Asymmetric random walk with unequal step size other than 1.

Solution 1:

The usual approach: let $q_n$ denote the probability of absorption starting from $n\geqslant0$, then $q_0=1$ and $$q_n=pq_{n-1}+(1-p)q_{n+a}$$ for every $n\geqslant1$. Furthermore, since the only negative steps are $-1$ steps, to hit $0$ starting from $n$, one must hit $n-1$ starting from $n$, then hit $n-2$ starting from $n-1$, and so on until $0$. Thus, $q_n=(q_1)^n$ for every $n\geqslant0$. Can you deduce the value of $q_1$?

Likewise, assume that $q_1=1$ and let $t_n$ denote the mean absorption time starting from $n\geqslant0$ (if $q_1\ne1$, the mean absorption time is infinite), then $t_0=0$ and $$t_n=1+pt_{n-1}+(1-p)t_{n+a}$$ for every $n\geqslant1$. Furthermore, since the only negative steps are $-1$ steps, the time to hit $0$ starting from $n$ is the sum of the time to hit $n-1$ starting from $n$, plus the time to hit $n-2$ starting from $n-1$, and so on until $0$. Thus, $t_n=nt_1$ for every $n\geqslant0$. Can you deduce the value of $t_1$?