Setup:
Let $X_n$ be a Gambler Ruin's game such that
$$X_n = x + \sum_{i =1 }^n Y_k$$
with $x \in \left\lbrace1,2,.., 99 \right\rbrace$ the initial value and $(Y_k)_{k \in \mathbb{N}}$ i.i.d with distribution given by $P(Y = -1) = \frac{3}{5}$ and $P(Y =1) = \frac{2}{5}$.
Let $T$ be the stopping time defined as
$$T = \min \left\lbrace n \geq 1: X_n = 0 \text{ or } X_n = 100\right\rbrace$$
We know that
$$M_n = X_n – \mu n$$
with $\mu = E[Y] $is a martingale adapted to $(X_n)$.
My question:
If I want to know the expected value of the Ruin time, which is defined as
$$T_0 = \min\left\lbrace n \geq 1: X_n = 0 \right\rbrace$$
If I suppose that it is the probability of ruin when there is no maximal gain. I need to show that $$|M_{\min(n, T_0)}|< \sum_{i = 1}^k a_i n^i$$
and that
\begin{align*}
E[T_0] < \infty
\end{align*}
But it seems to me that we dont even have control on the upper bound since:
By the strong law of law numbers, the process $(X_n)_{n \geq 0}$ diverges to –$\infty$ when $n \to \infty$. Therefore there exists almost surely a $N_0 \in \mathbb{N}$ such that $X_n \leq 0$ for all $n > N_0$. However, this $N_0$ is still random hence we cannot truly bound the stopped martingale.
Moreover, $T_0 < \infty$ almost surely but I cannot prove that its expected value is.
So, if anyone has any clue on how to solve that, or if it is possible given the theorems the teacher gave us, it would be very much appreciated.
(he realised the theorem he was teaching us was wrong 2 days before the final exam and now I'm not even sure if he knows what is the right version of the stopped time theorem).
Thank you!
Best Answer
Regarding the ruin time $T_0$:
Apply optional stopping to the martingale $M_n=X_n-\mu n=X_n+n/5$ at the bounded stopping time $T_0 \wedge n=\min(T_0,n)$:
$$x=E(M_0)=E(M_{T_0 \wedge n})=E(X_{T_0 \wedge n})+\frac15 E( T_0 \wedge n) \,. \tag{*} $$ Thus $E(T_0 \wedge n) \le 5x$, so by the monotone convergence theorem, $E(T_0 ) \le 5x \,.$ By the law of large numbers, $\lim_n X_{T_0 \wedge n}=0$ almost surely. Since $|X_{T_0 \wedge n}| \le x+T_0$ for all $n$, the dominated convergence theorem yields $\lim_n E(X_{T_0 \wedge n})=0$. Therefore, letting $n \to \infty$ in $(*)$, we conclude that $$E(T_0)=5n\,.$$