Suppose there is a gambler who has $n$ dollars gambles for infinitely many times. The payoff of the $t^{th}$ round is $X_t$, which is an integer beween $-1000$ and $1000$. We know that $\mathbb{E}[X_t|X_{t-1},\cdots,X_1]\leq c$. Here, $c$ is a negative constant. Suppose the probabliity of bankrupt is $p(n)$. Do we have $\lim_{n\rightarrow \infty}p(n)=0$?
I have thought this problem for a couple of days. I am not familiar with martingale theory so I do not know whether this problem is difficult.
Supplement: what about the case that $\mathbb{E}[X_t|X_{t-1},\cdots,X_1]\geq c$ where $c$ is a positive constant? Thank you very much!
$p(n)=1$ for each $n$.
Let $S_k= X_1+\ldots X_k$, so the gambler's fortune after $k$ rounds is $\max\{n+S_k,0\}$. Write $b=-c>0$. The hypothesis implies that $M_k=S_k+bk$ is a supermartingale, so by the Azuma-Hoeffding inequality
$$P(n+S_k >0)=P(M_k>bk-n) \le \exp\Bigl(\frac{-(bk-n)^2}{2 \cdot 1000^2 k}\Bigr)$$ which decays exponentially in $k$. By the Borel-Cantelli Lemma, the event $\{S_k>0\}$ will occur only finitely often almost surely, so the gambler will go bankrupt with probability 1.
https://en.wikipedia.org/wiki/Azuma%27s_inequality