CHAPTER 10

Stochastic Differential Equations

Consider a stochastic process (Xt) satisfying

dXt = b(t,Xt,Wt) dt + σt, Xt,Wt) dWt. (10.1)

Question.

(1) Can we obtain the existence and uniqueness theorem for (10.1)? What are the

properties of the solution?

(2) How to solve it?

10.1. Examples and some solution methods

Example 10.1. Solving

dXt = αXt dWt + σXt dt, (10.2)

where the initial condition X0 is given and α, σ are constant. Rewrite this equation as

dXt

Xt

= α dWt + σ dt,

and we can get ∫ t

0

dXu

Xu

=

∫ t

0

α dWu +

∫ t

0

σ du = αWt + σt. (10.3)

Due to the Ito’s formula,

df(t,Xt) =∂f

∂tf(t,Xt) dt +

∂f

∂xf(t,Xt) dXt +

1

2

∂2f

∂x2f(t,Xt) (dXt)

2,

237

238 10. STOCHASTIC DIFFERENTIAL EQUATIONS

we may take f(t, x) = f(x) = ln x, then

f ′(x) =1

x, f ′′(x) = − 1

x2.

Then

d ln(Xt) =1

Xt

dXt − 1

2

1

X2t

(dXt)2 =

1

Xt

dXt − α2

2dt,

i.e.,

ln(Xt) − ln(X0) =

∫ t

0

dXu

Xu

− α2t

2. (10.4)

Combining (10.3) and (10.4), we get

ln

(Xt

X0

)+

α2t

2=

∫ t

0

dXu

Xu

= αWt + σt.

Thus, the solution to (10.2) is given by

Xt = X0 exp

(αWt +

(σ − α2

2

)t

).

Definition 10.2. A stochastic process (Xt) of the form

Xt = X0 exp(αWt + μt)

is called the geometric Brownian motion.

Remark 10.3. (1) If (Wt) is independent of X0, then

E[Xt] = E[X0 exp

(αWt +

(σ − α2

2

)t

)]= E[X0]e

σt.

(2) (i) If σ > α2/2, then Xt −→ ∞ as t → ∞ P-a.s.

(ii) If σ < α2/2, then Xt −→ 0 as t → ∞ P-a.s.

(iii) If σ = α2/2, then Xt will fluctuate between arbitrary large and arbitrary

small values as t → ∞ P-a.s.

10.1. EXAMPLES AND SOME SOLUTION METHODS 239

Example 10.4 (Hull-White interest-rate model). Consider

dRt = (at − btRt) dt + σt dWt, with R0 = r, (10.5)

where at, bt, and σt are deterministic function. Then

dRt + btRt dt = at dt + σt dWt,

which implies that

d

(Rt exp

(∫ t

0

bu du

))= at exp

(∫ t

0

bu du

)dt + σt exp

(∫ t

0

bu du

)dWt.

Thus, the solution to (10.5) is given by

Rt = r exp

(−

∫ t

0

bu du

)+

∫ t

0

as exp

(−

∫ t

s

bu du

)ds

+

∫ t

0

σs exp

(−

∫ t

s

bu du

)dWs.

Example 10.5. Consider the stochastic differential equation

dXt = rXt(K − Xt) dt + βXt dWt, with X0 = x > 0. (10.6)

Rewrite the equation as

dXt

Xt

+ rXt dt = rK dt + β dWt.

Taking integration on the both sides, we have

∫ t

0

dXu

Xu

+ r

∫ t

0

Xu du = rKt + βWt.

Using a similar argument as in Example 10.1, we get

∫ t

0

dXu

Xu

= ln(Xt) − ln(X0) +1

2

∫ t

0

1

X2u

d〈X〉u

= ln

(Xt

x

)+

1

2

∫ t

0

1

X2u

β2X2u du = ln

(Xt

x

)+

1

2β2t.

240 10. STOCHASTIC DIFFERENTIAL EQUATIONS

Thus,

ln

(Xt

x

)+ r

∫ t

0

Xu du = βWt +

(rK − 1

2β2

)t,

which implies

Xt exp

(r

∫ t

0

Xu du

)= x exp

[βWt +

(rK − 1

2β2

)t

].

Integration with respect to t on the both sides, we obtain

x

∫ t

0

exp

[βWs +

(rK − 1

2β2

)s

]ds =

∫ t

0

Xs exp

(r

∫ s

0

Xu du

)ds

=

∫ t

0

exp

(r

∫ s

0

Xu du

)d

(∫ s

0

Xu du

)

=1

rexp

(r

∫ s

0

Xu du

)∣∣∣∣t

s=0

=1

r

[exp

(r

∫ t

0

Xu du

)− 1

].

Hence,

exp

(r

∫ t

0

Xu du

)= 1 + rx

∫ t

0

exp

[βWs +

(rK − 1

2β2

)s

]ds,

i.e., ∫ t

0

Xu du =1

rln

(1 + rx

∫ t

0

exp

[βWs +

(rK − 1

2β2

)s

]ds

).

Taking derivative with respect to t, we see that the solution to (10.6) is given by

Xt =1

r· rx exp

[βWt +

(rK − 1

2β2

)t]

1 + rx∫ t

0exp

[βWs +

(rK − 1

2β2

)s]

ds

=exp

[βWt +

(rK − 1

2β2

)t]

x−1 + r∫ t

0exp

[βWs +

(rK − 1

2β2

)s]

ds.

Example 10.6. Solving the stochastic differential equation

dXt = αt dt + btXt dWt.

Rewrite it as

dXt − btXt dWt = αt dt.

10.1. EXAMPLES AND SOME SOLUTION METHODS 241

����� Example 10.4 ����. ����� ρt such that

ρt dXt − btρtXt dWt = αtρt dt. (10.7)

By integration by parts,

d(ρtXt) = ρt dXt + Xt dρt + d〈ρ,X〉t.

Idea: Suppose Xt dρt = −btρtXt dWt.

(������, ���� idea, ������������.)

Hence, we want to find ρ such that

dρt

ρt

= −bt dWt.

By Ito formula,

−∫ t

0

bu dWu =

∫ t

0

dρu

ρu

= ln(ρt) − ln(ρ0) +1

2

∫ t

0

1

ρ2u

(dρu)2

= ln(ρt) − ln(ρ0) +1

2

∫ t

0

b2u du.

Hence,

ρt = ρ0 exp

(−

∫ t

0

bu dWu − 1

2

∫ t

0

b2u du

).

We may take ρ0 = 1. Thus,

d(ρtXt) = ρt dXt + Xt dρt + (dXt)(dρt)

= ρt dXt − btρtXt dWt + (αt dt + btXt dWt)(−btρtXt dWt)

= ρt dXt − btρtXt dWt − b2t ρtXt dt.

Plugging this result into (10.7), we have

d(ρtXt) + b2t ρtXt dt = αtρt dt.

242 10. STOCHASTIC DIFFERENTIAL EQUATIONS

Using integrating factor again, let Gt = exp

(∫ t

0

b2u du

), we get

d(GtρtXt) = Gt d(ρtXt) + ρtXt dGt = αtρtGt dt.

Set

F (t) = ρtGt = exp

(−

∫ t

0

bu dWu +1

2

∫ t

0

b2u du

).

Thus,

d(FtXt) = αtFt dt,

i.e.,

FtXt − F0X0 =

∫ t

0

αuFu du.

Hence, its solution is given by

Xt = F−1t X0 + F−1

t

∫ t

0

αuFu du

= X0 exp

(∫ t

0

bu dWu − 1

2

∫ t

0

b2u du

)+

∫ t

0

αu exp

(∫ t

u

bv dWv − 1

2

∫ t

u

b2v dv

)du.

Example 10.7. Solving the stochastic differential equation

LQ′′t + RQ′

t +1

CQt = Gt + αWt, (10.8)

where Wt is the white noise. Introduce the vector

Xt =

⎛⎜⎝ X

(1)t

X(2)t

⎞⎟⎠ =

⎛⎜⎝ Qt

Q′t

⎞⎟⎠ ,

then ⎧⎪⎨⎪⎩

(X(1)t )′ = X

(2)t ,

L(X(2)t )′ + RX

(2)t +

1

CX

(1)t = Gt + αWt.

Thus, we may rewrite (10.8) as

dXt = AXt dt + Ht dt + K dWt,

10.2. AN EXISTENCE AND UNIQUENESS RESULT 243

where (Wt) is a 1-dimensional Brownian motion,

Xt =

⎛⎜⎝ X

(1)t

X(2)t

⎞⎟⎠

A =

⎛⎜⎝ 0 1

− 1

CL−R

L

⎞⎟⎠

Ht =

⎛⎜⎝ 0

Gt

L

⎞⎟⎠

K =

⎛⎜⎝ 0

α

L

⎞⎟⎠ .

Thus,

d exp [exp(−At)Xt] = exp(−At)(Ht dt + K dWt).

The solution is of the form

Xt = exp(At)X0 + exp(At)

∫ t

0

exp(−As)(Hs ds + K dWs).

������ coordinate ��.

10.2. An existence and uniqueness result

Theorem 10.8 (Existence and uniqueness theorem for stochastic differential equa-

tion). Let T > 0 and let b, σ (b : [0, T ]×Rn → Rn, σ : [0, T ]×Rn → Rn×m) be measurable

functions satisfying

|b(t, x)| + |σ(t, x)| ≤ C(1 + |x|) for all x ∈ Rn, t ∈ [0, T ] (10.9)

244 10. STOCHASTIC DIFFERENTIAL EQUATIONS

for some constant C, where |σ|2 =n∑

i=1

m∑j=1

|σij|2, and

|b(t, x) − b(t, y)| + |σ(t, x) − σ(t, y)| ≤ D|x − y| for all x, y ∈ Rn, t ∈ [0, T ] (10.10)

for some constant D. Let Z be a random variable which is independent of FWT , the σ-

algebra generated by (Ws : 0 ≤ s ≤ T ), and E|Z|2 < ∞. Then the stochastic differential

equation

dXt = b(t,Xt) dt + σ(t,Xt) dWt, 0 ≤ t ≤ T, X0 = Z

has a unique t-continuous solution Xt with the properties that

(i) Xt is adapted to FWt ∨ σ(Z);

(ii) E[∫ T

0

|Xt|2 dt

]< ∞.

Remark 10.9. The conditions (10.9) and (10.10) are natural in view of the following

two simple examples from deterministic differential equations.

(a) Consider

dXt

dt= X2

t , X0 = 1, t ∈ [0,∞)

corresponding to b(t, x) = x2 (not satisfying (10.9)). This differential equation

has the (unique) solution

Xt =1

1 − t, 0 ≤ t < 1.

But no global solution in this case. More generally, (10.9) ensures that the

solution (Xt) does not explode, i.e., |Xt| does not tend to ∞ in a finite time.

(b) Consider the differential equation

dXt

dt= 3X

23t , X0 = 0 (10.11)

10.3. WEAK AND STRONG SOLUTIONS 245

which has more than one solution. For any a > 0, the function

Xt =

⎧⎪⎨⎪⎩

0 for t ≤ a

(t − a)3 for t > a

is the solution of (10.11).

In this case, b(x) = 3x23 does not satisfy the condition (10.10) at x = 0.

10.3. Weak and strong solutions

Definition 10.10. (1) A strong solution of the stochastic differential equation

dXt = b(t,Xt) dt + σ(t,Xt) dWt (10.12)

on a given probability space (Ω,F , P) and with respect to the fixed Brownian

motion motion W and initial value Z, is a stochastic process X with continuous

sample paths and with the following properties:

(i) X is adapted to the filtration (Ft);

(ii) P[X0 = Z] = 1;

(iii) P[∫ t

0

(|bi(s, xs)| + σ2ij(s,Xs)) ds < ∞

]= 1 for all 1 ≤ i ≤ d, 1 ≤ j ≤ r, and

0 ≤ t < ∞;

(iv) the integral version of (10.12):

Xt = X0 +

∫ t

0

b(s,Xs) ds +

∫ t

0

σ(s,Xs) dWs, 0 ≤ t < ∞.

(2) A weak solution of (10.12) is a triple (X, W ), (Ω,F , P), (Ft), where

(i) (Ω,F , P) is a probability space, and (Ft) is a filtration of sub-σ-algebra of

F satisfying the usual conditions.

246 10. STOCHASTIC DIFFERENTIAL EQUATIONS

(ii) X = (Xt,Ft)0≤t<∞ is a continuous, adapted Rn-valued process. W =

(Wt,Ft)0≤t<∞ is a standard Brownian motion, and (iii),(iv) of (1) are satis-

fied.

����, ��������� strong solution ����� weak solution.

Remark 10.11. There are stochastic differential equations which has no strong solu-

tion, but still has a weak solution.

Example 10.12 (Tanaka equation). Consider the stochastic differential equation

dXt = sign(Xt) dWt (10.13)

where

sign(x) =

⎧⎪⎨⎪⎩

1, if x ≥ 0

−1, if x < 0.

Note that σ(t, x) = sign(x) does not satisfy the Lipschitz condition (10.10). This implies

that we cannot apply Theorem 10.8.

(1) Claim that (10.13) has no strong solution.

Suppose X is a strong solution of (10.13). Then X is a Brownian motion by Levy

Theorem. Moreover,

dWt = sign(Xt) dXt.

This implies that the stochastic process (Wt) given by

Wt =

∫ t

0

sign(Xu) dXu

is a Brownian motion with respect to (FXt ). By the Tanaka equation ,

Wt = |Xt| − 2Lxt ,

10.4. FEYNMAN-KAC FORMULA 247

where Lxt is the local time of (Xt). This means that FW

t � FXt . This contradicts

to X is a strong solution.

(2) Find a weak solution of (10.13)

Choose (Xt) to be a Brownian motion (Bt). Then we can define (Bt) by

Bt :=

∫ t

0

sign(Bu) dBu =

∫ t

0

sign(Xu) dXu,

i.e., dBt = sign(Xt) dXt. Thus,

dXt = sign(Xt) dBt.

Hence, (Xt) is a weak solution.

10.4. Feynman-Kac formula

����������������.

Theorem 10.13 (Feynman-Kac formula). Consider the stochastic differential equation

dXt = β(t,Xt) dt + γ(t,Xt) dWt.

Let f be a Borel-measurable function. Fix T > 0 and let t ∈ [0, T ] be given. Define the

function

g(t, x) = Et,x[f(XT )] = E[f(XT )|Xt = x].

Assume that g(t, x) < ∞ for all (t, x). Then g(t, x) satisfies the partial differential equa-

tion

gt(t, x) + β(t, x)gx(t, x) +1

2γ2(t, x)gxx(t, x) = 0

with the terminal condition g(T, x) = f(x) for all x.

248 10. STOCHASTIC DIFFERENTIAL EQUATIONS

Remark 10.14. (g(t, Xt))0≤t≤T is a martingale.

Theorem 10.15 (Discounted Feynman-Kac formula). Consider the stochastic differ-

ential equation

dSt = rβ(t, St) dt + σγ(t, St) dWt

Let f be a Borel-measurable function and let r be constant. Fix T > 0 and let t ∈ [0, T ]

be given. Define the function

h(t, x) = Et,x[e−r(T−t)f(XT )] = E[e−r(T−t)f(XT )|Xt = x]

Assume that h(t, x) < ∞ for all (t, x). Then h(t, x) satisfies the partial differential equa-

tion

ht(t, x) + β(t, x)hx(t, x) +1

2γ2(t, x)hxx(t, x) = rh(t, x)

with the terminal condition h(T, x) = f(x) for all x.

� stochastic analysis ��� Feynmann-Kac Theorem �������. ����

����, ��������� finance �����.

Example 10.16. Suppose that (St) satisfies the stochastic differential equation

dSt = rSt dt + σSt dWt.

Let

f(t, x) = Et,x[h(St)].

Then f(t, x) satisfies the partial differential equation

ft(t, x) + rxfx(t, x) +1

2σ2x2fxx(t, x) = 0.

with terminal condition h(T, x) = f(x). This equation can be solved numerically.

10.4. FEYNMAN-KAC FORMULA 249

Theorem 10.17 (2-dimensional Feynman-Kac formula). Let Wt = (W 1t ,W 2

t ) be a

2-dimensional Brownian motion. Consider two stochastic differntial equations

dXt = β1(t,Xt, Yt) dt + γ11(t,Xt, Yt) dW 1t + γ12(t,Xt, Yt) dW 2

t ,

dYt = β2(t,Xt, Yt) dt + γ21(t,Xt, Yt) dW)t + γ22(t,Xt, Yt) dW 2

t .

Let f(x, y) be a Borel measurable function, define

g(t, x, y) = Et,x,y[f(XT , YT )],

h(t, x, y) = Et,x,y[e−r(T−t)f(XT , YT )].

Then g and h satisfy the partial differential equations

gt + β1gx + β2gy +1

2(γ2

11 + γ212)gxx + (γ11γ21 + γ12γ22)gxy +

1

2(γ2

21 + γ222)gyy = 0,

ht + β1hx + β2hy +1

2(γ2

11 + γ212)hxx + (γ11γ21 + γ12γ22)hxy +

1

2(γ2

21 + γ222)hyy = rh

with terminal conditions

g(T, x, y) = h(T, x, y) = f(x, y), for all x, y.