Partial Differential Equations : Wiener Process and Ito's Lemma

Consider a random variable r satisfying the stochastic differential
equation:



are positive constants and dX is a Wiener process. Let



for r into (-1,1) for ξ. Suppose the stochastic equation for the new
random variable ξ is in the form:

d ξ = a(ξ)dt + b(ξ)dX.

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a(ξ) and b(ξ) fulfill the conditions:



{a(-1) = 0, and {a(1) = 0,

{b(-1) = 0, and {b(1) = 0.

I am confused because I don’t know if you can use Ito’s lemma to
find the concrete expressions for a(ξ) and b(ξ). Also when taking the
partial derivative of ξ with respect to r, I do not know how to treat
the absolute value sign. Thanks.

Ito's Lemma

Recall that a Wiener process dW is normalized to have expectation E(dW 2 ) = dt.

(1)

A specialized form of Ito's lemma can be found for instance on web page

http://en.wikipedia.org/wiki/It%C5%8D's lemma

If we take the first two equations from this web page and change variables to those of the
assignment,
dS = a(S, t)dt + b(S, t)dX, (1.1)
and collect together the terms containing f /S, we obtain exactly the requested state-
ment
f f 1 2 2f
df = dS + + b dt. (1.2)
S t 2 S 2
An informal proof of equation (1.2) is further given on that web page. The key element of
the informal proof, in our notations here, is the replacement

dX 2 E(dX 2 ) = dt, (1.3)

in the limit dt 0, whose proof is too involved to be explained in a basic text.

The rest of the informal proof there consists of simple algebraic book-keeping. The same
book-keeping is done in our notations here in the next question.

(2)

If
dS = µSdt + SdX, (2.1)
µ 0, > 0, dX is a Wiener process, Pm > 0, and

S Pm
= =1- , (2.2)
S + Pm S + Pm

we use Taylor expansion and obtain

Pm Pm
d = 2
dS - dS 2 + O(dS 3 ) =
(S + Pm ) (S + Pm )3

(1 - )2 (1 - )3 2
= dS - 2
dS + O(dS 3 ). (2.3)
Pm Pm

1
In the same way as it is done for question (1) in the cited web page, we expand

dS 2 = µ2 S 2 dt2 + 2µS 2 dtdX + 2 S 2 dX 2 , (2.4)

make replacement (1.3) and note that the remaining terms in equation (2.2) are of higher
order than dt and so can be dropped, so that

dS 2 = 2 S 2 dt + o(dt). (2.5)

Using equations (2.1) and (2.5) and retaining only the expansion terms up to the order of
dt in equation (2.3), we obtain

(1 - )2 (1 - )3 2 2
d = (µSdt + SdX) - 2
S dt. (2.6)
Pm Pm

Noting from equation (2.2) that
Pm
S= (2.7)
1-
and substituting equation (2.7) into equation (2.6), we obtain

d = (1 - )(µdt + dX) - 2 (1 - ) 2 dt =

= (1 - )(µ - 2 )dt + (1 - )dX
= a()dt + b()dX, (2.8)
where
a() = (1 - )(µ - 2 ),
(2.9)
b() = (1 - ).
From equation (2.9) we see that a() and b() do indeed have the properties

a(0) = a(1) = b(0) = b(1) = 0. (2.10)




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