# Stopped martingale and stopped filtration

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.

## 1 Answer

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 $$X_T=X_{S \wedge T}+\sum_{k=S}^{T-1}(X_{k+1}-X_{k})=X_{S \wedge T}+\sum_{k=0}^{n-1}(X_{k+1}-X_{k})\mathbb{1}_{S\leq k Pick now any $$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 Finally, note that $$\mathbb{1}_{S\leq k k\}}$$ 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 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 $$\mathbb{E}[X_{t\wedge T}|F_{T \wedge s}]= X_T=X_{T \wedge s}$$

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