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!
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?