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Is there an example of a martingale in discrete time $X_0, X_1, X_2,\ldots$ and a stopping time $T$ so that $E(T) <\infty$ but $E(X_T) \neq E(X_0)$?

With added assumptions on how $X_n$ behaves, you can prove that $E(X_T)=E(X_0)$ For example if $|X_{n+1} - X_{n}| \leq c$ for some constant $c$, then we can show $E(X_T) = E(X_0)$.

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up vote 5 down vote accepted

How about this: let $Y_0,Y_1,\ldots$ be independent RVs, where $$ P(Y_n=2^n)=\frac{1}{2},\qquad P(Y_n=-2^n)=\frac{1}{2}. $$ For $n\geq 0$, let $X_n=\sum_{i=0}^{n}Y_i=X_{n-1}+Y_n$. This is a martingale: $$ \mathbb{E}[X_n\,\mid\,X_{n-1}]=X_{n-1}+\mathbb{E}[Y_n]=X_{n-1}, $$ since $X_{n-1}$ is measurable with respect to $Y_1,\ldots,Y_{n-1}$, while $Y_n$ is independent of these. Let $T$ be the minimum $n$ such that $X_n>0$, if such an $n$ exists, and $\infty$ otherwise.

Note that $T$ is also the first $n$ such that $Y_n>0$, since $2^n>1+2+\cdots+2^{n-1}$. So, $$ P(T=n)=\frac{1}{2^{n+1}}\qquad\text{and}\qquad\mathbb{E}[T]=\sum_{n=0}^{\infty}\frac{n}{2^{n+1}}=1. $$ Since $\mathbb{E}[T]<\infty$, $T$ is almost surely finite, and $X_T$ is almost surely well-defined. But, since $X_n\geq 1$ if $X_n>0$, it is always the case that $X_T\geq 1$ if $T<\infty$. Combining these, $\mathbb{E}[X_T]\geq1$, whereas $\mathbb{E}[X_0]=0$.

(It should be noted here that, as you would hope, our situation doesn't fit any of the various conditions for the optional stopping theorem; our steps aren't almost surely bounded, our stopping time isn't almost surely bounded, etc.)

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