I've a question about a proof in my lecture notes. We want to prove the following theorem.
$M=(M_t)_{t\ge 0}$ be a $(P,F)$-martingale, where $P$ is a probability measure and $F=(\mathcal{F}_t)$ a filtration (right continuous). Further we assume that $M$ has right continuous paths and $\sigma,\tau$ are $F$-stopping times with $\sigma\le \tau$. If $\tau$ is bounded or if $M$ is uniformly integrable, then $E[M_\tau|\mathcal{F}_\sigma]=M_{\sigma}$
The proof consists of 3 steps.
1) Here we conclude: We can assume that $M_t=E[M_N|\mathcal{F}_t]$ for all $t\le N$ with some fixed $N\in [0,\infty]$.
2) Since $\mathcal{F}_\sigma\subset \mathcal{F}_\tau$ and tower property of conditional expectation it's enough to prove: $E[M_N|\mathcal{F}_\tau]=M_\tau$.
3) Here we assume that $\tau$ has only countably many increasingly ordered values $t_n,n\in \mathbb{N}$ and use stopping theorem for discrete time to show $E[M_\tau]=E[M_N]$. For a general $\tau$ we approximate it from "above" and show similarly $E[M_\tau]=E[M_N]$
Now my questions:
to 2): Is this what is meant in 2): if $E[M_N|\mathcal{F}_\tau]=M_\tau$ is true, then
$$E[E[M_N|\mathcal{F}_\tau|\mathcal{F}_\sigma]=E[M_\tau|\mathcal{F}_\sigma]$$
The LHS is equal $E[M_N|\mathcal{F_\sigma}]$. Can I apply here 1) to get $E[M_N|\mathcal{F_\sigma}]=M_\sigma$? I'm not quiet sure, since in 1) we have a $t$, and here $\sigma$ is a stopping time.
to 3): Why does the Theorem follows if I just show $E[M_\tau]=E[M_N]$?
Thanks in advance for your help.
hulik
Best Answer
The second step says: if one proves $E[M_N|\mathcal F_\tau]=M_\tau$ for any bounded stopping time(so this is valid for $\sigma$) then for all stopping times $\sigma\leq \tau$ one has the statement of the theorem: $E[M_\tau|\mathcal F_\sigma]=M_\sigma$. So as you wrote $E[M_\tau|\mathcal F_\sigma]=E[E[M_N|\mathcal F_\tau]|\mathcal F_\sigma]=E[M_N|\mathcal F_\sigma]=|\mbox{our assumption!}|=M_\sigma$.
The third step seems very wierd, but you can finish yourself as the following, consider the sequence of ``simple'' stopping times $(\tau_n)_{n \geq 1}$ s.t. $\tau_n \to \tau+, n \to \infty$, so $M_{\tau_n}=E[M_N|\mathcal F_{\tau_n}]$ and take $A \in \mathcal F_\tau\subset \mathcal F_{\tau_n}$, hence $E[M_NI_A]=E[M_{\tau_n}I(A)]$. Now $(M_{\tau_n})_{n \geq 1}$ are uniformly integrable(think why!), so we need only take limit as $n\to\infty$ and we are done.
If you have further questions feel free to ask.