Existence and Uniqueness Theorem /Ito Formula

Existence and Uniqueness Theorem /Ito Formula

Stat 761 Winter 2015
Stochastic Processes
Instructor: A. Swishchuk
Lecture 31: Ito Formula
Outline
⇒ Ito Formula for One BM
⇒ General Ito Formula
⇒ Examples
1
1.1
o Formula
Itˆ
Itˆ
o Formula for one Brownian Motion
If f (x) and B(t) are differentiable functions, then the ordinary chain rule
would give
df (B(t)) = f 0 (B(t))B 0 (t)dt = f 0 (B(t))dB(t).
However, B(t) is not differentiable, and in particular has nonzero quadratic
variation, so the correct formula has an extra term, namely,
df (B(t)) = f 0 (B(t))B 0 (t)dt + 12 f 00 (B(t))dB(t)dB(t)
= f 0 (B(t))B 0 (t)dt + 21 f 00 (B(t))dt.
(since dB(t)dB(t) = dt.) This is Itˆ
o formula in differential form. Integrating this, we obtain Itˆ
o formula in integral form:
Z
Z t
1 t 00
0
f (B(u))du.
f (B(t)) = f (B(0)) +
f (B(u))dB(u) +
2 0
0
Itˆ
o formula in integral form is the mathematically meaningful form, because
we have solid definitions for both integrals apearing on the right-hand side:
the first, is an Itˆ
o integral and the second is a Riemann integral, the type
used in freshman calculus.
1.2
Derivation of Itˆ
o Formula
Consider f (x) = 21 x2 , so that f 0 (x) = x and f 00 (x) = 1. Taylor formula for
numbers xk , xk+1 implies
1
f (xk+1 ) − f (xk ) = (xk+1 − xk )f 0 (xk ) + (xk+1 − xk )2 f 00 (xk ).
2
In this case, Taylor’s formula to second order is exact because f is a quadratic
function.
Fix T > 0 and let Π = {t0 , t1 , ..., tn } be a partition of [0, T ]. Using Taylor’s
formula, we have:
f (B(T )) −
=
=
=
f (B(0)) = 21 B 2 (T ) − 21 B 2 (0)
Pn−1
[f (B(tk+1 )) − f (B(tk ))]
Pn−1
Pk=0
n−1
[B(tk+1 ) − B(tk )]2 f 00 (B(tk ))
[B(tk+1 ) − B(tk )]f 0 (B(tk )) + 21 k=0
k=0
Pn−1
P
n−1
1
2
k=0 [B(tk+1 ) − B(tk )]B(tk ) + 2
k=0 [B(tk+1 ) − B(tk )] .
We let ||Π|| → 0 to obtain
RT
f (B(T )) − f (B(0)) = 0 B(u)dB(u) + 12 < B > (T )
RT
RT
= 0 f 0 (B(u))dB(u) + 21 0 f 00 (B(u))du.
This is Itˆ
o formula in integral form for the special case
1
f (x) = x2 .
2
1.3
General Itˆ
o Formula
If function f (t, s) of two variables t and s has the following derivatives
0
ft0 (t, s), fs0 (t, s) and fss
(t, s), then the general Itˆ
o formula has the following look:
1
00
(t, s)]dt + σ(t, s)fs0 (t, s)dB(t),
df (t, s) = [ft0 (t, s) + a(t, s)fs0 (t, s) + σ 2 (t, s)fss
2
or
1
00
df (t, s) = ft0 (t, s) + fs0 (t, s)dS(t) + σ 2 (t, s)fss
(t, s)]dt,
2
where
dS(t) = a(t, S(t))dt + σ(t, S(t))dB(t),
1.4
S(0) = s.
Examples
Example 1: Geometric Brownian Motion (GBM) GBM is
1
2
1
2
S(t) = S(0)eσB(t)+(µ− 2 σ )t ,
where µ and σ > 0 are constant. Define
f (t, x) = S(0)eσx+(µ− 2 σ )t ,
so S(t) = f (t, B(t)). Then
1
ft = (µ − σ 2 )f,
2
fx = σf,
fxx = σ 2 f.
According to Itˆ
o formula,
dS(t) = df (t, B(t)) = ft dt + fx dB + 12 fxx dBdB
= (µ − 12 σ 2 )f dt + σf dB + 21 σ 2 f dt
= µS(t)dt + σS(t)dB(t).
Thus, GBM in differential form is
dS(t) = µS(t)dt + σS(t)dB(t),
and GBM in integral form is
Z
t
S(t) = S(0) +
Z
µS(u)du +
t
σS(u)dB(u)
0
0
a2
Example 2: Stochastic Exponent. Function X(t) := eaB(t)− 2 t satisfies the following equation:
dX(t) = aX(t)dB(t).
Proof. From Itˆ
o formula we get:
dX(t) = [−
a2
a2
X(t) + X(t)]dt + aX(t)dB(t) = aX(t)dB(t).
2
2
Example 3: E[B 4 (t)] = 3t2 . Indeed, by Itˆ
o formula for Zt := B 4 (t), we
have
dZt = 4B 3 (t)dB(t) + 6B 2 (t)dt,
or, in integral form,
Z
Zt − Z0 =
t
Z
3
4B (s)dB(s) + 6
t
B 2 (s)ds,
0
0
and, after taking E, we get E[B 4 (t)] = 3t2 .
Recommended Textbook: These Lecture Notes
Recommended Exercises:
1. Use Ito formula to prove that
Z t
Z t
1 3
2
Bs dBs = Bt −
Bs ds.
3
0
0
2. Solve the following SDE:
dXt = (m − Xt )dt + σdBt .
3. Use Ito formula to prove that
1
bk (t) = k(k − 1)
2
where bk (t) = E[Btk ], B0 = 0.
Z
t
bk−2 (s)ds,
0
Z0 = 0,