[944] Derivatives with W in the integral

[David Itkin]

Hi Lucas,

Remember when applying Ito's Lemma to a process [image: X_t] (in this
case [image:
X_t = \int_0^tW(u)^3-1du]) you define a function [image: f(t,x)] such
that [image:
f(t,W_t) = X_t]. So here we define [image: f(t,x) = \int_0^tW(u)^3-1du]
which has no x dependence so [image: \frac{\partial}{\partial x} f(t,x) = 0].
What may have confused you is that we don't replace [image: W(u)] -- we
would only replace [image: W(t)]'s with x.

Similarly for Assignment 4 define [image: f(t,x) = \int_0^xe^{-ts^2}ds] and
then you can apply regular calculus rules to obtain the derivatives you
need. Remember Brownian Motion is not differentiable so writing [image:
 \frac{\partial}{\partial t} \int_0^{W(t)}e^{-ts^2}ds] does not make sense.

Hope this helps,

David

On Sun, Nov 19, 2017 at 11:56 AM, Lucas Duarte Bahia <
lduarteb at andrew.cmu.edu> wrote:

> Hi,
>
> I am struggling here with some processes that have integral with W. Could
> you please give me some guidance if the way I am thinking is correct?
>
>
> 2015 midterm ex 1)
>
>
> [image: \frac{\partial}{\partial W(t)}\int_{0}^{t} (W(u)^3 -1) du]
>
>
> We get that this is 0. My intuition (after seeing the solution) is that
> this is zero because every increment in of W is independent of W(t).
>
> Assignment 4 -- ex 1 d)
>
>
> [image: \frac{\partial}{\partial t} \int_{0}^{W(t)} e^{-ts^2} ds]
>
> Here my understanding is that we should apply the Leibniz Rule and set
> W(t) as a constant when we do the partial derivative in relation to t.
>
> Thanks,
> *Lucas Duarte Bahia*
> MS. Computational Finance Student
> Carnegie Mellon University
> Telephone: (412) 378-1892
>
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