Question about asymmetric random walk and gambler's ruin Let $P(\xi_i = 1)=p$ and $P(\xi_i = -1)=1-p = q$. Let $0<p<1/2$ and $T_z = \text{inf }\{n:S_n=z\}$. If $a<0$ and $b>0$, then $P(\text{min}_n S_n\leq a)=P(T_a<\infty) = [(1-p)/p]^{-a}$. If it is known that $S_n -(p-q)n$ is a martingale, then $T_b\wedge n$ is a bounded stopping time and 
$$0=E(S_{T_b\wedge n}-(p-q)(T_b\wedge n))$$
Q1) Claim is that $b\geq S_{T_b\wedge n}\geq \text{min}_nS_n$ and $P(\text{min}_n S_n\leq a)= [(1-p)/p]^{-a}\implies E(\text{inf}_nS_n)>-\infty$. May I know why that implication holds? Is $E(\text{inf}_nS_n) = \sum_{a=0}^{\infty}P(\text{min}_n S_n\geq a)= \sum_{a=0}^{\infty} 1-[(1-p)/p]^{-a}$ but that is $=\infty$ (although $>-\infty)$?
Q2) Let the walk start at $x$. Use the martingale $(S_n-(p-q)n)^2-n(1-(p-q)^2)$ to show $\text{var }T_0 = cx$ for some constant $c$. My question is how to show $S_{T_0\wedge n}^2$ is dominated by an integrable RV? If $x<0$, then $0\geq S_{T_0\wedge n}\geq \text{min}_nS_n$ and I can perhaps claim $E(\text{inf }S_n)^2<\infty$ but if $x>0$, then $0\leq S_{T_0\wedge n}\leq \text{max}_nS_n$. Now I know that for $b>0$ and $a<x<b$, $P(T_b<\infty)=1=P(\text{max}_nS_n\geq b)$ and if I use $E(\text{sup}_nS_n) = \sum_{b=0}^{\infty}P(\text{max}_n S_n\geq b)= \sum_{b=0}^{\infty} 1=\infty$, let alone $E(\text{sup }S_n)^2<\infty$.
Thanks.
 A: Your problem is from Durreet's PTE five Theorem 4.8.9 and Exercise 4.8.5. You should note that in Theorem 4.8.9 (c) and (d), we assumed $\frac12<p<1$ rather than $0<p<\frac12$! Actually, if $0<p<\frac12$, $P(T_a<\infty)=1$ for all $a<0$.
Q1) I assume $\frac12<p<1$ in this part. Note that $\min_nS_n\leq 0$, and recall that if $X$ is a random variable taking values in $\mathbb N$ then $EX=\sum_{n=1}^\infty P(X\geq n)$, so in this case, let $X=-\min_nS_n$, we have
$$EX=\sum_{n=1}^\infty P(X\geq n)=\sum_{n=1}^\infty P(\min_n S_n\leq -n)=\sum_{n=1}^\infty \left(\frac qp\right)^n<\infty,$$
here we need to use $\frac12<p<1$ to get $\frac qp<1$. Hence $E(\inf_n S_n)=-EX>-\infty$.
Q2) In this part, we assume $0<p<\frac12$ and $x>0$, since the original exercise asks us to prove 
$$\text{var } T_0=x\cdot \frac{1-(p-q)^2}{(q-p)^3}=cx$$
with $c>0$.
Hence, $0\leq S_{T_0\wedge n}\leq\max_n S_n$. We want to show $E(\max_n S_n)^2<\infty$. For $b>x$, using the method in Theorem 4.8.9, we can get that $P_x(\max_n S_n\geq b)=P_x(T_b<\infty)=\left(\frac pq\right)^{b-x}$, so
$$E(\max_n S_n)^2=\sum_{n=1}^\infty (2n-1)P_x(\max_n S_n\geq n)<\infty.$$
I believe that you can move on now.
