# Non-symmetric simple random walk stopping time

Say there is a random walk $\{S_n\}$ with $S_0=0$ and $0<p=P(S_1=1)<\frac{1}{2}$. We know such a random walk would go to $-\infty$ eventually. Define the stopping time $T=\inf\{n: S_n=-\infty\}$, how can we argue that the stopping time is finite a.s., i.e., $P(T<\infty)=1$?

I know we definitely will use $0<p=P(S_1=1)<\frac{1}{2}$, and I know for such $p$, the returning time is not finite, which is transient. How can I apply this to show the stopping time $T$ is finite a.s.? Or is there another approach to this?

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## 1 Answer

You have not told us enough to reach your conclusions.

If you are saying each independent step is $+1$ with probability $p$ and $-1$ with probability $1-p$, with $p \lt \frac12$, then $S_n \not = -\infty$ for finite $n$, since $-n \le S_n \le n$. So $\Pr(T \lt \infty)=0$.

Since $p \lt \frac12$ you could conclude that $$\lim_{n \to +\infty} S_n = -\infty$$ with probability $1$ but that is not the same thing at all.

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Thanks for your answer! You are right about all the assumptions that I don't mention in my question (each independent step is +1 with probability p and −1 with probability 1−p, with p<1/2) From your argument, I think maybe we cannot reach the conclusion that the stopping time is finite a.s. The reason I asked this I want to check the conditions of Optional stopping theorem, so I can use it to reach another conclusion. – Qomo Apr 25 '12 at 23:30