betting in fair game over infinite horizon, is it possible to win?

Question

If a gambler were to play in a fair game, lets say he wins/loses 1 dollar with equal probability in each step. Let $X_i$ denote the amount of money he has after $i$ steps. And he plays until he either runs out of money or decides to stop, in which case he cashes in what he currently has. Let us denote this stopping time as $T$. Then starting with $n$ dollars, if $E(T)<\infty$ using optional stopping theorem one can show that $E(X_T)=n$. However this is no longer applicable if $E(T)=\infty$. Could there potentially be a strategy for the gambler such that $E(T)=\infty$ and $E(X_T)>n$?

Answer 1

Assuming that $T < \infty$ a.s., there is no way for $\mathbb{E}[X_T] > n$ if we enforce the no debt condition $X_m \ge 0$ for all $m$.

For any $m$ we have $\mathbb{E}[X_{T \wedge m}] = n$ by the optional stopping theorem (since $T \wedge m$ is a stopping time with finite expectation), and since $T < \infty$ a.s., $\lim_{m \rightarrow \infty} X_{T \wedge m} = X_T$ a.s. Since $X_{T \wedge m} \ge 0$ for all $m$, by Fatou's lemma we have \begin{align*} \mathbb{E}[X_T] = \mathbb{E}[\lim_{m \rightarrow \infty}X_{T \wedge m}] \le \liminf_{m \rightarrow \infty} \mathbb{E}[X_{T \wedge m}] \le n. \end{align*}

If we allow $\mathbb{P}(T = \infty) > 0$, it's unclear how to define $X_T$ on the event $\{T = \infty\}$.