Theoretical Tutorial Session 2

Xiaoming Song

Department of MathematicsDrexel University

July 27, 2016

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Outline

Itô’s formula

Martingale representation theorem

Stochastic differential equations

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Itô’s formula and martingale representation theorem

1. Using Itô’s formula to show that Mt = B3t − 3

∫ t0 Bsds is a

martingale.

Proof: Let f (x) = x3 ∈ C2(R). Then

f ′(x) = 3x2, and f ′′(x) = 6x .

Applying Itô’s formula to f (Bt ) we have

f (Bt ) = f (B0) +

∫ t

0f ′(Bs)dBs +

12

∫ t

0f ′′(Bs)ds,

that is,

B3t = 3

∫ t

0B2

s dBs + 3∫ t

0Bsds. (1)

Then

Mt = B3t − 3

∫ t

0Bsds = 3

∫ t

0B2

s dBs.

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Since

E(∫ t

0(B2

s )2ds)

= E(∫ t

0B4

s ds)

=

∫ t

03s2ds <∞

for all t ≥ 0, by the basic property of indefinite Itô integral, wecan show that

Mt = B3t − 3

∫ t

0Bsds = 3

∫ t

0B2

s dBs

is a martingale.

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2. Use Itô’s formula to show that

tBt =

∫ t

0Bsds +

∫ t

0sdBs. (2)

Proof: Let f (t , x) = tx . Then f ∈ C1,2(R+ × R) and

∂f∂t

(t , x) = x ,

∂f∂x

(t , x) = t ,

∂2f∂x2 (t , x) = 0.

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Applying Itô’s formula

f (t ,Bt ) = f (0,B0) +

∫ t

0

∂f∂t

(s,Bs)ds +

∫ t

0

∂f∂x

(s,Bs)dBs

+12

∫ t

0

∂2f∂x2 (s,Bs)ds,

we obtain

tBt =

∫ t

0Bsds +

∫ t

0sdBs.

Note that the above equation give us an integration by partsformula.

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3. Check if the process Xt = B3t − 3tBt is a martingale.

Solution: Using (1) and (2), we can write

Xt = B3t − 3tBt

= 3∫ t

0B2

s dBs + 3∫ t

0Bsds − 3

(∫ t

0Bsds +

∫ t

0sdBs

)=

∫ t

0(3B2

s − 3s)dBs.

We can also show that

E(∫ t

0(3B2

s − 3s)2ds)<∞, ∀t ≥ 0.

Therefore, the process Xt = B3t − 3tBt is a martingale.

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4. Find the stochastic integral representation on the timeinterval [0,T ] of the square integrable random variable B3

T .

Solution: Using (1) and (2) with t = T , we have

B3T = 3

∫ T

0B2

s dBs + 3∫ T

0Bsds

= 3∫ T

0B2

s dBs + 3

(TBT −

∫ T

0sdBs

)

= 3∫ T

0B2

s dBs + 3

(T∫ T

01dBs −

∫ T

0sdBs

)

=

∫ T

0(3B2

s + 3T − 3s)dBs.

The above is the integral representation for B3T since the

process 2Bs + 3T − 3s, s ∈ [0,T ] in in L2(P) and E(B3T ) = 0.

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5. Verify that the following processes are martingales:(a) Xt = t2Bt − 2

∫ t0 sBsds

(b) Xt = et/2 cos Bt

(c) Xt = et/2 sin Bt

(d) Xt = B1(t)B2(t), where B1 and B2 are two independentBrownian motion.

Solution 5(a): Let f (t , x) = t2x . Then f ∈ C1,2(R+ × R) and

∂f∂t

(t , x) = 2tx ,

∂f∂x

(t , x) = t2,

∂2f∂x2 (t , x) = 0.

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Applying Itô’s formula

f (t ,Bt ) = f (0,B0) +

∫ t

0

∂f∂t

(s,Bs)ds +

∫ t

0

∂f∂x

(s,Bs)dBs

+12

∫ t

0

∂2f∂x2 (s,Bs)ds,

we get

t2Bt =

∫ t

02sBsds +

∫ t

0s2dBs.

Hence, the process

Xt = t2Bt − 2∫ t

0sBsds =

∫ t

0s2dBs

is a martingale.

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Solution 5(b): Let f (t , x) = et/2 cos x . Then f ∈ C1,2(R+ × R)and

∂f∂t

(t , x) =12

f (t , x),

∂f∂x

(t , x) = − et/2 sin x ,

∂2f∂x2 (t , x) = − f (t , x).

Note that

∂f∂t

(t , x) +12∂2f∂x2 (t , x) = 0, and f (0,B0) = 1.

Then we apply Itô’s formula and show that the process

Xt = et/2 cos Bt = 1−∫ t

0es/2 sin BsdBs

is a martingale since E(∫ t

0 es sin Bs2ds) ≤

∫ t0 esds <∞ for all

t ≥ 0.11 / 36

Solution 5(c): Let f (t , x) = et/2 sin x . Then f ∈ C1,2(R+ × R)and

∂f∂t

(t , x) =12

f (t , x),

∂f∂x

(t , x) = et/2 cos x ,

∂2f∂x2 (t , x) = − f (t , x).

Note that

∂f∂t

(t , x) +12∂2f∂x2 (t , x) = 0, and f (0,B0) = 0.

Then we apply Itô’s formula and show that the process

Xt = et/2 sin Bt =

∫ t

0es/2 cos BsdBs (3)

is a martingale since E(∫ t

0 es cos Bs2ds) ≤

∫ t0 esds <∞ for all

t ≥ 0.12 / 36

Solution 5(d): For this exercise, we need to apply Itô’s formulain multidimensional case. Let f (x1, x2) = x1x2. Then f ∈ C2(R2)and

∂f∂x1

(x1, x2) = x2

∂f∂x2

(x1, x2) = x1

∂2f∂x2

1(x1, x2) = 0

∂2f∂x2

2(x1, x2) = 0

∂2f∂x1∂x2

(x1, x2) = 1

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Applying multidimensional Itô’s formula, one can obtain

f (B1(t)B2(t)) = f (B1(0)B2(0)) +

∫ t

0

∂f∂x1

(B1(s),B2(s))dB1(s)

+

∫ t

0

∂f∂x2

(B1(s),B2(s))dB2(s)

+12

∫ t

0

∂2f∂x2

1(B1(s),B2(s))ds

+12

∫ t

0

∂2f∂x2

2(B1(s),B2(s))ds

+

∫ t

0

∂2f∂x1∂x2

(B1(s),B2(s))dB1(s)dB2(s),

then noticing that dB1dB2 = 0, we can show that the process

Xt = B1(t)B2(t) =

∫ t

0B2(s)dB1(s) +

∫ t

0B1(s)dB2(s)

is a martingale, sinceE(∫ t

0 B1(s)2ds)

= E(∫ t

0 B2(s)2ds)

=∫ t

0 sds <∞, ∀t ≥ 0.14 / 36

6. If f (t , x) = eax− a22 t and Yt = f (t ,Bt ) = eaBt− a2

2 t where a is aconstant, then prove that Y satisfies the following linear SDE:

Yt = 1 + a∫ t

0YsdBs. (4)

Proof: Note that f (t , x) ∈ C1,2(R+ × R) and

∂f∂t

(t , x) = − a2

2f (t , x),

∂f∂x

(t , x) = af (t , x),

∂2f∂x2 (t , x) = a2f (t , x).

Note also that∂f∂t

(t , x) +12∂2f∂x2 (t , x) = 0.

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Applying Itô’s formula, we have

f (t ,Bt ) = f (0,B0) +

∫ t

0

∂f∂t

(s,Bs)ds +

∫ t

0

∂f∂x

(s,Bs)dBs

+12

∫ t

0

∂2f∂x2 (s,Bs)ds

= 1 +

∫ t

0

∂f∂x

(s,Bs)dBs,

that is

Yt = 1 + a∫ t

0YsdBs.

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Remark: i). Note that

E(∫ t

0|Ys|2ds

)= E

(∫ t

0e2aBs−a2sds

)=

∫ t

0ea2sE

(e2aBs− (2a)2

2 s)

ds

=

∫ t

0ea2sds <∞,

for all t ≥ 0. Hence, the Itô integral∫ t

0 YsdBs is well-defined.

ii). The solution of the stochastic differential equation

dYt = aYtdBt , Y0 = 1

is not Yt = eaBt , but Yt = eaBt− a22 t .

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7. Find the stochastic integral representation on the timeinterval [0,T ] of the following square integrable randomvariables:(a) F = BT

(b) F = B2T

(c) F = eBT

(d) F = sin BT

(e) F =∫ T

0 Btdt

(f) F =∫ T

0 tB2t dt

Solution 7(a): Since E(BT ) = 0, the stochastic integralrepresentation for BT is

BT =

∫ T

01dBt = E(BT ) +

∫ T

01dBt .

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Solution 7(b): Let f (x) = x2. Then f ∈ C2(R) and f ′(x) = 2xand f ′′(x) = 2. Using Itô’s formula and E(B2

T ) = T , we have

B2T = B2

0 +

∫ T

02BsdBs +

12

∫ T

02dt

=

∫ T

02BsdBs + T

= E(B2T ) +

∫ T

02BsdBs.

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Solution 7(c): We can calculate that

E(eBT ) = eT2

In fact, we can NOT apply Itô’s formula directly to get thestochastic integral representation, since if we choose f (x) = ex

and apply Itô’s formula to f (BT ), then we get

eBT = eB0 +

∫ T

0eBt dBt +

12

∫ T

0eBt dt

= 1 +

∫ T

0eBt dBt +

12

∫ T

0eBt dt .

We can not get rid of the integral with respect to dt .

Question: How can we get its stochastic integralrepresentation?

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In order to obtain the stochastic integral representation for eBT ,we will need the result (4) in Exercise 6 with a = 1 and t = T :

eBT− T2 = 1 +

∫ T

0eBt− t

2 dBt .

Multiplying eT2 on both sides of the above equation, we obtain

the following stochastic integral representation

eBT = eT2 + e

T2

∫ T

0eBt− t

2 dBt

= E(eBT ) +

∫ T

0eBt+

T−t2 dBt .

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Solution 7(d): Note that

E(sin BT ) = 0.

Since sin x and ex are closely related in

eix = cos x + i sin x ,

we can foresee the same problem if we apply Itô’s formuladirectly to f (BT ) = sin BT .

Instead, we make use of (3) and obtain

sin BT = e−T2

∫ T

0e

t2 cos BtdBt

= E(sin BT ) +

∫ T

0e−

T−t2 cos BtdBt .

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Solution 7(e): We have

E

(∫ T

0Btdt

)= 0.

Using (2) with t = T we get∫ T

0Btdt = TBT −

∫ T

0tdBt

= T∫ T

01dBt −

∫ T

0tdBt

=

∫ T

0(T − t)dBt

= E

(∫ T

0Btdt

)+

∫ T

0(T − t)dBt .

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Solution 7(f): Note that

E

(∫ T

0tB2

t dt

)=

∫ T

0tE(B2

t )dt =

∫ T

0t2dt =

T 3

3.

From Part 7(b), we know

B2t = 2

∫ t

0BsdBs + t , ∀t ≥ 0.

Using the above equation and then changing the order of theintegrals we have∫ T

0tB2

t dt =

∫ T

0t

(2∫ t

0BsdBs + t

)dt

=

∫ T

0t2dt + 2

∫ T

0

∫ t

0tBsdBsdt

=T 3

3+ 2

∫ T

0Bs

(∫ T

stdt

)dBs

=E

(∫ T

0tB2

t dt

)+

∫ T

0(T 2 − s2)BsdBs.

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8. Consider an n-dimensional Brownian motionB(t) = (B1(t),B2(t), · · · ,Bn(t)) and constants αi , i = 1, . . . ,n.Solve the following SDE:

dXt = rXtdt + Xt

n∑i=1

αidBi(t), X0 = x ,

where x ∈ R.

Solution: The coefficients in this SDE satisfy the Lipschitz andlinear growth conditions, so there exists a unique solution.

If α1 = α2 = · · · = αn = 0, then the above SDE becomes anODE

dXt = rXtdt , X0 = x .

and its unique solution is Xt = xert .

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If∑n

i=1 α2i 6= 0, then by using the standard method mentioned

in my first tutorial session we can show that the process

Bt =1√∑ni=1 α

2i

n∑i=1

αiBi(t)

is a Brownian motion.

Let Yt = e−rt . Then Y satisfies dYt = −rYtdt . Applyingmultidimensional Itô’s formula to f (x , y) = xy we have

d(XtYt ) = XtdYt + YtdXt + dXtdYt

= XtdYt + YtdXt

=Xt (−rYt )dt + Yt (rXtdt + Xt

n∑i=1

αidBi (t))

= XtYt

n∑i=1

αidBi (t)

=

√√√√ n∑i=1

α2i (XtYt )dBt . (5)

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Thus, XtYt satisfies the linear SDE (5). Note also thatX0Y0 = x . Then the solution to (5) is

XtYt = x exp

√√√√ n∑

i=1

α2i Bt −

t∑n

i=1 α2i

2

.

Therefore,

Xt = Y−1t x exp

√√√√ n∑

i=1

α2i Bt −

t∑n

i=1 α2i

2

= x exp

rt +

√√√√ n∑i=1

α2i Bt −

t∑n

i=1 α2i

2

= x exp

n∑

i=1

αiBi(t) +

(r −

∑ni=1 α

2i

2

)t

.

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9. Solve the following stochastic differential equations

dXt =1Xt

dt + αXtdBt , X0 = x > 0.

For which values of the parameter α the solution explodes?

Solution: Let Yt = e−αBt−α2t2 . Then Yt satisfies the following

SDE:dYt = −αYtdBt , Y0 = 1.

Using Itô’s formula, we have

d(XtYt ) = XtdYt + YtdXt + dXtdYt

= Xt (−αYtdBt ) + Yt

(1Xt

dt + αXtdBt

)− α2XtYtdt

=Yt

Xtdt − α2XtYtdt ,

which implies28 / 36

2(XtYt )d(XtYt ) = 2Y 2t dt − 2α2(XtYt )

2dt .

Then for each fixed ω, (Xt (ω)Yt (ω))2 solves the following linearODE:

y = 2Y 2t (ω)− 2α2y , y(0) = x2,

whose solution is given by

y(t) = e−2α2t(

x2 + 2∫ t

0e2α2sY 2

s (ω)

)ds.

Then

Xt = Y−1t e−α

2t

√x2 + 2

∫ t

0e2α2sY 2

s ds.

Since the trajectories of the process Y are continuous on [0,∞)almost surely, the above integral is well defined for all t ≥ 0,and hence Xt will not explode for any parameter α.

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10. Solve the following stochastic differential equations

dXt = X γt dt + αXtdBt , X0 = x > 0.

For which values of the parameters γ, α the solution explodes?

Solution: If α = 0, then the differential equation is an ODE

ddt

X = X γ , X0 = x > 0.

This is a separable equation and we know that the solutionexplodes when γ > 1.

If α 6= 0 and γ = 1, then this is a linear SDE and its solution isgiven by

Xt = eαBt+

(1−α2

2

)t, ∀t ≥ 0.

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If α 6= 0 and γ 6= 1, then we will use very similar steps as in

Problem 9 to obtain the solution. Let Yt = e−αBt−α2t2 . Then Yt

satisfies the following SDE:

dYt = −αYtdBt , Y0 = 1,

Using Itô’s formula, we have

d(XtYt ) = XtdYt + YtdXt + dXtdYt

= Xt (−αYtdBt ) + Yt(X γ

t dt + αXtdBt)− α2XtYtdt

=YtXγt dt − α2XtYtdt ,

which implies for each fixed ω ∈ Ω, y(t) = Xt (ω)Yt (ω) satisfiesthe following nonlinear ODE:

y = Yt (ω)1−γyγ − α2y , y(0) = x ,

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or equivalently,

y + α2y = Yt (ω)1−γyγ , y(0) = x .

Multiplying the above equation by eα2t and denoting z = eα

2ty ,we obtain

z =(

Yt (ω)eα2t)1−γ

zγ , z(0) = x .

We can separate the variables to solve this ODE as follows:

zzγ

=(

Yt (ω)eα2t)1−γ

= e−α(1−γ)Bt+α2(1−γ)t

2 , z(0) = x . (6)

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Note thatz(t) = Xt (ω)e−αBt (ω)+

12α

2t

and e−αBt (ω)+12α

2t is continuous in t . Then, Xt (ω) explodes ast ↑ T (ω) if and only if z(t) explodes when t ↑ T (ω).

Suppose that Xt (ω) explodes as t ↑ T (ω) for some T (ω) <∞.Then integrating (6) on both sides, we should get∫ ∞

x

dzzγ

=

∫ T

0e−α(1−γ)Bt+

α2(1−γ)t2 dt <∞, (7)

and hence, explosion might occur only if γ > 1.

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For γ > 1, then ∫ ∞x

dzzγ

=x1−γ

γ − 1.

Thus, if Xt (ω) explodes as t ↑ T (ω) for some T (ω) <∞, thefollowing equation holds∫ T

0e−α(1−γ)Bt+

α2(1−γ)t2 dt =

x1−γ

γ − 1.

Note also that for each t ≥ 0 we have

E(∫ t

0e−α(1−γ)Bs+

α2(1−γ)s2 ds

)=

∫ t

0e

α2(1−γ)2s2 +α2(1−γ)s

2 ds

=

∫ t

0e

α2(1−γ)(2−γ)s2 ds

=

t , if γ = 2,2

α2(1−γ)(2−γ)

(e

α2(1−γ)(2−γ)t2 − 1

), if γ 6= 2.

(8)

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So, for α 6= 0 and γ ≥ 2, we have

limt→∞

E(∫ t

0e−α(1−γ)Bs+

α2(1−γ)s2 ds

)=∞. (9)

Define

τ = inf

t ≥ 0,∫ t

0e−α(1−γ)Bs+

α2(1−γ)s2 ds =

x1−γ

γ − 1

.

That is, z (or equivalently, X ) explodes at τ .

Then τ is a stopping time, and moreover, from (9), we get

P(τ <∞) > 0,

that is, X explodes on τ <∞.

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For α 6= 0 and 1 < γ < 2, we get from (8)

limt→∞

E(∫ t

0e−α(1−γ)Bs+

α2(1−γ)s2 ds

)=

2α2(γ − 1)(2− γ)

.

If α and γ satisfy

2α2(γ − 1)(2− γ)

>x1−γ

γ − 1,

then we also getP(τ <∞) > 0,

that is, X explodes on τ <∞ in this case.

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