# Future of martingale after stopping time

Let $$(M_n)_{n\geq0}$$ be a non-negative martingale with filtration $$(\mathcal{F}_n)_{n\geq0}$$. Set $$T=\min\{n\geq0:M_n=0\}.$$ Show that $$M_n=0$$ for all $$n\geq T$$ almost surely.

As $$(M_n)_{n\geq0}$$ is non-negative, it will be enough to show that the expectation for all such $$n\geq T$$ is zero. But there is no necessary reason that the martingale is uniformly integrable, nor that the stopping time is bounded, so I can't use the Optional Stopping Theorem, and so I'm not sure how to show this. Any hints or advice would be greatly appreciated!

• What if $M_n\equiv 1$? $T=\infty$...
– user140541
Commented May 10, 2022 at 22:20
• I suppose that in the case the result holds trivially (as there are no $n\geq T$)? Commented May 10, 2022 at 23:20

As you observed, it suffices to check that $$\mathbb E\left[M_n\mathbf{1}_{n\geqslant T}\right]=0$$, since $$M_n\geqslant 0$$ will imply that $$M_n\mathbb{1}_{n\geqslant T}=0$$ almost surely.
To do so, we decompose $$\mathbf{1}_{n\geqslant T}$$ as the sum of $$\mathbf{1}_{T=k}$$ in order to get $$\mathbb E\left[M_n\mathbf{1}_{n\geqslant T}\right]=\sum_{k=0}^n\mathbb E\left[M_n\mathbf{1}_{T=k}\right]$$ and to reduce the problem to show that $$\mathbb E\left[M_n\mathbf{1}_{T=k}\right]=0$$ for each $$n\geqslant k$$.
Denote by $$\left(\mathcal F_n\right)_{n\geqslant -1}$$ the filtration associated to the martingale $$\left(M_n\right)_{n\geqslant 0}$$. We notice that $$\{T=k\}$$ belongs to $$\mathcal F_k$$, as $$\{T=k\}=\{M_k=0\}\cap\bigcap_{j=0}^{k-1}\{M_{j}>0\}$$ if $$k\geqslant 1$$ and $$\{T=0\}=\{M_0=0\}$$. As a consequence, $$\mathbb E\left[M_n\mathbf{1}_{T=k}\right]=\mathbb E\left[\mathbb{E}\left[M_n\mathbf{1}_{T=k}\mid\mathcal F_k\right]\right]= \mathbb E\left[\mathbf{1}_{T=k}\mathbb{E}\left[M_n\mid\mathcal F_k\right]\right]$$ and using the martingale property [until here, we only used $$\mathcal F_n$$-measurability of $$M_n$$], we get $$\mathbb E\left[M_n\mathbf{1}_{T=k}\right]= \mathbb E\left[\mathbf{1}_{T=k}M_k\right]$$ which is $$0$$ because $$\mathbf{1}_{T=k}M_k=0$$ almost surely.
• Thanks very much for your answer! Am I right in thinking that, before proving the result of the question, we actually can't say that, for $k\geq1$, $\{T=k\}=\{M_k=0\}\cap\{M_{k-1}>0\}$ but it should rather be $\bigcap_{0\leq j\leq k-1}\{M_j>0\}\cap\{M_k=0\}$? Not that this affects its inclusion in $\mathcal{F}_k$. Commented May 11, 2022 at 10:18