# expected hitting time of asymmetric random walk

$$X_1$$, $$X_2$$, ... are i.i.d. random variables with distribution P($$X_i$$ = 1) = $$\frac{2}{3}$$, P($$X_i$$ = -1) = $$\frac{1}{3}$$. Let $$S_n = \sum_{i=1}^n X_i$$ For each integer k > 0, define $$T_k = min \left\{ n\geq 1: S_n = k \right\}$$ Then $$E(T_k) = \frac{k}{2p-1} = 3k$$

But I do not know how the result comes from. Can anyone give me any idea about how should I solve this problem? Thanks a lot!

You can use renewal theory to solve this problem. The idea is to compute $$u(j)=E[T_k \mid S_0=j]$$ for each $$j=\dots,-1,0,1,\dots,k$$. Then $$u(0)$$ is what you want to find and $$u(k)=0$$. Then you also have the recursion relation $$u(j)=1+2/3 u(j+1)+1/3 u(j-1)$$ for $$j.
At the moment you still have a problem, because this is a second order recurrence relation and you only have one boundary condition. The solution is to approximate $$u$$ along a sequence $$u_n$$ with the artificial boundary condition $$u_n(-n)=0$$. These $$u_n$$'s can be computed and then you can send $$n \to \infty$$ to obtain the desired result.