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If $M$ is a martingale w.r.t. $(F_t)$, then by Doob's optional sampling theorem we can conclude that the stopped process $M^T$ is also a martingale w.r.t. the same filtration, whenever $T$ is an $(F_t)$-stopping time.

Is $M^T$ also a martingale w.r.t. to the filtration stopped at $T$, i.e. $(F_{t \wedge T})$?

I can see that $E(M^T_t | F_{s \wedge T}) = M^T_s$ whenever $s \leq t \leq T$ or $s \leq T \leq t$. But if $T \leq s \leq t$, then $E(M^T_t | F_{s \wedge T}) = E(M^T_T | F_{T})$. I don't know how to think about this last equality.

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Yes, this is true. In fact you have the more general result that if $T$ is a bounded stopping time and $S$ is any stopping time, then for $X_n$ a $F_n$-martingale, $\mathbb{E}[X_T|F_S]=X_S$ a.s. To see this, assume that $T\leq n$, and write $$ X_T=X_{S \wedge T}+\sum_{k=S}^{T-1}(X_{k+1}-X_{k})=X_{S \wedge T}+\sum_{k=0}^{n-1}(X_{k+1}-X_{k})\mathbb{1}_{S\leq k<T} $$ Pick now any $A \in F_{S}$. Then, $$ \mathbb{E}[X_T\mathbb{1}_A]=\mathbb{E}[X_{S\wedge T}\mathbb{1}_A]+\sum_{k=0}^{n-1}\mathbb{E}[(X_{k+1}-X_{k})\mathbb{1}_{S\leq k<T}\mathbb{1}_A] $$ Finally, note that $\mathbb{1}_{S\leq k <T}\mathbb{1}_A=\mathbb{1}_{(\{S\leq k\}\cap A) \cap \{T>k\}}$ and this set is in $F_k$ by definition of the stopped filtration. Hence, by the martingale property $$ \mathbb{E}[(X_{k+1}-X_{k})\mathbb{1}_{S\leq k<T}\mathbb{1}_A]=0 $$ for all k, and hence the result follows.

Regarding the case when $T\leq s\leq t$, just note that $X_{T\wedge t}=X_{T}=X_{T\wedge s}$, so $$ \mathbb{E}[X_{t\wedge T}|F_{T \wedge s}]= X_T=X_{T \wedge s} $$

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  • $\begingroup$ How did you get $E(X_{t \wedge T} | F_{s \wedge T}) = X_T$ when $T \leq s \leq t$? $\endgroup$
    – harisf
    Jun 27, 2023 at 8:11

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