LINKÖPING UNIVERSITY
Department of Mathematics
Mathematical Statistics
John Karlsson
TAMS29
Stochastic Processes with
Applications in Finance
7. Itô formula, stochastic differential equations
7.1
Let s ≤ t. We recall the fact that
Z t
f (s) dWs = 0
(1)
E
0
Rt
f (s)2 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
=
0
h(x)dWx
s
t
Z
h(x) dWx and
h(x) dWx are independent
=
s
"Z
2 #
s
=E
h(x) dWx
0
s
Z
+E
|
0
=
t
h(x) dWx ·
+E
Using the Itô isometry
Z
h(x) dWx · E
{z
} |
=0 by (1)
Z s
=E
s
t
h(x)dWx
{z
}
=0 by (1)
Z
2
h(x) 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 Itô isometry we get
"Z
2 #
Z t
Z t
t
Itô iso.
2
E
W (s) dWs
= E
W (s) ds =
E W (s)2 ds
0
0
Z
t
=
s ds =
0
0
2
t
.
2
b
Itô’s formula gives
d(Ws2 ) = ds + 2Ws dWs
1/5
and it follows
Z t
Z t
Z t
Z t
Ws dWs ⇐⇒
Ws dWs ⇐⇒ W (t)2 = t + 2
dW 2 =
ds +2
0
0
0
0
| {z } | {z }
=W (t)2 −W (0)2
=t
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
Itô’s formula gives
d(eWt ) =
eWt
dt + eWt dWt .
2
b
We have f (w, t) := cos(w) and thus
∂f
= 0,
∂t
Itô’s formula gives
∂2f
= − cos(w).
∂w2
∂f
= − sin(w),
∂w
d(cos(Wt )) = −
cos(Wt )
dt − sin(Wt )dWt .
2
7.4a
We study Yt := e−at Xt , Y0 = x0 . By Itô’s formula 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
= σe−at dWt .
It follows
Z
t
Yt − Y0 = σ
e−as dWs .
0
Using Xt = eat Yt yields
Xt = x0 eat + σeat
Z
t
e−as dWs ,
0
where X(0) = 0 gives x0 = 0.
b
2/5
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 ) =
CeAWt +Bt . From Itô’s formula it follows that
1 2 AWt +Bt
AWt +Bt
dt + CA · eAWt +Bt dWt
dX(t, Wt ) = C · B · e
+ A e
2
A2
= B · X(t, Wt ) +
X(t, Wt ) dt + A · X(t, Wt )dWt .
2
We see that (B + A2 /2) = a and A = σ i.e. we have
B =a−
A = σ,
σ2
A2
=a−
.
2
2
X0 = 1 gives C = 1 and we get
σ2
X(t, Wt ) = exp σWt + a −
t .
2
Alternative: We guess that Xt is some kind of exponential and we study ln Xt . We
have
1
1
d(ln Xt ) =
(dXt )2
dXt −
Xt
2Xt2
2
2 2
= dXt = aXt dt + σXt dWt =⇒ (dXt ) = σ Xt dt =
1
X2
Xt
Xt
= − σ 2 · t2 dt + a dt + σ dWt
2
Xt
Xt
Xt
σ2
= a−
dt + σdWt .
2
We get
Z t
a−
ln Xt =
0
σ2
2
Z
ds +
t
σdWt + C1 =
0
σ2
a−
t + σWt + C1
2
and it follows
σ2
Xt = C2 · exp σWt + a −
t .
2
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
3/5
Proof. Itô’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
σ2
1 0
g (Xt )g(Xt )dt + g(Xt )dWt =
dt + σ Xt dWt .
2
4
The above proposition states that a solution of
p
σ2
dXt =
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
σ 2 x2
Cσx
Integrate p
⇐⇒ 2 f = σx + C =⇒ f (x) =
+
+ C 2.
4
2
Thus
CσWt
σ 2 Wt2
+
+ C 2.
Xt =
4
2
X0 = 0 gives C = 0 and it follows
σ 2 Wt2
.
4
We check the answer by using Itô’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 =
Alternative: Make ansatz Xt = g(Wt ). Comparing Itô’s formula with the SDE we
get
1 00
σ2
g (Wt ) =
,
2
p4
p
g 0 (Wt ) = σ Xt = σ g(Wt ) =⇒
(2)
=⇒ g 0 (Wt )2 = σ 2 g(Wt )
(3)
From (2) we get
g(Wt ) = σ 2 Wt2 /4 + C1 · Wt + C2
and thus
g 0 (Wt )2 = (σ 2 Wt /2 + C1 )2 =
4/5
σ 2 Wt2
+ C1 σ 2 Wt + C12 .
4
Equation (3) now gives
σ 4 Wt2
(3)
+ C1 σ 2 Wt + C12 = σ 2 g(Wt ) .
4
With g(W0 ) = X0 = 0 it follows that C1 = 0 and thus
Xt = g(Wt ) =
5/5
σ 2 Wt2
.
4