We define the lazy random walk $S_n$ on $\mathbb{Z}$ which goes left with probability $\frac{1}{2}$, right with probability $\frac{1}{4}$, and stay in place with probability $\frac{1}{4}$. Assume that $S_0=0$ and $a,b \ge 1$ integers. Find the probability that $S_n$ hits $b$ before reaching $-a$.
I was thinking about applying the Doob's optional stopping theorem but can' figure out how. Can anyone help?
Way 1: define a time $\tau$ to hit the boundary, define $u(n)=P(S_\tau=b)$, define a recurrence relation for $u$ using the total probability formula, solve the resulting equation for all $u(n)$, plug in $n=0$. The setup of the recursion is $u(n)=(1/2) u(n+1) + (1/4) u(n) + (1/4) u(n-1)$ and then the boundary conditions are $u(b)=1,u(-a)=0$. This recurrence can be solved.
Way 2: define a time $\tau$ to hit the boundary, prove it satisfies one of the possible conditions of the optional stopping theorem. Then consider the martingale $M_n=S_n-n/4$. On the one hand $E[M_\tau]=E[M_0]$ while on the other hand $E[M_\tau]=b P(S_\tau=b) + (-a) P(S_\tau=-a)-E[\tau]/4$. Thus you can obtain the desired probability provided that you know $E[\tau]$; one way to get $E[\tau]$ is to follow a procedure very similar to Way 1.