1
3. Brownian motion
(February 16, 2012)
Introduction
The French mathematician and father of mathematical …nance Louis Bachelier initiated the mathematical equations of Brownian motion in his thesis
"Théorie de la Spéculation" (1900). Later, in the mid-seventies, the Bachelier
theory was improved by the American economists Fischer Black, Myron Scholes, and Robert Merton, which has had an almost indescribable in‡uence
on today’s derivative pricing and international economy. Here, Brownian
motion is still very important as it is in many other more recent …nancial
models.
The purpose of this chapter is to discuss some points of the theory of
Brownian motion which are especially important in mathematical …nance.
To begin with we show that Brownian motion exists and that the Brownian
paths do not possess a derivative at any point of time. Furthermore, we use
abstract Lebesgue integration to show the existence of a stochastic integral
Z T
f (t; !)dW (t)
0
with respect to Brownian motion W when f is progressively measurable and
Z T
f 2 (t)dt < 1:
E
0
For future applications in …nance, it is proved that every (W )-measurable
and square integrable mean-zero random variable may be represented as a
stochastic integral of the above type. We also indicate that the stochastic
integral above may be de…ned under the weaker integrability condition
Z T
f 2 (t)dt < 1 a.s.
0
In mathematical …nance change of variables or change of measures is a
standard proceeding. We give a detailed proof of the so called CameronMartin theorem, which deals with a deterministic change of variables and
the more general result by Girsanov is proved in a special case.
2
Finally in this chapter, we study some other di¤usion processes than
Brownian motion and, in particular, the so called square root process (X(t))t 0 ,
which solves the (Feller or CIR (after Cox, Ingersol, and Ross)) equation
p
dX(t) = {(
X(t))dt +
X(t)dW (t) (X(0); {; ; positive constants):
2
We prove that X(t) > 0 if {
=2 and derive its moment-generating
function.
The Itô formula and the Feynman-Kac formula will not be treated at all
in this chapter. Instead we hope it will increase the understanding of these
important results, which the reader has met in earlier and more calculus
inspired courses.
3.1 De…nition of Brownian motion
Recall that a Gaussian stochastic process (X(t))t2T is a stochastic process
such that each linear combination
n
X
ak X(tk )
k=0
(t0 ; :::; tn 2 T ; a0 ; :::; an 2 R and n 2 N)
is Gaussian. If, in addition, the expectation of each X(t) equals zero the
process is said to be centred.
A centred Gaussian process (W (t))t 0 ; starting at 0 at time 0; and with
the covariance function
E [W (s)W (t)] = min(s; t)
is called a standard Brownian motion. In this case, W (s) W (t) is a centred
Gaussian random variable with the second order moment
E (W (s)
W (t))2 = E W 2 (s)
=s
2W (s)W (t) + W 2 (t)
2 min(s; t) + t =j s
tj
W (t) 2 N (0; j s
t j):
and, thus
W (s)
3
Theorem 3.1.1. A Gaussian process X = (X(t))t 0 is a standard Brownian
motion if and only if the following conditions are true:
(i) X(0) = 0
(ii) X(t) 2 N (0; t); t
0
(iii) the increments of X are independent, that is, for any …nite times
0 t0 t1 ::: tn the random variables
X(t1 )
X(t0 ); X(t2 )
X(t1 ); :::; X(tn )
X(tn 1 )
are independent (or uncorrelated since X is Gaussian).
PROOF. First suppose X = (X(t))t 0 is a standard Brownian motion. Then
(i) holds and X is a Gaussian process such that X(t) 2 N (0; t): This proves
(ii): To prove (iii), let j < k < n to get
E [(X(tj+1 )
X(tj ))(X(tk+1 )
X(tk ))]
= E [X(tj+1 )X(tk+1 )] E [X(tj+1 )X(tk )] E [X(tj )X(tk+1 )]+E [X(tj )X(tk )]
= tj+1
tj+1
tj + tj = 0:
This proves (iii):
Conversely, assume X = (X(t))t 0 is a Gaussian process satisfying (i)
(iii). Then X(0) = 0 and if 0 s t,
E [X(s)X(t)] = E X(s)(X(t)
= E [X(s)(X(t)
= E [X(s)] E [X(t)
X(s)) + X 2 (s)
X(s)] + E X 2 (s)
X(s)] + E X 2 (s) = s:
From this follows that E [X(s)X(t)] = min(s; t) and X is a standard Brownian motion.
4
A stock price process S = (S(t))t 0 is called a geometric Brownian motion
if there exists a standard Brownian motion W = (W (t))t 0 such that
S(t) = S(0)e
for appropriate parameters
t+ W (t)
2 R och
; t
0
> 0: Since
E [S(t)] = S(0)e(
it is natural to introduce a new parameter
+
2
)t
2
de…ned by the equation
2
=
+
so that
S(t) = S(0)e(
2
2
)t+
2
W (t)
:
The parameters and are called the mean rate of return and the volatility
of S, respectively. If the time unit is years and = 0:25, S is said to have
the volatility 25 %.
Note that if t0 t1 ::: tn ; then the simple returns
S(t1 )
S(t0 )
1; ::::;
S(tn )
S(tn 1 )
1
are independent as are the log-returns
ln
S(t1 )
S(tn )
; ::::; ln
:
S(t0 )
S(tn 1 )
Again let W = (W (t))t 0 be a standard Brownian motion and a > 0: The
scaled process
1
X(t) = a 2 W (at); t 0
is a standard Brownian motion since the process is centred, Gaussian, and
E [X(s)X(t)] = a
1
min(as; at) = min(s; t):
Furthermore,
Y (t) = W (t + a)
W ( a); t
0
is a standard Brownian motion since the process is centred, Gaussian, and
E [Y (s)Y (t)] = E [(W (s + a)
W ( a))(W (t + a)
W ( a))]
5
= E [(W (s + a)(W (t + a)]
E [(W (s + a)W ( a)]
E [(W ( a)W (t + a)] + E(W ( a)W ( a)
= min(s + a; t + a)
a
a + a = min(s; t):
As a mnemonic rule we say that W starts afresh at each point of time.
Finally, the sign changed process
Z(t) =
W (t); t
0
is a standard Brownian motion.
Thus if a stock price process (S(t))t 0 is a geometric Brownian motion
and a > 0, (S(at))t 0 and (S(a + t))t 0 are geometric Brownian motions.
Moreover, (1=S(t))t 0 is a geometric Brownian motion.
We …nish this section with a useful de…nition. Suppose I is a subinterval
of [0; 1[ and 0 2 I: A centred Gaussian process (W (t))t2I ; starting at the
point 0 at time 0; and with the covariance function
E [W (s)W (t)] = min(s; t)
is called a standard Brownian motion with the time set I:
3.2 Existence of Brownian motion with continuous sample paths
In this section we will …rst show the existence of Brownian motion with
continuous paths as a consequence of the existence of Lebesgue measure.
The so called Wiener measure is the distribution law of real-valued Brownian
motion with continuous sample paths.
We …rst start with some Hilbert space theory. Suppose H is Hilbert
space. Two vectors f and g in H are said to be orthogonal (abbrev. f ? g)
if < f; g >= 0: Let n be a non-negative integer and set In = f0; :::; ng : A
sequence (ei )i2In of unit vectors in H is said to be an orthonormal sequence
if ei ? ej for all i 6= j; i; j 2 In : If (ei )i2In is an orthonormal sequence and
f 2 H; then
f
i2In hf; ei iei ? ej all j 2 In
6
and it follows that
kf
i2In hf; ei iei
k2 k f
i2In
i ei
k2 all real
1 ; :::;
n:
Moreover,
k f k22 =k f
i2In hf; ei iei
k22 + k
i2In hf; ei iei
k22
and we get
2
i2In hf; ei i
k f k22 :
If H is of …nite dimension we say that (en )n2In is an orthonormal basis
for H if it is an orthonormal sequence and
f=
i2In hf; ei iei
for every f 2 H:
A sequence (ei )1
i=0 in H is said to be an orthonormal sequence if (ei )i2In
is an orthonormal sequence for each non-negative integer n: In this case, for
each f 2 H;
1
2
k f k22
i=0 hf; ei i
and the series
1
i=0 hf; ei iei
converges since the sequence
(
n
1
i=0 hf; ei iei )n=0
of partial sums is a Cauchy sequence in H: The sequence (ei )1
i=0 is said to be
an orthonormal basis for H if it is an orthonormal sequence and
f=
1
i=0 hf; ei iei
for every f 2 H:
Theorem 3.2.1. An orthonormal sequence (ei )1
i=0 in H is an orthonormal
basis for H if
(hf; ei i = 0 for every i 2 N) ) f = 0:
PROOF. Let f 2 H and set
g=f
1
i=0 hf; ei iei :
7
Then, for any j 2 N;
1
i=0 hf; ei iei ; ej i
hg; ej i = hf
1
i=0 hf; ei ihei ; ej i
= hf; ej i
Thus g = 0 or
f=
The theorem is proved.
= hf; ej i
hf; ej i = 0:
1
i=0 hf; ei iei :
In order to show the existence of Brownian motion with continuous sample
paths we next construct an appropriate orthonormal basis of L2 ( ), where
is Lebesgue measure on the unit interval. Set
[ 12 ;1] (x); x 2 R
v(x) = [0; 1 [ (x)
2
Moreover, de…ne h00 (x) = 1; 0
x
1; :::; 2n 1 ,
n 1
hjn (t) = 2 2 v(2n 1 x
1; and for each n
j + 1); 0
x
1 and j =
1:
Stated otherwise, we have for each n 1 and j = 1; :::; 2n
8 n 1
j 12
j 1
>
2 ;
2
x
<
;
>
n
1
2
2n 1
>
>
>
<
n 1
j 12
hjn (x) =
2 ;
x 2nj 1 ;
2
n 1
>
2
>
>
>
>
:
0; elsewhere in [0; 1] :
1
Drawing a …gure it is simple to see that the sequence h00; hjn ; j = 1; :::; 2n 1 ;
n 1; is orthonormal in L2 ( ). We will prove that the same sequence constitute an orthonormal basis for L2 ( ): To this end, suppose f 2 L2 ( ) is
orthogonal to each of the functions h00; hjn ; j = 1; :::; 2n 1 ; n 1: Then for
each n 1 and j = 1; :::; 2n 1
Z
and, in particular
Z
j 1
2
2n 1
fd =
j 1
2n 1
j
2n 1
j 1
2n 1
fd =
Z
1
2n
1
j
2n 1
j 1
2
2n 1
Z
0
fd
1
fd = 0
8
since
Z
1
fd =
Z
1
f h00 d = 0:
0
0
Thus
Z
k
2n 1
f d = 0; 1
j
2n
k
1
j
2n 1
and we conclude that
Z b
Z 1
f d = 0; 0
1[a;b] f d =
a
b
1:
a
0
Accordingly from this, f = 0 and the sequence
(hk )1
k=0 = (h00; h11 ; h12 ; h22 ; h13 ; h23 ; h33 ; h43 ; :::)
is an orthonormal basis for L2 ( ), called the Haar basis.
Let 0 t 1 and de…ne for …xed k 2 N
Z t
Z 1
hk d
ak (t) =
[0;t] (x)hk (x)dx =
0
0
so that
[0;t]
Now, if 0
s; t
=
1
k=0 ak (t)hk
in L2 ( ):
1;
min(s; t) =
Z
1
[0;s] (x) [0;t] (x)dx
0
=
1
k=0 ak (s)hhk ;
Note that
t=
[0;t] i
=h
=
1
k=1 ak (s)hk ;
[0;t] i
1
k=0 ak (s)ak (t):
1
2
k=0 ak (t):
From now on suppose (Gk )1
k=0 is an i.i.d. where each Gk 2 N (0; 1) and
introduce the random series
1
k=0 ak (t)Gk
which converges in L2 (P ) (and a.s., see Corollary 1.4.1). Its sum de…nes a
Gaussian random variable denoted by W (t) and the process (W (t))0 t 1 is
a centred Gaussian stochastic process with the covariance
E [W (s)W (t)] = min(s; t):
9
:
Recall that
(h00; h11 ; h12 ; h22 ; h13 ; h23 ; h33 ; h43 ; :::) = (hk )1
k=0 :
We de…ne
(a00; a11 ; a12 ; a22 ; a13 ; a23 ; a33 ; a43 ; :::) = (ak )1
k=0
and
(G00; G11 ; G12 ; G22 ; G13 ; G23 ; G33 ; G43 ; :::) = (Gk )1
k=0 :
It is important to note that for …xed n;
Z 1
ajn (t) =
[0;t] (x)hjn (x)dx 6= 0 for at most one j:
0
Set
U0 (t) = a00 (t)G00
and
Un (t) =
2n 1
j=1 ajn (t)Gjn ;
n 2 N+ :
We know that
W (t) =
1
n=0 Un (t)
in L2 (P )
for …xed t:
Let C [0; 1] be equipped with the metric
d(f; g) =k f
g k1
where k f k1 = max0 t 1 j f (t) j : Recall that every f 2 C [0; 1] is uniformly
continuous and as R is separable, it follows that the Banach space C [0; 1] is
separable. Finally, if fn 2 C [0; 1] ; n 2N; and
1
n=0
k fn k1 < 1
the series
1
n=0 fn
converges since the partial sums
sn =
form a Cauchy sequence.
n
k=0 fk ;
k2N
10
We now de…ne
1
n=0
= f! 2 ;
Here
2 F since
k Un k1 < 1g :
k Un k1 = sup j Un (t) j
0 t 1
t2Q
for each n: Next we prove that n is a null set.
To this end let n 1 and note that
P k Un k1 > 2
n
4
maxn 1 (k ajn k1 j Gjn j) > 2
P
n
4
1 j 2
But
k ajn k1 =
:
1
2
n+1
2
and, hence,
P k Un k1 > 2
n
4
n 1
2
Since
x
1 ) P [j G00 j
x]
2
P k Un k1 > 2
n=0
P j G00 j> 2
1
ye
y 2 =2
x
we get
and conclude that
"1
X
E
1[kUn k1 >2
Z
h
n
4
#
] =
n
4
1
X
n=0
2n 1 e
n
+ 12
4
dy
p
x 2
i
:
e
x2 =2
2n=2
P k Un k1 > 2
n
4
< 1:
From this and the Beppo Levi Theorem (or the Borel-Cantelli Lemma)
P [ ] = 1:
The trajectory t ! W (t; !); 0
t
1; is continuous for every ! 2 :
Without loss of generality, from now on we can therefore assume that all
trajectories of W are continuous (by eventually replacing by ):
Finally let W1 and W2 be independent Brownian motions both with time
set [0; 1] and continuous sample paths. and de…ne
W (t) =
W1 (t); 0
W1 (1) + tW2 ( 1t )
t 1
W2 (1); t > 1:
11
It is readily seen that W = (W (t))t 0 is a standard Brownian motion with
continuous sample paths and time set [0; 1[.
From now on it is always assumed that we consider Brownian motions
with continuous sample paths and if not otherwise stated W denotes such a
process. Moreover, without loss of generality, it will always be assumed that
the underlying probability space is complete.
Exercises
1. Suppose 0 t1 < ::: < tn T < 1 and let I1 ; :::; In be open subintervals of the real line. The set
S(t1 ; :::; tn ; I1 ; :::; In ) = fx 2 C [0; T ] ; x(tk ) 2 Ik ; k = 1; :::; ng
is called an open n-cell in C [0; T ] : The -algebra generated by all open
cells in C [0; T ] is denoted by C: Prove that
C = B(C [0; 1]):
(The construction above shows that the map
W :
! C [0; T ]
which maps ! to the trajectory
t ! W (t; !); 0
t
T
is (F; C)-measurable. The image measure PW is called Wiener measure
in C [0; T ] :)
3.3 Non-di¤erentiability of Brownian paths
In a mathematical model with continuous time a stock price process S =
(S(t))t 0 should not have a derivative at any point of time and, as will shortly
be proved, a log-Brownian stock price will ful…l this desire.
12
Theorem 3.3.1. The function t ! W (t); t
point t a.s. :
0 is not di¤erentiable at any
PROOF. We may restrict ourselves to the unit interval 0
c; " > 0 and denote by B(c; ") the set of all ! 2 such that
j W (t)
W (s) j< c j t
s j if t 2 [s
t
1. Let
"; s + "] \ [0; 1]
for some s 2 [0; 1] : If the set
1 [
1
[
1
B(j; ):
k
j=1 k=1
is of probability zero we are done
From now on let c; " > 0 be …xed. It is enough to prove P [B(c; ")] = 0 :
To this end set
j
j+1
) W( ) j
Xn;k = max j W (
k j<k+3
n
n
for each integer n > 3 and k 2 f0; :::; n 3g : Furthermore, let n > 3 be so
large that
3
":
n
We claim that
6c
B(c; ")
min Xn;k
:
0 k n 3
n
To show this claim note that to every ! 2 B(c; ") there exists an s 2 [0; 1]
such that
j W (t)
W (s) j
and choose k 2 f0; :::; n
cjt
"; s + "] \ [0; 1]
3g such that
s2
If k
s j if t 2 [s
k k
3
; +
:
n n n
j < k + 3;
j W(
j+1
)
n
j
j+1
W( ) j j W(
)
n
n
W (s) j + j W (s)
j
W( ) j
n
13
6c
n
and, hence, Xn;k
6c
:
n
Now
B(c; ")
min Xn;k
0 k n 3
6c
n
and it is enough to prove that
lim P
n!1
0 k n 3
But
P
6c
n
min Xn;k
0 k n 3
= (n
6c
= 0:
n
min Xn;k
2)P Xn;0
1
= n(P j W ( ) j
n
6c
n
n 3
X
P Xn;k
6c
n
nP Xn;0
6c
n
k=0
6c 3
) = n(P (j W (1) j
n
6c
p )3
n
12c 3
n( p
):
2 n
where the right side converges to zero as n ! 1. The theorem is proved.
3.4 Stochastic integrals
Suppose W = (W (t))0 t T is a Brownian motion based on a complete probability space ( ; F; P ) and let F(t); 0 t T; be an increasing sequence of
-algebras contained in F such that
(i) each W (t) is F(t)-measurable
(ii) for 0 t u the increment W (u)
W (t) is independent of F(t):
14
Under these assumptions the family (F(t))0 t T is said to be a …ltration for
W: Moreover, a stochastic process X = (X(t))0 t T is said to be adapted if
each X(t) is F(t)-measurable. An adapted process such that X(t) 2 L1 (P )
and
0 s t T ) E [X(t) j F(s)] = X(s)
is called a martingale. If
0
s
t
T ) E [X(t) j F(s)]
X(s)
the process X is called a supermartingale. The minimum of two supermartingales is a new supermartingal. Moreover, if Xn ; n 2 N; are non-negative
supermartingales and
X(t) = lim Xn (t) a.s. for each t
n!1
then X = (X(t))0 t T is a supermartingale. In fact, if 0
A 2 F(s); by bounded convergence
E [min(X(t); k)
A]
= lim E [(min(Xn (t); k)
n!1
lim E [min(Xn (s); k)
n!1
A]
= E [min(X(s); k)
s
t
T and
A]
A]
and by letting k ! 1 the claim follows by monotone convergence.
In this section the purpose is to introduce a so called stochastic integral
Z T
f (t)dW (t)
0
under appropriate conditions on f = (f (t))0 t<T . It will always be assumed
that f is adapted but many more restrictions must be imposed.
A stochastic process f = (f (t))0 t T is said to be simple if the following
conditions hold:
(a) f is adapted and there exists a C 2 R such that j f (t; !) j C for
all 0 t T and ! 2
(b) there exists a partition 0 = t0 < t1 < :::: < tn = T such that
f (t) = f (tj 1 ) if t 2 [tj 1 ; tj [ for j = 1; :::; n:
15
For a simple integrand f we de…ne the stochastic integral in the natural
way
Z T
n
X
f (t)dW (t) =
f (tj 1 )(W (tj ) W (tj 1 ))
0
j=1
and have the so called Itô isometry
Z
Z T
2
f (t)dW (t)) = E
E (
T
f 2 (t)dt :
0
0
In addition, the process
Rt
Mf = e
0
f (s)dW (s)
1
2
Rt
f 2 (s)ds
0
0 t T
is a martingale. If we abide by the notion in (a) and (b) and let A 2 F(tn 1 )
E [Mf (T )
h h
E E ef (tn
1 )(W (tn )
A]
W (tn
= E [E [Mf (T )
1 ))
1 2
f (tn 1 )(tn
2
tn
= E [Mf (tn 1 )
A
1)
A]
j F(tn 1 )]]
i
j F(tn 1 ) Mf (tn 1 )
A
i
or
E [Mf (T ) j F(tn 1 )] = Mf (tn 1 ):
By iterating this line of reasoning we conclude that Mf is a martingale for
every simple f:
Often it is important to have a stochastic integral for less restrictive
conditions on the integrand f:
A stochastic process f = (f (t))0 t<T is said to be progressively measurable if for each t 2 [0; T ] ; the restriction map f (s; !); 0 s t; ! 2 ; is
(B([0; t] F(t))-measurable. If, in addition,
Z T
E
f 2 (t)dt < 1
0
the process f is said to belong to the class F:
Next suppose f 2 F is bounded and f ( ; !) continuous for every ! 2
Set for each positive integer n;
fn (t; !) = f (t; !) if
j
1
n
T
t<
j
T; j = 1; :::; n
n
:
16
and de…ne fn (t; !) to be left-continuous at the point t = T: Then, if
for Lebesgue measure on [0; T ] ; by dominated convergence
fn ! f in L2 (
P)
as n ! 1: Moreover, by the Itô isometry
Z T
Z T
Z T
2
E (
fm (t)dW (t)
fn (t)dW (t)) = E
(fm (t)
0
stands
0
fn (t))2 dt
0
RT
2
and it follows that the sequence ( 0 fn (t)dW (t))1
n=1 converges in L (P ) and
RT
the limit is denoted by 0 f (t)dW (t): Clearly, the Itô isometry holds for f:
In the next step we assume f 2 F is bounded and de…ne for each postive
integer n;
Z
t
f (s; !)ds if 0
fn (t; !) = n
(t
t
T:
1 +
)
n
The function fn ( ; !) is continuous. Moreover, by the Lebesgue di¤erentiation
theorem, for every …xed ! 2 ;
lim fn ( ; !) = f ( ; !) a.s. [ ]
n!0
and hence, by dominated convergence
Z T
(fn (t)
lim E
n!1
f (t))2 dt = 0:
0
RT
As above the Itô isometry implies that the sequence ( 0 fn (t)dW (t))1
n=1 conRT
2
verges in L (P ) and the limit is denoted by 0 f (t)dW (t): Furthermore, the
Itô isometry holds for f:
The next step is simple. If f 2 F, de…ne fn = f [jf j n] for every n 2 N+
and use dominated convergence to conclude that
fn ! f in L2 (
P ) as n ! 1:
RT
As above we can now de…ne the stochastic integral 0 f (t)dW (t) such that
the Itô isometry holds.
If f is simple the stochastic integral is de…ned path-wise and the map
Z t
f (s)dW (s); 0 t T
0
17
is continuous and, in addition, it is readily seen to be a martingale. The
de…nitions above imply the martingale property of the stochastic integral for
every f 2 F: To prove continuity we need Doob’s maximal inequality.
Suppose f 2 F and choose simple functions fn such that
fn ! f in L2 (
P)
as n ! 1 and note that for each 0 t T;
Z t
Z t
f (s)dW (s) in L2 (P )
fn (s)dW (s) !
0
0
as n ! 1: Furthermore, choose integers 1 n1 < n2 < ::: such that
Z T
(fnk (t) fnk+1 (t))2 dt
2 3k :
E
0
If we note that the square of a bounded martingale in discrete time is a
non-negative submartingale, the Doob maximal inequality and Itô isometry
yield
#
"
Z m
Z m
j=2
j=2
sup j
P
0 j
P
"
fnk (s)dW (s)
2m
sup j
0 j
0
0
2m
Z
2
k
2k
#
fnk+1 (s)dW (s) j
j=2m
fnk+1 (s))dW (s) j2
(fnk (s)dW (s)
0
2k
2 E
Z
2
T
(fnk (s)
fnk+1 (s))2 ds
2
k
0
and, hence,
P
sup j
0 t T
Z
t
fnk (s)dW (s)
0
2k
2 E
Z
t
0
Z
21
fnk+1 (s)dW (s) j
T
(fnk (s)
fnk+1 (s))2 ds
2
k
:
0
Now the series
1 Z
X
(
1
0
fnk (s)dW (s)
Z
0
fnk+1 (s)dW (s))
k
18
converges absolutely in the Banach space C [0; T ] and is therefore convergent
with probability one. Accordingly from this,
Z
lim
fnk (s)dW (s) =def X
k!1
0
exists and is a continuous function with probability one. But for every t;
Z t
f (s)dW (s) a.s.
X(t) =
0
Rt
and we conclude that the process ( 0 f (s)dW (s))0
version.
t T
possesses a continuous
Example
R t 3.4.1. Consider the is progressively measurable random function
f (t) = 0 cos s dW (s); 0 t T: We want to compute E [X 2 ] if
X=
Z
T
(f (t) +
Z
t
f (s)dW (s))dW (t):
0
0
To this end …rst note that
2
E f (t) =
Z
t
cos2 s ds
0
s=t
s + 21 sin 2s
=
2
s=0
1
1
= (t + sin 2t) 2 L2 ( ):
2
2
From this we conclude that f 2 F since
Z T
Z T
2
E f 2 (t) dt < 1:
E
f (t)dt =
0
0
Thus the stochastic integral
and by the Itô isometry
Rt
0
f (s)dW (s) is well de…ned for every …xed t
Z t
E ( f (s)dW (s))2
0
=E
Z
0
t
2
f (s)ds =
Z
0
t
1
1
(s + sin 2s)ds
2
2
19
Moreover,
R
1 t2
= (
2 2
0
1
1
cos 2t + ) 2 L2 ( ):
4
4
f (s)dW (s) 2 F since
Z
E
T
0
Z
=
Z t
( f (s)dW (s))2 dt
0
Z t
E ( f (s)dW (s))2 dt < 1:
T
0
0
Accordingly from this the stochastic integral
de…ned and by the Itô isometry
Z T Z t
( f (s)dW (s))dW (t))2
E (
0
0
=E
Z
RT Rt
( 0 f (s)dW (s))dW (t) is well
0
T
0
Z t
Z
2
( f (s)dW (s)) dt =
0
T
0
1 t2
(
2 2
1
1
cos 2t + )dt
4
4
T
1 T3 1
= (
sin 2T + ):
4 3
4
2
Moreover, by using by the Itô isometry twice and that f (s) is a mean-zero
random variable, we have
Z
E
T
f (t)dW (t)
Z
0
0
=E
Z
T
(f (t)
0
=
Z
=
Z
T
0
=
Z
0
T
E
Z t
( f (s)dW (s))dW (t)
0
Z
t
f (s)dW (s))dt
0
T
E f (t)
0
T
Z
t
f (s)dW (s) dt
0
Z
0
t
cos sdW (s)
Z
t
f (s)dW (s) dt
0
Z t
Z T Z t
( E [(cos s) f (s)] ds)dt =
( 0ds)dt = 0:
0
Finally, noting that
0
0
20
Z
X=
T
f (t)dW (t) +
0
we have
Z
0
T
Z t
( f (s)dW (s))dW (t)
0
E X2
1 T2 1
1
1 T3 1
T
= (
cos 2T + ) + (
sin 2T + )
2 2
4
4
4 3
4
2
3
2
T
T
T
1 1
1
=
+
+ +
cos 2T
sin 2T:
12
4
8
8 8
16
Our next theorem shows that stochastic integration with respect to Brownian motion leads to several new martingales and supermartingales.
Rt
Theorem 3.4.1. Suppose f 2 F. Then ( 0 f (s)dW (s))0
and
Rt
R
1 t 2
Mf = e 0 f (s)dW (s) 2 0 f (s)ds
t T
is a martingale
0 t T
a supermartingale.
In particular, if C 2 R and j f j C;
h Rt
i
C2
E e 0 f (s)dW (s)
e 2 t; 0
t
T;
and Mf is a martingale.
PROOF. Suppose f 2 F and choose simple functions fn such that
fn ! f in L2 (
P)
as n ! 1 and note that for each 0 t T;
Z t
Z t
fn (r)dW (r) !
f (r)dW (r) in L2 (P )
0
0
as n ! 1: But if 0 s t T and A 2 Fs
Z t
Z s
E
fn (r)dW (s) A = E
fn (r)dW (s)
0
0
A
21
and by letting n ! 1 we have
Z t
E
f (r)dW (s)
A
Z
=E
0
s
f (r)dW (s)
A
:
0
Rt
Thus ( 0 f (s)dW (s))0 t T is a martingale.
By eventually passing to a subsequence we may assume that
Z T
(fn f )2 d ! 0 a.s.
0
and it follows thar there exists a set 0 of measure one such that
Z t
Z t
2
fn2 (s)ds on 0 for every t 2 [0; T ] :
fn (s)ds !
0
0
Moreover, we may assume
Z
Z
fn (s)dW (s) !
0
f (s)dW (s) a.s.
0
in C [0; T ] (to see this use the Doob maximal theorem and Itô isometry as
above). Since the process
Rt
Mfn = e
0
1
2
fn (s)dW (s)
Rt
0
fn2 (s)ds
0 t T
is a non-negative martingale it follows that the limit process
Rt
e
0
f (s)dW (s)
1
2
Rt
0
f 2 (s)ds
0 t T
is a supermartingale.
Next suppose f 2 F and j f j C for an appropriate real number C: As
h Rt
i
R
f (s)dW (s) 12 0t f 2 (s)ds
0
1
E e
and we get
h
h Rt
E e 0 f (s)dW (s)
Rt
E e
0
f (s)dW (s)
i
e
i
1
; 0
t
1 2
C t
2
C2
t
2
T:
22
Now we claim that the process Mf is a martingale. In fact, by Itô’s lemma
dMf (t) = f (t)Mf (t)dW (t):
Note here that
(f (t)Mf (t))0
2 F:
t T
Indeed, by using the …rst part of the theorem already proved
Z T
R
Rt
2
1 t 2
E
f (t)e 0 f (s)dW (s) 2 0 f (s)ds
0
Z
2
C E
C
2
Z
T
e2
0
T
Mf (t) = 1 +
0
f (s)dW (s)
dt
2
e2C t dt < 1:
0
Hence
Rt
Z
t
f (s)Mf (s)dW (s)
0
and we are done.
Example 3.4.1. Suppose f is simple and suppose " > 0: Next we will show
how it is possible to control the tail probability
Z T
f (t)dW (t) j> "
P j
0
RT
from the tail probability of the random variable 0 f 2 (t)dt:
Given N > 0 we will show that
Z T
Z T
N
P j
f (t)dW (t) j> "
P
f 2 (t)dt > N + 2 :
"
0
0
To see this suppose
0 = t0 < t1 < ::: < tn
1
< tn = T
and de…ne
f (t) = f (tj 1 ); tj
1
t < tj ; j = 1; :::; n:
23
Moreover, let m be the largest integer m n such that
Z tm
m
X1
2
f (t)dt =
f 2 (tj 1 )(tj tj 1 ) N:
0
j=1
Note that
[tm
Next we de…ne
t] 2 Ft :
f (t) if t < tm ;
0 otherwise in [0; T ]
fN (t) =
and have that fN 2 F is bounded and
Z T
m
X1
2
fN (t)dt =
f 2 (tj 1 )(tj
0
Thus
tj 1 )
N:
j=1
Z
E (
T
fN (t)dW (t))2
N
0
and furthermore
Z
f (t) = fN (t) for every t < T if
T
f 2 (t)dt
N:
0
Now
P j
Z
0
T
f (t)dW (t) j> "
+P
Z
P j
T
Z
T
fN (t)dW (t) j> "
0
f 2 (t)dt > N :
0
But
P j
Z
0
T
fN (t)dW (t) j> "
1
E
"2
" Z
2
T
fN (t)dW (t)
0
and we are done.
A progressively measurable function such that
Z T
f 2 (t)dt < 1 a.s.
0
#
N
"2
24
is
R Tsaid to belong to the class Floc : It is possible to de…ne the stochastic integral
f (t)dW (t) for each f 2Floc such that the process
0
Z t
( f (s)dW (s))0
t T
0
has a continuous version (in this context the Example 3.4.1 is useful (see
e.g. A. Friedman, Stochastic Di¤erential Equations and Applications, Vol 1,
Academic Press 1975)) .
If f is a smooth function and
f 0 (t; W (t)) 2 F (as a function of (t; !)).
the Itô lemma gives
1 00
(t; W (t))dt:
df (t; W (t)) = ft0 (t; W (t))dt + fx0 (t; W (t))dW (t) + fxx
2
However, the same formula turns out to be true under the much weaker
condition
f 0 (t; W (t)) 2 Floc (as a function of (t; !)).
If f 2Floc and
we de…ne
is any random variable with values in the interval [0; T ]
Z
f (t)dW (t) = I( )
0
where
I(t) =
Z
t
f (s)dW (s):
0
Furthermore, if is a stopping time so that the level set f
measurable for every t 2 [0; T ] it is possible to show that
Z
0
Exercises
f (t)dW (t) =
Z
0
T
1[t
] f (t)dW (t):
tg is Ft -
25
1. Show that the function W (t; !); 0
measurable.
t
T; ! 2
; is progressively
2. Suppose 0 t0 a b T and f (t; !) = 1[a;b] (t)W (t0 ; !); 0
! 2 : Show that f is progressively measurable.
t
T;
3. Find E [X 2 ] if
X=
Z
T
0
Z t
( sin(u + t)dW (u))dW (t):
0
4. Suppose G 2 N (0; 1) and
p
Hn (x; y) = E [(x + i yG)n ] ; n 2 N
for every x 2 R and y
0: Show that H0 (x; y) = 1; H1 (x; y) = x;
H2 (x; y) = x2 y; H3 (x; y) = x3 3xy and H4 (x; y) = x4 6x2 y + 3y 2 :
In addition, show that
@
Hn (x; y) = nHn 1 (x; y); n = 1; 2; :::
@x
@
1 @2
Hn (x; y) +
Hn (x; y) = 0; n = 2; 3; :::
@y
2 @x2
and
e
x
1
2
2y
=
1
X
n=0
5. Prove that
M (t) =
1
X
n=0
n
n!
Hn (x; y);
2 R:
n
n!
Hn (W (t); t);
> 0:
6. Use the Itô lemma to show that
Z t
Hn (W (t); t) = n
Hn 1 (W (s); s)dW (s); n = 1; 2; ::: :
0
26
3.5 Martingale representation
Theorem 3.5.1. (Itô representation, one dimension) Suppose F(t) =
(W (s); s t) and X 2 L2 ( ; F(T ); P ): There exists a 2 F such that
Z T
(t)dW (t):
X = E [X] +
0
PROOF. Suppose f 2 L2 ( ) and
Rt
Mf (t) = e
Recall that
E
RT
0
Z
0
f (t)dW (t) 2 N (0;
RT
0
ZT
T
(f (t)Mf (t))2 dt =
0
0
=
ZT
Rt
f 2 (t)e
0
0
1
2
f (s)dW (s)
f 2 (s)ds
Rt
0
f 2 (s)ds
; t 2 [0; T ] :
f 2 (t)dt): By the Tonelli theorem
h Rt
f 2 (t)E e2 0 f (s)dW (s)
RT
dt
e
0
f 2 (s)ds
ZT
Rt
0
f 2 (s)ds
f 2 (t)dt < 1
0
and it follows that f Mf 2 F: Moreover, the Itô lemma yields
dMf (t) = f (t)Mf (t)dW (t)
and
Mf (T ) = 1 +
Z
T
f (t)Mf (t)dW (t):
0
Set
and
L0 = Y ; Y 2 L2 ( ; F(T ); P ) and E [Y ] = 0
Lint =
Z; Z =
Z
T
(t)dW (t) where
0
2F :
Here L0 and Lint are closed subspaces of L2 ( ; F(T ); P ) and
Lint
L0 :
i
dt
27
Now suppose Y 2 L0 \ L?
int : If we can prove Y = 0, then Lint = L0 and we
are done.
To start with Y ? Mf (T ) for every f 2 L2 ( ). To de…ne an appropriate
f choose 0 = t0 < t1 < ::: < tn = T and 1 ; 2 ; ::: n 2 R and de…ne
f (t) =
k;
t 2 [tk 1 ; tk [ ; k = 1; :::; n:
Since
Mf (T ) = e
1 W (t1 )+ 2 (W (t2 )
where
1
2
a=
is a constant,
E Ye
1 W (t1 )+ 2 (W (t2 )
W (t1 ))+:::+
Z
W (tn
n (W (tn )
1 ))+a
T
f 2 ( )d
0
W (t1 ))+:::+
n (W (tn )
W (tn
1 ))
= 0:
Stated otherwise, with
Y+ = max(0; Y )
and
Y = max(0; Y )
we have
E Y+ e
1 W (t1 )+ 2 (W (t2 )
W (t1 ))+:::+
n (W (tn )
W (tn
1 ))
E Y e
1 W (t1 )+ 2 (W (t2 )
W (t1 ))+:::+
n (W (tn )
W (tn
1 ))
E Y+ es(
1 W (t1 )+ 2 (W (t2 )
W (t1 ))+:::+
n (W (tn )
W (tn
1 )))
=
:
Thus
E Y es(
1 W (t1 )+ 2 (W (t2 )
W (t1 ))+:::+
n (W (tn )
W (tn
=
1 )))
for every real s and by analytic continuation
E Y+ ei(
E Y ei(
1 W (t1 )+ 2 (W (t2 )
1 W (t1 )+ 2 (W (t2 )
W (t1 ))+:::+
W (t1 ))+:::+
n (W (tn )
n (W (tn )
W (tn
W (tn
1 )))
1 )))
=
:
Now the uniqueness theorem for the Fourier transform yields that
(Y+ P )((W (t1 ); W (t2 )
W (t1 ); (W (tn )
W (tn 1 )) 2 B)
(Y P )((W (t1 ); W (t2 )
W (t1 ); (W (tn )
W (tn 1 )) 2 B)
28
for every B 2 B(Rn ): Since F(T ) = (W (s); s T ); 0 t T; we conclude
that Y P = Y+ P or Y P = 0 that is Y = 0: This completes the proof of the
theorem.
Corollary 3.5.1. If M = (M (t))0 t T is a square integrable martingale
with respect to the …ltration F(t) = (W (s); s t); 0 t T; there exists
a 2 F such that
M (t) = E [M (0)] +
Zt
(s)dW (s); 0
t
T:
0
Exercises
1. Find the martingale representation of the random variable j W (T ) j :
3.6 Cameron-Martin-Girsanov theory
If
2 L2 ( ) we already know from the previous section that the process
M
(t) =def Z (t) = e
Rt
(s)dW (s)
0
1
2
Rt
0
2 (s)ds
; 0
t
T
is a martingale. Moreover, we have
Theorem 3.6.1 (Cameron-Martin) If
2 L2 ( ) the process (Z (t))0
is a martingale and the process
Z t
~
W (t) = W (t) +
(s)ds; 0 t T
0
is a Brownian motion with respect to the probability measure
P~ = Z (T )P:
t T
29
PROOF. For short write Z = Z: Fix 0 s t T and A 2 F(s): Now for
any real ;
h
i
~ (t) W
~ (s))
i (W
E e
A Z(T )
h
i
Rt
= E ei (W (t) W (s)+ s (r)dr) )) A Z(T )
i
h
R
i (W (t) W (s)+ st (r)dr)
=E e
A Z(t)
i
i
h h
Rt
Rt
R
1 t 2
= E E ei (W (t) W (s)+ s (r)dr) s (r)dW (r) 2 s (r)dr j F(s) A Z(s) :
Rt
Here as the random variable i (W (t) W (s))
(r)dW (r) is a complex
s
linear combination of members of a Gaussian process we get
h
i
Rt
Rt
R
1 t 2
E ei (W (t) W (s)+ s (r)dr) s (r)dW (r) 2 s (r)dr j F(s)
h
= E ei
(W (t) W (s))
1
= e 2 E [(i
2
=e
2
h
Rt
s
(r)dr+ 21
Rt
s
Rt
s
~ (t) W
~ (s))
i (W
E e
Now if 0
h
Pn
E e
k=1
i
t0
:::
k (W (tk )
~ (tk
W
~
(r)dW (r)
s
(W (t) W (s))
(t s) i
Thus
Rt
i
j F(s) ei
(r)dW (r))2 jF (s)] i
e
2 (r)dr
ei
Rt
s
Rt
1
2
(r)dr
s
1
2
1
2
(r)dr
s
(r)dr
i
A Z(T ) = E [
Rt
Rt
s
2 (r)dr
2
A Z(s)] e
2
Rt
2 (r)dr
s
Rt
2 (r)dr
s
2
=e
(t s)
2
(t s)
:
:
tn ; we have
1 ))
i
h Pn 1
Z(T ) = E e k=1 i
= ::: = E [Z(t0 )] e
1
2
Pn
2
k=1 k (tk
tk
1)
~
k (W (tk )
=e
1
2
~ (tk
W
Pn
1 ))
i
Z(tn 1 ) e
2
k=1 k (tk
tk
1)
2
n (t
n
2
:
This proves the Cameron-Martin theorem.
Suppose the distribution measure of W on the Banach spase C [0; T ] is
denoted by and let a (A) = (A a) if a 2 C [0; T ] and A 2 B(C [0; T ]): The
set H of all absolutely continuous functions f :[0; T ] ! R such that f (0) = 0
tn
1)
30
and f 0 2 L2 ( ) is called the Cameron-Martin space of
shows that
a
and Theorem 3.6.1
if a 2 H:
<<
It can be proved that
H = fa 2 C [0; T ] ;
a
<< g :
(see e.g. H.-H. Kuo, Gaussian measures in Banach spaces, Springer (1975)).
If B(a; r) denotes the closed ball in C [0; T ] with centre a and radius r; we
have
R
1 T
0
2
(B(a; r))
e 2 0 (a (t)) dt ; if a 2 H
=
lim
r!0+ (B(0; r))
0; if a 2 C [0; T ] H
(see my paper A note on Gauss measures which agree on small balls, Ann.
H. Poincaré, section B, 13, 231-238, (1977)). Thus
H=
a 2 C [0; T ] ; lim
r!0+
(B(a; r))
>0 :
(B(0; r))
The quantity (B(0; r)) can be computed explicitly using the separation
of variables method (see Section 4.5).
Theorem 3.6.2. Suppose
2 Floc .
(a) The process
Z (t) = e
Rt
0
(s)dW (s)
1
2
Rt
0
2 (s)ds
; 0
is a martingale if and only if
E [Z (T )] = 1:
(b) (Novikov’s condition) If
h 1 RT
E e2 0
2 (s)ds
i
< 1;
t
T
31
then E [Z (T )] = 1:
(c) (Girsanov’s Theorem) If E [Z (T )] = 1; the process
~ (t) = W (t) +
W
Z
t
(s)ds; 0
t
T
0
is a Brownian motion with respect to the probability measure
P~ = Z (T )P:
We will not prove Theorem 3.6.2 here (for a proof see J. M. Steele, Stochastic Calculus and Financial Applications, Springer 2000 ; for a slightly
weaker result than 3.6.5 see A. Friedman, Stochastic Di¤erential Equations
and Applications, Vol 1, Academic Press 1975)). Let us just remark that
the case with a simple bounded
follows exactly as the Cameron-Martin
theorem.
Example 3.6.1. It is possible to …nd a
2 Floc such that Z
martingale. A well known example is the following.
Set
= min t 2 [0; 1] ; W 2 (t) = 1 t
and
(t) =
2W (t)
1[t
(1 t)2
]:
Then
P [0 <
and
Z
0
1
2
(t)dt = 4
Z
0
< 1] = 1
W 2 (t)
dt < 1 a.s.
(1 t)4
Now by the Itô lemma, for t < 1;
d
W 2 (t)
2W 2 (t)
2W (t)
1
=
dt +
dW (t) +
dt
2
3
2
(1 t)
(1 t)
(1 t)
(1 t)2
is not a
32
from which
Z
1
(t)dW (t)
0
Z
Z
1
2
1
2
(t)d(t)
0
Z 1
2W (t)
2W 2 (t)
=
1
dW
(t)
1[ t] dt
[ t]
t)2
t)4
0 (1
0 (1
Z
Z
Z
Z
W 2 (t)
2W 2 (t)
1
2W 2 (t)
d
=
+
dt
+
dt
dt
(1 t)2
t)3
t)2
t)4
0
0 (1
0 (1
0 (1
Z
1
1
1
W 2( )
+
2W 2 (t)(
)+
)dt
=
2
3
4
(1
)
(1 t)
(1 t)
(1 t)2
0
Z
1
1
+
dt = 1:
1
t)2
0 (1
1
Hence
E [Z (1)] < e 1 :
Example 3.6.2. Let C be the Banach space of all real-valued continuous
functions on the interval [0; 1] and equip C with the Borel …eld B: We claim
that there is a probability measure P~ :B ! [0; 1] such that under P~ ; the map
f :C ! C de…ned by
f (x) =
t ! x(t) +
Z tp
0
j x(s) jds; 0
t
1
is a standard Brownian motion.
To prove the claim let P :B ! [0; 1] be Wiener measure and put W (x) =
x; x 2 C: Note that under P; the map W is a real-valued standard Brownian
motion. Set
Z tp
~
W (t) = W (t) +
j W (s) jds; 0 t 1
0
and
dP~ = e
R1p
0
jW (s)jdW (s)
1
2
R1
0
jW (s)jds
If the Novikov condition
h
E e
1
2
R1
0
jW (s)jds
i
<1
dP:
33
~ is a Brownian motion. But
is ful…lled we know that under P~ ; the process W
Z 1
j W (s) j ds max j W (s) j
0 s 1
0
and for given y
0; we use the Bachelier double law
P max W (s)
y = 2P [W (1)
0 t 1
to get
Z
P
y]
1
0
j W (s) j ds
P max W (s)
P max j W (s) j
y
y +P
0 s 1
2P max W (s)
min W (s)
y =
0 s 1
y = 4P [W (1)
0 s 1
y
0 s 1
y]
2e
y2
2
Hence
h
E e
1
2
Z
1
2
1
R1
0
P
0
jW (s)jds
Z
0
i
Z
1
P
0
1
j W (s) j ds
Z
0
1
j W (s) j ds
y
2
y e dy
Z
1
0
2 ln u du =
e
y2
2
y
e 2 dy < 1
and we are done.
The de…nitions and results in this section extend to Brownian motion in
several dimensions and the generalization is often only a question of notation
(cf the Shreve book).
Example 3.6.3.
Let T 2 ]0; 1[ and let W = (W1 (t); W2 (t))0 t T be a
2-dimensional Brownian motion with the natural …ltration Ft = (W (s);
s t); 0 t T: We want to …nd a constant C such that the measure P~
de…ned by the equation
dP~ = CeW1 (T )W2 (T ) dP on FT
34
is a probability measure and, moreover, we want to …nd a martingale (Lt ; Ft )0
such that
dP~ = Lt dP on Ft
t T
for every 0 t T:
With this aim, …rst note that
Lt =C = E eW1 (T )W2 (T ) j Ft =
eW1 (t)W2 (t) E e(W1 (T )
W1 (t))(W2 (T ) W1 (t))+W2 (t)(W1 (T ) W1 (t))+W1 (t)(W2 (T ) W2 (t))
h
eW1 (t)W2 (t) E e(T
p
p
t)XY +b T tX+a T tY
i
j Ft =
j(a;b)=(W1 (t);W2 (t))
where X; Y 2 N (0; 1) are independent. Moreover,
i
h
p
p
(T t)XY +b T tX+a T tY
=
E e
Z
1
(x2 +y 2
2
e
R2
p
p
2(T t)xy)+b T tx+a T ty dxdy
2
Z
e
p
1
(u2 +v 2 )+b
2
T t(u+ p
=
1
Hence
1 2
b ((T
a2 (T
(T
1
p
1
(T
t)2
e 2(1
eW1 (t)W2 (t)
Lt =C = p
1
(T
t)2
e
C=
p
Lt = p
1
T2
1
(T
2
t)2
t)2
T t
(b(T
2(1 (T t)2 )
e
p
dudv
1
p
1
(T
t)+a)2 )
t)
ab(T t)2
b2 (T
+
+
t)2 )
1 (T t)2
2(1 (T
(T
W1 (t)2 (T t)
+
2(1 (T t)2 )
and it follows that
and
t)+
e2
t)2
(T
t)y
t)2 y
p
T t
v)+ p a T t 2 v
1 (T t)2
1 (T t)
R2
p
1
u =px (T
v = 1 (T
t)
t)2 )
=
:
t)2 W1 (t)W2 (t)
W (t)2 (T t)
+ 2
1 (T t)2
2(1 (T t)2 )
T2
W1 (t)2 (T t)
W (t)W2 (t)
W (t)2 (T t)
+ 1
+ 2
2(1 (T t)2 )
1 (T t)2
2(1 (T t)2 )
:
=
=
35
3.7 Stochastic di¤erential equations
Partial di¤erential equations and stochastic di¤erential equations are basic
for most models in option pricing and we …nd it natural to complement the
Shreve book in various ways.
To start with let (W (t))0 t T be a standard Brownian motion and F(t) =
(W (s); 0 s t); 0 t T: We will consider the stochastic di¤erential
equation
dX(t) = a(t; X(t))dt + b(t; X(t))dW (t); 0
t
T
with the initial value x0 2 R or, equivalently, the stochastic integral equation
Z t
Z t
b(s; X(s))dW (s); 0 t T;
a(s; X(s))ds +
X(t) = x0 +
0
0
where it is assumed that the process X = (X(t))0 t T is adapted, that is the
random variable X(t) is F(t)-measurable for every t 2 [0; T ] : Throughout it
will be assumed that the functions
a : [0; T ]
R!R
b : [0; T ]
R!R
and
are continuous but, in general, more restrictions will be imposed.
Theorem 3.7.1. Suppose there is a constant K such that
j a(t; x)
a(t; y) j
Kjx
y j; 0
t
T; x; y 2 R;
j b(t; x)
b(t; y) j
Kjx
y j; 0
t
T; x; y 2 R;
j a(t; x) j
K(1+ j x j); 0
t
T; x 2 R
j b(t; x) j
K(1+ j x j); 0
t
T; x 2 R:
and
36
Then the di¤erential equation
dX(t) = a(t; X(t))dt + b(t; X(t))dW (t); 0
t
T
equipped with the initial condition X(0) = x0 possesses an adapted solution
(X(t))0 t T with continuous trajectories a.s. and such that
sup E X 2 (t) < 1:
0 t T
If (Y (t))0
t T
is another solution with these properties, then
P [X(t) = Y (t) for every 0
t
T ] = 1:
For short we only prove uniqueness in Theorem 3.7.1 and refer to A.
Friedman, Stochastic Di¤erential Equations and Applications, Vol 1, Academic Press 1975, for the remaining part.
PROOF OF UNIQUENESS. If (X(t))0 t T and (Y (t))0 t T are solutions,
Z t
X(t) Y (t) =
(a(s; X(s)) a(s; Y (s)))ds
0
+
Z
t
(b(s; X(s))
b(s; Y (s)))dW (s):
0
But
( + )2
2(
2
+
2
);
;
2R
and we get
E (X(t)
2
Y (t))
Z t
2E ( (a(s; X(s))
a(s; Y (s)))ds)2
0
Z t
+2E ( (b(s; X(s))
b(s; Y (s)))dW (s))2 :
0
By applying the Cauchy-Schwarz inequality we get
Z t
Z t
2
E ( (a(s; X(s)) a(s; Y (s)))ds)
tE
(a(s; X(s))
0
0
a(s; Y (s)))2 ds
37
and since
a(s; Y (s)))2
(a(s; X(s))
it follows that
Z t
E ( (a(s; X(s))
K 2 (X(s)
2
tK
a(s; Y (s)))ds)
2
Z
Y (s))2
t
Y (s))2 ds:
E (X(s))
0
0
Moreover,
b(s; Y (s)))2
(b(s; X(s))
K 2 (X(s)
Y (s))2
and
E
Z
t
(b(s; X(s))
2
b(s; Y (s))) ds
K
2
0
Z
t
E (X(s)
Y (s))2 ds
0
2K 2 t( sup E X 2 (t) + sup E Y 2 (t) ) < 1:
0 t T
0 t T
Hence
(b(s; X(s))
b(s; Y (s)))0
t T
2F
and
Z t
E ( (b(s; X(s))
2
b(s; Y (s)))dW (s))
=E
0
Z
t
(b(s; X(s))
0
K
2
Z
t
Y (s))2 ds:
E (X(s)
0
From the above, if C = 4K 2 max(T; 1),
E (X(t)
2
Y (t))
C
Z
t
Y (s))2 ds:
E (X(s)
0
Consequently, by introducing the function
f (t) = E (X(t)
we get
f (t)
C
Z
0
Y (t))2 ; 0
t
t
f (s)ds; 0
t
T
T
b(s; Y (s)))2 ds
38
Next let
M = sup f (s)
0 s T
and use induction to conclude
f (t)
M Cn n
t ; n 2 N:
n!
From these inequalities it follows that f = 0 or, stated otherwise,
P [X(t) = Y (t)] = 1; for every 0
t
T:
Hence
P [X(t) = Y (t) for every rational t 2 [0; T ]] = 1
and as the processes (X(t))0 t T and (Y (t))0 t T possess continuous sample
paths with probability one the uniqueness part of the theorem is proved.
Example 3.7.1. Suppose ; ; ; and x0 are real numbers. The equation
dX(t) = ( + X(t))dt + X(t)dW (t); t
0
with the initial condition X(0) = x0 possesses a unique solution. In the
special case = 0 and x0 = 1 the solution equals
2
)t+
2
U (t) = e(
W (t)
:
To treat the general case set
Y (t) =
X(t)
= X(t)V (t)
U (t)
where
V (t) = e(
+
2
)t
2
W (t)
:
From
dV (t) = V (t)((
+
2
)dt
dW (t))
and the product rule for di¤erentiation
d(X(t)V (t)) = V (t)dX(t) + X(t)dV (t) + dX(t)dV (t)
39
= V (t) f( + X(t))dt + X(t)dW (t)g
+X(t)V (t) (
+
2
)dt
2
dW (t)
Thus
X(t)V (t) = x0 +
Z
X(t)V (t)dt = V (t)dt:
t
V (u)du
0
or, equivalently,
(
X(t) = e
2
)t+
2
W (t)
(x0 +
Z
t
e(
+
2
)u
2
W (u)
du):
0
Note that X(t) > 0 if x0 > 0 and > 0:
For positive parameters x0 ; {; ; and ;the process (X(t))t
by the equation
dX(t) = {(
0
governed
X(t))dt + X(t)dW (t)
with the initial condition X(0) = x0 describes the volatility in the so called
Garch di¤usion model. The same equation also describes the short rate in
the so called Brennan-Schwartz model.
Example 3.7.2 (The stochastic Verhulst equation). Let s0 ; > 0 be positive
parameters and and real numbers. The stochastic di¤erential equation
dS(t) = S(t)(
S(t))dt + S(t)dW (t); t
S(0) = s0 ; S(t) > 0; 0 t T
0
does not meet the assumptions in Theorem 3.7.1. To solve the equation put
X(t) =
1
S(t)
and rewrite the stochastic di¤erential equation in the following equivalent
form
dX(t) =
1
S 2 (t)
=
fS(t)(
( X(t)
S(t))dt + S(t)dW (t)g +
)dt
X(t)dW (t) +
2
1
S 3 (t)
X(t)dt
(dS(t))2
40
2
( +(
)X(t))dt
X(t)dW (t):
Now by the previous example,
(
X(t) = e
2
2
)t
W (t)
(X(0) +
Z
t
2
)u+
2
e(
W (u)
du)
0
and we get
e(
S(t) =
X(0) +
or
2
)t+
2
Rt
0
e(
2
s0 e( 2
S(t) =
Rt
1 + s0 0 e(
W (t)
2
)u+
2
W (u)
du
)t+ W (t)
2
)u+
2
W (u)
:
du
Example 3.7.3. Suppose we want to solve the stochastic di¤erential equation
dX(t) = a(X(t))dt + b(X(t))dW (t); 0 t T
with the initial value X(0) = x0 2 R: To this end we start with a partition
0 = t0 < t1 < ::: < tn = T
and rewrite the integral equation
Z t
Z t
X(t) = x0 +
a(X(s))ds +
b(X(s))dW (t); 0
0
in the equivalent form
(
X(tk ) = X(tk 1 ) +
t
T
0
R tk
tk
X(0) = x0
R tk
a(X(s))ds
+
b(X(s))dW (t); k = 1; :::; n:
tk 1
1
The standard Euler scheme reads as follows:
X (n) (0) = x0
X (n) (tk ) = X (n) (tk 1 )+a(X (n) (tk 1 )) tk +b(X (n) (tk 1 )) W (tk ); k = 1; :::; n
where
tk = tk
tk 1 ; k = 1; :::; n
41
and
W (tk ) = W (tk )
W (tk 1 ); k = 1; :::; n:
In the so called Milstein scheme the term
Z tk
b(X(s))dW (t)
tk
1
is approximated in a slightly di¤erent way. First
db(X(s)) = b0 (X(s))dX(s) +
b2 (X(s)) 00
b (X(s))ds
2
b2 (X(s)) 00
b (X(s)))ds + b(X(s)b0 (X(s))dW (s)
2
= (a(X(s))b0 (X(s)) +
and from this
+
b(X(s)) = b(X(tk 1 )
Z
s
(a(X(u))b0 (X(u)) +
tk
1
+
Z
s
b(X(u)b0 (X(u))dW (u):
tk
1
X(tk 1 ) =
Z
Now
X(tk )
b2 (X(u)) 00
b (X(u)))du
2
tk
tk
a(X((s))ds +
Z
tk
tk
1
b(X(s))dW (t)
1
= a(X(tk 1 ) tk + b(X(tk 1 ) W (tk ) + Rk
where
Rk =
Z
tk
tk
1
"Z
#
s
b(X(u)b0 (X(u))dW (u) dW (s) + Rk0 :
tk
1
Next we approximate the expression
"Z
Z
tk
tk
#
s
b(X(u)b0 (X(u))dW (u) dW (s)
1
tk
1
by
0
b(X(tk 1 )b (X(tk 1 ))
Z
tk
tk
1
Z
(
s
tk
dW (u))dW (s)
1
42
where
Z
tk
tk
1
Z
(
=
s
dW (s))dW (s) =
tk
Z
tk
(W (s)
tk
1
W (tk 1 ))dW (s)
1
tk
W (s)dW (s)
tk
W 2 (tk )
=
2
Z
W (tk 1 )(W (tk )
W (tk 1 ))
1
tk
W 2 (tk 1 )
2
tk
1
W (tk 1 )(W (tk )
1
= (( W (tk ))2
2
W (tk 1 ))
tk ):
If we drop Rk0 we get the Milsteins schema:
X (n) (0) = x0
and
X (n) (tk ) = X (n) (tk 1 ) + a(X (n) (tk 1 )) tk + b(X (n) (tk 1 )) W (tk )
1
+ b(X (n) (tk 1 )b0 (X (n) (tk 1 ))(( W (tk ))2
2
tk ); k = 1; :::; n:
Exercises
1. Suppose X is the Garch di¤usion process. Show that 1=X solves the
stochastic Verhulst equation.
2. Suppose x0 is a real constant. Solve the di¤erential equation
p
1
dX(t) = X(t)dt + 1 + X 2 (t)dW (t); t
2
with the initial value X(0) = x0 :
0
43
(Answer: X(t) = sinh(W (t) + ln(x0 +
p
1 + x20 )))
3.8 A Dirichlet problem in one space dimension
In mathematical …nance the connection between Brownian motion and partial di¤erential equation is of fundamental importance and requires a course
on its own. In spite of this, in this section we will make a minor occursion
into this important …eld without beeing able to give complete mathematical
proofs of every statement.
Consider two continuously di¤erentiable functions a and b on a compact
interval [ ; ] and assume min b > 0: The boundary problem
b2 (x) 00
u
2
+ a(x)u0 = 0
u( ) = A; u( ) = B
is simple to solve for any real A and B using the integrating factor
Rx
e
2a(z)
dz
b2 (z)
:
The purpose of this section is to give a probabilistic solution of the solution u of the above Dirichlet problem. To this end consider smooth bounded
extensions of a and b de…ned on R and, again, denote these extensions by
a and b; respectively. Furthermore, choose x 2 ] ; [ and denote by X the
solution of the stochastic di¤erential equation
dX(t) = a(X(t))dt + b(X(t))dW (t); 0
t<1
with the initial value X(0) = x (Hint: Use Theorem 3.7.1 for each the time
interval 0 t n; n 2 N+ ).
Next consider a smooth extension of u de…ned on R: For simplicity, this
extension is denoted by u and without loss of generality we may assume that
u0 is bounded. By Itô’s lemma, we have
1
du(X(t)) = u0 (X(t))dX(t) + u00 (X(t))(dX(t))2
2
1
= (u0 (X(t))a(X(t) + u00 (X(t))b2 (X(t))dt + u0 (X(t))b(X(t))dW (t):
2
44
Hence
u(X(t)) = u(x) +
Z
t
0
+
Z
1
(u0 (X(s)a(X(s) + u00 (X(s))b2 (X(s))ds
2
t
u0 (X(s))b(X(s))dW (s):
0
Thus if is the …rst time X hits the boundary of [ ; ] and T is a strictly
positive number,
Z min( ;T )
1
(u0 (X(s)a(X(s) + u00 (X(s))b2 (X(s))ds
u(X(min( ; T ))) = u(x) +
2
0
+
Z
min( ;T )
u0 (X(s))b(X(s))dW (s)
0
and
u(X(min( ; T ))) = u(x) +
Z
min( ;T )
u0 (X(s))b(X(s))dW (s):
0
Since the expectation of the stochastic integral in the right-hand side equals
zero, we have
u(x) = E [u(X(min( ; T )))]
and letting T ! 1 yields the formula
u(x) = E [u(X( ))]
since P [ = 1] = 0 (see I. Karatzas and S. E. Shreve, "Brownian Motion and
Stochastic Calculus", Chapter 5). To empasize that X(0) = x we sometimes
write
u(x) = E [u(X( )) j X(0) = x]
or
u(x) = Ex [u(X( ))] :
We now have
u(x) = APx [ = ] + BPx [ = ]
where the notation Px indicates that the process starts at the point x:
45
In the special case A = 1 and B = 0 we get
Rx
u(x) = 1
R
and
Px [ = ] = 1
Ry
e
Ry
e
Rx
e
R
e
2a(z)
dz
b2 (z)
2a(z)
dz
b2 (z)
Ry
Ry
dy
dy
2a(z)
dz
b2 (z)
2a(z)
dz
b2 (z)
dy
:
dy
Suppose ; {; are strictly positive parameters and consider the square
root process (X(t))t 0 governed by the CIR equation
p
dX(t) = {(
X(t))dt +
X(t)dW (t)
with the initial condition X(0) = x0 > 0. Moreover, assume
2
{
If 0 <
2
:
it is readily seen that
<x<
Px ( = ) ! 0 as
! 0:
Hence the CIR equation has a strictly positive solution for every t 0 and
x > 0:
The CIR equation is used in interest rate theory and, moreover, it represents the volatility process in Heston’s option pricing model and the following
result is of great interest.
Theorem 3.8.1. If
Ex e
X(t)
> 0;
1
=
(1 +
2
(1
2{
e
{t ))
2{
2
exp(
e
1+
2
2{
{t
(1
x
e
{t )
):
For a proof of Theorem 3.8.1 in a special case, see Exercise 6.6 in the
Shreve book "Stochastic Calculus for Finance II. Continuous-Time Models".
46
PROOF. Set
2
d2 f
+ {(
2 dx2
Lf =
and
x
1
u(t; x) =
(1 +
2
(1
2{
e
{t ))
2{
2
x)
exp(
df
dx
e
2
1+
2{
{t
(1
x
e
{t )
):
By direct computation we …nd that the function u(t; x) satis…es the equation
Lu =
@u
if t > 0; x > 0
@t
with the initial condition u(0; x) = exp( x):
Now suppose v(t; x) is a smooth function and use the Itô lemma to conclude that
v(min(t; ); X(min(t; )))
Z min(t; )
Z min(t; )
p
@v
@v
= v(0; x) +
(s; X(s)) X(s)dW (s) +
(
+ Lv)ds
@x
@t
0
0
if is any stopping time. Moreover, choose a …xed number t0 > 0 and
v(t; x) = u(t0 t; x) for t < t0 : Then
u(t0 min(t; ); X(min(t; ))) = u(t0 ; x)+
Z
min(t; )
0
Now if 0 <
that
Noting that
p
@u
(t0 s; X(s)) X(s)dW (s):
@x
< x and
is the …rst time the process X hits
Ex [u(t0
min(t; ); X(min(t; )))] = u(t0 ; x):
! 1 if
we conclude
! 0 we use dominated convergence to get
Ex [u(t0
t; X(t))] = u(t0 ; x):
Finally, letting t ! t0 the theorem follows.
let
If we di¤erentiate with respect to
! 0+ , it follows that
Ex [X(t)] = e
{t
in the formula in Theorem 3.8.1 and
x + (1
e
{t
x):
47
(for a strict proof A. Friedman, Stochastic Di¤erential Equations and Applications, Vol 1, Academic Press 1975, is useful). In particular, the process
p
( X(t))0 t T 2 F:
In addition, it is known that
Ex [X n (t)] < 1 if n 2 N:
In fact much more is true. To explain this, set
X(t)
z( ) = Ex e
where
z( ) =
1
2
(1
(1
2{
{t ))
e
2{
2
if
exp(
<0
e
{t
x
2
1
(1
2{
e
2{
e
{t )
{t )
)
is de…ned in the domain
=
2 C; Im 6= 0 if Re
2 (1
:
We already know that
z(n) (0) = Ex [X n (t)] :
Now if
0
z( ) =
<
1
X
z(n) (0)
0
n
n!
= Ex e
In applications below
we asume
X(t)
:
2
2{
2 (1
e
2
{t )
>1
and
Ex e
X(t)
n
will always be strictly smaller than 1 and since
{
it follows that
2{
;
e {t )
1
X
Ex [X n (t)]
=
n!
0
2 (1
is analytic for Re <
2 (1
2{
e
{t )
: