[944] Last Example From TA Session Friday

[David Itkin]

Hi Richard,

Indeed [image: e^{2W(u)}] is not a martingale and you have to manually
compute the conditional expectation (as we did) to determine [image:
E[e^{2W(u)}|\mathcal{F}_s]], but Brownian motion by itself is a martingale.
Keep in mind that functions of martingales need not be martingales.

David

On Sat, Nov 18, 2017 at 7:30 AM, Richard Rosenbaum <rrosenba at andrew.cmu.edu>
wrote:

> Hi all,
>
> For the last problem in the TA session, David narrowed it down to
> E[exp(2W) | Fs].
>
> Since we are integrating from s to t, my first thought was to assume that
> exp(2W) was a martingale when conditioned on Fs. Is this incorrect because
> of the BM being in the exponent?
>
> For example, if we had this same situation but instead of exp(2W), we just
> had W, I believe we could assume martingale, plug in s, and integrate. Is
> this thought process correct?
>
> Thanks again.
>
> Rich
>
>
> --
> Richard Rosenbaum
> Master of Science in Computational Finance Candidate
> Carnegie Mellon University - Class of Dec. 2018
>
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