TITLE: Stopping time on an asymmetric random walk
QUESTION [8 upvotes]: Suppose we have an asymmetric random walk whose step $x_i$ is distributed as  $P(\xi_i = 1) = p$ and $P(\xi_i = -1) = 1-p = q$, where $p >1/2$. The hitting time, $T_x$ is defined as $\inf{\{n : S_n = x\}}$ where $S_n$ represents the simple walk $S_n = \sum_{i \leq n} \xi_i$. It can be shown that $$\Bbb E T_1 = (p-q)^{-1}$$ However, how can we deduce from this that $\Bbb ET_b = b(p-q)^{-1}$ for all $b>0$? I tried to use Wald's equation, but does not seem to work.

REPLY [3 votes]: Let's see what we can do for $T_2$. We have
$$E[T_2]=E[E[T_2\mid T_1]]$$
Now what is $E[T_2\mid T_1=t]$? We have $$T_2\mid\{ T_1=t\}=\inf\{n:S_{t+n}=1\}=t+\inf\{n:S_n=1\}=2t$$ Here I used the fact that your random walk is a homogeneous Markov chain. So $E[T_2\mid T_1]=2T_1$, thus $$E[T_2]=2E[T_1]=2(p-q)^{-1}$$
Then you can prove the rest by induction.