Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.