Solutions lesson 7

¨
LINKOPING
UNIVERSITY
Department of Mathematics
Mathematical Statistics
VT1-2014
TAMS29
ˆ formula, stochastic differential equations
7. Ito
7.1
Let s ≤ t. We recall the fact that
Z t
f (s) dWs = 0
(1)
E
0
Rt
2
f (s) ds < ∞ i.e. we have E[Xt ] = 0. It follows
Z s
Z t
h(x) dWx ·
h(x)dWx − 0 · 0
CX (s, t) = E[Xs · Xt ] − E[Xs ] · E[Xt ] = E
0
0
Z s
Z s
Z t
=E
h(x) dWx ·
h(x)dWx +
h(x)dWx
0
0
s
"Z
#
Z
Z
for any f such that
0
2
s
=E
s
h(x) dWx
0
0
s
Z
t
h(x) dWx ·
+E
Z
h(x)dWx
s
t
=
h(x) dWx and
h(x) dWx are independent =
0
s
"Z
#
Z t
2
Z s
s
h(x)dWx
=E
h(x) dWx
+E
h(x) dWx · E
0
0
s
{z
} |
{z
}
|
=
Using the Itˆ
o isometry
=0 by (1)
Z s
=E
=0 by (1)
Z
h(x)2 dx + 0 · 0 =
0
s
h(x)2 dx.
0
We get a similar result for t ≤ s and we conclude
Z min(s,t)
CX (s, t) =
h(x)2 dx.
0
7.2a
Using the Itˆ
o isometry we get
"Z
2 #
Z t
Z t
t
Itˆ
o iso.
2
E
W (s) dWs
= E
W (s) ds =
E W (s)2 ds
0
0
Z
=
t
0
2
s ds =
0
t
.
2
b
We note the identity
(2)
2a(b − a) = 2ab − 2a2 = −(a − b)2 − a2 + b2 = b2 − a2 − (b − a)2 .
1/6
Now by the definition of the Itˆo integral we have the following (where the limit is
understood in L2 sense)
Z t
n
X
t(j − 1)
tj
t(j − 1)
W (s) dWs = 2 · lim
W
· W
−W
2
n→∞
n
n
n
0
j=1
2
2 2
n
X
tj
tj
t(j − 1)
t(j − 1)
(2)
= lim
− W
W
−W
−W
n→∞
n
n
n
n
j=1 |
{z
}
2
Telescoping sum
n X
2
= lim W (t) − W (0) −
n→∞
W
j=1
|
tj
n
−W
t(j − 1)
n
2
{z
}
−→ t, in L2 *
n→∞
(3)
= W (t)2 − t.
NOTE: Problem 5.2(c) showed that
2
n X
tj
t(j − 1)
a.s.
W
−W
−→ t,
n
n
j=1
while we need to show the limit in L2 . We do this by the following calculation
where we denote
tj
t(j − 1)
t
∆Wj := W
−W
∼ N 0,
.
n
n
n
We have




2 
2


n
n
n
X
X


 X

E 
∆Wj2  − t  = E 
∆Wj2  − 2t 
∆Wj2  + t2 
j=1
j=1
j=1

2 


n
n
X
X


∆Wj2   − 2tE 
∆Wj2  + E t2
= E 
j=1
j=1


n X
n
n
X
X
=E
∆Wi2 ∆Wj2  − 2t
E ∆Wj2 +t2
| {z }
i=1 j=1
j=1
t
=n
=
n
X
i,j=1
t
E ∆Wi2 ∆Wj2 −2t · n · + t2
n
|
{z
}
2
2
t
3t
=n
2 ,i6=j; = n2 ,i=j
t2
3t2
+
n
·
− 2t2 + t2
n2
n2
t2
3t2
= t2 − +
− 2t2 + t2
n
n
2t2
=
−→ 0.
n n→∞
= n(n − 1) ·
2/6
It follows
"Z
"
2 #
2 #
t
1 W (t)2 − t
(3)
= E W (t)4 − 2tW (t)2 + t2
W (s) dWs
= E
E
2
4
0
=
t 1 3t2
1 t2
t2
t2
E W (t)4 − E W (t)2 + E t2 =
− +
= .
4 | {z } 2 | {z } 4
4
2
4
2
=t
=3t2
Alternative: Itˆ
o’s formula gives us
d(Ws2 ) = ds + 2Ws dWs
and it follows
Z t
Z t
Z t
Z t
ds +2
Ws dWs ⇐⇒ W (t)2 = t + 2
Ws dWs ⇐⇒
dW 2 =
0
0
0
0
| {z } | {z }
=t
=W (t)2 −W (0)2
Z
⇐⇒ 2
t
Ws dWs = W (t)2 − t.
0
7.3a
We have f (w, t) := ew and thus
∂f
∂2f
=
= ew .
∂w
∂w2
∂f
= 0,
∂t
Itˆ
o’s formula gives
d(eWt ) =
eWt
dt + eWt dWt .
2
b
We have f (w, t) := cos(w) and thus
∂f
= 0,
∂t
Itˆ
o’s formula gives
∂f
= − sin(w),
∂w
d(cos(Wt )) = −
∂2f
= − cos(w).
∂w2
cos(Wt )
dt − sin(Wt )dWt .
2
7.4a
We study Yt := e−at Xt , Y0 = x0 . We have
dYt = −ae−at Xt dt + e−at dXt = −ae−at Xt dt + ae−at Xt dt + σe−at dWt
|{z}
aXt dt+σdWt
−at
= σe
dWt .
3/6
It follows
Z
t
e−as dWs .
Yt − Y0 = σ
0
Using Xt = eat Yt yields
Xt = x0 eat + σeat
Z
t
e−as dWs ,
0
where X(0) = 0 gives x0 = 0.
b
If
dXt = aXt dt + σXt dWt , X0 = 0,
then the process Xt will stay at 0 since the increments are proportional to Xt .
We will therefore assume that X0 = 1 instead. We make the ansatz X(t, Wt ) =
CeAw+Bt . It follows
1
dX(t, Wt ) = C · B · eAWt +V t + A2 eAWt +V t dt + CA · eAWt +bt dWt
2
2
A
= B · X(t, Wt ) +
X(t, Wt ) + A · X(t, Wt )dWt .
2
We see that (B + A2 /2) = a and A = σ i.e. we have
B =a−
A = σ,
A2
σ2
=a−
.
2
2
X0 = 1 gives C = 1 and we get
σ2
t .
X(t, Wt ) = exp σWt + a −
2
Alternative: We guess that Xt is some kind of exponential and we study ln Xt . We
have
1
1
d(ln Xt ) =
dXt −
(dXt )2
Xt
2Xt2
= dXt = aXt dt + σXt dWt =⇒ (dXt )2 = σ 2 Xt2 dt =
1
X2
Xt
Xt
= − σ 2 · t2 dt + a dt + σ dWt
2
Xt
Xt
Xt
2
σ
= a−
dt + σdWt .
2
We get
Z t
ln Xt =
0
σ2
a−
2
Z
ds +
0
t
σ2
σdWt + C1 = a −
t + σWt + C1
2
and it follows
σ2
Xt = C2 · exp σWt + a −
t .
2
4/6
We note that X0 = 1 gives C2 = 1.
c
We formulate the following proposition
Proposition 7.1. If g ≡ g(f ) is a function of f and f ≡ f (w) is a solution
of the ordinary differential equation df = g(f (w))dw i.e. f 0 (w) = g(f (w)), then
Xt = f (Wt ) is a solution of the stochastic differential equation
dXt =
1 0
g (Xt )g(Xt )dt + g(Xt )dWt .
2
Proof. Itˆ
o’s lemma gives
f 00 (Wt )
dt + f 0 (Wt )dWt
2
, f 0 (W ) = g(f (W )) = g(X )
,
t
t
t
=
0
f 00 (Wt ) = g f (Wt )
= g 0 (f (Wt )) · f 0 (Wt ) = g 0 (Xt ) · g(Xt )
dXt =
=
1 0
g (Xt )g(Xt )dt + g(Xt )dWt .
2
√
Letting g(x) := σ x gives g 0 (x)g(x) = σ 2 /2. It follows
p
1 0
σ2
g (Xt )g(Xt )dt + g(Xt )dWt =
dt + σ Xt dWt .
2
4
The above proposition states that a solution of
dXt =
p
σ2
dt + σ Xt dWt
4
is given by Xt = f (Wt ) where f is a solution of f 0 (x) = g(f (x)) or equivalently
df = g(f (x))dx. We have
p
Sep. of variables df
√ = σdx
df = g(f (x))dx ⇐⇒ df = σ f (x)dx
⇐⇒
f
2 2
p
σ x
Cσx
Integrate
⇐⇒ 2 f = σx + C =⇒ f (x) =
+
+ C 2.
4
2
Thus
σ 2 Wt2
CσWt
Xt =
+
+ C 2.
4
2
X0 = 0 gives C = 0 and it follows
σ 2 Wt2
.
4
We check the answer by using Itˆo’s formula and get
p
1 2σ 2
2σ 2 Wt
σ2
σWt
σ2
dXt =
dt +
dWt =
dt + σ
dWt =
dt + σ Xt dWt .
2 4
4
4
2
4
Xt =
5/6
Alternative: Make ansatz Xt = g(Wt ). We get
1 00
σ2
g (Wt ) =
,
2
p 4
p
g 0 (t) = σ Xt = σ g(Wt ) =⇒
(4)
=⇒ g 0 (t)2 = σ 2 g(Wt )
(5)
From (4) we get
g(Wt ) = σ 2 Wt2 /4 + C1 · Wt + C2
(6)
and thus
g 0 (Wt )2 = (σ 2 Wt2 /2 + C1 )2 =
σ 4 Wt4
+ C1 σ 2 Wt + C12 .
4
Equation (5) now gives
2
2
σ 4 Wt4
(5)
(6) σ Wt
+ C1 σ 2 Wt + C12 = σ 2 g(Wt ) =
+ C1 · Wt + C2
4
4
from which it follows C1 = C2 = 0 and thus
σ 2 Wt2
.
Xt = g(Wt ) =
4
6/6

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