Stopping time on an asymmetric random walk

Question

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.

Answer 1

You can write $$T_2=T_1+(T_2-T_1)$$ where $T_1$ and $T_2-T_1$ have the same distribution by the Strong Markov Property. Hence $$\Bbb ET_2=\Bbb E[T_1+T_2-T_1]=\Bbb E[T_1]+\Bbb E[T_2-T_1]=2(p-q)^{-1}$$

Answer 2

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.