Lecture 7: Stochastic Differential Equations

Lecturer: Phạm Thị Hồng Thắm

Foundations of Mathematical Finance

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Table of Contents

1 Ito’s Lemma

2 Stochastic Differential Equations

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 Ito’s Lemma

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Ito’s Lemma

Suppose Xt is a stochastic process satisfying the following stochastic

differential equation (SDE)

dXt = µ(X , t)dt + σ(X , t)dWt

where µ(X , t) and σ(X , t) are smooth functions in terms of X and t.

If we define the stochastic process Yt to be a function of X and t

Yt = f (Xt , t)

where f is a smooth function in X and t then a natural question

arises which is if Yt also satisfies a similar SDE.

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Ito’s Lemma

That is if there exist µ̂(X , t) and σ̂(X , t) such that

dYt = µ̂(X , t)dt + σ̂(X , t)dWt

To answer this question, we consider the Taylor’s expansion of f (X , t).

f (Xt+h , t + h) = f (Xt , t) + fx (Xt , t)(Xt+h − Xt ) + ft (Xt , t)h

1 1
+ fxx (Xt , t)(Xt+h − Xt )2 + fxt (Xt , t)(Xt+h − Xt )h + ftt (X
2 2
Therefore,
1 1
dYt = fx dXt + ft dt + fxx (dXt )2 + fxt dXt dt + ftt (dt)2 .
2 2

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Ito’s Lemma

Note that
dXt = µ(X , t)dt + σ(X , t)dWt
and using the fact that

dWt dt = 0, (dWt )2 = dt, (dt)2 = 0

we deduce
dXt dt = 0.
Therefore,
1
dYt = fx dXt + ft dt + fxx (dXt )2 .
2

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Ito’s Lemma

Substituting
dXt = µ(X , t)dt + σ(X , t)dWt
we obtain
 
1 2
dYt = fx µ(Xt , t) + ft + fxx σ (Xt , t) dt + fx σ(Xt , t)dWt .
2

This is the content of Ito’s lemma.

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Ito’s Lemma

Theorem
If Xt satisfies the SDE

dXt = µ(X , t)dt + σ(X , t)dWt

and Yt = f (Xt , t) then

1
dYt = fx dXt + ft dt + fxx (dXt )2 .
2
More explicitly,
 
1 2
dYt = fx µ(Xt , t) + ft + fxx σ (Xt , t) dt + fx σ(Xt , t)dWt .
2

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Ito’s Lemma

Corollary
If Yt = f (Xt ) only, i.e., f only depends on X and not depend on t then
1
dYt = fx dXt + fxx (dXt )2 .
2

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Example

Example
Show that Yt = Wt2 satisfies the SDE

dYt = 2Wt dWt + dt.

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Solution:
Let f (X , t) = X 2 . Then it is readily verified that

fx = 2X , ft = 0, fxx = 2, fxt = 0, ftt = 0.

Let Xt = Wt . Then

µ(Xt , t) = 0, σ(Xt , t) = 1.

Apply Ito’s lemma we have

 
1 2
d(Yt ) = fx µ(Xt , t) + ft + fxx σ (Xt , t) dt + fx σ(Xt , t)dWt
2
= dt + 2Wt dWt .

In other words,
d Wt2 = 2Wt dWt + dt.

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Example

Example
Suppose Xt is a geometric Brownian motion
dXt
= µdt + σdWt .
Xt
Let Yt = log(Xt ) then

dYt = µ − 0.5σ 2 dt + σdWt .

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Integration by Parts

Proposition
Suppose we have two diffusion processes Xt and Yt . Then

d(Xt Yt ) = Yt dXt + Xt dYt + dXt dYt .

Equivalently,
Z t Z t Z t
Xt Yt = X0 Y0 + Ys dXs + Xs dYs + dXs dYs .
0 0 0

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Example

Let Xt = Yt = Wt . Then

d(Wt2 ) = Wt dWt + Wt dWt + (dWt )2

= 2Wt dWt + t.

Equivalently, Z t
1
Ws dWs = (Wt2 − t).
0 2

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Example

Example
Suppose Xt and Bt satisfy
dXt
= µdt + σdWt
Xt
dBt = rdt

where µ ,σ and r are constants. Find the SDE satisfied by

Xt /Bt .

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 Stochastic Differential Equations

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Stochastic Differential Equations

In this section we present some techniques used to solve some simple

SDEs.
The SDE in consideration will have the following form

dXt = µ(Xt , t)dt + σ(Xt , t)dWt .

That means for h > 0

Xt+h − Xt − µ(Xt , t)h − σ(Xt , t)(Wt+h − Wt )

is a random variable with vanishing mean and variance as h → 0.

Let us first consider the deterministic case. That is

σ(Xt , t) ≡ 0.

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Solving ODEs

Then our SDE becomes an ordinary differential equation (ODE).

dXt = µ(Xt , t)dt

Some elementary techniques to solve ODEs include

▶ Separation of variables,
▶ Using integrating factors.

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Example

Example
Solve the ODE
dXt = µXt dt
where µ is a constant.

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Solution:

This is an example of a separable differential equation.

We separate Xt from t to obtain
dXt
= µdt
Xt
Integrating both sides yields

log(Xt ) = µt + c

where c is a constant.
Hence, let A = e c .
Xt = e c+µt = Ae µt .
where A is any positive constant.

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Example

Example
Solve the ODE  
3
dXt = − Xt + 1 dt
t

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Solution:

This is an example of a first order linear differential equation.

To solve this equation, we rewrite it as
dXt 3
+ Xt = 1.
dt t
The integrating factor corresponds to this equation is given by
Z t 
3
I (t) = exp ds = t 3 .
0 s

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Multiplying by I (t) yields

dXt
t3 + 3t 2 Xt = t 3 .
dt
which is equivalent to
d 3

t Xt = t 3 .
dt
Therefore
t4
t 3 Xt = +c
4
for some constant c.
As a result, the solution of the ODE is
t c
Xt = + .
4 t3

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Finding the Expected value of Xt
Consider the SDE

dXt = µ(Xt , t)dt + σ(Xt , t)dWt

That implies for h > 0

Xt+h − Xt − µ(Xt , t)h − σ(Xt , t)(Wt+h − Wt )

is a random variable with vanishing mean and variance as h → 0.

Taking expectation,

E (Xt+h − Xt ) − µ(Xt , t)h − E(σ(Xt , t))E(Wt+h − Wt ) + o(h)

which is simplified to
E(Xt+h ) − E(Xt ) o(h)
= µ(Xt , t) +
h h
Taking limit we obtain the ODE for E(Xt ).
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Proposition
Suppose Xt satisfies the SDE

dXt = µ(Xt , t)dt + σ(Xt , t)dWt .

Then E(Xt ) satisfies the ODE

dE(Xt )
= E(µ(Xt , t)).
dt

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Example

Example
Show that
dE(Wt4 )
= 6E(Wt2 ).
dt
Then deduce that
E(Wt4 ) = 3t 2 .
Use a similar method to find E(Wt6 ).

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Example

Example
Suppose Xt satisfies the SDE
dXt
= µdt + σdWt
Xt

a) Given that X0 = 1, show that E(Xt ) = e µt .

b) Let Zt = Xt e −µt . Show that

dZt = σZt dWt

c) By considering d(log(Zt )), solve for Zt and hence deduce a formula

for Xt .

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Solution:

a) Let Yt = E(Xt ). Then

dYt
= µYt .
dt
This is a separable ODE. We separate variables as follows
dYt
= µdt
Yt
Integrating both sides yields

log(Yt ) = µt + c ⇒ Yt = Ae µt .

Since X0 = 1, we deduce

E(Xt ) = e µt .

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b) Now that we know E(Xt ) = e µt , define

Zt = Xt e −µt .

By the product rule

d(Zt ) = d Xt e −µt

= Xt (−µe −µt dt) + e −µt dXt + dXt (−µe −µt dt)

= σZt dWt .

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c) Consider log(Zt ). By Ito’s lemma
 2
dZt 1 dZt
d(log(Zt )) = −
Zt 2 Zt
1
= σdWt − (σdWt )2
2
= σdWt − 0.5σ 2 dt.

Integrate both sides,

log(Zt ) = σWt − 0.5σ 2 t.

As a result,
2 )t+σW
Xt = e µt Zt = e (µ−0.5σ t
.

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Example
Consider the SDE
2 /2
dXt = tXt dt + e t dWt

a) Given that X0 = 1, show that

2 /2
E(Xt ) = e t .
2 /2
b) Let Zt = Xt e −t . Find Zt and hence solve the SDE.

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