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On page 217 of the book Probability Essentials By Jean Jacob and Philip Protter found an issue that can not do, is about martingales and stopping times. If anyone has any tips on how to do, I'd appreciate it a lot:

$$E \{ E (Y\mid F_T)\mid F_S\} = E \{ E (Y\mid F_S)\mid F_T\} = E (Y\mid F_{T \wedge S})$$

$T$, $S$ are stopping times.

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What did you try? – Did Sep 14 '12 at 18:13
tips: with $Y_T = \mathbb E(Y | F_T)$ etc. $Y_{T\wedge S},Y_T$ is martingale. Also $\lbrace T < S \rbrace \in F_S$ and of course $F_T$ – mike Sep 15 '12 at 10:35

I have a method but it assumes T and S are bounded stopping times. So let T and S be bounded stopping times. In Jacod and Protter, the theory is for Discrete time martingales and so T and S take on non negative integer values. (Note: All random variable equalities and inequalities are almost sure ones).

Let $M_t = E[Y|F_t]$

Then we need to show $$ E[M_T|F_S] = M_{S\wedge T}$$

Let $$A= \{\omega : S(w) \leq T(\omega)\}$$ $$ T_1 = T\vee S$$ $$ T_2 = T \wedge S $$ Note that $T_1$ and $T_2$ are stopping times. And $A \in F_S \cap F_T$. These are all exercises in Jacod and Protter.

$\therefore M_{S \wedge T} = M_S1_A + M_T1_{A^C}$

Now $E[M_T|F_S] = E[M_T 1_A + M_T1_{A^C}|F_S]$.

$E[M_T 1_A|F_S] = E[M_{T_1}1_A|F_S] = E[M_{T_1}|F_S]1_A$

But $T_1\geq S$. By Doob's Optional Sampling Theorem, and since $T, S, T_1$ and $T_2$ are bounded stopping times, $E[M_{T_1}|F_S]1_A = M_S1_A \quad (1)$

Similarly $E[M_T1_{A^C}|F_S]=E[M_{T_2}1_{A^C}|F_S]=E[M_{T_2}|F_S]1_{A^C}$

But $T_2 \leq S$. Therefore $F_{T_2} \subseteq F_S$.

$\Rightarrow E[M_T1_{A^C}|F_S] = M_{T_2}1_{A^C} = M_T1_{A^C} \quad (2)$

From (1) and (2)

$E[M_T|F_S] = M_S1_A + M_T1_{A^C} = M_{S \wedge T} $

You can interchange T and S and the proof goes through with the corresponding changes. As for the unbounded case, a limiting argument would be required. See if you can get it using the above, replacing T and S with $T\wedge N$ and $S \wedge N$, letting N go to $\infty$. Be careful though.

Also in "Stochastic Processes" by Richard Bass, the above is true even in the continuous time case.

Edit: Finally proved it for the unbounded case. But had to resort to Doob's Martingale convergence and Levy's zero one law, both available on Wiki.

Let T and S be two stopping times not necessarily bounded. Consider $$ T_n = T\wedge n $$ $$S_n = S\wedge n$$ Now $T_n$ and $S_n$ are bounded stopping times. Moreover $T_n \uparrow T$ and $S_n \uparrow S$. Consider the following: $$E[M_{T_n}|F_{S_m}] = M_{T_n \wedge S_m} \quad (3)$$ which is what we proved before. Now let us fix m. Let $T_n\wedge S_m = U_n$ and $T\wedge S_m = U$. Note $U_n \uparrow U$.

Claim: $M_{U_n}$ is a discrete martingale wrt $F_{U_n}$. And it satisfies the condition for Doob's martingale convergence.

Proof: Since $E[M_{U_n}] = E[M_0] = E[Y] < \infty$, and $E[M_{U_k}|F_{U_m}] = M_{U_m}$ if $m<k$ by Doobs OST and $\sup_{n \geq 1}E[M_{U_n}] = \sup_{n \geq 1}E[Y] <\infty$, the claim is proved.

Therefore $$\lim_{n \rightarrow \infty} M_{U_n} = M_U$$ Applying this to (3), we get $$\lim_{n \rightarrow \infty} E[M_{T_n}|F_{S_m}] = \lim_{n \rightarrow \infty}M_{T_n \wedge S_m} = M_{T\wedge S_m}\quad (4)$$

Now by the same logic, $M_{T_n} \rightarrow M_T$ and $\sup_{n \geq 1} E[M_{T_n}] = E[Y] < \infty $. By Conditional DCT, $$\lim_{n \rightarrow \infty} E[M_{T_n}|F_{S_m}] = E[M_T|F_{S_m}] \quad (5) $$ By (4) and (5) $$E[M_T|F_{S_m}] = M_{T\wedge S_m}$$ Now let $m \rightarrow \infty$. The RHS goes to $M_{T\wedge S}$ by Martingale convergence. Now Le'vy's conditional convergence (aka his zero-one law) states that if $X \in L^1$ and $F_m$ is a filtration then $$E[X|F_m] \rightarrow E[X|F_\infty] = X$$. Using this, we get $$ E[M_T|F_{S_m}] \rightarrow E[M_T|F_{S_\infty}] = E[M_T|F_S]$$


Note: Le'vy's zero one law is a corollary of Doob's martingale convergence. So we as good as used that only and conditional DCT.

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