# Expectation of infimum of asymmetric 1D random walk greater than -$\infty$

I'm reading Durrett's book on Probability and in the example of the asymmetric 1D random walk with $P(X_1=1)=p>1/2$, when trying to compute the expectation of the hitting time $T_{b}:=\inf\{n: S_n=b\}$ using optional stopping, in order to use the dominated convergence theorem to prove that $E[S_{T_b\wedge n}]\to E[S_{T_b}]$ he makes the following statement: $$E[\min_{n} S_n]>-\infty.$$

First of all, I think the $\min$ should be replaced by $\inf$ and intuitively I agree with the result, but I can't seem to be able to prove it.

I know that my random walk $S_n$ converges almost surely to $\infty$ by the law of large numbers and the fact that $p>1/2$ and I was thinking that this should imply that there are finitely many negative $S_n$, but this finite number is $\omega$-dependent (i.e. random). My idea was then to say that $$E[\inf_{n}S_n]\geq \sum_{k\in\mathbb{N}} (-k)P(\inf_{n}S_n=-k)$$ and then to use the fact the the above probabilities should be zero for all but finitely many $k$, but I don't know how to get rid of the $\omega$-dependance.

Any hints or (other) ideas?

Let $q$ be the probability that starting from $0$ the walk ever reaches $-1$. Then the probability $a_n$ that starting from $0$ the walk ever reaches $-n$ satisfies
$$a_{n+1}=qa_n\;,$$
since the walk reaches $-(n+1)$ if and only if it first reaches $-n$ and then with probability $q$ reaches $-(n+1)$. Thus $a_n=q^n$. We have $q\ne1$, so the desired expectation can be computed as a finite expression in terms of $q$.
• Thank you for your answer, but I don't unserstand how I can proceed from there. You are saying that $P(\exists j: S_j=-n)=q^n$. How does this say anything about $P(\inf_j S_j=-n)$? – DCPC Jul 13 '16 at 12:56
• @DCPC: $P(\inf_jS_j=-n)=P(\exists j: S_j=-n)-P(\exists j: S_j=-(n+1))$. But more directly, $E[\inf_jS_j]=-\sum_{n=1}^\infty P(\exists j: S_j=-n)$. – joriki Jul 13 '16 at 13:50
• Thank you. I agree with your reasoning except one thing. Your initial claim. If you consider $P(\exists j: S_j=-1)$, this is equal to $P(\cup_{j} \{S_j=-1\}) = \sum_{j} P( T_{-1}=j)=P(T_{-1}<\infty)\geq q$, where $T_{-1}$ is the hitting time of $-1$. Then, by the same reasoning, $P(\exists j: S_j=-n)= P(T_{-1}<\infty)^n$. – DCPC Jul 14 '16 at 8:15
• @DCPC: I don't quite understand that comment. Why do you write $\ge q$ and not $=q$? And I'm not sure what you're getting at -- are you saying that the conclusion you reach in the end is wrong, and hence my argument must be wrong? If so, why is the conclusion wrong? – joriki Jul 14 '16 at 10:24
• I'm saying it's greater than $q$ because $q=P(S_1=-1)$, but $P(S_3=-1, S_1>-1, S_2>-1)=P(T_{-1}=3)>0$ and similarly $P(T_{-1}=2k+1)= c_k p^{k}q^{k+1}>0$, where $c_k$ is the k-th Catalan number. – DCPC Jul 14 '16 at 13:23