# Stopping theorem for continuous martingale

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 $E[M_\tau]=E[M_N]$?

hulik

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It is difficult to understand some steps in your post. OK, The first your question: Can I apply here 1) to get $E[M_N|F_\sigma]=M_\sigma$? I'm not quiet sure, since in 1) we have a , and here σ is a stopping time. Here you do not need martingale property, you just use tower property that you stated once again. I do not understand the conlcusion in 3). Yo can proceed as the following: take $\tau_n$ - simple''stopping times that converge to $\tau$ above'' and use the fact that $(M_{\tau_n})_{n \geq 1}$ are uniformly integrable. If you have further questions about this, please ask. –  Tarasenya Apr 9 '12 at 16:26
@ Tarasenya: Thanks for your comment. Can yout turn your comment into an asnwer? I do not understand your answer for my first question, i.e. : "You just use tower property that you stated once again." I used tower property, but why do I know that $E[M_N|F_\sigma]=M_\sigma$? Also your answer for point 3) is not clear for me. More information, as in an answer, would be appreciated. Thanks for your help! –  user20869 Apr 21 '12 at 13:44

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.
@ Tarasenya: Thanks for the answer! now I get point 2. But still, why is it enough to prove $E[M_\tau]=E[N]$, to conclude the theorem? Additionally, why are $(M_{\tau_{n}})$ uniformly integrable? THanks for your help! –  user20869 May 6 '12 at 12:25
If $Y \in L^1$ then $\{ E[Y|\mathcal{G}], \mathcal{G} \subseteq \mathcal{F}] \}$, where $\mathcal{G}$ ranges over all sub $\sigma$ algebras of $\mathcal{F}$, is uniformly integrable. This shows that $(M_{\tau_n})$ is uniformly integrable since $M_{\tau_n} = E[M_N | \mathcal{F}_\tau]$. –  nullUser Feb 7 at 5:50