3The stochastic integral
III Stochastic Calculus and Applications
3.5 Itˆo formula
We now prove the equivalent of the integration by parts and the chain rule, i.e.
Itˆo’s formula. Compared to the world of usual integrals, the difference is that
the quadratic variation, i.e. “second order terms” will crop up quite a lot, since
they are no longer negligible.
Theorem
(Integration by parts)
.
Let
X, Y
be a continuous semi-martingale.
Then almost surely,
X
t
Y
t
− X
0
Y
0
=
Z
t
0
X
s
dY
s
+
Z
t
0
Y
s
dX
s
+ hX, Y i
t
The last term is called the Itˆo correction.
Note that if
X, Y
are martingales, then the first two terms on the right are
martingales, but the last is not. So we are forced to think about semi-martingales.
Observe that in the case of finite variation integrals, we don’t have the
correction.
Proof. We have
X
t
Y
t
− X
s
Y
s
= X
s
(Y
t
− Y
s
) + (X
t
− X
s
)Y
s
+ (X
t
− X
s
)(Y
t
− Y
s
).
When doing usual calculus, we can drop the last term, because it is second order.
However, the quadratic variation of martingales is in general non-zero, and so
we must keep track of this. We have
X
k2
−n
Y
k2
−n
− X
0
Y
0
=
k
X
i=1
(X
i2
−n
Y
i2
−n
− X
(i−1)2
−n
Y
(i−1)2
−n
)
=
n
X
i=1
X
(i−1)2
−n
(Y
i2
−n
− Y
(i−1)2
−n
)
+ Y
(i−1)2
−n
(X
i2
−n
− X
(i−1)2
−n
)
+ (X
i2
−n
− X
(i−1)
2
−n
)(Y
i2
−n
− Y
(i−1)2
−n
)
Taking the limit
n → ∞
with
k
2
−n
fixed, we see that the formula holds for
t
a
dyadic rational. Then by continuity, it holds for all t.
The really useful formula is the following:
Theorem
(Itˆo’s formula)
.
Let
X
1
, . . . , X
p
be continuous semi-martingales, and
let
f
:
R
p
→ R
be
C
2
. Then, writing
X
= (
X
1
, . . . , X
p
), we have, almost surely,
f(X
t
) = f(X
0
) +
p
X
i=1
Z
t
0
∂f
∂x
i
(X
s
) dX
i
s
+
1
2
p
X
i,j=1
Z
t
0
∂
2
f
∂x
i
∂x
j
(X
s
) dhX
i
, X
j
i
s
.
In particular, f(X) is a semi-martingale.
The proof is long but not hard. We first do it for polynomials by explicit
computation, and then use Weierstrass approximation to extend it to more
general functions.
Proof.
Claim. Itˆo’s formula holds when f is a polynomial.
It clearly does when
f
is a constant! We then proceed by induction. Suppose
Itˆo’s formula holds for some f . Then we apply integration by parts to
g(x) = x
k
f(x).
where x
k
denotes the kth component of x. Then we have
g(X
t
) = g(X
0
) +
Z
t
0
X
k
s
df(X
s
) +
Z
t
0
f(X
s
) dX
k
s
+ hX
k
, f (X)i
t
We now apply Itˆo’s formula for f to write
Z
t
0
X
k
s
df(X
s
) =
p
X
i=1
Z
t
0
X
k
s
∂f
∂x
i
(X
s
) dX
i
s
+
1
2
p
X
i,j=1
Z
t
0
X
k
s
∂
2
f
∂x
i
∂x
j
(X
s
) dhX
i
, X
j
i
s
.
We also have
hX
k
, f (X)i
t
=
p
X
i=1
Z
t
0
∂f
∂x
i
(X
s
) dhX
k
, X
i
i
s
.
So we have
g(X
t
) = g(X
0
) +
p
X
i=1
Z
t
0
∂g
∂x
i
(X
s
) dX
i
s
+
1
2
p
X
i,j=1
Z
t
0
∂
2
g
∂x
i
∂x
j
(X
s
) dhX
i
, X
j
i
s
.
So by induction, Itˆo’s formula holds for all polynomials.
Claim.
Itˆo’s formula holds for all
f ∈ C
2
if
|X
t
(
ω
)
| ≤ n
and
R
t
0
|
d
A
s
| ≤ n
for
all (t, ω).
By the Weierstrass approximation theorem, there are polynomials
p
k
such
that
sup
|x|≤k
|f(x) − p
k
(x)| + max
i
∂f
∂x
i
−
∂p
∂x
i
+ max
i,j
∂
2
f
∂x
i
∂x
j
−
∂p
k
∂x
i
∂x
j
≤
1
k
.
By taking limits, in probability, we have
f(X
t
) − f(X
0
) = lim
k→∞
(p
k
(X
t
) − p
k
(X
0
))
Z
t
0
∂f
∂x
i
(X
s
) dX
i
s
= lim
k→∞
∂p
k
∂x
i
(X
s
) dX
i
s
by stochastic dominated convergence theorem, and by the regular dominated
convergence, we have
Z
t
0
∂f
∂x
i
∂x
j
dhX
i
, X
j
i
s
= lim
k→∞
Z
t
0
∂
2
p
k
∂x
i
∂x
j
dhX
i
, X
j
i.
Claim. Itˆo’s formula holds for all X.
Let
T
n
= inf
t ≥ 0 : |X
t
| ≥ n or
Z
t
0
|dA
s
| ≥ n
Then by the previous claim, we have
f(X
T
n
t
) = f(X
0
) +
p
X
i=1
Z
t
0
∂f
∂x
i
(X
T
n
s
) d(X
i
)
T
n
s
+
1
2
X
i,j
Z
t
0
∂
2
f
∂x
i
∂x
j
(X
T
n
s
) dh(X
i
)
T
n
, (X
j
)
T
n
i
s
= f(X
0
) +
p
X
i=1
Z
t∧T
n
0
∂f
∂x
i
(X
s
) d(X
i
)
s
+
1
2
X
i,j
Z
t∧T
n
0
∂
2
f
∂x
i
∂x
j
(X
s
) dh(X
i
), (X
j
)i
s
.
Then take T
n
→ ∞.
Example.
Let
B
be a standard Brownian motion,
B
0
= 0 and
f
(
x
) =
x
2
. Then
B
2
t
= 2
Z
t
0
B
S
dB
s
+ t.
In other words,
B
2
t
− t = 2
Z
t
0
B
s
dB
s
.
In particular, this is a continuous local martingale.
Example.
Let
B
= (
B
1
, . . . , B
d
) be a
d
-dimensional Brownian motion. Then
we apply Itˆo’s formula to the semi-martingale
X
= (
t, B
1
, . . . , B
d
). Then we
find that
f(t, B
t
) − f(0, B
0
) −
Z
t
0
∂
∂s
+
1
2
∆
f(s, B
s
) ds =
d
X
i=1
Z
t
0
∂
∂x
i
f(s, B
s
) dB
i
s
is a continuous local martingale.
There are some syntactic tricks that make stochastic integrals easier to
manipulate, namely by working in differential form. We can state Itˆo’s formula
in differential form
df(X
t
) =
p
X
i=1
∂f
∂x
i
dX
i
+
1
2
p
X
i,j=1
∂
2
f
∂x
i
∂x
j
dhX
i
, X
j
i,
which we can think of as the chain rule. For example, in the case case of Brownian
motion, we have
df(B
t
) = f
0
(B
t
) dB
t
+
1
2
f
00
(B
t
) dt.
Formally, one expands
f
using that that “(d
t
)
2
= 0” but “(d
B
)
2
= d
t
”. The
following formal rules hold:
Z
t
− Z
0
=
Z
t
0
H
s
dX
s
⇐⇒ dZ
t
= H
t
dX
t
Z
t
= hX, Y i
t
=
Z
t
0
dhX, Y i
t
⇐⇒ dZ
t
= dX
t
dY
t
.
Then we have rules such as
H
t
(K
t
dX
t
) = (H
t
K
t
) dX
t
H
t
(dX
t
dY
t
) = (H
t
dX
t
) dY
t
d(X
t
Y
t
) = X
t
dY
t
+ Y
t
dX
t
+ dX
t
dY
t
df(X
t
) = f
0
(X
t
) dX
t
+
1
2
f
00
(X
t
) dX
t
dX
t
.