Partial Differential Equations with
Applications to Finance
Seminar 1: Proving and applying Dynkin’s
formula
Group 4: Bertan Yilmaz, Richard Oti-Aboagye and Di Liu
May 20, 2015
Chapter 1
Proving Dynkin’s formula
We start by defining two crucial building blocks for the proof of Dynkin’s
formula.
Definition 1.1 (Diffusion): A diffusion is a stochastic process X which
is the solution to the stochastic differential equation (SDE):
dXt = µ(Xt )dt + σ(Xt )dWt ,
X0 = x0 for t ≥ 0
(1.1)
where µ, σ and x0 satisfy the conditions of the existence and uniqueness
theorem for SDEs and Wt denotes a Wiener process.
Definition 1.2 (Infinitesimal generator): Let X = (Xt )t≥0 be a diffusion. The infinitesimal generator L of Xt is defined as:
Lf (x) = lim+
t→0
EP [f (Xt )] − f (x)
,
t
x∈R
(1.2)
given that this limit exists. The set of functions f : Rn → R such that the
above limit exists at x is denoted by DA (x), while DA denotes the set of
functions for which the limit exists ∀x ∈ Rn .
Let Xt = Xtx now be a 1-dimensional Itô process of the form Xtx (ω) =
Rt
Rt
x + u(s, ω) ds + v(s, ω) dBs (ω) where B is a 1-dimensional Brownian
0
0
motion, x is a constant and f ∈ C02 (R), i.e. f is in C 2 (R) and has compact
support in R. Assume that u(t, ω) and v(t, ω) are ”nice” stochastic processes.
We moreover assume that u(t, ω) and v(t, ω) are bounded on the set of (t, ω)
such that X(t, ω) belongs to the support of f . Define Z = f (X) and apply
1
Itô’s formula. We obtain:
dZ =
∂f
1 ∂ 2f
1 ∂ 2f
∂f
∂f
2
2
(X)dX +
dt
+
(vdB).
(X)dX
=
u
(vdB)
+
∂x
2 ∂x2
∂x
2 ∂x2
∂x
We have that (vdB)2 = v 2 (dB)2 = v 2 dt, and thus:
Zt
!
Zt
∂f
1 2 ∂ 2f
∂f
u
+ v
ds + v dB.
2
∂x 2 ∂x
∂x
f (Xt ) = f (X0 ) +
0
0
Consequently:
ExP [f (Xτ )] = f (x)+ExP
" Zτ
#
! #
" Zτ
∂f
1 2 ∂ 2f
∂f
u (Xs )+ v
(Xs ) ds +ExP
v (Xs )dB .
∂x
2 ∂x2
∂x
0
0
(1.3)
To advance in our proof of Dynkin’s formula, we are in the need of defining
Borel measurable functions.
Definition 1.3 (Borel algebra and Borel measurable function): A
Borel algebra B(R) can be defined as B(R) = σ{open sets} where σ denotes
a σ-algebra. A real-valued function f : R → R which is measurable with
respect to the Borel algebra B(R) is said to be Borel measurable.
Proposition 1.1: Every continuous function f : Rn → R is Borel measurable.
Let g be a bounded Borel measurable function, i.e. ∃M ∈ R such that
|g| ≤ M. Then, for all finite T ∈ R we have:
ExP
"Z
τ ∧T
#
g(Xs )dBs = ExP
" ZT
0
#
1{s<t} g(Xs )dBs = 0
0
since g(Xs ) and 1{s<t} are both Fs -measurable. Here, τ ∧ T = min(τ, T )
and 1{s<t} denotes the indicator function defined as:
(
1 if s < t
1{s<t} =
0 if s ≥ t.
2
Furthermore, using Itô’s symmetry, we obtain:
ExP
τ ∧T
Z
" Zτ
g(Xs )dBs −
0
!2 #
= ExP
g(Xs )dBs
" Zτ
#
g 2 (Xs )ds .
τ ∧T
0
Now we have that:
#
" Zτ
#
" Zτ
T →∞
M2 ds = ExP [M2 (τ −τ ∧T )] = M2 ExP [τ −τ ∧T ] −−−→ 0
g 2 (Xs )ds ≤ ExP
ExP
τ ∧T
τ ∧T
T →∞
since τ ∧ T −−−→ τ . Hence:
" Zτ
#
"Z
#
τ ∧T
g(Xs )dBs = 0
ExP
g(Xs )dBs = lim ExP
T →∞
0
(1.4)
0
Combining (1.3) and (1.4) we now obtain:
ExP [f (Xτ )] = f (x) + ExP
" Zτ
! #
∂f
1 2 ∂ 2f
u (Xs ) + v
(Xs ) ds .
∂x
2 ∂x2
(1.5)
0
Using (1.5) with τ = t, u = µ and v = σ together with the definition of
L in (1.2), we arrive at Lemma 1.1.
Lemma 1.1: Let X = (Xt )t≥0 be a diffusion and the unique solution to
the SDE:
dXt = µ(Xt )dt + σ(Xt )dWt , X0 = x0 , t ≥ 0.
If f ∈ C02 (R), then the limit in (1.2) exists and:
Lf (x) = µ(x)
∂f (x) 1 2 ∂ 2 f (x)
+ σ (x)
.
∂x
2
∂x2
(1.6)
We consequently remark that the integrand in (1.5) effectively is the
expression for the generator of a diffusion Lf (Xs ) with u = µ(x) and v =
σ(x). Hence, we finally arrive at Dynkin’s formula.
Theorem 1.1 (Dynkin’s formula): Let f ∈ C02 (Rn ) and suppose that
τ is a stopping time such that EP [τ ] < ∞. Then:
" Zτ
#
ExP [f (Xτ )] = f (x) + ExP
Lf (Xs ) ds .
(1.7)
0
3
Chapter 2
Applications of Dynkin’s
formula
Here we present a few useful and interesting applications of the powerful
Dynkin formula. We start by defining the (first) exit time from an open set.
Definition 2.1 (First exit time): Let D be an open set and X x =
denote a diffusion starting at X0 = x ∈ D. The (first) exit time τ (x)
out of D is defined as:
(Xtx )t≥0
τ (x) = inf{t > 0 : Xtx ∈
/ D}.
2.1
The expectation of the exit time from a
bounded set
In this section, we are aiming to prove that the expectation of the (first)
exit time from a bounded set is finite with the aid of Dynkin’s formula. We
are proving the case for a 1-dimensional domain, but the proof can easily be
extended to the multidimensional case.
Let X = (Xt )t≥0 denote a 1-dimensional diffusion at an interval (a, b),
starting at x ∈ (a, b) and is thus the unique solution to the SDE:
(
dXt = µdt + σdWt ,
X0 = x
where σ and µ are constants and σ > 0.
4
t≥0
Integrating from 0 to τ , we first obtain:
Zτ
Xτ − x = µτ + σ
dWt
0
where τ denotes the earlier defined (first) exit time. Dynkin’s formula
reads:
" Zτ
#
ExP [f (Xτ )] = f (x) + ExP
Lf (Xs ) ds .
0
But since we do not know if E[τ ] < ∞, we cannot use Dynkin’s formula
directly. Instead, we take τ ∧ T for some finite T ∈ R and we know that
E[τ ∧ T ] < ∞. We obtain:
"Z
τ ∧T
EP [Xτ ∧T ] = EP [x] + µEP [(τ ∧ T )] + σEP
#
dWt .
0
Since in our interval (a, b), a < b, we have that:
EP [a] ≤ EP [Xτ ∧T ] ≤ EP [b].
Since a, b and x are constants, we have that EP [a] = a, EP [b] = b and
EP [x] = x. We then obtain:
"Z
τ ∧T
#
dWt = 0.
EP
0
Hence:
b−x
a−x
≤ EP [(τ ∧ T )] ≤
, ∀T > 0.
µ
µ
Now, from the monotone convergence theorem, we have that:
EP [τ ] = EP [ lim (τ ∧ T )] < ∞.
T →∞
Hence we conclude that the expectation of the (first) exit time from a
bounded set in one dimension is finite.
5
2.2
Application 1
We want to show that:
Z1
sdB(s) ∼ N (0, 1/3)
J=
0
by finding its moment generating function m(u) = E[euJ ] and using Dynkin’s
formula.
To solve this, we consider the Itô integral:
Zt
X(t) = sdB(s), t ≤ 1
0
and J = X1 . We have the Itô process X(t) where:
dX(t) = tdB(t) and f (X(t), t) = f (x) = eux .
We apply Lemma 1.1 to above and hence obtain:
Lf (x) =
t2 2 ux
ue .
2
Therefore, by Dynkin’s formula:
uX(t)
E[e
1
] = 1 + u2
2
Zt
s2 E[euX(s) ]ds,
0
∂f
∂t
where
= 0 and f (X(0), 0) = 1. Denote h(t) = E[euX(t) ]. Differentiation with respect to t thus yields:
h0 (t) =
t2 2
u h(t)
2
and thus:
h0 (t)
t2
= u2
h(t)
2
which after integration yields:
Zt
1 2 t3
1 2
ln h(t) = u
s2 ds ⇒ h(t) = e 2 u 3
2
0
3
which corresponds to the normal distribution N (0, t3 ). Thus, X(t) =
Rt
0
3
sdB(s) ∼ N (0, t3 ) and J = X1 ∼ N (0, 13 )
6
.
2.3
Application 2
Consider the n-dimensional standard Brownian motion B(t) with B(0) = 0
and:
τR = inf{t > 0 : |B(t)| = R}.
Find E0 [τR ].
For a n-dimensional standard Brownian motion we have:
1
Lf (B(t)) = ∆f (B(t)).
2
Since it is not known whether E0 [τR ] < ∞, we can not apply Dynkin’s
formula directly. So we define γR = τR ∧ T for some finite T ∈ R. We
thereafter choose f ∈ C02 such that f (x) := |x|2 . By Dynkin’s formula we
now have:
#
" ZγR
#
" ZγR
1
∆f (B(s))ds = 0+E0
n·ds = n·E0 [γR ].
E0 [f (B(γR ))] = f (0)+E0
2
0
0
Now from the monotone convergence theorem, we have that:
E0 [τR ] = E0 [ lim γR ] = lim E0 [γR ] ≤ M < ∞
T →∞
T →∞
for some finite M ∈ R. Now, since f (B(τR )) = |B(τR )|2 and E0 [B(τR )] = R,
we finally obtain:
R2
E0 [τR ] =
.
n
2.4
Application 3
This example of an application of Dynkin’s formula is similar to the previous
application but is nevertheless an interesting and highlighting example of
how Dynkin’s formula can be applied.
Consider a flea market in the form of a rectangle with sides a and b, with
a mother who has lost her son walking around inside furiously looking for
her lost son. We denote the rectangle domain of the market by D and the
boundary of the market by Γ = ∂D. The son is in fact waiting for his mother
at a point at the boundary of the market, and once his mother reaches any
point on the boundary of the market it is just a matter of time before they
meet. The question now is, how long time, at most, is it expected that the
7
poor son will have to wait for his mother to exit the flea market and find
him? Hopefully it is not infinite. Since the woman is hectically and stressfully
walking around looking for her son, her movements can be well-described by
a 2-dimensional Brownian motion Wt . We consider the searching process to
start when the mother is at W0 = (x, y) ∈ R2 . We define:
δ = sup kx − xΓ k,
xΓ ∈Γ
i.e. the maximum distance from the initial point of the Brownian motion
describing the searching process of the mother to the boundary of the rectangular shaped flea market. We realize that
√ δ has to be smaller than the
diagonal length of the flea market which is a2 + b2 .
The generator of a Brownian motion is:
1
L = ∆.
2
2
We now choose a function f ∈ C0 (R2 ) such that Lf (x, y) = constant and
we consequently choose f (x, y) = x2 + y 2 . Hence:
1
Lf = (2 + 2) = 2.
2
But since we do not know if EP [τ ] < ∞ we cannot use Dynkin’s formula
directly. Instead, we take τ ∧T for some finite T and we know that EP [τ ∧T ] <
∞. Hence, we obtain from Dynkin’s formula:
EP [f (Wτ ∧T )] = x2 + y 2 + EP
"Z
τ ∧T
#
2 ds .
0
2
2
We realize that EP [f (Wτ ∧T )] < a + b since EP [Wτ ] <
√
x2 + y 2 + 2EP [τ ∧ T ] < a2 + b2
yielding:
a2 + b 2 − x 2 − y 2
EP [τ ∧ T ] <
.
2
We know from the monotone convergence theorem that:
EP [τ ] = EP [ lim (τ ∧ T )] < ∞.
T →∞
Hence:
a2 + b 2 − x 2 − y 2
a2 + b 2
EP [τ ] <
≤
2
2
8
a2 + b2 and so:
and the expected time until the mother and her son are reunited is luckily
finite.
9