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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.

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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.

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  • $\begingroup$ Thank you. Could you provide a reference for this result? Never mind. I found it. Thanks again. $\endgroup$
    – Calculon
    Mar 20, 2016 at 22:29

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