# Fail of optional sampling theorem

Could anyone help me see why the optional sampling theorem ($E(M_{\tau}\mid\mathcal{F}_{\sigma})=M_{\sigma}$ a.s.) fails for certain stopping times $\sigma\leq\tau$ for the not uniformly integrable martingale $$e^{aW_{t}-a^2t/2}$$ where $a$ is not 0 and $W$ is a Brownian Motion.

Why does $E(M_{\tau}\mid\mathcal{F}_{\sigma})=M_{\sigma}$ not hold for this martingale?

Really having trouble with this one so help is very much appreciated!

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have you proved the martingale inequality question you asked yesterday? if not, i am not dishing out hints again when I think you are not actually following them. –  Lost1 Apr 14 '13 at 10:16
Yes, I have as a matter of fact. I needed a bit of a push into the right direction and it took a while but I finally did manage. So thank you for your help on that question I do appreciate it as I had tried so many approaches but non seemed to work. For this question I can only say I've come quite far by proving its not integrable and that it satisfies the martingale property but on this last part again I need a bit of a push. However I realize this is non of your concern, so there is no need to help me out if you'd rather not. –  user70267 Apr 14 '13 at 10:36
okay, sorry. good. I am glad. too often i find people don't actually try the question themselves when I/others only gives hints rather than complete solutions. I would just not waste time replying to them in the future. as for this question, have you understood what i wrote? –  Lost1 Apr 14 '13 at 10:42
Well i'd again take $\sigma=0$ so on the righthandside one would get the value 1 in the equation. And then for $\tau$ i'd like a stopping time that tends to infinity so that the almnost sure limit of the martingale sets in and the left side becomes 0. But I'm a bit confused as there is an extra $t$ in the martingale besides from the brownian motion one. For a brownian motion It's rather easy. –  user70267 Apr 14 '13 at 11:09
I had tried so many approaches but non seemed to work... For your questions to come, you might want to include in them a description of these "so many approaches" you tried. So far you are remarkably silent on these. –  Did Apr 14 '13 at 11:24

Optional sampling theorem only works for bounded stopping times in general. For any arbitary $\tau$ and $\sigma$, we can actually define bounded stopping time by $\tau\wedge t$ and $\sigma\wedge t$. Now. If we apply optional sampling theorem to $\tau\wedge t$ and $\sigma\wedge t$, we will get

$\mathbb{E}(M_{\tau\wedge t}|\mathcal{F}_{\sigma\wedge t})=M_{\sigma\wedge t}$

assuming $\tau\geq\sigma$ almost surely.

In order to arrive at your expression. what we do is to take $t\rightarrow\infty$ on both handside. In order to do that on the left, we need uniform integrability.

Okay so, are you aware if you take a Brownian motion, and take $\tau$ to be the first time it hits 1 and $\sigma = 0$, you will get a contradiction? because 1 side would be 1, the other would be 0.

The exponential martingle you wrote down converges to 0 almost surely. can you try something similiar?

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