# Stopping time and random walk: Proof that Stopping time of reaching a certain value is finite a.s. [duplicate]

This is a basic question but I was wondering if there was a simple proof (I have a rather complex one).

The problem is: Given a symmetric random walk ($P(X_i=1)=P(X_i=-1)=0.5$) and $S_n = \sum_{i=1}^n X_i$, define $T$ as $T=\inf\{n\geq0:S_n=a\}$, $S_0=0$. Prove that $T$ is almost surely finite, $P(T<\infty)=1$.

Thanks.

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## marked as duplicate by Nate Eldredge, Byron Schmuland, Did, Henry, Asaf KaragilaJul 23 '12 at 22:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

This is a multi-duplicate (and typical (homework)). You could try to compute $P(T_a\lt T_{-b})$, and let $b\to+\infty$. –  Did Jul 23 '12 at 8:53

## 1 Answer

The symmetric random walk is a recurrent and irreducible Markov Chain, thus by recurrence theorem $f_{ij}=1$ where $f_{ij}$ is the probability that starting from $i$ we have $X_n=j$ for some $n\geq 1$, that is $P(T<\infty)=1$.

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In other words: $P(T\lt\infty)=1$ because $P(T\lt\infty)=1$. –  Did Jul 23 '12 at 16:29