LR − c)u(X(s))ds


0
0
199
193
is a martingale relative to FτR ∧t . But (LR − c)u ≤ 0 in [0, τR ].
Therefore
τR ∧t
τR ∧t
Z
Z
1
|b|2 ds)
u(X(τR ∧ t)) exp(−c(τR ∧ t) +
hb, dXi −
2
0
0
is a supermartingale relative to FτR ∧t , P x . Therefore u(X(τR ∧ t)e−c(τR ∧t)
is a supermartingale relative to QRx (optional sampling theorem). Therefore
R
E Qx (u(X(tR ∧ t))e−c(τR ∧t) ≤ u(x);
letting t → ∞, we get, using Fatou’s lemma,
R
E xQ (u(X(τR )e−cτR ) ≤ u(x).
Therefore
R
E Qx (e−cτR ) ≤
Thus
E Px (e−cσR ) ≤
u(x)
.
inf |u(y)|
|y|=R
u(x)
inf |u(y)|
|y|=R
(by change of variable). Let R → ∞ to get Lt
R→∞
P x {w : B(w) < ∞} = 0.
R
e−cσR dP x = 0, i.e.
Sketch of proof for Part (ii).
By using the same technique as for Part (i), show that u(X(t))e−ct is
a submartingale upto time τR relative to QRx , so that
E Px (e−cσR ) ≥
u(x)
u(x)
≥
> 0;
sup |u(y)| ||u||∞
|y|=R
let R → ∞ to get the result.
194
∂2
1
∂
+ x3 , there is explosion. (Hint: take
2 ∂x
∂x
−1 2
u = etan (x ) and show that Lu ≥ u).
Exercise. Show that if L =
25. Construction of a
Diffusion Process
Problem . Given a : [0, ∞) × Rd → S d+ , bounded measurable and b : 195
[0, ∞) × Rd → Rd bounded measurable, to find (Ω, Ft , P, X) where Ω is
a space, {Ft }t≥0 an increasing family of σ-algebras on Ω, P a probability
measure on the smallest σ-algebra containing all the Ft ’s. and X :
[0, t) × Ω → Rd , a progressively measurable function such that X(t, w) ∈
I[b(t, Xt ), a(t, Xt )].
Let Ω = C[0, ∞); Rn ), β(t, ·) = n-dimensional Brownian motion,
Ft = σ{β(s) : 0 ≤ s ≤ t}, P the Brownian measure on Ω and a and b as
given in the problem. We shall show that thee problem has a solution,
under some special conditions on a and b.
Theorem . Assume that there exists σ : [0, ∞) × Rd → Md×n (Md×n =
set of all d × n matrices over the reals) such that σσ∗ = a. Further let
X
i, j
X
i, j
X
j
|σi j (t, x)| ≤ C,
X
j
|b j (t, x)| ≤ C,
|σi j (t, x1 ) − σi j (t, x2 )| ≤ A|x1 − x2 |,
|b j (t, x1 ) − b j (t, x2 )| ≤ A|x1 − x2 |.
201
25. Construction of a Diffusion Process
202
Then the equation
(1)
ξ(t, ·) = x +
Zt
hσ(s, ξ(s, ·)), dβ(s, ·)i +
0
196
Zt
b(s, ξ(s, ·))ds
0
has a solution. The solution ξ(t, w) : [0, ∞) × Ω → Rd can be taken
to be such that ξ(t, ·) is progressively measurable and such that ξ(t, ·) is
continuous for a, a.e. If ξ, η are progressively measurable, continuous
(for a.a.e) solutions of equation (l), then ξ = n for a.a.w.
Proof. The proof proceeds in several steps.
Lemma 1. Let Ω be any space with (Ft )t≥0 an increasing family of σalgebras. If 0 ≤ T ≤ ∞ then there exists a σ-algebra A0 ⊂ A =
B[0, T ) × FT such that a function f : [0, T ] × Ω → R is progressively
measurable if and only if f is measurable with respect to A0 .
Proof. Let A0 = {A ∈ A : χA is progressively measurable}. Clearly
[0, T ] × Ω ∈ A0 , and if A ∈ A0 , Ac ∈ Ac . Thus A0 is an algebra. As
increasing limits (decreasing limits) of progressively measurable functions are progressively measurable, A0 is a monotone class and hence a
σ-algebra.
Let f : [0, T ] × Ω → R be progressively measurable; in fact, f + =
f − |f|
f + 1 = f| −
,f =
. Let g = f + . Then
2
2
gn =
n2n
X
i−1
i=1
197
2n
χg−1 [ i−1n ,
2
i
)
2n
+ nχg−1 [n,)
is progressively measurable. Hence nVgn is progressively measurable,
i.e. nχg−1 [n,∞) is progressively measurable. Similarly φg−1 [ i−1n , in ) is pro2 2
gressively measurable, etc. Therefore, by definition, gn is measurable
with respect to A0 . As g = f + is the pointwise limit of gn , f + is measurable with respect to A0 . Similarly f − is A0 -measurable. Thus f is
A0 -measurable.
203
Let f : [0, T ] × Ω → R be measurable with respect to A0 . Again, if
g = f+
n2n
X
i−1
gn =
χ −1 i−1 i + nχg−1 [n,∞)
2n g [ 2n , 2n )
i=1
i−1 i
is A0 -measurable. Since g−1 [n, ∞), . . . g−1 [ n , n ) ∈ A0 . So gn
2
2
is progressively measurable. Therefore g is progressively measurable.
Hence f is progressively measurable. This completes the proof of the
Lemma.
To solve (1) we use the standard iteration technique.
Step 1. Let ξ0 (t, w) = x,
ξn (t, w) = x +
Zt
hσ(s, ξn−1 (s, w)), dβ(s, w)i +
0
Zt
b(s, ξn−1 (s, w))ds.
0
By induction, it follows that ξn (t, w) is progressively measurable.
Step 2. Let ∆n (t) = E(|ξn+1 (t) − ξn (t)|2 ). If 0 ≤ t ≤ T , ∆n (t) ≤ C ∗
Rt
0
∆n−1
(s)ds and ∆0 (t) ≤ C ∗ t, where C ∗ is a constant depending only on T .
Proof.
∆0 (t) = E(|ξ(t) − x|2 )

 t
Zt

 Z


2
= E | h(s, x), dβ(s, x)i + b(s, x)ds| 


0
0

 t

 Z


2
≤ 2E | hσ(s, x), dβ(s, x)i|  +


0
 t

 Z



2
+ 2E | b(s, x)ds| 


(use the fact that |x + y|2
0
≤ 2(|x|2 + |y|2 ) ∀x, y ∈ Rd )
25. Construction of a Diffusion Process
204
 t

 t

Z

 Z




∗ 
2


= 2E  T r σσ ds = 2E | b(s, x)ds|  .




0
198
or
0

 t

 t

 Z

Z



2 
∗ 


∆0 (t) ≤ 2E  tr σσ ds + 2E t |b(s, x)| ds




0
0
(Cauchy-Schwarz inequality)
≤ 2nd C 2 (1 + T )t.
∆n (t) = E(|ξn+1 (t) − ξn (t)|2 )
 t
 Z

= E | hσ(s, ξn (s, w)) − σ(s, ξn−1 (s, w))dβi+

0
+
Zt
0



b(s, ξn (s, w)) − b(s, ξn−1 (s, w))ds|2 


 t

 Z


≤ 2E | hσ(s, ξn (s, w)) − (s, ξn−1 (s, w)), dβ(s, w)i|2  +


0
+ 2E(|
Zt
(b(s, ξn (s, w)) − b(s, ξn−1 (s, w))ds|2 )
0
Zt
≤ 2E( tr[(σ(s, ξn (s, w)) − σ(s, ξn−1 (s, w))]×
0
∗
× [σ (s, ξn (s, w)) − σ∗ (s, ξn−1 (s, w))]ds]+
 t

 Z



+ 2E t |b(s, ξn (s, w)) − b(s, ξn−1 (s, w))|2 ds


0
≤ 2dn A
2
Zt
0
2
∆n−1 (s)ds + 2tA n
Zt
0
∆n−1 (s)ds
205
2
≤ 2dn A (1 + T )
ZT
∆n−1 (s)ds.
0
199
This proves the result.
Step 3. ∆n (t) ≤
(C ∗ t)n+1
∀n in 0 ≤ t ≤ T , where
(n + 1)!
C ∗ = max{2nd C 2 (1 + T ),
and
A2 (1 + T )}.
Proof follows by induction on n.
Step 4. ξn |[0,T ]×Ω is Cauchy in L2 ([0, T ] × Ω, B([0, T ] × Ω), µ × P), where
µ is the Lebesgue measure on [0, T ].
Proof. ∆n (t) ≤
(C ∗ t)n+1
implies that
(n + 1)!
||ξn+1 − ξn ||22 ≤
(C ∗ T )n+2
.
(n + 2)!
Here || · ||2 is the norm in L2 ([0, T ] × Ω). Thus
∞
X
n=1
||ξn+1 − ξn ||2 < ∞,
proving Step (4).
Step 5. (4) implies that ξn |[0,T ]×Ω is Cauchy in L2 ([0, T ] × Ω, A0 , µ ×
P) where A0 is as in Lemma 1. Thus ξn |[0,T ]×Ω converges to ξT in
L2 ([0, T ] × Ω) where ξT is progressively measurable.
Step 6. If ξn |[0,T 2 ]×Ω ξT 2 in L2 ([0, T 2 ] × Ω) and
ξn |[0,T 1 ]×Ω ξT 1
in
L2 ([0, T 1 ] × Ω),
then ξT 2 |[0,T ]×Ω = ξT 1 a.e. on [0, T 1 ] × Ω, T 1 < T 2 .
1
This follows from the fact that if ξn → ξ in L2 , a subsequence of
(ξn ) converges pointwise a.e. to ξ.
200
25. Construction of a Diffusion Process
206
Step 7. Let ξ be defined on [0, ∞) × Ω by ξ|[0,T ]×Ω = ξ T . We now show
that
ξ(t, w) = x +
Zt
hσ(s, ξ(s, ·)), dβ(s, ·)i +
0
Zt
b(s, ξ(s, ·))ds.
0
Proof. Let 0 ≤ t ≤ T . By definition,
ξn (t, w) = x +
Zt
hσ(s, ξn−1 (s, ·)), dβ(s, ·)i +
Zt
b(s, ξn−1 (s, ·))ds.
0
0
 t
2 
Z
 



E  h(σ(s, ξn (s, ·)) − σ(s, ξ(s, w))), dβ(s, w)i 


0
= E(
ZT
tr[(σ(s, ξn (s, w)) − σ(s, ξ(s, w)))(σ(s, ξn (s, w)) − σ(s, ξ(s, w)))∗ ds
0
≤ dn A
ZT Z
0
|ξn (s, w) − ξ(s, w)|2 ds → 0
as n → ∞
Ω
(by Lipschitz condition on σ).
Therefore
Zt
hσ(s, ξn−1 (s, w), dβ(s, w)i →
0
201
Zt
hσ(s, ξ(s, w)), dβ(s, w)i
0
in L2 (Ω, P). Similarly,
Zt
b(s, ξn (s, w))ds →
0
Zt
b(s, ξ(s, w))ds,
in
0
Thus we get
(*)
ξ(t, w) = x +
Zt
0
hσ(s, ξ(s, w)), dβ(s, w)i +
L2 .
207
+
Zt
b(s, ξ(s, w))ds
a.e. in
t, w.
0
Step 8. Let ξ(t, w) ≡ the right hand side of (∗) above. Then ξ(t, w) is
almost surely continuous because the stochastic integral of a bounded
progressively measurable function is almost surely continuous. The re∞
S
sult follows by noting that [0, ∞) = [0, n] and a function on [0, ∞) is
n=1
continuous iff it is continuous on [0, n], ∀n.
Step 9. Replacing ξ by ξ in the right side of (∗) we get a solution
Zt
ξ(t, w) = x +
hσ(s, ξ), dβi +
0
Zt
b(s, ξ(s, w))ds
0
that is a.s. continuous ∀t and a.e.
Uniqueness. Let ξ and η be two progressively measurable a.s. continuous functions satisfying (1). As in Step 3,
Zt
Zt
∗
E(|ξ(t, w) − x| ) ≤ 2(E( tr σσ ds) + 2E(t b|2 ds)
2
0
0
ZT
ZT
∗
≤ 2E( tr σσ ds + 2E(T
|b|2 ds),
0
if 0 ≤ t ≤ T
0
< ∞.
202
Thus
E(|ξ(t, w)|2 )
is bounded in 0 ≤ t ≤ T . Therefore
φ(t) = E(|ξ(t, w) − η(t, w)|2 )
≤ 2E(|ξ(t, w)|2 ) + 2E(|η(t, w)|2 )
25. Construction of a Diffusion Process
208
and so φ(t) is bounded in 0 ≤ t ≤ T . But
2
φ(t) ≤ 2dn A (1 + T )
Zt
φ(s)ds
0
as in Step 2; using boundedness of φ(t) in 0 ≤ t ≤ T we can find a
constant C such that
φ(t) ≤ Ct
and
φ(t) ≤ C
Zt
φ(s)ds,
0 ≤ t ≤ T.
0
By iteration φ(t) ≤
(Ct)n (CT )n
≤
. Therefore
n!
n!
φ=0
on
[0, T ],
i.e. ξ(t, w) = η(t, w) a.e. in [0, T ]. But rationals being dense in R we
have
ξ = η a.e. and ∀t.
It is now clear that ξ ∈ I[b, a].
Remark. The above theorem is valid for the equation
ξ(t, w) = x0 +
Zt
t0
hσ(s, ), dβi +
Zt
b(s, ξ)ds,
t0
∀t ≥ t0 .
This solution will be denoted by ξt0 ,x0 .
203
Proposition . Let φ : C[(0, ∞); Rn ) → C([t0 , ∞); Rd ) be the map sending w to ξt0 ,x0 , P the Brownian measure on C([0, ∞); Rn ). Let Pt0 ,x0 =
Pφ−1 be the measure induced on C([t0 , ∞); Rd ). Define X : [t0 , ∞) ×
C)[t0 , ∞); Rd ) by X(t, w) = w(t). Then X is an Itö process relative to
(C([t0 , ∞); Rd ), t0 , Pt0 ,x0 ) with parameters
[b(t, Xt ), a(t, Xt )].
The proof of the proposition follows from
209
Exercise. Let (Ω, F t , P), (Ω, Ft , P) be any two measure spaces with X,
Y progressively measurable on Ω, Ω respectively. Suppose λ : Ω → Ω
is such that λ is (F t , Ft )-measurable for all t, and Pλ−1 = P. Let
X(t, w) = Y(t, λw), ∀w ∈ Ω. Show that
(a) If X is a martingale, so is Y.
(b) If X ∈ I[b(t, Xt ), a(t, Xt )] then
Y ∈ I[b(t, Yt ), a(t, Yt )].
P
Lemma . Let f : R2 → R be (Ω, P)-measurable, a sub - σ - algebra
P
of F . Let X : Ω → R and Y : Ω → R be such that X is -measurable
P
and Y is -independent. If g(w) = f (X(w), Y(w)) with E(g(w)) < ∞,
then
X
E(g|
)(w) = E( f (x, Y)| x=X(w) ,
i.e.
E( f (X, Y)|P )(w)
=
Z
f (X(w), Y(w′ ))dP(w′ ).
Ω
Proof. Let A and B be measurable subsets in R. The result is trivially 204
verified if f = XA×B. The set
A = {F ∈ R : the result is true for XF }
is a monotone class containing all measurable rectangles. Thus the
Lemma is true for all characteristic functions. The general result follows by limiting procedures.
26. Uniqueness of Diffusion
Process
IN THE LAST section we proved that
ξ(t, w) = x0 +
Zt
205
hσ(s, ξ(s, w)), dβ(s, w) +
t0
Zt
b(s, (s, w)ds
t0
has a solution under certain conditions on b and σ where σσ∗ = a. The
was constructed on (C([t0 , ∞); Rd ), Ft0 ) so that
measure Pt0 ,x0 = Pξt−1
0 ,x0
the map X(t, w) = w(t) is an Itô process with parameters b and a. We
now settle the uniqueness question, about the diffusion process.
Theorem . Let
(i) a : [0, ∞) × Rd → S d+ and b : [0, ∞) × Rd → Rd be bounded
measurable functions;
(ii) Ω = C([0, ); Rd );
(iii) X : [0, ∞) × Ω → Rd be defined by X(t, w) = w(t);
(iv) Xt = σ{X(s) : 0 ≤ s ≤ t};
(v) P be any probability measure on


[ 

= σ  Xt 
t≥0
211
26. Uniqueness of Diffusion Process
212
such that P{X(0) = x0 } = 1 and X is an Itô process relative to
(Ω, Xt , P) with parameters b(t, Xt ) and a(t, Xt );
(vi) σ : [0, ∞) × Rd → Md×n be a bounded measurable map into the
set of all real d × n matrices such that σσ∗ = a on [0, ∞) × Rd .
206
Then there exists a generalised n-dimensional Brownian motion β
P
on (Ω, t , Q) and a progressively measurable a.s. continuous map ξ :
[0, ∞) × Ω → Rd satisfying the equation
(1)
ξ(t, w) = x0 +
Zt
0
hσ(s,
Z
(s, w)), dβ(s, w)i +
Zt
b(s, ξ(s, w))ds
0
with Qξ −1 = P, where ξ : Ω → Ω is given by (ξ(w))(t) = ξ(t, w).
Roughly speaking, any Itô process can be realised by means of a
diffusion process governed by equation (1) with σσ∗ = a.
Proof. Case (i). Assume that there exist constants m, M > 0 such that
mI ≤ a(t, x) ≤ MI and σ is a d × d matrix satisfying σσ∗ = a. In this
P
case we can identify (Ω, t , Q) with (Ω, Ft , P). Since D(t, ·) is an Itô
process,
exphθ, X(t)i −
Zt
1
hθ, b(s, X(s, ·))ids −
2
0
Zt
hθ, a(s, X(s, ·))θids
0
is a (Ω, Ft , P)-martingale. Put
Y(t, w) = X(t, w) −
Zt
b(s, X(s, w))ds − x0 .
0
Clearly Y(t, w) is an Itô process corresponding to the parameters
[0, a(s, X s )],
213
so that
1
exphθ, Y(t, w)i −
2
Zt
hθ, a(s, X(s, ·))θids
0
is a (Ω, Ft , P)-martingale. The conditions m ≤ a ≤ M imply that σ−1
exists and is bounded. Let
η(t) =
Zt
−1
σ dY =
0
Zt
σ−1 (s, X(s, ·))dY(s, ·),
0
so that (by definition of a stochastic integral) η is a (Ω, Ft , P)-Itô pro- 207
cess with parameters zero and σ−1 a(σ−1 )∗ = 1. Thus η is a Brownian
motion relative to (Ω, Ft , P). Now by change of variable formula for
stochastic integrals,
Zt
0
σdη =
Zt
σσ−1 dY
0
= Y(t) − Y(0) = Y((t),
since Y(0) = 0. Thus
X(t) = x0 +
Zt
σ(s, X(s, ·))d +
0
Zt
b(s, X(s, ·))ds.
0
Taking Q = P we get the result.
Case (ii). a = 0, b = 0, x0 = 0, σ = 0 where σ ∈ Md×n . Let (Ω∗ , Ft∗ , P∗ )
be an n-dimensional Brownian motion. Define
X
(Ω, , Q) = (Ω × Ω∗ , Ft × Ft∗ , P × P∗ ).
t
If β is the n-dimensional Brownian motion on (Ω∗ , Ft∗ , P∗ ), we define β on Ω by β(t, w, w∗ ) = β(t, w∗ ). It is easy to verify that β is an
P
n-dimensional Brownian motion on (Ω, t , Q). Taking ξ(t, w, w∗) = x0
we get the result.
26. Uniqueness of Diffusion Process
214
Before we take up the general case we prove a few Lemmas.
Lemma 1. Let σ : Rn → Rd be linear σσ∗ = a : Rd → Rd ; then 208
there exists a linear map which we denote by σ−1 : Rd → Rn such that
σ−1 aσ−1∗ = πNσ⊥ , where π denotes the projection and Nσ null space of
σ.
Proof. Let Rσ = range of σ. Clearly σ : Nσ⊥ → R is an isomorphism.
Let τ : Rσ → Nσ⊥ be the inverse. We put
⊥
σ−1 = τ ⊕ 0 : Rσ ⊕ R⊥
σ → Nσ ⊕ Nσ .
Lemma 2. Let X, Y be martingales relative to (Ω, Ft , P) and (Ω, F t , P)
respectively. Then Z given by
Z(t, w, w) = X(t, w)Y(t, w)
is a martingale relative to
(Ω × Ω, Ft × F t , P × P).
Proof. From the definition it is clear that for every t > s
Z
Z
Z(t, w, w)d(P × P)|Fs ×Fs =
Z(s, w, w)d(P × P)
A×A
A×A
if A ∈ Fs and A ∈ F s . The general case follows easily.
As a corrollary to Lemma 2, we have
Lemma 3. Let X be a d-dimensional Itô process with parameters b and
a relative to (Ω, Ft , P) and let Y be a d-dimensional Itô process relative
to (Ω, F t , P) relative to b and a. Then Z(t, w, w) = (X(t, w), Y(t, w))h is ai
(d + d)-dimensional Itô process with parameters B = (b, b), A = a0 a0
relative to (Ω × Ω, Ft × F t , P × P).
215
209
Lemma 4. Let X be an Itô process relative to (Ω, Ft , P) with parameRt
ters 0 and a. If θ is progressively measurable such that E( |θ|2 , ds) <
0
Rt
∞, ∀t and θaθ∗ is bounded, then hθ, dXi ∈ I[0, θaθ∗ ].
0
Proof. Let θn be defined by



θi ,
θni = 

0,
if |θ| ≤ n,
otherwise;
Rt
Then hθn , dXi ∈ I[0, θn aθn∗ ]. Therefore
0
Xn (t) = exp(λ
Zt
λ2
hθn , dXi −
2
Zt
hθn , aθn ids
0
0
is a martingale converging pointwise to
 t

Zt
 Z

2
λ


X(t) = exp λ hθ, dXi −
hθ, aθids .


2
0
0
Rt
To prove that hθ, dXi is an Itô process we have only to show that
0
Xn (t) is uniformly integrable. Without loss of generality we may assume
that λ = 1. Let [0, T ] be given

 t

Zt

 Z




E(Xn2 (t, w)) = E exp 2 hθn , dXi − hθn , aθn ids



0
0

 t
Zt

 Z


= E exp 2 hθn , dXi − 2 hθn , aθn ids


0
+
Zt
0



hθn , aθn ids .

0
26. Uniqueness of Diffusion Process
216
≤ eT sup hθn , aθn i.
0≤t≤T
210
But hθ, aθ∗ i is bounded and therefore hθn , aθn i is uniformly bounded
in n. Therefore (Xn ) are uniformly integrable. Thus X(t, ·) is a martingale.
Case (iii). Take d = 1, and assume that
Zt
a−1 χ(a>0) ds < ∞, ∀t
0
with a > 0; let σ = +ve squareroot of a. Define 1/σ = 1/σ if σ > 0,
and 1/σ = 0 if σ = 0. Let
Y(t) = X(t) − x0 −
Zt
b(s, X(s))ds.
0
Denote by Z the one-dimensional Brownian motion on (Ω∗ , Ft∗ , P∗ )
where Ω∗ = C([0, ∞), R). Now
Y ∈ I[0, a(s, X(s, ·))], Z ∈ I[0, 1].
By Lemma 3,
!!
a 0
.
(Y, Z) ∈ I (0, 0);
0 1
If
∗
η(t, w, w ) =
Zt
0
!
1
χ(σ>0) , χσ=0 , d(Y, Z)i
h
σ(s, X(s, ·))
then Lemma 4 shows that
η ∈ I[0, 1].
Therefore η is a one-dimensional Brownian motion on Ω = (Ω ×
Ω∗ , Ft × Ft∗ , P × P∗ ). Put
Y(t, w, w∗ ) = Y(t, w) and
X(t, w, w∗) = X(t, w);
217
211
then
Zt
σdη =
Zt
=
Zt
0
1
σ χ(σ>0) dY +
σ
Zt
σχ(σ=0) dZ
0
0
χ(σ>0) dY.
0
Since
 t

 t
2 
Z

Z
 






2



E  χ(σ=0) dY   = E  σ χ(σ=0) ds = 0,






0
0
it follows that
Zt
σdη =
0
Zt
dY = Y(t) = Y(t, w, w∗ ).
0
Thus,
∗
X(t, w, w ) = x0 +
Zt
σ(s, X(s, w, w∗ )dη+
Zt
b(s, X(s, w, w∗ )ds
0
+
0
with X(t, w, w∗) = X(t, w). Now
−1
(P × P∗ )X (A) = (P × P∗ )(A × Ω∗ ) = P(A).
Therefore
(P × P∗ )X
−1
=P
or
QX
−1
= P.
26. Uniqueness of Diffusion Process
218
Case (iv). (General Case). Define
Y(t, ·) = X(t, ·) − x0 −
Zt
b(s, X(s, ·))ds.
0
Therefore Y ∈ I[0, a(s, X(s, ·))] relative to (Ω, Ft , P). Let Z be the 212
n-dimensional Brownian motion on (Ω∗ , Ft∗ , P∗ ) where
Ω∗ = C([0, ∞); Rn ).
"
"
##
a(s, X s ), 0
(Y, Z) I (0, 0);
0
I
Let σ be a d × n matrix such that σσ∗ = a on [0, ∞) × Rd . Let
η(t, w, w ) =
Zt
=
Zt
∗
−1
σ (s, X(s, w))dY(s, w) +
0
Zt
rNσ (s, Z(s, w∗))dZ(s, w∗ )
0
h(σ−1 (s, X(s, w)), πNσ (s, Z(s, w∗ ))), d(Y, Z)i.
0
Therefore η is an Ito process with parameters zero and
!
∗!
a 0 σ−1
−1
A = (σ , πN )
0 I πNσ∗
= σ−1 a(σ−1 )∗ + πNσ πNσ∗ .
= πNσ + πNσ
(for any projection PP∗ = PP = P)
= IRn .
Therefore η is n-dimensional Brownian motion on
(Ω, Ft , P) = (Ω × Ω∗ , Ft × Ft∗ , P × P∗ ).
Zt
σ(s, X(s, w))dη(s, w, w∗ )
0
219
=
Zt
=
Zt
σσ dY +
=
Zt
πRσ dY, since σσ−1 = πRσ and σπNσ = 0,
=
Zt
σ(s, X(s, w, w∗ ))dη(s, w, w∗ ), where X(s, w, w∗) = X(s, w),
0
−1
0
Zt
σπNσ dZ.
0
0
(1 − πRσ )dY.
0
213
Claim.
Rt
πRσ dY = 0.
0
For
 t
2 
Zt
Z
  Zt




E  πRσ dY   =
aπRσ ds =
σσ∗ πRσ ds
 

0
0
=
Zt
0
σ(0)ds = 0.
0
Therefore we obtain
Zt
∗
σ(s, X(s, w))dη(s, w, w ) =
0
Zt
dY = Y(t) − Y(0) = Y(t)
0
putting Y(t, w, w∗) = Y(t, w), one gets
∗
X(t, w, w ) = x0 +
Zt
0
σ(s, X(s, w, w∗ ))dη(s, w, w∗ )
26. Uniqueness of Diffusion Process
220
+
Zt
b(s, X(s, w, w∗ ))ds.
0
As in Case (iii) one shows easily that
(P × P∗ )X
−1
= P.
This completes the proof of the theorem.
Corollary . Let a : [0, ∞) × Rd → S d+ , and b : [0, ∞) × Rd → Rd
be bounded, progressively measurable functions. If for some choice of
a Lipschitz function σ : [0, ∞) × Rd → Md×n , σσ∗ = a then the Itô
process corresponding to [b, a) is unique.
214
To state the result precisely, let P1 and P2 be two probability measures on C([0, ∞); Rd ) such that X(t, w) = w(t) is an Itô process with
parameters b and a. Then P1 = P2 .
Proof. By the theorem, there exists a generalised n-dimensional BrowP
nian motion βi on (Ωi , it , Qi ) and a map ξi : Ωi → Ω satisfying (for
i = 1, 2)
(t,w)
i
= x0 +
Zt
σ(s, ξi (s, w))dβi (s, w) +
0
Zt
b(s, ξi (s, w))ds.
0
and Pi = Qi ξi−1 .
Now σ is Lipschitz so that ξi is unique but we know that the iterations converge to a solution. As the solution is unique the iterations
converge to ξi . Each iteration is progressively measurable with respect
to
i
t = σ{βi (s); 0 ≤ s ≤ t} so that ξi is also progressively
measurable with respect to Fti . Thus we can restate the result as follows:
There exists (Ωi , Fti , Qi ) and a map ξi : Ωi → Ω satisfying
ξi (t, w) = x0 +
Zt
0
σ(s, ξi (s, w))dβi (s, w)
221
+
Zt
b(s, ξi (s, w))ds,
0
and Pi = Qi ξi−1 .
(Ωi , Fti , Qi , βi ) can be identified with the standard Brownian motion
∗
(Ω , Ft∗ , Q, β). Thus P1 = Qξ −1 = P2 , completing the proof.
27. On Lipschitz Square
Roots
Lemma . Let f : R → R be such that f (x) ≥ 0, f (x) ∈ C 2 and | f ′′ (x)| ≤ 215
A on (−∞, ∞); then
√
p
| f ′ (x)| ≤ f (x) 2A.
Proof.
0 ≤ f (y) = f (x) + (y − x) f ′ (x) +
≤ f (x) + Z f ′ (x) +
(y − x)2 ′′
f (ξ)
2
Z 2 ′′
f (ξ)
2
where Z = y − x, or f (y) ≤ f (x) + Z f ′ (x) +
AZ 2
. Therefore
2
AZ 2
+ Z f ′ (x) + f (x) ≥ 0, ∀Z ∈ R
2
| f ′ (x)|2 ≤ 2A f (x).
So
| f ′ (x)| ≤
√
2A f (x).
Note. If we take f (x) = x2 , we note that the constant is the best possible.
Corollary . If f ≥ 0, | f ′′ | ≤ A, then
√
√
√
| ( f (x1 )) − ( f (x2 )) ≤ (A/2)|x1 − x2 |.
223
27. On Lipschitz Square Roots
224
√
Proof. Let ǫ > 0, then ( f (x) + ǫ) is a smooth function.
√
f ′ (x)
( f (x) + ǫ)′
( ( f (x) + ǫ))′ = √
= √
.
2 ( f (x) + ǫ) 2 ( f (x) + ǫ)
Therefore
√
√
√
|( ( f (x) + ǫ))′ | ≤ (2A/2) ≤ (A/2),
or
√
√
√
| ( f (x1 ) + ǫ) − ( f (x2 ) + ǫ)| ≤ (A/2)|x1 − x2 |.
216
Let ǫ → 0 to get the result.
We now consider the general case and give conditions on the matrix
a so that σ defined by σσ∗ = a is Lipschitz.
Theorem . Let a : Rn → S d+ be continuous and bounded C 2 -function
such that the second derivative is uniformly bounded, i.e. ||D s Dr ai j || ≤
d
). If σ : Rn → S d+ is
M, where M is independent of i, j, r, s; (Dr ≡
dxr
the unique positive square root of a, then
||σ(x1 ) − σ(x2 )|| ≤ A|x1 − x2 |, ∀x1 , x2 , A = A(M, d).
Proof.
Step 1. Let A ∈ S d+ be strictly positive such that ||I − A|| < 1. Then
√
√
A = (I − (I − A))
∞
X
Cr
(I − A)r ,
=
r!
r=0
√
so that on the set {A : ||I − A|| < 1} the map A → A is C ∞ (in fact
analytic).
Now assume that A is any positive definite matrix. Let λ1 , . . . , λn
be the eigen values so that λ j > 0, j = 1, 2 . . . n. Therefore I − ǫA is
225
aymmetric with eigen values 1 − ǫλ j . By choosing ǫ sufficiently small
we can make
||I − ǫA|| = max{1 − ǫλ1 , . . . , 1 − ǫλn } < 1.
Fixing such an ǫ we observe that
√
1 √
1 √
A = √ (ǫA) = √ (I − (I − ǫA)).
ǫ
ǫ
So the map A →
definite matrices.
√
217
A is smooth on the set of symmetric positive
√
Step 2. Let n = 1, σ(t0 ) = (a(t0 ) where a(t0 ) is positive definite. Assume a(t0 ) to be diagonal so that σ(t0 ) is also diagonal.
X
σi j (t)σ jk (t) = aik (t).
j
Differentiating with respect to t at t = t0 we get
X
X
σ′i j (t0 )σ jk (t0 ) = a′ik (t0 )
σi j (t0 )σ′jk (t0 ) +
j
j
or
√
√
aii (t0 )σ′ik (t0 ) + akk (t0 )σ′ik (t0 ) = a′ik (t0 )
or
σ′ik (t0 )
a′ik (t0 )
√
.
=√
(aii (t0 )) + (akk (t0 ))
Since the second derivatives are bounded by 4M and aii −2ai j +a j j ≥
0, we get
√
√
|a′ii (t) + 2a′i j (t) + a′j j (t)| ≤ (8M) (aii (t) + 2ai j (t) + a j j (t))
√
√ √
≤ (8M) 2 (aii + a j j )(t)
or
(1)
√ √
√
|a′ii (t) + 2a′i j (t) + a′j j (t)| ≤ 4 M( aii + a j j ).
27. On Lipschitz Square Roots
226
Since a is non-negative definite,
√
√
|a′ii (t)| ≤ (2M) (aii (t)), ∀i.
substituting this in (1), we get
√ √
√
|a′i j (t)| ≤ 4 M( aii + a j j ),
and hence
√
|σ′i j (t0 )| ≤ 4 M.
Step 3. Let a(t0 ) be positive definite and σ its positive definite square
root. There exists a constant unitary matrix α such that αa(t0 )α−1 = b(t0 )
is a diagonal positive definite matrix. Let λ(t0 ) be the positive square
root of b(t0 ) so that
λ(t0 ) = ασ(t0 )α−1 .
Therefore σ′ (t0 ) = (α−1 λ′ α)(t0 ) where (σ′ (t0 ))i j = σ′i j (t0 ) and
a′′ (t0 ) = (α−1 b′′ α)(t0 ).
Since α is unitary.
||λ|| = ||σ||, ||a′′ || = ||b′′ ||, ||λ′ || = ||σ′ ||.
By hypothesis, ||b′′ || = ||a′′ || ≤ C(d) · M. Therefore
√
||λ′ || ≤ 4 (MC(d)),
i.e.
√
||σ′ || ≤ 4 (MC(d)).
Thus ||σ(t1 ) − σ(t2 )|| ≤ |t1 − t2 |C(M, d).
Step 4. Let a : R → S d+ and σ be the unique non-negative definite
square root of a. For each ǫ > 0 let aǫ = a + ǫI, σǫ = unique positive
square root of aǫ . Then by step 3,
||σǫ (t1 ) − σǫ (t2 )|| ≤ C(M, d)|t1 − t2 |.
218
227
219
If a is diagonal then it is obvious that σǫ → σ as ǫ → 0. In the
general case reduce a to the diagonal form and conclude that σǫ → σ.
Thus
||σ(t1 ) − σ(t2 )|| ≤ C(M, d)|t1 − t2 |.
Step 5. Let a : Rn → S d+ and σ2 = a, with ||Dr D s ai j || ≤ M, ∀x, i, j; r,
s × ǫRn . Choose x1 , x2 ∈ Rn . Let x1 = y1 , y2 , . . . , yn+1 = x2 be (n + 1)
points such that yi and yi+1 differ almost in one coordinate. By Step 4,
we have
||σ(yi ) − σ(yi+1 )|| ≤ C|yi − yi+1 |.
(*)
The result follows easily from the fact that
||x||1 =
n
X
i=1
|xi | and
||x||2 = (x1 + · · · + xn )1/2
are equivalent norms.
This completes the proof of the theorem.
28. Random Time Changes
LET
X
1X
∂
∂2
L=
+
bj
ai j
2 i, j
∂xi ∂x j
∂x j
j
with a : Rb → S d+ and b : Rd → Rd bounded measurable funcitons. Let
X(t, ·), given by X(t, w) = w(t) for (t, w) in [0, ∞) × C([0, ∞) : Rd ) be an
Itô process corresponding to (Ω, Ft , Q) with parameters b and a where
Ω = C([0, ∞); Rd ). For every constant c > 0 define



 1 X
2
X
∂
∂

+
bj
ai j
Lc ≡ c 

2 i, j
∂xi ∂x j
∂x
j
j
Define Qc by Qc = PT c−1 where (T c w)(t) = w(ct). Then one can
show that X is an Itô process corresponding to (Ω, Ft , Qc ) with parameters cb and ca [Note: We have done this in the case where ai j = δi j ].
Consider the equation
∂u
= Lc u with
∂t
u(0, x) = f (x).
∂u
This can be written as
= Lu with u(0, x) = f (x) when τ = ct.
∂τ
Thus changing time in the differential equation is equivalent to stretching time in probablistic language.
So far we have assumed that c is a constant. Now we shall allow c
to be a function of x.
229
220
28. Random Time Changes
230
Let φ : Rd → R be any bounded measurable function such that
0 < C1 ≤ φ(x) < C2 < ∞,
∀x ∈ Rd
and suitable constants C1 and C2 . If
" X
#
X
1
∂
∂2
L≡
+
bj
ai j
2
∂xi ∂x j
∂x j
221
we define
" X
#
X
1
∂
∂2
Lφ = φL ≡ φ
+
bj
.
ai j
2
∂xi ∂x j
∂x j
In this case we can say that the manner which time changes depends
on the position of the particle.
Define T φ : Ω → Ω by
(T φ w)(t) = w(τt (w))
where τt (w) is the solution of the equation
Zτt
ds
= t.
φ(w(s))
0
1
1
≤ t ≤ τt . When φ ≡ c a
C1
C2
constant, then τt = ct and T φ coincides with T c .
As
Zλ
1
0 < C1 ≤ φ ≤ C2 < ∞,
ds
φ(w(s))
As C1 ≤ φ ≤ C2 it is clear that τt
0
is continuous and increases strictly from 0 to ∞ as λ increases, so that
τt exists, is unique, and is a continuous function of t for each fixed w.
Some properties of T φ .
(i) If l is the constant function taking the value 1 then it is clear that
T l = identity.
231
222
(ii) Let φ and ψ be two measurable funcitons such that 0 < a ≤ φ(x),
ψ(x) ≤ b < ∞, ∀x ∈ Rd . Then T φ ◦ T = T φψ = T ψ ◦ T φ .
Proof. Fix w. Let τt be given by
Zτt
1
ds = t.
φ(w(s))
0
Let w∗ (t) = w(τt ) and let σt be given by
Zσt
0
1
ds = t.
φ(w∗ (s))
Let w∗∗ (t) = w∗ (σt ) = w(τσt ). Therefore
((T ψ ◦ T φ )w)(t) = (T φ w∗ )(t) = w∗ (σt )
= w∗∗ (t) = w(τσt ).
Hence to prove the property (ii) we need only show that
Zτσt
1
1
ds = t.
φ(w(s)) ψ(w(s))
0
Since
Zτt
1
ds = t,
φ(w(s))
dt
1
=
dτt φ(w(τt ))
0
and
1
1
dt
=
=
dσt ψ(w∗ (σt )) ψ(w(τσt ))
Therefore
dτσt dσt
dτσt
=
= φ(w(τσt ))φ(w∗ (σt ))
−
dt
dσt
dt
= φ(w(τσt )ψ(w(τσt ))
28. Random Time Changes
232
= (φψ)(w(τσt )).
223
Thus
Zτσt
1
ds = t.
(φψ)(w(s))
0
This completes the proof.
(iii) From (i) and (ii) it is clear that T φ−1 = T φ−1 where φ−1 =
1
.
φ
(iv) (τt ) is a stopping time relative to τt . i.e.




Zλ






1


ds
≥
r
w
:
∈ λ for each λ ≥ 0.






φ(w(s))




0
(v) T φ (w)(t) = w(τt w) = Xτt (w).
Thus T φ is (Ft − Fτt )-measurable, i.e. T φ−1 (Ft ) ⊂ Fτt .
Since X(t) is an Itô process, with parameters b, a, ∀ f ∈ C0∞ (Rd ),
Rt
f (X(t)) − (L f )(X(s))ds is a martingale relative to (Ω, Ft , P). By the
0
optional sampling theorem
f (Xτt ) −
Zτt
(L f )(X(s))ds
0
is a martingale relative to (Ω, Fτt , P), i.e.
f (Xτt ) −
Zt
0
(L f )(X(τ s ))dτ s
233
is a martingale relative to (Ω, Fτt , P). But
f (X(τt )) −
Zt
dτ s
= φ. Therefore
dt
(L f )(Xτs )φ(Xτs )ds
0
is a martingale.
Put Y(t) = Xτt and appeal to the definition of Lφ to conclude that
f (Y(t)) −
Zt
224
(Lφ f )(Y(s))ds
0
is a martingale. Y(t, w) = Xτt (w) = (T φ w)(t). Let F t = σ{Y(s) : 0 ≤ s ≤
t}. Then F t = T φ−1 (Ft ) ⊂ Fτt . Thus
f (Y(t)) −
Zt
(Lφ f )(Y(s))ds
0
is a martingale relative to (Ω, F t , P). Define Q = PT φ−1 so that
f (X(t)) −
Zt
(Lφ f )(X(s))ds
0
is an (Ω, Ft , Q)-martingale, i.e. Q is an Itô process that corresponds
to the operator φL. Or, PT φ−1 is an Itô process that corresponds to the
operator φL.
We have now proved the following theorem.
Theorem . Let Ω = C([0, ∞); Rd ); X(t, w) = w(t);
X
∂
∂2
1X
+
bj
.
ai j
L=
2 i, j
∂xi ∂x j
∂x j
Suppose that X(t) is an Itô process relative to (Ω, Ft , P) that corresponds to the operator L. Let 0 ≤ C1 ≤ φ ≤ C2 where φ : Rd → R 225
is measurable. If Q = PT φ−1 , then X(t) is an Itô process relative to
(Ω, Ft , Q) that corresponds to the operator φL.
28. Random Time Changes
234
As 0 < C1 ≤ φ ≤ C2 , we get 0 < 1/C2 ≤ 1/φ < 1/C1 with
T φ−1 ◦ T φ = I. We have thus an obvious corollary.
Corollary . There exists a probability measure P on Ω such that X is an
Itô process relative to (Ω, Ft , P) that corresponds to the operator L if
and only if there exists a probability measure Q on Ω such that X is an
Ito process relative to (Ω, Ft , Q) that corresponds to the operator φL.
Remark . If C2 ≥ φ ≥ C1 > 0 then we have shown that existence and
uniqueness of an Itô process for the operator L guarantees existence and
uniqueness of the Itô process for the operator φL. The solution is no
longer unique if we relax the strict positivity on C1 as is illustrated by
the following example.
1 ∂2
Let φ ≡ a(x) = |x|α ∧ 1 where 0 < α < 1 and let L = a . Define
2 ∂x
δ0 on {C([0, ∞); R)} by



1, if θ ∈ A, ∀Aǫ,
δ0 (A) = 

0, if θ < A,
where θ is the zero function on [0, ∞).
Claim. δ0 is an Itô process with parameters 0 and a. For this it is enough
to show that, ∀ f ∈ C0∞ (R)
f (X(t)) −
Zt
(L f )(X(s))ds
0
226
is a martingale, using a(0) = 0, it follows easily that
Z Zt
A
(L f )(X(σ))dσdδ0 = 0
0
∀ Borel set A of C([0, ∞); R) and
Z
A
R
f (X(t))dδ0 = 0 if θ < A and
A
f (X(t))dδ0 = f (0)
235
if θ ∈ A, and this is true ∀t, showing that X(t, w) = w(t) is an Itô process
relative to δ0 corresponding to the operator L.
Next we shall define T a (as in the theorem); we note that T a cannot
be defined everywhere (for example T a (θ) is not defined). However T a
is defined a.e. P where P = P0 is the Brownian motion.
 t

Z
 Zt Z∞
−y
1
1
1


P
√
e 2s dy ds < ∞
E 
ds =
α
α
 |X(s)|

y (2πs)
0
0
0
since 0 < α < 1. Thus by Fubini’s theorem,
Zt
0
1
ds < ∞
|w(s)|α
a.e.
Taking t = 1, 2, 3 . . ., there exists a set Ω∗ such that P(Ω∗ ) = 1 and
Zt
0
1
ds < ∞,
|w(s)|α
Observe that
Zt
0
∀t,
∀w ∈ Ω∗
1
ds < ∞
|w(s)|α
implies that
Zt
0
1
ds < ∞,
|w(s)|α ∧ 1
for
227
Zt
=
Z
[0,t]{|w(s)|α >1}
ds
=
|w(s)|α ∧ 1
0
Z
ds
+
|w(s)|α ∧ 1
{|w(s)|α ≤1}[0,t]
ds
,...
|w(s)|α
28. Random Time Changes
236
α
≤ m{(|w(s)| > 1)[0, t]} +
Zt
0
1
ds < ∞
|w(s)|α
(m = Lebesgue measure)
Thus T a is defined on the whole of Ω∗ . Using the same argument
as in the theorem, it can now be proved that X is an Itô process relative
to Q corresponding to the operator L. Finally, we show that Q{θ} = 0.
Q{θ} = PT a−1 {θ}. Now: T a−1 {θ} = empty. For, let w ∈ T a−1 {θ}. Then
w(τt ) = 0, ∀t, w ∈ Ω∗ . Since |τt − τ s | ≤ |t − s|, one finds that τt is a
continuous function of t. Further τ1 > 0, and w = 0 on [0, τ1 ] gives
Zτ1
0
1
ds = ∞.
|w(s)|α ∧ 1
This is false unless T a−1 {θ} = empty. Thus Q{θ} = 0 and Q is different from δ0 .
29. Cameron - Martin Girsanov Formula
LET US REVIEW what we did in Brownian motion with drift.
Let (Ω, Ft , P) be a d-dimensional Brownian motion with
P{w : w(0) = x} = 1.
Let b : Rd → Rd be a bounded measurable function and define

 t
Zt

Z
1


|b|2 ds .
Z(t) = exp  hb, dxi −


2
0
0
Then we see that Z(t, ·) is an (Ω, Ft , P)-martingale. We then had a
probability measure Q given by the formula
dQ = Z(t, ·).
dP Ft
We leave it as an exercise to check that in effect X is an Itô process
relative to Q with parameters b and I. In other words we had made a
transition from the operator ∆/2 to ∆/2 + b · ∇. We now see whether
such a relation also exists for the more general operator L.
Theorem . Let a : Rd → S d+ be bounded and measurable such that
a ≥ CI for some C > 0. Let b : Rd → Rd be bounded, Ω = ([0, ∞); Rd ),
X(t, w) = w(t), P any probability measure on Ω such that X is an Itô
237
228
29. Cameron - Martin - Girsanov Formula
238
process relative to (Ω, Ft , P) with parameters [0, a]. Define Qt on Ft
by the rule
dQt dP
229
F
t
 t

Zt
Z

1


−1
−1
= Z(t, ·) = exp  ha b, dXi −
hb, a bids .


2
0
0
Then
(i) {0t }t ≥ 0 is a consistent family.
(ii) there exists a measure Q on σ(||Ft ):
Q = Qt .
Ft
(iii) X(t) is an Itô process relative to (Ω, Ft , Q) with parameters [b, a],
i.e. it corresponds to the operator
X
∂
∂2
1X
+
bj
.
ai j
2 i, j
∂xi ∂x j
∂x j
j
Proof.
Rt
(i) Let A(t) = ha−1 b, dXi. Then A ∈ I[0, hb, a−1 bi].
0
Therefore Z(t) is a martingale relative to (Ω, Ft , P) hence {Qt }t≥0
is a consistent family.
(ii) Proof as in the case of Brownian motion.
(iii) We have to show that
exp[hθ, X(t, ·)i − hθ,
Zt
1
bdsi −
2
0
is a martingale relative to (Ω, Ft , Q).
Zt
0
hθ, aθids]
239
Now for any function θ which is progressively measurable and bounded
Zt
Zt
1
hθ, aθids]
exp[ hθ, dXi −
2
0
0
230
is an (Ω, Ft , P)-martingale. Replace θ by θ(w) = θ + (a−1 b)(χ(s, w)),
where θ now is a constant vector. Then
Zt
Zt
1
−1
hθ + a−1 b, aθids
exp[ hθ + a b, dXi −
2
0
0
is an (Ω, Ft , P)-martingale, i.e.
1
exp[hθ, X(t)i −
2
Zt
1
hθ, aθids −
2
0
Zt
1
ha−1 b, aθi −
2
0
Zt
hθ, bi]
0
is an (Ω, Ft , Q)-martingale, and
ha−1 b, aθi = ha∗ a−1 b, θi
= haa−1 b, θi
(since a = a∗ )
= hb, θi.
Thus
1
exp[hθ, X(t)i −
2
Zt
hθ, aθids −
0
Zt
hθ, bids]
0
is an (Ω, Ft , Q)-martingale, i.e. X is an Itô process relative to (Ω, Ft , Q)
with parameters [b, a]. This proves the theorem.
We now prove the converse part.
Theorem . Let
L1 =
∂2
1X
i, jai j
2
∂xi ∂x j
29. Cameron - Martin - Girsanov Formula
240
and
L2 ≡
231
X
∂
1X
∂2
+
bj
,
ai j
2 i, j
∂xi ∂x j
∂x
j
j
where a : Rd → S d+ is bounded measurable such that a ≥ CI for some
C > 0; b : Rd → Rd is bounded and measurable. Let Ω = C([0, ∞); Rd )
with Ft as usual. Let θ be a probability measure on σ(∪Ft ) and X a
progressively measurable function such that X is an Itô process relative
to (Ω, Ft , Q) with parameters [b, a] i.e. X corresponds to the operator
L2 . Let
Zt
Zt
1
Z(t) = exp[− ha−1 b, dXi +
hb, a−1 bids].
2
0
0
Then
(i) Z(t) is an (Ω, Ft , Q)-martingale.
(ii) If Pt is defined on Ft by
dPt = Z(t),
dQ Ft
Then there exists a probability measure P on σ(∪Ft ) such that
P = Pt
Ft
(iii) X is an Itô process relative to (Ω, Ft , P) corresponding to parameters [0, a], i.e. X corresponds to the operator L1 .
Proof.
(i) Let
A(t) =
Zt
h−a−1 b, dXi.
0
Then A(t) is an Itô process with parameters [h−a−1 b, bi, ha−1 b, bi].
241
Thus
Zt
exp[A(t) −
1
h−a b, bids −
2
−1
Zt
ha−1 b, bids]
0
0
is an (Ω, Ft , Q)-martingale, i.e. Z(t) is an (Ω, Ft , Q) martingale.
(ii) By (i), Pt is a consistent family. The proof that there exists a 232
probability measure P is same as before.
Since X is an Itô process relative to Q with parameters b and a,
Zt
Zt
Zt
1
hθ, aθids]
exp[ hθ, dXi − hθ, bids −
2
0
0
0
is a martingale relative to Q for every bounded measurable θ. Replace θ
by θ(w) = θ − (a−1 b)(X(s, w)) where θ now is a constant vector to get
Zt
exp[hθ, X(t)i −
−1
ha b, dXi −
0
−
Zt
hθ, bi +
0
1
2
Zt
Zt
ha−1 b, bids−
0
hθ − a−1 b, aθ − bids]
0
is an (Ω, Ft , Q) martingale, i.e.
exp[hθ, Xi −
Zt
−1
ha b, dXi −
0
1
−
2
Zt
Zt
hθ, bids +
0
1
hθ, aθids −
2
0
Zt
1
ha b, bids +
2
−1
0
+
Zt
0
ha−1 b, aθids]
Zt
ha−1 b, bids−
Zt
hθ, bids+
0
0
29. Cameron - Martin - Girsanov Formula
242
is an (Ω, Ft , Q) martingale. Let θ ∈ Rd , so that
1
exp[hθ, Xi −
2
Zt
1
hθ, bids −
2
0
Zt
1
hθ, aθids +
2
0
Zt
ha−1 b, aθids]Z(t)
0
is an (Ω, Ft , Q)-matringale and
ha−1 b, ai = hb, θi
233
Therefore
1
exp[hθ, Xi −
2
(since a = a∗ ).
Zt
hθ, aθids]Z(t)
0
is an (Ω, Ft , Q) martingale.
dP Using the fact that
= Z(t), we conclude that
dQ Ft
1
exp[hθ, Xi −
2
Zt
hθ, aθids]
0
is a martingale relative to (Ω, Ft , P), i.e. X ∈ I[0, a] relative to
(Ω, Ft , P).
This proves the theorem.
Summary. We have the following situation
L1 , Ω, Ft , Ω = C([0, ∞); Rd ), L2 , Ω, Ft .


P a probability measure 
 X is an Itô process relative

 to a probability measure Q
such that X is an Itô Pro- 


cess relative to P corre-  =⇒  corresponding to L2 . Q is



dQ sponding to the operator 

given by
= Z(t, ·)
L1 .
dP Ft
X is an Itô process relative
to P corresponding to L1
dP 1
where
F =
dQ t Z(t, ·)






⇐=

 X is an Itô process relative
 to Q corresponding to L .
2



243
Thus existence and uniqueness for any system guarantees the existence and uniqueness for the other system.
Application. (Exercise).
Take d = 1, a : R → R bounded and measurable with 0 < C1 ≤
∂
a ∂2
+ b . Show that there exists a unique
a < C2 < ∞. Let L =
2
2 ∂x
∂x
probability masure P on Ω = C([0, ∞); R) such that X(t) is Itô relative
to P corresponding to L. (X(t, w) ≡ w(t)) for any given starting point.
234
30. Behaviour of Diffusions
for Large Times
LET L2 = ∆/2 + b · ∇ WITH b : Rd → Rd measurable and bounded 235
on each compact set. We assume that there is no explosion. If P x is the
d-dimensional Brownian measure on Ω = C([0, ∞); Rd ) we know that
there exists a probability measure Q x on Ω such that
 t

Zt
Z

1
dQ x 
2 
|b| ds
= exp  hb, dXi −

dP x t
2
0
0
Let K be any compact set in Rd with non-empty interior. We are
interested in finding out how often the trajectories visit K and whether
this ‘frequency’ depends on the starting point of the trajectory and the
compact set K.
Theorem . Let K be any compact set in Rd having a non-empty interior.
Let
K
E∞
= {w : w revisits K for arbitrarily large times}
= {w : there exists a sequence t1 < t2 < · < ∞
with tn → ∞ such that w(tn ) ∈ K}
Then,
K
either Q x (E∞
) = 0, ∀x, and ∀K,
or
K
Q x (E∞
) = 1, ∀x, and ∀K.
245
30. Behaviour of Diffusions for Large Times
246
Remark.
1. In the first case lim |X(t)| = +∞ a.e. Q x , ∀x, i.e. almost
t→+∞
all trajectories stay within K only for a short period.
These trajectories are called transient. In the second case almost
all trajectories visit K for arbitrary large times. Such trajectories
are called recurrent.
236
2. If b = 0 then Q x = P x . For the case d = 1 or d = 2 we know
that the trajectories are recurrent. If d ≥ 3 the trajectories are
transient.
Proof.
Step 1. We introduce the following sets.
E0K = {w : X(t, w) ∈ K for some t ≥ 0},
EtK0 = {w : X(t, w) ∈ K for some t ≥ t0 }, 0 ≤ t0 < ∞.
Then clearly
K
=
E∞
∞
\
EnK =
n=1
\
EtK0 .
t0 ≥0
Let
K
ψ(x) = Q x (E∞
), F = χE∞K .
K
)
E Qx (F|Ft ) = E Qx (χE∞K |Ft ) = QX(t) (E∞
by the Markov property,
= ψ(X(t)) a.e. Q x .
Next we show that ψ(X(t)) is a martingale relative to Q x . For, if
s < t,
E Qx (ψ(X(t))|Fs )
= E Qx (E Qx (F|Ft )|Fs )
= E Qx (F|Fs )
= ψ(X(s)).
247
237
Equating the expectations at time t = 0 and time t one gets
Z
ψ(X(t))dQ x
ψ(x) =
=
Ω
Z
ψ(y)q(t, x, y)dy,
where q(t, x, A) = Q x (Xt ∈ A), ∀A Borel in Rd .
We assume for the present that ψ(x) is continuous (This will be
shown in Lemma 4 in the next section). By definition 0 ≤ ψ ≤ 1.
Step 2. ψ(x) = 1, ∀x or ψ(x) = 0, ∀x.
Suppose that ψ(x0 ) = 0 for some x0 . Then
Z
0 = ψ(x0 ) =
ψ(y)q(t, x0 , y)dy.
As q > 0 a.e. and ψ ≥ 0 we conclude that ψ(y) = 0 a.e. (with respect to
Lebesgue measure). Since ψ is continuous ψ must vanish identically.
If ψ(x0 ) = 1 for some x0 , we apply the above argument to 1 − ψ
to conclude that ψ = 1, ∀x. We now show that the third possibility
0 < ψ(x) < 1 can never occur.
Since K is compact and ψ is continuous,
0 < a = inf ψ(y) ≤ sup ψ(y) = b < 1.
y∈K
y∈K
From an Exercise in the section on martingales it follows that
ψ(X(t)) → χE∞K
a.e.
Qx
as t → +∞.
Therefore lim ψ(X(t))(1 − ψ(X(t))) = 0 a.e. Q x . Now
t→∞
K
ψ(x0 ) = Q x0 (E∞
) = Q x0 {w : w(t) ∈ K for arbitrary large time}
≤ Q x0 {w : a ≤ ψ(X(t, w)) ≤ b for arbitrarily large times}
≤ Q x0 {w : (1 − b)a ≤ ψ(X(t))[1 − ψ(X(t)] ≤ b(1 − a)
for arbitrarily large times}
30. Behaviour of Diffusions for Large Times
248
= 0.
238
Thus ψ(x) = 0 identically, which is a contradiction. Thus for the
given compact set K,
either
K
Q x (E∞
) = 0, ∀x,
K
Q x (E∞
) = 1, ∀x.
or
◦
K0
Step 3. If Q x (E∞
) = 1 for some compact set K0 (K 0 , ∅) and ∀x, then
K
Q x (E∞ ) = 1, ∀ compact set K with non-empty interior.
K0
We first given an intuitive argument. Suppose Q x (E∞
) = 1, i.e.
almost all trajectories visit K0 for arbitrarily large times. Each time a
trajectory hits K0 , it has some chance of hitting K. Since the trajectory
visits K0 for arbitrarily large times it will visit K for arbitrarily large
times. We now give a precise arguent. Let
τ0 = inf{t : X(t) ∈ K0 }
τ1 = inf{t ≥ t0 + 1
...
...
...
...
X(t) ∈ K0 }
τn = inf{t ≥ tn−1 + 1 : X(t) ∈ K0 }
...
...
...
...
Clearly τ0 < τ1 < . . . < and τn ≥ n.
Q x (EnK ) ≥ Q x {X(t) ∈ K for t ≥ τn }
∞
[
[τ j , τ j + 1]}
≥ Q x {X(t) ∈ K for t ∈
j=n








\

= 1 − Qx 
X(t)
<
K
for
t
∈
[τ
,
τ
+
1]

j
j




 j≥n









\

≥ 1 − Qx 
X(τ
+
1)
∈
K

j




 j≥n

249
239
We claim that
\
Q x ( X(τ j + 1) < K) = 0,
j≥n
K ) = 1, completing the
so that Q x (EnK ) = 1 for every n and hence Q x (E∞
proof of the theorem.
Now
◦
q(1, x, K) ≥ q(1, x, K) > 0,
◦
∀x, K
interior of
K.
It is clear that if xn → x, then
◦
◦
lim q(1, xn , K) ≥ q(1, x, K).
◦
Let d = inf q(1, x, K). Then there exists a sequence xn in K0 such
x∈K0
◦
that d = Lt q(1, xn , K). K0 being compact, there exists a subsequence
n→∞
yn of xn with yn → x in K0 , so that
◦
◦
◦
d = lim q(1, x, K) = lim q(1, yn , K) ≥ q(1, x, K) > 0.
n→∞
Thus
inf q(1, x, K) ≥ d > 0.
x∈K0
Now


N
Y

Q x 
X(τ j + 1) < K|FτN 
j=n
=
N−1
Y
j=n
χ(X(τ j + 1) < K)Q x (X(τN + 1) ∈ K|FτN ) because
τ j + 1 ≤ τN
=
for
j < N,
N−1
Y
(χX(τ j +1) < K)QX(τN ) (X(1) < K) by the strong
j=n
30. Behaviour of Diffusions for Large Times
250
Markov property,
=
N−1
Y
q(1, X(τN ), K c ) χ(X(τ j +1)<K) .
j=1
240
Therefore


N

\
Q x  X(τ j + 1) < K 
j=n
N
\
= E (Q x ( X(τ j + 1) < K|τN )))
Qx
j=n


N−1

Y

c
Qx 
= E  (χ[X(τ j +1)<K] )q(1, X(τN ), K )
j=n
Since K0 is compact and X(τN ) ∈ K0 ,
q(1, X(τN ), K c ) = 1 − q(1, X(τN ), K) ≤ 1 − d
Hence




N
\

N−1

\



Q x  X(τ j + 1) < K  ≤ (1 − d)Q x 
X(τ j + 1) < K  .
j=n
j=n
Iterating, we get


N

\
Q x  X(τ j + 1) < K  ≤ (1 − d)N−n+1 , ∀N.
j=n
241
Let N → ∞ to get


\

Q x  X(τ j + 1) < K  = 0,
j=n
since 0 ≤ 1 − d < 1. Thus the claim is proved and so is the theorem.
251
◦
K ) = 1 if and only if
Corollary . Let K be compact, K , ∅. Then Q x (E∞
Q x (E0K ) = 1, ∀x.
K ) = 1; then Q (E K ) = 1 because E K E K . SupProof. Suppose Q x (E∞
x 0
∞ 0
pose Q x (E0K ) = 1, then
Q x (EnK ) = E Qx (E Qx (χEnK |Fn ))
= E Qx (QX(n) (E0K ))
= E Qx (1)
= 1, ∀n.
K ) = 1.
Therefore Q x (E∞
K ) = 0 then it need not imply that
Remark. If Q x (E∞
Q x (E0K ) = 0.
Example . Take b = 0 and d = 3. LEt K = S 1 = {x ∈ R3 such that
|x| ≤ 1}. Define



1, for |x| ≤ 1,
ψ(n) = 

 1 , for |x| ≥ 1.
|x|
K ) = 0. In fact, P (E K ) = ψ(x) (Refer
P x (E0K ) , constant but P x (E∞
x 0
Dirichlet Problem).
31. Invariant Probability
Distributions
Definition. Let {P x } x∈Rd be a family of Markov process on
Ω = C([0, ∞); Rd )
indexed by the starting points x, with homogeneous transition probability p(t, x, A) = P x (Xt ∈ A) for every Borel set A in Rd . A probability
measure µ on the Borel field of Rd is called an invariant distribution if,
∀A Borel in Rd .
Z
p(t, x, A)dµ(x) = µ(A).
Rd
We shall denote dp(t, x, y) by p(t, x, dy) or p(t, x, y)dy if it has a density.
Proposition . Let L2 = ∆/2 + b · ∇ with no explosion. Let Q x be the
associated measure. If {Q x } has an invariant measure µ then the process
is recurrent.
Proof. It is enough to show that if K is a compact set with non-empty
interior then
K
Q x (E∞
)=1
for some x. Also Q x (EtK ) ≥ Q x (Xt ∈ K) = q(t, x, K). Therefore
Z
Z
µ(K) =
q(t, x, K)dµ(x) ≤
Q x (EtK )dµ(x).
253
242
31. Invariant Probability Distributions
254
K ) as t → ∞.
Now, 0 ≤ Q x (EtK ) ≤ 1 and Q x (EtK ) decreases to Q x (E∞
Therefore by the dominated convergence theorem
Z
K
µ(K) ≤
Q x (E∞
)dµ(x).
243
Sn
If the process were transient, then Q x (E∞
) = 0, ∀n, where S n =
d
{x ∈ R : |x| ≤ n}, i.e. µ(S n ) = 0, ∀n. Therefore µ(Rd ) = 0, which is
false. Thus the process is recurrent.
The converse of this proposition is not true as is seen by the following example.
1 ∂2
so that we are in a one-dimensional situation (BrowLet L =
2 ∂x2
nian motion). Then
Z
−(x−y)2
1
1
√
e 2t dy ≤ √
λ(K),
p(t, x, K) =
(2πt)
(2πt)
K
where λ denotes the Lebesgue measure on R. If there exists an invariant
distribution µ, then
Z
Z
1
λ(K)
µ(K) =
p(t, x, K)dµ(x) ≤ √
λ(K) dµ(x) = √
(2πt)
(2πt)
Letting t → ∞, we get µ(K) = 0 ∀ compact K, giving µ = 0, which
is false.
Theorem . Let L = ∆/2 + b · ∇ with no explosion. Assume b to be
C ∞ . Define the formal adjoint L∗ of L by L∗ = ∆/2 − ∇ · b (i.e. L∗ u =
1
∆u − ∇ · (bu)). Suppose there exists a smooth function φ(C 2 - would
2
do) suchR that L∗ φ = 0, φ ≥ 0, inf φdx = 1. If one defines µ by the rule
µ(A) = φ(y)dy, then µ is an invariant distribution relative to the family
A
{Q x }.
244
Proof. We assume the following result from the theory of partial differential equations.
255
If f ∈ C0∞ (G) where G is a bounded open set with a smooth boundary ∂G and f ≥ 0, then there exists a smooth function UG : [0, ∞)×G →
[0, ∞) such that
∂UG
= LUG on (0, ∞) × G,
∂t
UG (0, x) = f (x) on {0} × G,
∀x ∈ G.
UG (t, x) = 0,
Let t > 0. As UG , φ are smooth and G is bounded, we have
Z
Z
Z
∂
∂
UG φds =
φLUG dx
UG (t, x)φ(x)dx =
∂t
∂t
G
G
G
Using Green’s formula this can be written as
#
Z
Z
Z "
∂
∂
1
∂UG
∗
− UG
UG (t, x)φ(x)dx =
UG L φ −
φ
dS +
∂t
2
∂n
∂n
G
G
∂G
Z
+ hb · niUG (t, x)φ(x)dS
∂G
Here n is assumed to be the inward normal to ∂G. So,
Z
Z
∂UG
1
∂
(x)
(t, x)dS
UG (t, x)φ(x)dx = −
∂t
2
∂n
G
∂G
(Use the equation satisfied by φ and the conditions on UG ). Now UG (t, x)
≥ 0, ∀x ∈ G, UG (t, x) = 0, ∀x in ∂G, so that
∂UG
(t, x) ≥ 0.
∂n
This means that
∂
∂t
245
Z
G
UG (t, x)φ(x)dx ≤ 0, ∀t > 0,
31. Invariant Probability Distributions
256
i.e.
R
UG (t, x)φ(x)dx is a monotonically decresing function of t. There-
G
fore
Z
Z
UG (t, x)φ(x)dx ≤
G
G
Z
=
G
Z
=
UG (0, x)φ(x)dx
f (x)φ(x)dx
f (x)φ(x)dx.
Rd
Next we prove that if U : [0, ∞)×Rd → [0, ∞) is such that
∀t > 0 and U(0, x) = f (x), then
Z
Z
U(t, x)φ(x)dx ≤
f (x)φ(x)dx.
Rd
∂U
∂t
= LU,
Rd
The solution UG (t, x) can be obtained by using Itô calculus and is
given by
Z
UG (t, x) =
f (X(t))χ{τG>t } dQ x .
We already know that
U(t, x) =
Therefore
Z
Now
U(t, x)φ(x)dx =
"
Z
f (X(t))dQ x .
"
f (X(t))φ(x)DQ x dx.
f (X(t))χ{τG >t} dQ x φ(x)dx
Z
Z
UG (t, x)φ(x)dx ≤
f (x)φ(x)dx.
Rd
257
246
Letting G increase to Rd and using Fatou’s lemma, we get
"
Z
f (X(t))φ(x)dQ x dx ≤
f (x)φ(x)dx
This proves the assertion made above.
Let
Z
µ(A) =
φ(X)dx,
ν(A) =
Z
A
Q x (Xt ∈ A)dµ(x) =
Z
q(t, x, A)dµ(x).
Let f ∈ C0∞ (G), f ≥ 0, where G is a bounded open set with smooth
boundary. Now
Z
"
f (y)dν(y) =
f (y)q(t, x, y)dµ(x)dy
"
=
f (X(t))dQ x dµ(x)
Z
=
U(t, x)dµ(x)
Z
=
U(t, x)φ(x)dx
Z
Z
≤
f (x)φ(x)dx =
f (x)dµ(x).
Thus, ∀ f ≥ 0 such that f ∈ C0∞ ,
Z
Z
f (x)dν(x) ≤
f (x)dµ(x).
This implies that ν(A) ≤ µ(A) for every Borel set A. (Use mollifier 247
s and the dominated convergence theorem to prove the above inequality
for χA when A is bounded). Therefore ν(Ac ) ≤ µ(Ac ), or 1 − µ(A) ≤ 1 −
31. Invariant Probability Distributions
258
µ(A), since µ, ν are both probability measures. This gives µ(A) = ν(A),
i.e.
Z
µ(A) =
q(t, x, A)dµ(x), ∀t,
i.e. µ is an invariant distribution.
We now see whether the converse result is true or not. Suppose there
exists a probability measure µ on Rd such that
Z
Q x (Xt ∈ A)dµ(x) = µ(A), ∀A Borel in Rd and ∀t.
R
The question we have in mind is whether µ(A) = φdx for some
A
R
smooth φ satisfying L∗ φ = 0, φ ≥ 0, φ(x)dx = 1. To answer this we
proceed as follows.
R
By definition µ(A) = q(t, x, A)dµ(x). Therefore
"
f (X(t))dQ x dµ(x)
"
=
f (y)q(t, x, y)dy dµ(x)
Z
=
f (y)dµ(y)∀ f ∈ C0∞ (Rd )|| f ||∞ ≤ 1.
(1)
Since X is an Itô process relative to Q x with parameters b and I,
f (X(t)) −
Zt
(L f )(X(s))ds
0
248
is a martingale. Equating the expectations at time t = 0 and time t we
obtain

 t

Z


E Qx ( f (X(t)) = f (x) + E Qx  (L f )(X(s))ds


0
Integrating this expression with respect to µ gives
"
f (X(t))dQ x dµ(x) =
Z
f (x)dµ(x)
" Zt
Rd
0
(L f )(X(s))ds dQ x dµ.
259
Using (1), we get
0=
Z Z Zt
Rd Ω
(L f )(X(s))dQ x ds dµ(x)
0
Applying equation (1) to the function L f we then get
0=
Z Zt
Rd
=t
Z
(L f )(y)dµ(y)ds
0
(L f )(y)dµ(y),
∀t > 0.
Rd
Thus
0=
Z
(L f )(y)dµ(y),
∀ f ∈ C0∞ (Rd ).
Rd
In the language of distributions this just means that L∗ µ = 0.
From the theory of partial differential equations it then follows that
there exists a smooth function φ such that ∀A Borel in Rd ,
Z
µ(A) =
φ(y)dy
A
with L∗ φ = 0. As µ ≥ 0, φ ≥ 0 and since
Z
µ(Rd ) = 1,
φ(x)dx = 1.
Rd
249
We have thus proved the following (converse of the previous) theorem.
Theorem . Let µ be an invariant distribution with respect to the family
{Q x } with b : Rd → Rd being C ∞ . Then there exists a φ ∈ L′ (Rd ), φ ≥ 0,
φ smooth such that
Z
L∗ φ = 0,
φ(y)dy = 1
31. Invariant Probability Distributions
260
and such that
µ(A) =
Z
φ(y)dy,
∀A
Borel in Rd .
A
Theorem (Uniqueness). Let φ1 , φ2 be smooth on Rd such that
Z
Z
φ1 , φ2 ≥ 0, 1 =
φ1 dy =
φ2 dy, L∗ φ1 = 0 = L∗ φ2 .
Rd
Rd
Then φ1 = φ2 .
Proof. Let f (x) = φ1 (x) − φ2 (x),
Z
µi (A) =
φi (x)dx,
i = 1, 2.
A
Then µ1 , and µ2 are invariant distributions. Therefore
Z
Z
q(t, x, y)φi (x)dx =
q(t, x, y)dµi (x)
= φi (y),
(a.e.),
Taking the difference we obtain
Z
q(t, x, y) f (x)dx = f (y),
i = 1, 2.
a.e.
Now
Z
Z Z
| f (y) dy =
| q(t, x, y) f (x)dx|dy
"
≤
q(t, x, y)| f (x)|dx dy
Z
Z
=
| f (x)|dx q(t, x, y)dy
Z
=
| f (x)|dx.
261
250
(*)
Thus
"
Z Z
| f (x)|q(t, x, y)dx dy =
| q(t, x, y) f (x)dx|dy ∀t.
We show that f does not change sign,R i.e. f ≥ 0 a.e. or f ≤ 0 a.e.
The result then follows from the fact that f (x)dx = 0. Now
Z
Z
| q(1, x, y) f (x)dx| ≤
q(1, x, y)| f (x)|dx
and (∗) above gives
Z Z
"
| q(1, x, y) f (x)dx|dy =
q(1, x, y)| f (x)|dx dy.
Thus
|
Z
q(1, x, y) f (x)dx| =
Z
q(1, x, y)| f (x)|dx a.e. y,
i.e.
|
Z
q(1, x, y) f (x)dx +
E−
=
Z
E+
Z
q(1, x, y) f (x)dx|
E−
q(1, x, y) f (x)dx −
Z
q(1, x, y) f (x)dx a.e. y,
E−
where
E + = {x : f (x) > 0},
E − = {x : f (x) < 0},
E 0 = {x : f (x) = 0}.
Squaring both sides of the above equality, we obtain




 Z
Z
 q(1, x, y) f (x)dx  q(1, x, y) f (x)dx = 0,
(**)

 

E+
a.e.
y.
E−
251
31. Invariant Probability Distributions
262
Let A be a set of positive Lebesgue measure; then p(1, x, A) =
P x (X(1) ∈ A) > 0. Since Q x is equivalent to P x on Ω we have Q x (X(1) ∈
A) = q(1, x, A) > 0. Therefore q(1, x, y) > 0 a.e. y for each x. By
Fubini’s theorem q(1, x, y) > 0 a.e. x, y. Therefore for almost all y,
q(1, x, y) > 0 for almost all x. Now pick a y such that (∗∗) holds for
which q(1, x, y) > 0 a.e. x.
We therefore conclude from (∗∗) that either
Z
q(1, x, y) f (x)dx = 0, in which case f ≤ 0 a.e.,
E+
or
Z
q(1, x, y) f (x)dx = 0,
in which case
E−
f ≥0
a.e.
Thus f does not change its sign, which completes the proof.
Remark. The only property of the operator L we used was to conclude
q > 0. We may therefore expect a similar result for more general operators.
R
Theorem . Let L∗ φ = 0 where φ ≥ 0 is smooth and φ(x)dx = 1. Let K
be any compact set. Then
Z
sup |q(t, x, y) − φ(y)|dy → 0 as t → +∞.
x∈K
Lemma 1. Let b be bounded and smooth. For every f : Rd → Rd that is
bounded and measurable let u(t, x) = E Qx ( f (X(t)). Then for every fixed
t, u(t, x) is a continuous function of x. Further, for t ≥ ǫ > 0,
|u(t, x) −
252
Z
1
|x − y|2
u(t − ǫ, y) √
exp
−
dy|
2ǫ
(2πǫ)d
√
≤ || f ||∞ (ect (ecǫ − 1)),
where c is a constant depending only on ||b||∞ .
263
Proof. Let
(T t f )(x) = E Qx ( f (X(t)) = E Px ( f (X(t))Z(ǫ, t))+
(1)
+ E Px ( f (X(t))(Z(t) − Z(ǫ, t))),
where

 t
Zt

Z
1


|b|2 ds ,
Z(t) = exp  hb2 , dxi −


2
0
0
 t

Zt
Z

1


Z(ǫ, t) = exp  hb, dxi −
|b(X(s))|2 ds .


2
ǫ
ǫ
E Px ( f (X(t))Z(ǫ, t)) = E Px (E PX ( f (X(t))Z(ǫ, t)|ǫ ))
= E Px (E PX ǫ)( f (X(t − ǫ))Z(t − ǫ)))
(by Markov property),
Px
= E (u(t − ǫ, X(ǫ)).
(2)
=
Z
"
#
1
|(x − y)|2
u(t − ǫ, y) √
exp −
dy.
2ǫ
( (2πǫ))d
Now
(E Px (|Z(t) − Z(ǫ, t)|)) =
= E Px (|Z(ǫ)Z(ǫ, t) − Z(ǫ, t)|))2
= E Px (Z(ǫ, t)Z(ǫ) − 1|))2
≤ (E Px ((Z(ǫ) − 1)2 ))(E Px (Z 2 (ǫ, t)))
(by Cauchy Schwarz inequality),
Px
≤ E (Z 2 (ǫ) − 2Z(ǫ) + 1)E Px (Z 2 (ǫ, t))
≤ E Px (Z 2 (ǫ) − 1)E Px (Z 2 (ǫ, t)), (since E Px (Z(ǫ)) = 1),
Zt
Zt
Zt
2
2
Px 2
Px
|b| ds + |b|2 ds))
≤ E (Z (ǫ) − 1)E (exp(2 hb, dXi −
2
ǫ
ǫ
ǫ
31. Invariant Probability Distributions
264
≤ E Px (Z 2 (ǫ) − 1)ect ,
253
using Cauchy Schwarz inequality and the fact that
Px
E (exp(2
Zt
22
hb, dXi −
2
ǫ
Zt
|b|2 ds)) = 1.
ǫ
Thus
E Px (|Z(t) − Z(ǫ, t)||2 ≤ (ecǫ − 1)ect
where c depends only on ||b||∞ . Hence
(3)
|E Px ( f (X(t))(Z(t) − Z(ǫ, t))| ≤ || f ||∞ E Px (|Z(t) − Z(ǫ, t)|)
√
≤ || f ||∞ ((ecǫ − 1)ect ).
Substituting (2) and (3) in (1) we get
"
#
Z
−|x − y|2
1
exp
dy
|u(t, x) − u(t − ǫy) · √
2ǫ
( (2πǫ)d )
√
≤ || f ||∞ ((ecǫ − 1)ect )
Note that the right hand side is independent of x and as ǫ → 0 the
right hand side converges to 0. Thus to show that u(t, x) is a continuous
function of x (t fixed), it is enough to show that
"
#
Z
−|x − y|2
1
exp
dy
u(t − ǫ, y) √
2ǫ
( (2πǫ)d
254
is a continuous function of x; but this is clear since u is bounded. Thus
for any fixed tu(t, x) is continuous.
Lemma 2. For any compact set K ⊂ Rd , for r large enough so that
K ⊂ {x : |x| < r},
x → Q x (τr ≤ t)
is continuous on K for each t ≥ 0, where
τr (w) = inf{s : |w(s)| ≥ r}.
265
Proof. Q x (τr ≤ t) depends only on the coefficient b(x) on |x| ≤ r. So
modifying, if necessary, outside |x| ≤ r, we can very well assume that
|b(x)| ≤ M for all x. Let
τǫr = inf{s : s ≥ ǫ, |w(s)| ≥ r}.
Q x (τǫr ≤ t) = E Qx (u(X(ǫ))),
where
u(x) = Q x (τr ≤ t − ǫ)
As b and u are bounded, for every fixed ǫ > 0, by Lemma 1, Q x (τr ≤
t) is a continuous function of x. As
|Q x (τǫr ≤ t) − Q x (τr ≤ t)| ≤ Q x (τr ≤ ǫ),
to prove the lemma we have only to show that
limit sup Q x (τr ≤ ǫ) = 0
ǫ→0 x∈K
Now
Q x (τr ≤ ǫ) =
Z
Z()dP x
{τ ≤ǫ}
r
Z
√
≤ ( (Z(ǫ))2 dP x )1/2 · P x (τr ≤ ǫ),
by Cauchy-Schwarz inequality. The first factor is bounded because b is 255
bounded. The second factor tends to zero uniformly on K because
sup P x (τr ≤ ǫ) ≤ P( sup |w(s)| > δ)
x∈K
0≤s≤ǫ
where
δ = inf |(x − y)|.
y∈K
|x|=r.
Lemma 3. Let K be compact in Rd . Then for fixed t, Q x (τr ≤ t) monotically decreses to zero as r → ∞ and the convergence is uniform on
K.
31. Invariant Probability Distributions
266
Proof. Let fr (x) = Q x (τr ≤ t). As {τr ≤ t} decreases to the null set, fr (x)
decreases to zero. As K is compact, there exists an r0 such that for r ≥
r0 , fr (x) is continuous on K, by Lemma 2. Lemma 3 is a consequence
of Dini’s theorem.
Lemma 4. Let b : Rr → Rd be smooth (not necessarily bounded). Then
E Qx ( f (X(t))) is continuous in x for every fixed t, f being any bounded
measurable function.
Proof. Let br be any bounded smooth function on Rd such that br ≡ b
on |x| ≤ r and Qrx the measure corresponding to br . Then by Lemma 1,
E Qx ( f (X(t))) is continuous in x for all r. Further,
r
|E Qx ( f (X(t))) − E Qx ( f (X(t)))| ≤ 2|| f ||∞ · Q x (τr ≤ t).
256
The result follows by Lemma 3.
Lemma 5. With the hypothesis as the same as in Lemma 1, (S 1 ) is an
equicontinuous family, where
S 1 = { f : Rd → R, f bounded measurable, || f ||∞ ≤ 1}
Proof. For any f in S 1 , let U(x) = U(t, x) ≡ E Qx ( f (X(t))) and
#
"
Z
−|x − y|2
1
dy.
exp
Uǫ (x) = Uǫ (t, x) =
U(t − ǫ, y) √
2ǫ
( (2πǫ)d )
By Lemma 1,
|U(x) − Uǫ (x)| ≤ (((ecǫ − 1)ǫ ct ))1/2
|U(x) − U(y)| ≤ |U(x) − Uǫ (x)| + |Uǫ (y) − U(y)| + |Uǫ (x) − Uǫ (y)|
√
≤ 2 ((ecǫ − 1)ect ) + |Uǫ (x) − Uǫ (y)|.
The family {Uǫ : f ∈ S 1 } is equicontinuous because every U occuring in the expression for Uǫ is bounded by 1, and the exponential factor
is uniformly continuous. Thus the right hand side is very small if ǫ is
small and |x − y| is small. This proves the lemma.
267
Lemma 6. Let b be smooth and assume that there is no explosion (b
is not necessarily bounded). Then (S 1 ) is an equi-continuous family
∀t > 0.
Proof. Let r > 0 be given. Define br ∈ C ∞ such that br = 0 on |x| > r+1,
br = b on |x| ≤ r, br : Rd → R. By Lemma 2, we have that
r
{E Qx ( f (X(t))) : f ∈ S 1 }
is equicontinuous, where Qrx is the probability measure corresponding 257
to the function br .
r
E Qx ( f (X(t))χ{τr >t} )E Qx ( f (X(t))χ{τr >t} ).
(1)
Therefore
r
|E Qx ( f (X(t))) − E Qx ( f (X(t))|
= |E Qx ( f (X(t))χ{τr >t} ) + E Qx ( f (X(t))χ{τr ≤t} )
r
r
− E Qx ( f (X(t))χ{τr >t} ) − E Qx ( f (X(t)))χ{τr ≤t} )
r
= |E Qx ( f (X(t)χ{τr ≤t} ) − E Qx ( f (X(t))χ{τr ≤t} )|
r
≤ || f ||∞ (E Qx (χ{τr ≤t} ) + E Qx (χ{τr ≤t} )
≤ l[E Qx (χ(τr ≤t) ) + E Qx (χ(τr ≤t) )](use (1) with f = 1)
= 2E Qx (χ(τr ≤t) ).
Thus
r
sup sup |E Qx ( f (X(t)) − E Qx ( f (X(t))| ≤ 2 sup(χ{τr ≤t} ).
x∈K
x∈K || f ||∞ ≤1
By Lemma 3,
sup E 0x (τr ≤ t) → 0
x∈K
for every compact set K as n → ∞, for every fixed t.
The equicontinuity of the family (S 1 ) now follows easily. For fixed 258
r
x0 , put ur (x) = E Qx ( f (X(t))) and u(x) = E Qx ( f (X(t))) and let K =
s[x0 , 1] = {x : |x − x0 | ≤ 1}. Then
|u(x) − u(x0 )| ≤ |u(x) − ur (x)| + |u(x0 ) − ur (x0 )| + |ur (x) − ur (x0 )|
31. Invariant Probability Distributions
268
≤ 2 sup E Qy (χ(τr ≤|t) ) + |ur (x) − ur (x0 )|
y∈K
By the previous lemma {ur } is an equicontinuous family and since
sup E Qy (χ(τr ≤1) ) → 0, {u : || f ||∞ ≤ 1} is equicontinuous at x0 . This
y∈K
proves the Lemma.
Lemma 7. T r ◦ T s = T t+s , ∀s, t ≥ 0.
Remark. This property is called the semigroup property.
Proof.
T r (T s f )(x)
"
=
f (z)q(s, y, z)q(t, x, y)dy dz.
Thus we have only to show that
Z
q(t, x, y)q(s, y, A)dy = q(t + s, x, A).
q(t + s, x, A) = E Qx (X(t + s) ∈ A)
= E Qx (X(t + s) ∈ A|t ))
= E Qx (E Qx X(t)(X(s) ∈ A))),
by Markov property
Qx
= E (q(s, X(t), A))
Z
=
q(t, x, y)q(s, y, A)dy,
259
which proves the result.
As a trivial consequence we have the following.
Lemma 8. Let ǫ > 0 and let S 1 be the unit ball in B(Rd ). Then
is equicontinuous.
S
t≥ǫ
T t (S 1 )
269
Proof.
S
t≥ǫ>0
T t (S 1 ) = T (
S
T t (S 1 )) (by Lemma 7) T ǫ (S 1 ).
t≥0
The result follows by Lemma 6.
Lemma 9. Let u(ttx) = E Qx ( f (X(t))) with || f ||∞ ≤ 1. Let ǫ > 0 be given
and K any compact set. Then there exists a T 0 = T 0 (ǫ, K) such that
∀T ≥ T 0 and ∀x1 , x2 ∈ K,
|u(T, x1 ) − u(T, x2 )| ≤ ǫ.
Proof. Define q∗ (t, x1 , x2 , y1 , y2 ) = q(t, x1 , y1 )q(t, x2 , y2 ) and let Q(x1 ,x2 )
be the measure corresponding to the operator
1
L = (∆x1 + ∆x2 ) + b(x1 ) · ∇x1 + b(x2 ) · ∇x2
2
i.e., for any u : Rd × Rd → R,
1 X ∂2 u X
∂u
+
bi (x1 , . . . , xd ) 2 +
2
2 i=1 ∂xi
∂xi
i=1
2d
Lu =
+
t
d
X
bi (xd+1,...,x2d )
i=1
∂u
.
∂xi+d
Then Q(x1 ,x2 ) will be a measure on C([0, ∞); Rd × Rd ). We claim that
Q(x1 ,x2 ) = Q x1 × Q x2 . Note that
C([0, ∞); Rd × Rd ) = C([0, ∞); Rd ) × C[(0, ∞); Rd )
and since C([0, ∞); Rd ) is a second countable metric space, the Borel 260
field of C([0, ∞)Rd × Rd ) is the σ-algebra generated by
B = (C([0, ∞); Rd )) × B(C[0, ∞); Rd ).
By going to the finite-dimensional distributions one can check that
P(x1 ,x2 ) = P x1 × P x2 .

 t
Zt

Z
dQ(x1 ,x2 ) 1

(1) 2 
(1)
|b
|
ds
hb
,
dX
i
−
=
exp
 ×

1
dP(x1 ,x2 ) Ft
2
0
0
31. Invariant Probability Distributions
270
 t

Zt
Z

1

(2) 2 
(2)
× exp  hb , dX2 i −
|b | ds ,


2
0
where
0
b(1) (x1 . . . xd ) = b(x1 . . . , xd ) · b(2) (xd+1 . . . x2d ) = b(xd+1 , . . . , x2d ),
so that Q(x1 ,x2 ) = Q x1 × Q x2 .
It is clear that if φ defined an invariant measure for the process Q x ,
i.e.
Z
Z
φ(x)dx =
φ(y)Qy(Xt ∈ A)dy,
A
then φ(y1 )φ(y2 ) defines an invariant measure for the process Q(x1 ,x2 ) .
Thus the process Q(x1 ,x2 ) is recurrent.
Next we show that u(T − t, X1 (t)) is a martingale (0 ≤ t ≤ T ) for any
fixed T on C([0, T ]; Rd ).
E Qx (u(T − t, X(t)|Fs ))
Z
= [ u(T − t, y)q(t − s, x, dy)]x=X(s)
"
=[
f (z)q(T − t, y, dz)q(t − s, x, dy)]x=X(s)
Z
= [ f (z)q(T − s, x, dz)]x=X(s)
= u(T − s, X(s)),
s < t.
261
It now follows that u(T − t, X1 (t)) is a martingale on C([0, ∞); Rd ) ×
C([0, ∞); Rd ). Hence u(T − t, X1 (t)) − u(T − t, X2 (t)) is a martingale
relative to Q(X1 ,x2 ) .
Let V = S (0, δ/2) ⊂ Rd × Rd with δ < 1/4. If (x1 , x2 ) ∈ V, then
|x1 − x2 | ≤ |(x1 , 0) − (0, 0)| + |(0, 0) − (0, x2 )| < δ.
271
Claim 1. Q(x1 ,x2 ) (τV ≤ T ) → 1 as T → ∞, where τV is the exit time
from Rd − V.
Proof. If w is any trajectory starting at some point in V, then τV =
0 ≤ T , ∀T . If w starts at some point outside V then, by the recurrence
property, w has to visit a ball with centre 0 and radius δ/2; hence it must
get into V at some finite time. Thus {τV ≤ T } ↑ to the whole space as
T ↑ ∞. Next we show that the convergence is uniform on compact sets.
If x1 , x2 ∈ K, (x1 , x2 ) ∈ K × K (a compact set). Put gT (x1 , x2 ) =
Q(x1 ,x2 ) (τV ≤ T ). Then gT (x1 , x2 ) ≥ 0 and gT (x1 , x2 ) increases to 1 as T
tends to ∞.
gT (x1 , x2 ) = Q(x1 ,x2 ) (τV ≤ T )
Q(x1 ,x2 ) (τ1V ≤ T ),
where
262
τ1V
= inf{t ≥ 1 : (x1 , x2 ) ∈ V}.
Therefore
gT (x1 , x2 ) ≥ E Q (x1 , x2 )(E Q (x1 , x2 )((τ1V ≤ T )|1 ))
= E Q (x1 , x2 )(Q(X1 (1),X2 (1)) {τ1V ≤ T )})
= E Q (x1 , x2 )(ψT (X1 (1), X2 (1))),
where ψT is a bounded non-negative function. Thus, if
hT (x1 , x2 ) = Q(x1 ,x2 ) (τ1V ≤ T ) =
= E Q (x1 , x2 )(ψT (X1 (1), X2 (1))),
then by Lemma 4, hT is continuous for each T , gT ≥ hT and hT increases
to 1 as T → ∞. Therefore, hT converges uniformly (and so does gT ) on
compact sets.
Thus given ǫ > 0 chose T 0 = T 0 (ǫ, K) such that if T ≥ T 0 ,
sup sup Q(x1 ,x2 ) (τV ≥ T − 1) ≤ ǫ.
x2 ∈K x1 ∈K
31. Invariant Probability Distributions
272
By Doob’s optional stopping theorem and the fact that
u(T − t, X1 (t)) − u(t − t, X2 (t))
is a martingale, we get, on equating expectations,
|u(T, x1) − u(T, x2 )|
= |E Q(x1 ,x2 ) [u(T − 0, X1 (0) − u(T − 0, X2 (0)]|
= |E Q(x1 ,x2 ) [u(T − (τv ∧ (T − 1)), X1(T − (τv ∧ (T − 1))−
− u(T − (τv ∧ T (−1)), X2(T − (τv ∧ (T − 1)]|
Z
[u(1, X1(1)) − u(1, X2(1))]dQ(x1,x2 ) +
|
{τv ≥T −1}
Z
+
[u(T − τv , X1 (T − τv )) − u(T − τv ), X2 (T − τv ))dQ(x1 ,x2 ) |.
{τv <(T −1)}
263
Therefore
|u(T, x1) − u(T, x2)|
Z
|[u(1, X1(1)) − u(1, X2 (1))]|dQ(x1,x2 ) +
≤
{τv ≥(T −1)}
+|
Z
[u(T − τv , X1 (T − τv )) − u(T − τv , X2 (T − τv ))dQ(x1 ,x2 ) |
{τv <(T −1)}
≤ 2ǫ + |
Z
[u(T − τv , X1 (T − τv )) − u(T − τv , X2 (T − τv ))]dQ(x1 ,x2 ) |,
{τv <(T −1)}
since u is bounded by 1.
The second integration is to be carried out on the set {T − v ≥ 1}.
S
Since
T t (S 1 ) is equicontinuous we can choose a δ > 0 such that
t≥1
whenever x1 , x2 ∈ K such that |x1 − x2 | < δ
|u(t, x1 ) − u(t, x2 )| ≤ ǫ,
∀t ≥ 1.
264
Thus |u(T, x1 ) − u(T, x2 )| ≤ 3ǫ whenever x1 , x2 ∈ K and T ≥ T 0 .
This proves the Lemma.
273
Corollary to Lemma 9. sup
x1 ,x2 ∈K
R
|q(t, x1 , y)|dy converges to 0 as t → ∞.
Proof. Since the dual of L1 is L∞ , we have
Z
|q(t, x1 , y) − q(t, x2 , y)|dy
Z
= sup | [q(t, x1 , y) − q(t, x2 , y)] f (y)dy|
|| f ||∞ ≤1
and the right side converges to 0 as t → ∞, by Lemma 9.
We now come to the proof of the main theorem stated before Lemma
1. Now
Z
|q(t, x, y) − φ(y)|dy
Z
Z
=
|q(t, x, y) − φ(x1 )q(t, x1 , y)dx1 |dy
(by invariance property)
Z Z
Z
1
1
=
| q(t, x, y)φ(x )dx − φ(x1 )q(t, x1 , y)dx1 |dy
Z
(since
φ(x1 )dx1 = 1)
"
≤
|q(t, x, y) − q(t, x1 , y)|φ(x1 )dx1 dy (since φ ≥ 0)
Z
Z
1
1
=
φ(x )dx
|q(t, x, y) − q(t, x1 , y)|dy
Since
Z
1
φ(x )dx = Lt
n→∞
|x1 |≤n
choose a compact set L K such that
Z
Z
1
φ(x1 )dx1
Z
R
φ(x1 )dx1 ,
φ(x1 )dx1 < ǫ. Then
Rd −L
|q(t, x, y) − q(t, x1 , y)|dy
265
31. Invariant Probability Distributions
274
=
Z
1
Z
1
φ(x )dx
L
+
Z
1
1
φ(x )dx
Z
|q(t, x, y) − q(t, x1 , y)|dy
"
1
1
Z
|q(t, x, y) − q(t, x1 , y)|dv + 2ǫ.
Rd −L
≤
|q(t, x, y) − q(t, x1 , y)|dy+
L
φ(x )dx
Chose t0 such that whenever t ≥ t0 ,
Z
|q(t, x, y) − q(t, x1 , y)|dy ≤
1+
R
ǫ
φ(x1 )dx1
L
∀x, x1 in L. (Corollary to Lemma 9). Then
Z
|q(t, x, y) − φ(y)|dy ≤ 3ǫ,
if t ≥ t0 ∀x ∈ K completing the proof of the theorem.
32. Ergodic Theorem
Theorem . Let f : Rd → R be bounded and measurable with || f ||∞ ≤ 1. 266
If φ is an invariant distribution for the family {Q x }, x ∈ Rd then
Z
Qx
lim E ( f (X(t1 )) f (X(t2 ))) = [ f (y)φ(y)dy]2
t1 →∞
0≤t2 −t1 →∞
Proof.
E Qx [( f (X(t1 ) f (X(t2 ))]
= E Qx (E Qx [ f (X(t1 )) f (X(t2 ))|Ft1 ])
= E Qx ( f (X(t1 ))(E Qx [ f (X(t2 ))|Ft1 ]))
Z
Qx
f (y)q(t2 − t1 , X(t1 ), y))dy), t2 > t1
= E ( f (X(t1 ))
(1)
=
Z
f (z)q(t1 , x, z)dz
Z
(by Markov property),
f (y)q(t2 − t1 , z, y)dy
does any bounded an measurable f . By theorem of § 31,
Z
sup |
f (y)[q(t, x, y) − φ(y)]dy| → 0
x∈K
as t → +∞. We can therefore write (1) in the form
E Qx [( f (X(t1 )) f (X(t2 ))] =
Z
Z
Z
= ( f (z)q(t1 , x, x)dz)
fφ +
f (z)q(t1 , x, z)A(t2 − t1 , z)dz,
275
32. Ergodic Theorem
276
267
where A(t2 −t1 , z) converges to 0 (uniformly on compact sets as) t2 −t1 →
+∞.
To prove the theorem we have therefore only to show that
Z
f (z)q(t1 , x, z)A(t2 − t1 , z)dz → 0
R
R
as t1 → +∞ and t2 − t1 → ∞ (because f (z)q(t1 , x, z)dz → f φ). Now
Z
|
f (z)q(t1 , x, z)A(t2 − t1 , z)dz|
Z
≤ || f ||∞ q(t1 , x, z)|A(t2 − t1 , z)|dz
Z
(2)
≤
q(t1 , x, z)|A(t2 − t1 , z)|dz
Let K be any compact set, then
Z
Z
Z
q(t1 , x, z)dz =
χK q(t1 , x, z)dz →
χK φ(z)dz
K
at t1 → ∞. Given ǫ > 0, let K be compact so that
Z
| χK c φ(z)dz| ≤ ǫ;
R
then | χK c q(t1 , x, z)dz| ≤ 2ǫ if t1 ≫ 0. Using (2) we therefore get
Z
|
f (z)q(t1 x, z)A(t2 − t1 , z)dz|
Z
Z
≤
q(t1 , x, z)|A(t2 − t1 , z)|dz + q(t1 , x, z)|A(t2 − t1 , z)|dz
≤
K
Z
Kc
q(t1 , x, z)|A(t2 − t1 , z)|dz + 2
K
≤
Z
K
Z
q(t1 , x, z)dz,
Kc
since |A(t2 − t1 , z)| ≤ 2,
q(t1 , x, z)|A(t2 − t1 , z)|dz + 2ǫ, if t1 ≫ 0.
277
268
The theorem now follows from the fact that
lim sup |A(t2 − t1 , z) = 0.
t2 −t1 →∞ z∈K
Weak Ergodic Theorem.


 Zt


Q x  1
lim E |
f (X(s))ds − f (x)φ(x)dx| > ǫ  = 0.
t→∞
 t

0
Proof.


Z
 Zt

1


E Qx |
f (X(s))ds −
f (x)φ(x)dx| > ǫ 
 t

0


Z
 Zt

1 Qx  1

≤ 2 E |
f (X(s))ds −
f (y)φ(y)dy|2  ,
 t

ǫ
0
by Tchebychev’s inequality. We show that the right side → 0 as t → ∞.
Now


Z

 Zt
1


f (X(s))ds −
f φ|2 
E Qx |

 t
0
1
= E [| 2
t
Qx
−2
1
t
Zt Zt
Z
f (X(σ1 )) f (X(σ2 ))dσ1 dσ2 + ( f φdy)
0
0
Zt
f (X(σ))dσ
Z
f φdy]
0
Also
Qx
sup |E [ f (X(t)) −
x∈K
Z
f (y)φ(y)dy]|
32. Ergodic Theorem
278
= sup |
Z
Z
q(t, x, y) f (y)dy −
f (y)φ(y)dy|
Z
≤ || f ||∞ sup |q(t, x, y) − φ(y)|dy;
x∈K
x∈K
269
the right hand side tends to 0 as t tends to +∞. Consider
Z
Zt
1
f (X(σ))dσ −
f (y)φ(y)dy)|
|E (
t
0
 T

Z
Zt
 Z

1
1

Qx 
= |E 
f (X(σ))dσ −
f (y)φ(y)dy +
f (X(σ))dσ , 0 ≤ T ≤ t,
t

t
Qx
0
≤
1
|
t
ZT
0
E Qx f (X(σ))dσ − T
0
Z
f (y)φ(y)dy|+


 Zt

t − T Z
1


+ |E Qx 
f (y)φ(y)dy |.
f (X(σ))dσ −
t

t
T
Given ǫ > 0 choose T large so that
Z
|E Qx ( f (X(σ)) −
f (y)φ(y)dy| ≤ ǫ,
(σ ≥ T ).
Then
|E
Qx
≤|
1 Z
t
1
t
0
T
Z
t
f (X(σ))dσ −
Z
E Qx [ f (X(σ))] −
0
T
t
f (y)φ(y)dy| ≤
Z
f (y)φ(y)dy]| +
≤ 2ǫ
provided t is large. Thus
1
lim E [
t→+∞
t
Qx
Zt
0
f (X(σ))dσ] =
Z
f φdy.
t−T
ǫ
t
279
To prove the result we have therefore only to show that


#2
 "Z
 Zt Zt
1

Qx 


f (y)φ(y)dy
f (X(σ1 )) f (X(σ2 ))dσ1 dσ2  =
lim E  2

t
t→+∞
0
0
270
POR is the region σ2 ≥ t0 , σ1 − σ2 ≥ t0 .
Let I


!
 Z
 Zt Zt

1
f φdy
f (X(σ1 )) f (X(σ2 ))dσ1 dσ2  −
= E Qx 

t
0 0
!2 
Z 
Z

2
 Qx
= 2
f (y)φ(y)dy  dσ1 dσ2
E ( f (X(σ1 )) f (X(σ2 ))) −
t
0 ≤ σ2 ≤ σ1 ≤ t.
Then
2
|I| ≤ 2
t
Z
∆PQR
Qx
|E ( f (X(σ1 )) f (X(σ2 ))) −
Z
!2
f (y)φ(y)dy |dσ1 dσ2
32. Ergodic Theorem
280
+
2
· 2|| f ||2∞
2
t
[area of OAB − area of PQR]
By the Ergodic theorem the integrand of the first term on the right
can be made less than ǫ/2 provided t0 is large (see diagram). Therefore

!2 
 2
(t − 2t0 ) 
ǫ 2
4
2  t
|I| ≤ · 2 area of PQR + 2 || f ||∞  −

2 t
2
2
t
ǫ 2|| f ||2∞
+
[4tt0 − 4t02 ].
2
t2
<ǫ
≤
271
if t is large. This completes the proof of the theorem.
33. Application of Stochastic
Integral
LET b BE A bounded function. For every Brownian measure P x on 272
Ω = C([0, ∞); Rd ) we have a probability measure Q x on (Ω, F ).
Problem. Let q(t, x, A) = Q x (Xt ∈ A) · q(t, x, ·) is a probability measure
on Rd . We would like to know if q(t, x, ·) is given by a density function
on Rd and study its properties.
Step (i). q(t, x, ·) is absolutely continuous with respect to the Lebesgue
measure.
For, p(t, x, A) = P x (X(t) ∈ A) is given by a density function. Therefore p(t, x, ·) ≫ md (Lebesgue measure). Since
Q x ≪ P x on Ft ,
q(t, z, ·) ≤ Md on Ft .
Step (ii). Let q(t, x, y) ≥ 0 be the density function of q(t, x, ·) and write
p(t, x, y) for the density of p(t, x, ·). Let 1 < α < ∞. Put
q(t, x, y)
.
r(t, x, y) =
p(t, x, y)
Z
Z
qα dy =
rα pα dy
Rd
=
Z
α 1/α
r p
P
α−1
α
dy ≤
281
Z
!1/α
×
r pdy
α2
33. Application of Stochastic Integral
282
×
273
Z
p
!α−1/α
dy
Z
q(t, x, y)dy
α+1
Step (iii).
Q x (X(t) ∈ A) =
=
=
Therefore
Z
Z
r(t, x, y)p(t, x, y)dy
r(t, x, y)P x (Xt ∈ dy).
dQ x t
= r(t, x, y)
dP x t
Therefore
!1/α Z
dQ x t α
α2
= r pdy
dP x t α2 ,Px
dQ x α
, since Ftt ⊂ Ft ,
≤ dP x t α2 ,Px
2
{E Px [Z(t)α ]}1/α
Zt
Zt
α2
Px
2
|b|2 ds)]}1/α
= {E [exp(α
hb, dXi −
2
Px
= {E [exp(α
2
0
0
Zt
Zt
0
α4
hb, dXi −
2
α4 − α2
|b| ds +
2
2
0
Zt
|b|2 ds)]}1/α ,
0
i.e.,
Z
α2
r pdy
!1/α
1/α



t
Zt



 4


4 Z
2

α
α
−
a
 P x 



2
2



hb,
dXi
−
|b|
ds
ct
+
α
≤
E
exp














 2
2


0
0
where c is such that |b|2 ≤ c. Using Schwarz inequality we then get
!#1/α
!1/α "
Z
α4 − α2
α2
ct
.
≤ exp
r pdy
2
283
274
Hence
Z
qα dy ≤ exp
"
α4 − α2
ct
2
#!1/α Z
!α−1/α
Pα+1 dy
Significance. Pure analytical objects like q(t, x, y) can be studied using
stochastic integrals.
33. Application of Stochastic Integral
284
Appendix
Language of Probability
275
Definition. A probability space is a measure space (Ω, B, P) with P(Ω)
= 1. P is called a probability measure or simply a probability. Elements
of B are called events. A measurable function X : (Ω, B) → Rd is
called d-dimensional random variable. Given the random variable X,
define F : Rd → R by
F((a1 , . . . an )) = P{w : Xi (w) < ai , for i = 1, 2, . . . , d}
where X = (X1 , X2 , . . . , Xd ). Then F is called the distribution
funcR
tion
variable X. For any random variable X, X dP =
R of the random
R
( X1 dP, . . . , Xd dP), if it exists, is called mean
of X or expectation of
R
X and is denoted by E(X). Thus E(X) = XdP = µ. E(X n ), where
X n = (X1n , X2n , . . . , Xdn ) is called the nth moment about zero. E((X − µ)n )
is called the nth central moment. The 2nd central moment is called variance and is denoted by σ2 we have the following.
Tchebyshev’s Inequality.
Let X be a one-dimensional random variable with mean and variance
µ. Then for every ǫ > 0, P{w : |X(w) − µ| ≥ ǫ} ≤ σ2 /ǫ 2 .
276
Generalised Tchebyshev’s Inequality. Let f : R → R be measurable
such that f (u) = f (−u), f is strictly positive and increasing on (0, ∞).
Then for any random variable X : Ω → R,
P(w : |X(w)| > ǫ) ≤
E( f (X))
f (ǫ)
for every ǫ > 0.
For any random variable X : Ω → Rd , φ(t) = E(eitX ) : Rd → C is
called the characteristic function of X. Here t = (t1 , . . . , td ) and tX =
t1 X1 + t2 X2 + · · · + td Xd .
285
Independence. Events E1 , . . . , En are called independent if for every
{i1 , . . . , ik } ⊂ {1, 2, . . . , n} we have
P(Ei1 ∩ . . . ∩ Eik ) = P(Ei1 )P(Ei2 ) . . . P(Eik ).
An arbitrary collection of events {Eα : α ∈ I} is called independent
if every finite sub-collection is independent. Let {Fα : α ∈ I} be a
collection of sub-σ-algebras of B. This collection is said to be independent if for every collection {Eα : α ∈ I}, where Eα ∈ Fα , of events
is independent. A collection of random variables {Xα : α ∈ I} is said
to be independent if {σ(Xα ) : α ∈ I} is independent where σ(Xα ) is the
σ-algebra generated by Xα .
Theorem . Let X1 , X2 , . . . , Xn be random variables with F X1 , . . . , F Xn
as their distribution functions and let F be distribution function of X =
(X1 , . . . , Xn ), φX1 , . . . , φXn the characteristic functions of X1 , . . . , Xn and
φ that of X = (X1 , . . . , Xn ). X1 , . . . , Xn are independent if and only if
F((a1 , . . . , an )) = F X1 (a1 ) . . . F Xn (an ) for all a1 , . . . , an , iff φ((t1 , . . . , tn ))
= φX1 (t1 ) . . . φXn (tn ) for all t1 , . . . , tn .
277
Conditioning.
Theorem . Let X : (Ω, B, P) → Rd be a random variable, with E(X)
finite, i.e. if X = (X1 , . . . , Xd ), E(Xi ) is finite for each i. Let C be a subd
σ-algebraR of B. Then
R there exists a random variable Y : (Ω, C ) → R
such that YdP = XdP for every C in C .
C
C
If Z is any random variable with the same properties then Y = Z
almost everywhere (P).
Definition . Any such Y is called the conditional expectation of X with
respect to C and is denoted by E(X|C ).
If X = χA , the characteristic function of A in B, then E(χA |C ) is
also denoted by P(A|C ).
Properties of conditional expectation.
1. E(1|C ) = 1.
33. Application of Stochastic Integral
286
2. E(aX + bY|C ) = aE(X|C ) + bE(Y|C ) for all real numbers a, b and
random variables X, Y.
3. If X is a one-dimensional random variable and X ≥ 0, then
E(X|C ) ≥ 0.
4. If Y is a bounded C -measurable real valued random variable and
X is a one-dimensional random variable, then
E(Y X|C ) = Y E(X|C ).
5. If D ⊂ C ⊂ B are σ-algebras, then
E(E(X|C )|D) = E(X|D).
278
6.
R
Ω
|E(X|D)|d(P|D) ≤
R
E(X|C )d(P|C ).
Ω
Exercise 1. Let (Ω, B, P) be a probability space, C a sub-σ-algebra of
B. Let X(t, ·)Y(t, ·) : Ω → R be measurable with respect to B and C
respectively where t ranges over the real line. Further let E(X(t, ·)|C ) =
Y(t, ·) for each t. If f is a simple C -measurable function then show that
Z
Z
X( f (w), w)d(P|C ) =
Y( f (w)w)dP
C
C
for every C in C .
[Hint. Let A1 , . . . , An be a C -measurable partition such that f is constant on each Ai . Verify the equality when C is replaced by C ∩ Ai .]
Exercise 2. Give conditions on X, Y such that exercise 1 is valid for all
bounded C -measurable functions and prove your claim.
The next lemma exhibits conditioning as a projection on a Hilbert
space.
Lemma . Let (Ω, B, P) be any probability space C a sub-σ-algebra of
B. Then
287
(a) L2 (Ω, C , P) is a closed subspace of L2 (Ω, B, P).
(b) If π : L2 (Ω, B, P) → L2 (Ω, C , P) is the projection,then π( f ) =
E( f |C ).
(a) is clear, because for any f ∈ L1 (Ω, C , P)
Z
Z
f d(P|C ) =
f dP
Proof.
Ω
Ω
(use simple function 0 ≤ s1 ≤ . . . ≤ f , if f ≥ 0) and L2 (Ω, C , P) 279
is complete.
(b) To prove this it is enough to verify it for characteristic functions
because both π and f → E( f |C ) are linear and continuous.
Let A ∈ B, C ∈ C then π(χC ) = χC . As π is a projection
Z
Z
χA π(χC )d(P|B),
π(χA )χC d(P|B) =
i.e.
Z
π(χA )d(P|B) =
C
Z
XA d(P|B).
C
Since π(χA ) is C -measurable,
Z
Z
π(χA )d(P|B) =
π(χA )d(P|C )
C
C
Therefore
Z
C
π(χA )d(P|C ) =
Z
χA d(P|B), ∀C in C .
C
Hence
π(χA ) = E(χA |C ).
33. Application of Stochastic Integral
288
Kolmogorov’s Theorem.
280
Statement. Let A be any nonempty set and for each finite ordered subset
(t1 , t2 , . . . , tn ) of A [i.e. (t1 , . . . , tn ) an ordered n-tuple with ti in A], let
P(t1 ,...,tn ) be a probability on the Borel sets in Rdn = Rd × Rd × · · · Rd .
Assume that the family P(t1 ,...,tn ) satisfies the following two conditions
(i) Let τ : {1, 2, . . . , n} → {1, 2, . . . , n} be any permutation and fτ :
Rdn → Rdn be given by
fτ ((x1 , . . . , xn )) = (xτ(1) , . . . , xτ(n) ).
We have
P
(E)
(tτ(1) ,...,tτ(n) )
= P(t1 ,...,tn ) ( fτ−1 (E))
for every Borel set E of Rdn . In short, we write this condition as
Pτt = Pt τ−1 .
(ii) P (E) = P(t1 ,t2 ,...,tn ,tn+1 ,...tn+m ) (E × Rdm ) for all Borel sets E of Rdn
(t1 ,...,tn )
and this is true for all t1 , . . . , tn , tn+1 , . . . , tn+m of A.
Then, there exists a probability space (Ω, B, P) and a collection of
random variable {Xt : t ∈ A} : (Ω, B) → Rd such that
P (E) = P{w : (Xt1 (w), . . . , Xtn (w)) ∈ E}
(t1 ,...,tn )
for all Borel sets E of Rdn .
281
Proof. Let Ω = π{Rdt : t ∈ A} where Rdt = Rd for each t. Define
Xt : Ω → Rd to be the projection given by Xt (w) = w(t). Let B0 be
the algebra generated by {Xt : t ∈ A} and B the σ-algebra generated by
{Xt : t ∈ A}. Having got Ω and B we have to construct a probability P
on (Ω, B) satisfying the conditions of the theorem.
Given t1 , . . . , tn define
π(t1 ,...,tn ) : Ω → Rd × Rd × · · · × Rd (n times)
289
by
π(t1 ,...,tn ) (w) = (w(t1 ), . . . , w(tn )).
It is easy to see that every element of B0 is π−1
(t1 ,...,tn ) (E) for suitable
dn
t1 , . . . , tn in A and a suitable Borel set E of R . Define P on B0 by
P(π−1
(t1 ,...,tn ) (E)) = P(t1 ,...,tn ) (E). Conditions (1) and (2) ensure that P is
a well-defined function on B0 and that, as P(t1 ,...,tn ) are measures, P is
finitely additive on B0 .
Claim . Let C1 ⊃ C2 ⊃ . . . ⊃ Cn ⊃ . . . be a decreasing sequence in B0
with limit P(Cn ) ≥ δ > 0. Then ∩Cn is non-empty. Once the claim is
n→∞
proved, by Kolmogorov’s theorem on extension of measures, the finitely
additive set function P can be extended to a measure P on B. One easily
sees that P is a required probability measure.
Proof of the Claim. As Cn ∈ B0 , we have
Cn = π−1(n)
(n)
(t1 ,...,tk(n)
)
(En ) for suitable ti(n) in A
and Borel set En in Rdk(n) . Let
(n)
)
T n = (t1(n) , . . . , tk(n)
and
(n)
}.
An = {t(n) , . . . , tk(n)
We can very well assume that An is increasing with n. Choose a compact 282
subset En′ of En such that
PT n (En − En′ ) ≤ δ/2n+1 .
′
′
n+1 . If C ′′ = C ′ ∩C ′ ∩. . .∩C ′
If Cn′ = π−1
n
n
T n (E n ), then P(C n −C n ) ≤ δ/2
1
2
′′
′
′′
then Cn ⊂ Cn ⊂ Cn , Cn is decreasing and
P(Cn′′ ) ≥ P(Cn ) −
n
X
i=1
P(Ci − Ci′ ) ≥ δ/2.
We prove ∩Cn′′ is not empty, which proves the claim.
Choose wn in Cn′′ . As πT 1 (wn ) is in the compact set E1 for all n,
choose a subsequence
(1)
n(1)
1 , n2 , . . . of 1, 2, . . . such that πT 1 (wnk (1))
33. Application of Stochastic Integral
290
converges as k → ∞. But for finitely many n(1)
k ’s, πT 2 (ωnm (1)) is in
(1)
′
the compact set E2 . As before choose a subsequence n(2)
k of nk such
that πT 1 (ωnk (2)) converges as k → ∞. By the diagonal process obtain a
subsequence, w∗n of wn such that πT m (w∗n ) converges as n → ∞ for all m.
Thus, if t is in
∞
[
Am , then limit w∗n (t) = xt
n→∞
m=1
exists. Define w by w(t) = 0 if t ∈ A
easily sees that w ∈
283
Martingales.
∞
T
n=1
∞
S
m=1
Am , w(t) = xt if t ∈
∞
S
Am . One
m=1
Cn′′ , completing the proof of the theorem.
Definition. Let (Ω, F , P) be a probability space, (T, ≤) a totally ordered
set. Let (Ft )t∈T be an increasing family of sub-σ-algebras of F . A
collection (Xt )t∈T of random variables on Ω is called a martingale with
respect to the family (Ft )t∈T if
(i) E(|Xt |) < ∞, ∀t ∈ T ;
(ii) Xt is Ft -measurable for each t ∈ T ;
(iii) E(Xt |Fs ) = X s a.s. for each s, t in T with t ≥ s. (Markov property).
If instead of (iii) one has
(iii)′ E(Xt |Fs ) ≥ (≤)X s a.s.,
then (Xt )t∈T is called a submartingale (respectively supermartingale).
From the definition it is clear that (Xt )t∈T is a submartingale if
and only if (−Xt )t∈T is a supermartingale, hence it is sufficient to
study the properties of only one of these. T is usually any one of
the following sets
[0, ∞), N, Z, {1, 2, . . . , n}, [0, ∞]
or
N ∪ {∞}.
291
Examples. (1) Let (Xn )n=1,2... be a sequence of independent random
variables with
E(Xn ) = 0.
Then Yn = X1 + · · · + Xn is a martingale with respect to (Fn )n=1,2,... 284
where
Fn = σ{Y1 , . . . , Yn } = σ{X1 , . . . , Xn }.
Proof. By definition, each Yn is Fn -measurable.
E(Yn ) = 0.
E((X1 + · · · + Xn + Xn+1 + · · · + Xn+m )|σ{X1 , . . . , Xn })
= X1 + · · · + Xn + E((Xn+1 + · · · + Xn+m )|σ{X1 , . . . , Xn })
= Yn + E(Xn+1 + · · · + Xn+m ) = Yn .
(2) Let (Ω, F , P) be a probability space, Y a random variable with
E(|Y|) < ∞. Let Ft ⊂ F be a σ-algebra such that ∀t ∈ [0, ∞)
Ft ⊂ Fs
if t ≤ s.
If Xt = E(Y|Ft ), Xt is a martingale with respect to (Ft ).
Proof.
(i) By definition, Xt is Ft -measurable.
(ii) E(Xt ) = E(Y) (by definition) < ∞.
(iii) if t ≥ s,
E(Xt |Fs ) = E(E(Y|Ft )|Fs ) = E(Y|Fs ) = X s
Exercise 1. Ω = [0, 1], F = σ-algebra of all Borel sub sets of Ω, P =
Lebesgue measure.
Let Fn =-algebra generated by the sets
h 2n − 1 i
h 1 h 1 2 ,1 .
0, n n , n ; . . . ,
2 2 2
2n
33. Application of Stochastic Integral
292
Let f ∈ L′ [0, 1] and define

n
n −1
Zj/2
Z1
2X

n
Xn (w) = 2 
f dy + χ[ 2n n−1 ,1]
χ j−1 j
2
 j=1 [ 2n , 2n )
j−1/2n
2n −1/2n
Show that (Xn ) is a martingale relative to (Fn ).
285



f dy

Exercise. Show that a submartingale or a supermartingale {X s } is a martingale iff E(X s ) = constant.
Theorem . If (Xt )t∈T , (Yt )t∈T are supermartingales then
(i) (aXt + bYt )t∈T is a supermartingale, ∀a, b ∈ R+ = [0, ∞).
(ii) (Xt ∧ Yt )t∈T is a supermartingale.
Proof.
(i) Clearly Zt = aXt + bYt is Ft -measurable and E(|Zt |) ≤
aE(|Xt |) + bE(|Yt |) < ∞.
E(aXt + bYt |Fs ) = aE(Xt |Fs ) + bE(Yt |Fs )
≤ aX s + bY s = Z s ,
if t ≥ s.
(ii) Again Xt ∧ Yt is Ft -measurable and E(|Xt ∧ Yt |) < ∞,
E(Xt ∧ Yt |Fs ) ≤ E(Xt |Fs ) ≤ X s .
286
Similarly
E(Xt ∧ Yt |Fs ) ≤ E(Yt |Fs ) ≤ Y s ,
if
t ≥ s.
Therefore
E(Xt ∧ Yt |Fs ) ≤ X s ∧ Y s .
Jensen’s Inequality. Let X be a random variable in (Ω, B, P) with
E(|X|) < ∞ and let φ(x) be a convex function defined on the real line
such that E(|φ0 X|) < ∞. Then
φ(E(X|C )) ≤ E(φ0 X|C ) a.e.
where C is any sub-σ-algebra of B.
293
Proof. The function φ being convex, there exist sequences a1 , a2 , . . . an ,
. . . , b1 , b2 , . . . of real numbers such that φ(x) = sup(an x + bn ) for each x.
n
Let Ln (x) = an x + bn . Then
Ln (E(X|C )) = E(Ln (X)|C ) ≤ E(φ(X)|C )
for all n so that
φ(E(X|C )) ≤ E(φ(X)|C ).
Exercise. (a) If {Xt : t ∈ T } is a martingale with respect to {Ft : t ∈
T } and φ is a convex function on the real line such that E(|φ(Xt )|) <
∞ for every t, then {φ(Xt )} is a sub martingale.
(b) If (Xt )t∈T is a submartingale and φ(x) is a convex function and
nondecreasing and if E(|φ0 Xt |) < ∞, ∀t then {φ(Xt )} is a submartingale. (Hint: Use Jensen’s inequality).
Definition . Let (Ω, B, P) be a probability space and (Ft )t∈[0,∞) an in- 287
creasing family of sub-σ-algebras of F . Let (Xt )t∈[0,∞) be a family of
random variables on Ω such that Xt is Ft -measurable for each t ≥ 0.
(Xt ) is said to be progressively measurable if
X : [0, t] × Ω → R
defined by
X(s, w) = X s (w)
is measurable with respect to the σ-algebra B[0, t] × Ft for every t.
Stopping times. Let us suppose we are playing a game of chance, say,
tossing a coin. The two possible outcomes of a toss are H (Heads) and
T (Tails). We assume that the coin is unbiased so that the probability
of getting a head is the same as the probability of getting a tail. Further
suppose that we gain +1 for every head and lose 1 for every tail. A game
of chance of this sort has the following features.
1. A person starts playing with an initial amount N and finishes with
a certain amount M.
33. Application of Stochastic Integral
294
2. Certain rules are specified which allow one to decide when to stop
playing the game. For example, a person may not have sufficient
money to play all the games, in which case he may decide to play
only a certain number of games.
288
It is obvious that such a game of chance is fair in that it is neither advantageous nor disadvantageous to play such a game and on the average
M will equal N, the initial amount. Furthermore, the stopping rules that
are permissible have to be reasonable. The following type of stopping
rule is obviously unreasonable.
Rule. If the first toss is a tail the person quits at time 0 and if the first
toss is a head the person quits at time t = 1.
This rule is unreasonable because the decision to quit is made on
the basis of a future event, whereas if the game is fair this decision
should depend only on the events that have already occured. Suppose,
for example, 10 games are played, then the quitting times can be 0,
1, 2, . . . , 10. If ξ1 , . . . , ξ10 are the outcomes (ξi = +1 for H, ξi = −1 for
T ) then the quitting time at the 5th stage (say) should depend only on
ξ1 , . . . , ξ4 and not any of ξ5 , . . . , ξ10 . If we denote ξ = (ξ1 , . . . , ξ10 ) and
the quitting time τ as a function of ξ then we can say that {ξ : τ = 5
depends only ξ1 , . . . , ξ4 }. This leads us to the notion of stopping times.
Definition. Let (Ω, F , P) be a probability space, (Ft )t∈[0,∞) an increasing family of sub-σ-algebras of F . τ : Ω → [0, ∞] is called a stopping
time or Markov time (or a random variable independent of the future) if
{w : τ(w) ≤ t} ∈ Ft
for each
t ≥ 0.
Observe that a stopping time is a measurable function with respect
to σ(∪Ft ) ⊂ F .
289
Examples.
1. τ = constant is a stopping time.
2. For a Brownian motion (Xt ), the hitting time of a closed set is
stopping time.
T
Exercise 2. Let Ft+ ≡
Fs ≡ Ft .
Def s>t
295
[If this is satisfied for every t ≥ 0, Ft is said to be right continuous]. If
{τ < t} ∈ Ft for each t ≥ 0, then τ is a stopping time. (Hint: {τ ≤ t =
∞
T
{τ < t + 1/n} for every k).
n=k
S
Ft . If τ is a
We shall denote by F∞ the σ-algebra generated by
t∈T
stopping time, we define
Fτ = {A ∈ F∞ : A ∩ {τ ≤ t} ∈ Ft , ∀t ≥ 0}
Exercise 3.
(a) Show that Fτ is a σ-algebra. (If A ∈ Fτ ,
Ac ∩ {τ ≤ t} = {t ≤ t} − A ∩ {τ ≤ t}).
(b) If τ = t (constant) show that Fτ = Ft .
Theorem . Let τ and σ be stopping times. Then
(i) τ + σ, τvσ, τ ∧ σ are all stopping times.
(ii) If σ ≤ τ, then Fσ ⊂ Fτ .
(iii) τ is Fτ -measurable.
(iv) If A ∈ Fσ , then A ∩ {σ = τ} and A ∩ {σ ≤ τ} are in Fσ∧τ ⊂
Fσ ∩ Fτ . In particular, {τ < σ}, {τ = σ}, {τ > σ} are all in 290
Fτ ∩ Fσ .
(v) If τ′ is Fτ -measurable and τ′ ≥ τ, then τ′ is a stopping time.
(vi) If {τn } is a sequence of stopping times, then lim τn . lim τn are also
stopping times provided that Ft+ = Ft , ∀t ≥ 0.
(vii) If τn ↓ τ, then Fτ =
Proof.
∞
T
n=1
Fτn provided that Ft+ = Ft , ∀t ≥ 0.
(i)
{σ + τ} > t} = {σ + τ > t, τ ≤ t, σ ≤ t} ∪ {τ > t} ∪ {σ > t};
{σ + τ > t, σ ≤ t} = τ ≤ A
33. Application of Stochastic Integral
296
[
=
r∈Q
0≤r≤t
{σ > r > t − τ, τ ≤ t, σ ≤ t}
(Q = set of rationals)
{σ > r > t − τ, τ ≤ t, σ ≤ t} = {t ≥ σ > r} ∩ {t ≥ τ > t − r}
= {σ ≤ t} ∩ {σ ≤ r}c ∩ {τ ≤ t} ∩ {τ ≤ t − r}c .
The right side is in Ft . Therefore σ + τ is a scopping time.
{τVσ ≤ t} = {τ ≤ t} ∩ {σ ≤ t}
{τ ∧ σ > t} = {τ > t} ∩ {σ > t}
(ii) Follows from (iv).
(iii) {τ ≤ t}{τ ≤ s} = {τ ≤ t ∧ s} ∈ Ft∧s ⊂ Fs , ∀s ≥ 0.
(iv) A ∩ {σ < τ} ∩ {σ ∧ τ ≤ t} = [A ∩ {σ ≤ t < τ}]
U[A ∩ U {σ ≤< τ} ∩ {τ ≤ t}] ∈ Ft .
r∈Q
0≤r≤t
291
A ∩ {σ ≤ τ} ∩ {σ ∧ τ ≤ t} = A ∩ {σ ≤ τ} ∩ {σ ≤ t}.
It is now enough to show that (σ ≤ τ) ∈ Fσ ; but this is obvious
because (τ < σ) = (σ ≤ τ)c is in Fσ∧τ ⊂ Fσ . Therefore A ∩{σ ≤
τ} ∈ Fσ∧τ and (iv) is proved.
(v) {τ′ ≤ t} = {τ′ ≤ t} ∩ {τ ≤ t} ∈ Ft as (τ′ ≤ t) ∈ Fτ . Therefore τ′ is
a stopping time.
(vi) lim τn ≡ sup inf τk
n k≥n
= sup inf inf{τn , τn+1 , . . . , τn+ℓ }.
n
ℓ
By (i), inf{τn , τn+1 , . . . , τn+ℓ } is a stopping time. Thus we have
only to prove that if τn ↑ τ or τn ↓ τ where τn are stopping times,
∞
T
then τ is a stopping time. Let τn ↑ τ. Then {τ ≤ t} =
{τn ≤ t}
n=1
so that τ is a stopping time. Let τn ↓ τ. Then
∞
\
{τ ≥ t} =
{τn ≥ t}.
n=1
297
By Exercise 3, τ is a stopping time. That lim τn is a stopping time
is proved similarly.
(vii) Since τ ≤ τn , ∀n, Fτ ⊂
∞
T
n=1
Fτn . Let A ∈
A ∩ (τn < t) ∈ Ft , ∀n. A ∩ (τ < t) =
∞
T
∞
T
n=1
Fτn . Therefore
(A ∩ (τm < t)) ∈ Ft .
m=1
Therefore A ∈ Fτ .
Optional Sampling Theorem. (Discrete case). Let {X1 , . . . , Xk } be a
martingale relative to {F1 , . . . , Fk }. Let {τ1 , . . . , τ p } be a collection of
stopping times relative to {F1 , . . . , Fk } such that τ1 ≤ τ2 ≤ . . . ≤ τ p 292
a.s. and each τi takes values in {1, 2, . . . , k}. Then {Xτ1 , . . . , Xτ p } is a
martingale relative to {Fτ1 , . . . , Fτ p } where for any stopping time τ,
Xτ (ω) = Xτ(w) (ω).
Proof. It is easy to see that each Xτi is a random variable. In fact Xτm =
k
P
Xi χ{τm =i} . Let τ ∈ {1, 2, . . . , k}. Then
i=1
E(|Xτ |) ≤
k Z
X
j=1
|X j |dP < ∞.
Consider
(Xτ j ≤ t) ∩ (τ j ≤ s) =
that
\
(Xℓ ≤ t) ∈ Fs .
ℓ≤s
Then (Xτ j ≤ t) is in Fτ j , i.e. Xτ j is Fτ j -measurable. Next we show
E(Xτ j |Fτk ) ≤ Xτk ,
(*)
(*) is true if and only if
Z
Z
Xτ j dP ≤
Xτk dP
A
if
j ≥ k.
for every
A ∈ Fτk .
A
The theorem is therefore a consequence of the following
33. Application of Stochastic Integral
298
Lemma . Let {X1 , . . . , Xk } be a supermartingale relative to
{F1 , . . . , Fk }.
If τ and σ are stopping times relative to {F1 , . . . , Fk } taking values in
{1, 2, . . . , k} such that τ ≤ σ then
Z
Z
Xτ dP ≥
Xσ dP for every A ∈ Fτ .
A
293
A
Proof. Assume first that σ − τ ≤ 1. Then
Z
(Xτ − Xσ )dP =
k
X
=
k
X
A
Z
j=1 [A∩(τ= j)∩(τ<σ)]
Z
j=1 [A∩(τ= j)]
(Xτ − Xσ )dP
(X j − X j+1 )dP
A ∈ Fi . Therefore A ∩ (τ = j) ∈ F j . By supermartingale property
Z
(X j − X j+1 )dP ≥ 0.
[A∩(τ= j)]
Therefore
Z
(Xτ − Xσ )dP ≥ 0.
A
Consider now the general case τ ≤ σ. Define τn = σ ∧ (τ + n).
Therefore τn ≥ τ. τn is a stopping time taking values in {1, 2, . . . , k},
τn+1 ≥ τn , τn+1 − τn ≤ 1, τk = σ.
R
R
Therefore Xτn dP ≥ Xτn+1 dP, ∀A ∈ Fτn . If A ∈ Fτ then A ∈
A
A
Fτn , ∀n. Therefore
Z
Z
Z
Xτ2 dP ≥ . . . ≥
Xτk dP,
Xτ1 dP ≥
A
A
A
∀A ∈ Fτ .
299
Now τ1 − τ ≤ 1. τ ≤ τ1 . Therefore
Z
Z
Z
Xτ dP ≥
Xτ1 dP ≥
Xσ dP.
A
A
A
This completes the proof.
N.B. The equality in (*) follows by applying the argument to
{−X1 , . . . , −Xk }.
Corollary 1. Let {X1 , X2 , . . . , Xk } be a super-martingale relative to
{F1 , . . . , Fk }.
If τ is any stopping time, then
E(Xk ) ≤ E(Xτ ) ≤ E(X1 ).
Proof. Follows from the fact that {X1 , Xτ , Xk } is a supermartingale relative to {F1 , Fτ , Fk }.
Corollary 2. If {X1 , X2 , . . . , Xk } is a super-martingale relative to
{F1 , . . . , Fk }
and τ is any stopping time, then
E(Xτ ) ≤ E(|X1 |) + 2E(Xk− ) ≤ 3 sup E(|Xn |)
1≤n≤k
where for any real x, x− =
|x| − x
.
2
|Xk | − Xk
, so 2E(Xk− ) = E(|Xk |) − E(Xk ).
Proof. Xk− =
2
By theorem {Xτ ∧0, Xk ∧0} is a super-martingale relative to {Fτ , Fk }.
Therefore E(Xk ∧ 0|Fτ ) ≤ E(Xτ ∧ 0). Hence
E(Xk− ) ≥ E(Xτ− ) =
E(|Xτ |) − E(Xτ )
.
2
294
33. Application of Stochastic Integral
300
Therefore
E(|Xτ |) ≤ 2E(Xk− ) + E(Xτ )
≤ 2E(Xk− ) + E(X1 ) ≤ 3 sup E(|Xn |).
1≤n≤k
295
Theorem . Let (Ω, F , P) be a probability space and (Ft )t≥0 on increasing family of sub-σ-algebras of F . Let τ be a finite stopping time, and
(Xt )t≥0 a progressively measurable family (i.e. X : [0, ∞) × Ω → R
defined by X(t, w) = Xt (w) is progressively measurable). If Xτ (w) =
Xτ(w) (w), then Xτ is Fτ -measurable.
Proof. We show that {w : X(τ(w), w) ≤ t, τ(w) ≤ s} ∈ Ft for every t.
Let Ω s = {w : τ(w) ≤ s}; Ω s ∈ Fs and hence the σ-algebra induced by
Fs on Ω s is precisely
{A ∩ Ω s : A ∈ Fs } = {A ∈ Fs : A ⊂ Ω s }.
Since τ(w) is measurable,
w → (τ(w), w)
of Ω s → [0, s] × Ω s
is (Fs , B[0, s] × Fs )-measurable. Since X is progressively measurable,
X
[0, s] × Ω s −
→ R is measurable.
Therefore {w : X(τ(w), w) ≤ t, τ(w) ≤ s} ∈ σ-algebra on Ω s . Therefore Xτ is Fτ measurable.
The next theorem gives a condition under which (Xt )t≥0 is progressively measurable.
Theorem . If Xt is right continuous in t, ∀w and Xt is Ft -measurable,
∀t ≥ 0 then (Xt )t≥0 is progressively measurable.
Proof. Define
!
[nt] + 1
[nt] + 1
Xn (t, w) = X
,w ·
↓ t.
n
n
301
Then
Lt Xn (t, w) = X(t, w) (by right continuity)
n→∞
296
Step 1. Suppose T is rational, T = m/n where m ≥ 0 is an integer. Then
{(t, w) : 0 ≤ t < T, Xn (t, w) ≤ α}


−1

[ 


h i i + 1 Xi+1

=
,
(−∞,
α]
X




 n n

n
0≤i≤m−1
Thus if T = m/n, Xn |[0,T )×Ω is B[0, T ] × FT -measurable. Now
km
T =
. Letting k → ∞, by right continuity of X(t) one gets X|[0,)×Ω is
kn
[0, T ] × FT -measurable. As X(T ) is FT -measurable, one gets X|[0,T ]×Ω
is [0, T ] × FT -measurable.
Step 2. Let T be irrational. Choose a sequence of rationals S n increasing
to T .
{(t, w) : 0 ≤ t ≤ T, X(t, w) ≤ α}
∞
[
=
{(t, w) : 0 ≤ t ≤ S n , X(t, w) ≤ α} ∪ {T } × XT−1 (−∞, α]
n=1
The countable union is in B[0, T ]×FT by Step 1. The second member is also in B[0, T ] × FT as X(T ) is FT -measurable. Thus X|[0,T ]×Ω
is B[0,T ] × FT -measurable when T is irrational also.
Remark. The technique used above is similar to the one used for proving that a right continuous function f : R → R is Borel measurable.
Theorem . Let {X1 , . . . , Xk } be a supermartingale and λ ≥ 0. Then
R
(1) λP( sup Xn ≥ λ) ≤ E(X1 ) −
Xk dP
1≤n≤k
sup Xn <λ
1≤n≤k
≤ E(X1 ) + E(Xk− ).
297
33. Application of Stochastic Integral
302
(2) λP( inf Xn ≤ −λ) ≤ −
1≤n≤k
R
Xk dP
{inf Xn ≤−λ}
≤ E(Xk− ).
Proof. Define
τ(w) = inf{n : Xn ≥ λ} if
= k,
if
sup Xn < λ.
sup Xn ≥ λ,
n
Clearly τ ≥ 0 and τ is a stopping time. If τ < k, then Xτ (w) ≥ λ for
each w.
Z
Z
E(Xτ ) =
Xτ dP +
Xτ dP
(sup Xn <λ)
(sup Xn ≥λ)
≥ λP(sup Xn ≥ λ) +
Z
Xk dP.
(sup Xn <λ)
Therefore
E(X1 ) ≥ λP(sup Xn ≥ λ) +
Z
Xk dP,
(sup Xn <λ)
λP(sup Xn ≥ λ) ≤ E(X1 ) −
Z
Xk dP ≤ E(X1 ) + E(Xk− )
(sup Xn <λ)
The proof of (2) is similar if we define



inf{n : Xn ≤ −λ}, if inf Xn ≤ −λ,
τ(w) = 

k,
if inf Xn > −λ.
298
Kolmogorov’s Inequality (Discrete Case). Let {X1 , . . . , Xk } be a finite
sequence of independent random variables with mean 0. Then
!
1
2
P sup (|X1 + · · · + Xn | ≥ λ) ≤ 2 E((X1 + X2 + · · · + Xk ) )
λ
1≤n≤k
303
Proof. If S n = X1 + · · · + Xn , n = 1, 2, . . . , k, then {S 1 , . . . , S k } is a
martingale with respect to {F1 , . . . , Fk } where Fn = σ{X1 , . . . , Xn }.
Therefore S 12 , . . . , S k2 is a submartingale (since x → x2 is convex). By
the previous theorem,
λ2 P{inf −S n2 ≤ −λ2 } ≤ E((−S 2K )− )
Therefore
P{sup |S n | ≥ λ} ≤
=
E((−S k2 )− )
λ2
=
E(S k2 )
λ2
1
E((X1 + X2 + · · · + Xk )2 ).
λ2
Kolmogorov’s Inequality (Continuous case). Let {X(t) : t ≥ 0} be a
continuous martingale with E(X(0)) = 0. If 0 < T < ∞, then for any
ǫ>0
n
1
P w : sup |X(s, w)| ≥ ǫ ≤ 1 E((X(T ))2 ).
ǫ
0≤s≤T
Proof. For any positive integer k define Y0 = X(0),
!
T T 2T
Y1 = X k − X(0), Y2 = X k − X k , . . . , Y2 k
2
2
2
!
!
k
k
2T
(2 − 1)
T .
=X
−X
2k
2k
By Kolmogorov inequality for the discrete case, for any δ > 0.


nT 
 1

P  sup |X k | > δ ≤ 2 E((X(T ))2 ).
δ
2
0≤n≤2k
nT By continuity of X(t), Ak = {w : sup |X k | > δ} increases to 299
2
0≤n≤2k
{ sup |X(s)| > δ} so that one gets
0≤s≤T
(1)
!
1
P sup |X(s)| > δ ≤ 2 E((X(T ))2 ).
δ
0≤s≤T
33. Application of Stochastic Integral
304
Now
!
1
P sup |X(s)| ≥ ǫ ≤ limit P sup |X(s)| > ǫ −
m→∞
m
0≤s≤T
0≤s≤T
≤ limit
m→∞
!
1
E((X(T ))2 ),
(ǫ − 1/m)2
by (1).
= 1/ǫ 2 E((X(T ))2 ).
This completes the proof.
Optional Sampling Theorem (Countable case). Let {Xn : n ≥ 1} be
a supermartingale relative to {Fn : n ≥ 1}. Assume that for some
X∞ ∈ L1 , Xn ≥ E(X∞ |Fn ). Let σ, τ be stopping times taking values in
N ∪ {∞}, with σ ≤ τ. Define Xτ = X∞ on {σ = ∞} and Xτ = X∞ on
{σ = ∞}. Then E(Xτ |Fσ ) ≤ Xσ .
Proof. We prove the theorem in three steps.
Step 1. Let X∞ = 0 so that Xn ≥ 0. Let τk = τ ∧ k, σk = τ ∧ k. By
optional sampling theorem for discrete case E(Xτk ) ≤ E(Xk ) ≤ E(X1 ).
By Fatou’s lemma, E(Xτ ) < ∞. Again by optional sampling theorem
for the discrete case,
E(Xτk |Fσk ) ≤ Xσk . . . , (0).
Let A ∈ Fσ . Then A ∩ {σ ≤ k} ∈ Fσk , and by (0)
Z
Z
Z
Z
Xτ dP ≤
Xτk dP ≤
Xσk dP ≤
300
A∩(τ≤k)
A∩{τ≤k}
Letting k → ∞,
(1)
Z
A∩(σ≤k)
Xτ dP ≤
A∩(τ,∞)
Z
A∩(σ≤k)
Xσ dP.
A∩(σ,∞)
Clearly
(2)
Z
A∩(τ=∞)
Xτ dP =
Z
A
X∞ dP =
Z
A∩(σ=∞)
Xσ dP
Xσ dP.
305
By (1) and (2), inf X dP ≤
A
R
Xσ dP, proving that
A
E(Xτ |Fσ ) ≤ Xσ .
Step 2. Suppose Xn = E(X∞ |Fn ). In this case we show that Xτ =
E(X∞ |Fτ ) for every stopping time so that E(Xτ |Fσ ) = Xσ . If A ∈ Fτ ,
then
Z
Z
Xτ dP =
X∞ dP for every k.
(τ≤k)
(1)
A∩(τ≤K)
Letting k → ∞,
Z
A∩(τ,∞)
(2)
Z
Z
Xτ dP =
X∞ dP,
A∩(τ,∞)
Xτ dP =
Z
A
A∩(τ=∞)
X∞ dP =
Z
X∞ dP
A∩(τ=∞)
The assertion follows from (1) and (2).
Step 3. Let Xn be general. Then
301
Xn = Xn − E(X∞ |Fn ) + E(X∞ |Fn ).
Apply Step (1) to Yn = Xn − E(X∞ |Fn ) and Step (2) to
Zn = E(X∞ |Fn )
to complete the proof.
Uniform Integrability.
Definition . Let (Ω, B, P) be any probability space, L1 = L1 (Ω, B, P).
A family H ⊂ L1 is calledR uniformly integrable if for every ǫ > 0 there
|X|dP < ǫ for all X in H.
exists a δ > 0 such that
(|X|≥δ)
33. Application of Stochastic Integral
306
Note. Every uniformly integrable family is a bounded family.
Proposition . Let Xn be a sequence in L1 and let Xn → X a.e. Then
Xn → X in L1 iff {Xn : n ≥ 1} is uniformly integrable.
Proof. is left as an exercise.
As {Xn : n ≥ 1} is a bounded family, by Fatou’s lemma X ∈ L1 .
Let ǫ > 0 be given. By Egoroff’s theorem there exists a set F such that
P(F) < ǫ and Xn → X uniformly on F.
Z
Z
|Xn − X|dP
|Xn − X|dP ≤ ||Xn − X||∞,Ω−F+
≤ ||Xn − X||∞,Ω−F +
Z
≤ ||Xn − X||∞,Ω−F +
+
Z
F
|Xn |dP +
|X|dP
Z
|Xn |dP +
|X|dP+
F∩(|X|≥δ)
Z
|Xn |dP +
Z
Z
F
F
F∩(|Xn |≥δ)
F∩{|Xn |≤δ}
≤ ||Xn − X||∞,Ω−F +
Z
XdP
F∩(|X|≤δ)
|Xn |dP +
(|Xn |≥δ)
Z
|X|dP + 2δǫ
(|X|≥δ)
302
The result follows by uniform integrability of {X, Xn : n ≥ 1}.
Corollary . Let C be any sub-σ-algebra of B. If Xn → X a.e. and Xn is
uniformly integrable, then E(Xn |C ) → E(X|C ) in L1 (Ω, C , P).
Proposition . Let H ⊂ L1 . Suppose there exists an increasing convex
function G : [0, ∞) → [0, ∞) such that
limit
t→∞
G(t)
= ∞ and
t
sup E(G(|X|)) < ∞.
X∈H
Then the family H is uniformly integrable.
307
Example . G(t) = t2 is a function satisfying the conditions of the theorem.
Proof. (of the proposition). Let
M = sup E(G(|X|)).
X∈H
Let ǫ > 0 be given. Choose δ > 0 such that
G(t) M
≥
t
ǫ
Then for X in H
Z
Z
ǫ
|X|dP ≤
M
(|X|≥δ)
for
t ≥ δ.
G(|X|)dP ≤
(|X|≥δ)
ǫ
M
Z
G(|X|)dP ≤ ǫ
G
Remark. The converse of the theorem is also true.
Exercise . Let H be a bounded set in L∞ , i.e. there exists a constant M
such that ||X||∞ ≤ M for all X in H. Then H is uniformly integrable.
Up Crossings and Down Crossings.
Definition . Let a < b be real numbers; let s1 , s2 , . . . , sk be also given
reals. Define i1 , i2 , . . . , ik as follows.



inf{n : sn < a},
i1 = 

k, if no si < a;



inf{n > i1 : sn > b},
i2 = 

k, if sn ≤ b for each n > i1 ;



inf{n > i2 : sn < a},
i3 = 

k, if sn ≥ a for each n > i2 ;
and so on
303
308
33. Application of Stochastic Integral
Let t1 = si1 , t2 = si2 , . . .. If (t1 , t2 ), (t3 , t4 ), . . ., (t2p−1 , t2p ) are the
only non-empty intervals and (t2p+1 , t2p+2 ), . . . are all empty, then p is
called the ******** of the sequence s1 , . . . , sk for the interval [a, b] and
is denoted by U(s1 , . . . , sk ; [a, b]).
Note. U (the up crossing) always takes values in {0, 1, 2, 3, . . .}.
Definition. For any subset S of reals define
U(S ; [a, b]) = sup{U(F; [a, b]) : F is a finite subset of S }
The number of down crossings is defined by
D(S ; [a, b]) = U(−S ; [−b, −a]).
For any real valued function f on any set S we define
U( f, S , [a, b]) = U( f (S ), [a, b]).
If the domain of S is known, we usually suppress it.
304
Proposition . Let a1 , a2 , . . . be any sequence of real numbers and S =
{a1 , a2 , . . .}. If U(S , [a, b]) < ∞ for all a < b, then these sequence {an }
is a convergent sequence.
Proof. It is clear that if T ⊂ S then U(T, [a, b]) ≤ U(S , [a, b]). If
the sequence were not convergent, then we can find a and b such that
lim inf an < a < b < lim sup an . Choose n1 < n2 < n3 . . .; m1 < m2 <
. . . such that ani < a and ami > b for all i. If T = {an1 , am1 , an2 , am2 , . . .},
then U(S ; [a, b]) ≥ U(T ; [a, b]) = ∞; a contradiction.
Remark. The converse of the proposition is also true.
Theorem . (Doob’s inequalities for up crossings and down crossings).
Let {X1 , . . . , Xk } be a submartingale relative to {F1 , . . . , Fk } a < b.
Define U(w, [a, b]) = U(X1 (w), . . . , Xk (w); [a, b]) and similarly define
D(w, [a, b]). Then
(i) U, D are measurable functions;
309
(ii) E(U(·, [a, b])) ≤
E((Xk − a) + 1) − E((X1 − a)+ )
;
b−a
(iii) E(D(·, [a · b])) ≤ E((Xk − b)+ )/(b − a).
Proof.
(i) is left as an exercise.
(ii) Define Yn = (Xn − a)+ ; there are submartingales. Then clearly
Yn ≤ 0 if and only if Xn ≤ a and Yn ≥ b − a iff Xn ≥ b, so that
UY1 (w), . . . , Yk (w); [0, b − a]) = U(X1 (w), . . . , Xk (w); [a, b])
305
Define
τ1 = 1



inf{n : Yn = 0}
τ2 = 

k, if each Yn = 0



inf{n > τ2 : Yn > b − a,
τ3 = 

k, if Yn < b − a for each n > τ2 ;
τk+1 = k.
As {Y1 , . . . , Yk } is a submartingale, by optional sampling theorem
Yτ1 , . . . , Yτk+1 is also a submartingale. Thus
E(Yτ2 − Yτ1 ) + E(Yτ4 − Yτ3 ) + · · · ≥ 0.
(1)
Clearly
[(Yτ3 − Yτ2 ) + (Yτ5 − Yτ4 ) + · · · ](w) ≥ (b − a) ∪ (Y1 (w), . . . Yk (w);
[0, b − a]) = (b − a) ∪ (w, [a, b]).
Therefore
(2)
E(Yτ3 − Yτ2 ) + E(Yτ5 − Yτ4 ) + · · · ≥ (b − a)E(U(·, [a, b])).
33. Application of Stochastic Integral
310
By (1) and (2),
E(Yk − Y1 ) ≥ (b − a)E(U(·, [a, b]))
giving the result.
(iii) Let Yn = (Xn − a)+ so that
D(Y1 (w), . . . Yk (w); [0, b − a]) = D(X1 (w), . . . , Xk (w); [a, b])
306
Define
τ1 = 1;



inf{n : Yn ≥ b − a},
τ2 = 

k, if each Yn < b − a;



inf{n > τ2 : Yn = 0},
τ3 = 

k, if each Yn > 0 for each n > τ2 ;
τk+1 = k.
By optional sampling theorem we get
0 ≥ E(Yτ2 − Yτ3 ) + E(Yτ4 − Yτ5 ) + · · · .
Therefore
0 ≥ (b − a)E(D(Y1 , . . . , Yk ; [0, b − a])) + E((b − a) − Yk ).
Hence
E(D(·, [a, b])) ≤ E((Xk − a)+ − (b − a))/(b − a)
≤
E((Xk − b)+ )
, for (c − a)+ − (b − a) ≤ (c − b)+
(b − a)
for all c.
311
Corollary . Let {X1 , . . . , Xk } be a supermartingale. U, D as in theorem.
Then
(i) E(D(·, [a, b])) ≤
(ii) E(U(·, [a, b])) ≤
E(X1 ∧ b) − E(Xk ∧ b)
.
b−a
E((Xk − b)− )
.
b−a
(i) E(D(·, [a, b])) = E(U(−X1 (w), . . . , −Xk (w), [−b, −a])
Proof.
≤
≤
E((−Xk + b)+ − (−X1 + b)+ )
,
b−a
E((b ∧ Xk ) − (b ∧ X1 ))
,
b−a
by above theorem,
since for
since for all a, b, c, (b − c)+ − (b − a)+ ≤ (b ∧ a) − (b ∧ c).
(ii) E(U(·, [a, b])) =
= E(D(−X1 (w), . . . , −Xk (w); [−b, −a]))
E((−Xk + a)+ )
, by theorem,
b−a
E((Xk − b)− )
≤
,
b−a
(since (−Xk + a)+ ≤ (Xk − b)− ,
≤
307
Theorem . Let {Xn : n = 1, 2, . . .} be a supermartingale relative to {Fn :
n = 1, 2, . . .}. Let (Ω, F , P) be complete.
(i) If sup E(Xn− ) < ∞, then Xn converges a.e. to a random variable
n
denoted by X∞ .
(ii) if {Xn : n ≥ 1} is uniformly integrable, then also X∞ exists. Further, {Xn : n = 1, 2, . . . , n = ∞} is a supermartingale with the
natural order.
(iii) if {Xn : n ≥ 1} is a martingale, then {Xn : n ≥ 1, n = ∞} is a
martingale.
33. Application of Stochastic Integral
312
Proof.
(i) Let U(w[a, b]) = U(X1 (w), X2 (w), . . . , [a, b]). By the corollary to Doob’s inequalities theorem,
E(U(·, [a, b]) ≤ sup E((Xn − b)− ) < ∞
n
308
for all a < b. Allowing a, b to vary over the rationals alone we
find that the sequence Xn is convergent a.e.
(ii) Sup E(Xn− ) ≤ sup E(|Xn |) < ∞ so that X∞ exists. As Xn → X∞ in
n
n
L1 we get that {Xn : n ≥ 1, n = ∞} is a supermartingale.
(iii) follows from (ii).
Proposition . Let {Xt : t ≥ 0} be a supermartingale relative to {Ft : t ≥
0}. I = [r, s], a < b and S any countable dense subset. Let U(w, S ∩
I, [a, b]) = U(·, {Xt (w) : t ∈ S ∩ I}, [a, b]). Then
E(U(·, S ∩ I, [a, b)) ≤
E((X s − b)− )
.
b−a
Proof. Let S ∩ I be an increasing union of finite sets Fn : then
E(U(·, Fn , [a, b])) ≤
E((Xmax Fn − b)− ) E((X s − b)− )
≤
.
b−a
b−a
The result follows by Fatou’s lemma.
Exercise . If further Xt is continuous i.e. t → Xt (w) is continuous for
each w, then prove that
E(U(·, I, [a, b])) ≤
E((X s − b)− )
b−a
Theorem . Let (Ω, F , P) be complete and {Xt : t ≥ 0} a continuous
supermartingale.
309
(i) If sup E(Xt− ) < ∞, then Xt converges a.e. to a random variable
t≥0
X∞ .
313
(ii) If {Xt : t ≥ 0} is uniformly integrable then also X∞ exists and
{Xt : t ≥ 0, t = ∞} is a supermartingale.
(i) E(U(·, [0, n], [a, b])) ≤ E((Xn − b)− )/(b − a) so that
Proof.
limit E(U(·, [0, n], [a, b])) ≤ sup
n→∞
0≤s
E((X s − b)− )
b−a
for all a < b. Thus {Xt (w) : t > 0} converges a.e. whose limit in
denoted by X∞ which is measurable.
(ii) As E(Xt− ) ≤ E(|Xt |) by (i) X∞ exists, the other assertion is a consequence of uniform integrability.
Corollary . Let {Xt : t ≥ 0} be a continuous uniformly integrable martingale. Then {Xt : 0 ≤ t ≤ ∞} is also a martingale.
Exercise. Let {Xt : t ≥ 0} be a continuous martingale such that for some
Y with 0 ≤ Y ≤ 1 E(Y|Ft ) = Xt show that Xt → Y a.e.
Lemma . Let (Ω, F , P) be a probability space, F1 ⊃ F2 ⊃ F3 . . . be
sub-σ-algebras. Let X1 , X2 , . . . be a real valued functions measurable
with respect to F1 , . . . , Fn , . . . respectively. Let
(i) E(Xn−1 |Fn ) ≤ Xn
(ii) sup E(Xn ) < ∞.
n
Then {Xn : n ≥ 1} is uniformly integrable.
310
Proof. By (i) E(Xn ) is increasing. By (ii) given ǫ > 0, we can find n0
such that if n ≥ n0 then E(Xn ) ≤ E(Xn0 ) + ǫ. For and δ > 0,
Z
|Xn |dP
n ≥ n0
= E(Xn ) +
Z
(|Xn |≥δ)
(Xn ≤−δ)
−Xn dP −
Z
(Xn <δ)
Xn dP
33. Application of Stochastic Integral
314
≤
≤ǫ+
Z
Z
−Xn0 dP −
(Xn ≤−δ)
Xn0 dP −
(Xn ≥δ)
Z
Z
Xn0 dP + E(Xn )
by (i)
(Xn <δ)
(because E(Xn ) ≤ E(Xn0 ) + ǫ)
Xn0 dP
(Xn ≤−δ)
≤ǫ+
Z
|Xn0 |dP
(|Xn |≥δ)
Thus to show uniform integrability we have only to show P(|Xn | ≥
δ) → 0 uniformly in n as δ → ∞. Now
E(|Xn |) = E(Xn + 2Xn− )
≤ E(Xn ) + 2E(|X1 |) by (i)
≤ M < ∞ for all n by (ii)
The result follows as P(|Xn | ≥ δ) ≤ M/δ.
311
Optional Sampling Theorem. (Continuous case).
Let {Xt : t ≥ 0} be a right continuous supermartingale relative to
{Ft : t ≥ 0}. Assume there exists an X∞ ∈ L′ (Ω, F , P) such that Xt ≥
E(X∞ |Ft ) for t ≥ 0. For any stopping time τ taking values in [0, ∞], let
Xτ = X∞ on {τ = ∞}. Then
(i) Xτ is integrable.
(ii) If σ ≤ τ are stopping times, then
E(Xτ |Fσ ) ≤ Xσ .
Proof. Define
[2n σ] + 1
[2n σ] + 1
,
τ
=
.
n
2n
2n
Then σn , τn are stopping times, σn ≤ τn , σ ≤ σn , τ ≤ τn . σn ,
τn take values in Dn = {∞, 0, 1/2n , 2/2n , . . . , 1/2n , . . .} so that we have
E(Xτn |Fσn ) ≤ Xσn . Thus if, A ∈ Fσ ⊂ Fσn , then
Z
Z
(*)
Xτn dP ≤
Xσn dP
σn =
A
A
315
As σ1 ≥ σ2 ≥ . . ., by optional sampling theorem for the countable
case, we have
E(Xσn−1 |Fσn ) ≤ Xσn .
Further
Fσ1 ⊃ Fσ2 ⊃ . . . ;
E(Xσn |F0 ) ≤ X0 .
By the lemma {Xσn }, {Xτn } are uniformly integrable families. By
right continuity Xσn → Xσ pointwise and Xτn → Xτ pointwise. Letting
n → ∞ in (*) we get the required result.
Lemma (Integration by Parts). Let M(t, ·) be a continuous progres- 312
sively measurable martingale and A(t, w) : [0, ∞) × Ω → R be of
bounded variation for each w. Further, assume that A(t, w) is Ft -measurable for each t. Then
Y(t, ·) = M(t, ·)A(t, ·) =
Zt
M(s, ·)dA(s, ·)
0
is a martingale if
E( sup |M(s, ·)| ||A(·)||t ) < ∞
0≤s≤t
for each t, where ||A(w)||t is the total variation of A(s, w) in [0, t].
Proof. By hypothesis,
n
X
i=0
M(s, ·)(A(si+1 , ·) − A(si , ·))
converges to
Zt
M(u, ·)dA(u, ·) in L1 as n → ∞
s
and as the norm of the partition s = s0 < s1 < . . . < sn+1 = t converges
to zero. Hence it is enough to show that
E([M(t, ·)A(t, ·) −
n
X
i=0
M(si+1 , ·)(A(si+1 , ·) − A(si , ·))]|Fs )
33. Application of Stochastic Integral
316
= M(s, ·)A(s, ·).
But the left side = E(M(sn+1 , ·)A(sn+1 , ·)−
−
n
X
i=0
M(si+1 , ·)(A(si+1 , ·) − A(si , ·))|Fs )
= M(s, ·)A(s, ·).
313
Taking limits as n → ∞ and observing that
sup |(si+1 − si )| → 0,
0≤i≤n
we get
E(M(t, ·)A(t, ·) −
Zt
M(u, ·)dA(u, ·)|Fs )
0
= M(s, ·)A(s, ·) −
Zs
M(u, ·)dA(u, ·).
0
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