Stopped martingale in discrete time

For a discrete martingale and stopping time $T$,The indicator is measurable at time $n$, so the stopped process is a martingale. If it is uniformly bounded and $T<\infty$ almost surely, bounded convergence gives $\mathbb{E}{M}_T=\mathbb{E}{M}_0$ directly.

For an almost surely finite stopping time $T$, if $\mathbb{E}|M_T|<\infty$ and $\mathbb{E}[|M_n|1_{\{T>n\}}]\to 0$, then $M_{n\wedge T}\to M_T$ with convergence in L1. Thus L1 convergence implies uniform integrability makes the stopped martingale uniformly integrable.

Let an integrable process $(M_n)$ be adapted to $({\mathcal{F}}_n)$. It is a martingale if and only if $\mathbb{E}[M_{n\wedge \tau }]=\mathbb{E}[M_0]$ for every $n$ and every stopping time $\tau$. For the reverse implication, fix $A\in {\mathcal{F}}_n$ and take $\tau =n$ on $A$ and $\tau =n+1$ on $A^c$. Comparing its stopped expectation at $n+1$ with the expectation obtained from the deterministic stopping time $n+1$ gives $\mathbb{E}[1_A(M_{n+1}-M_n)]=0$. Since this holds for every $A\in {\mathcal{F}}_n$, it is exactly $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$.

For simple symmetric random walk stopped on hitting $-a$ or $b$ and $0<z<1$, choose $w>1$ byThen $z^n{w}^{\pm X_n}$ are martingales. Solving the two boundary equations gives