Let $(X_n)_{n\geq 1}$ be a martingale with respect to $(Y_n)_{n\geq 1}$, i.e., the martingale condition $$ \mathbb{E}[X_n|Y_1, \ldots, Y_{n-1}] = X_{n-1} $$ holds. From this condition, it follows that the process $(X_n)$ has constant mean $\mathbb{E}[X_n] = \mathbb{E}[X_1]$ for all $n$. Under some assumptions, we then also have $\mathbb{E}[X_T] = \mathbb{E}[X_1]$, where $T$ is a stopping time. My question is, does this optional stopping theorem also apply to processes that have constant mean, but do not fulfill the martingale condition above? An example for such a process was given here: Example of a sequence of r.v.'s with constant stopping time that is not a Martingale


1 Answer 1


No, it does not hold. In fact, an equivalent definition of a martingale is an integrable adapted process that satisfies the optional stopping theorem.

As a counter-example, let your process be $X_n$ where $\mathbb{P}[X_i = 1] = 1/2$ and $\mathbb{P}[X_i = -1] = 1/2$ and $X_i$ are i.i.d. Let $\tau=\inf\{n: X_n =1\}$, then $\mathbb{E}[X_{\tau}]=1$.

Note that I am NOT taking the process to be a random walk. It's just a process which at each time $n$ takes the value $-1$ or $1$.

  • $\begingroup$ Thanks! The first statement seems pretty strong. Is there a reference where this is discussed? $\endgroup$
    – TriSSSe
    Jan 20, 2015 at 19:25
  • 1
    $\begingroup$ I first saw it in Continuous martingales and Brownian motion by Revuz and Yor. I don't have my copy to hand right now I'm afraid, but let me know if you can't find it and I'll look it up when I get a chance. $\endgroup$
    – user180850
    Jan 20, 2015 at 20:23

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