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!
Best Answer
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.