It is known that if $T$ is a stopping time such that $E[T] < \infty$ and $(M_n)$ is a martingale with bounded increments, i.e. $\lvert M_n - M_{n-1}\rvert \leq K < \infty$ for every $n$, almost surely, then $E[M_T] = E[M_0]$.
My question is as follows. Suppose $M_n = \sum_{i=1}^nX_i$ where $X_1,X_2,\ldots$ is an integrable iid sequence with zero mean. Does the result above still hold? I don't have bounded increments anymore but I have a particular structure for the martingale. Is this structure powerful enough to compensate for lack of bounded increments in this case?
There is such a thing as Wald's equation apparently. I just found out about this. It comes very close to this problem.
Yes, this follows directly from Wald's (first) equation. There is no need to assume finite variance.
Wald's First Equation: Let $T$ be a stopping time with first moment and $X_i$ be an iid sequence with a first moment. Let $M_n=\sum_{i=1}^n X_i$. Then $\mathbb E M_T=\mathbb E X_1\mathbb E T$.
By plugging in $\mathbb E X_1=0$, we get $\mathbb E M_T=0$ as desired.