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Theoretical Tutorial Session 2
Xiaoming Song
Department of MathematicsDrexel University
July 27, 2016
1 / 36
Outline
Itô’s formula
Martingale representation theorem
Stochastic differential equations
2 / 36
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.
4 / 36
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.
5 / 36
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.
6 / 36
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.
7 / 36
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
13 / 36
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 .
17 / 36
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 .
18 / 36
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.
19 / 36
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?
20 / 36
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 .
21 / 36
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 .
22 / 36
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 .
23 / 36
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.
24 / 36
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 .
25 / 36
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)
26 / 36
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
.
27 / 36
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.
29 / 36
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.
30 / 36
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 ,
31 / 36
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)
32 / 36
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.
33 / 36
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)
34 / 36
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 <∞.
35 / 36
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.
36 / 36