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Say, an asymmetric random walk, at each step it goes left by 1 step with chance $p$, and goes right by $a$ steps with chance $1-p$. (where $a$ is positive constant).

The chain stops whenever it reaches 0 or any state which is negative (in this particular case it cannot reach negative state).

Is there an easy way to calculate the absorption probability? Is there an easy way to calculate the mean hitting time before absorption?

How can we verify them?

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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 $t_1$?

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the issue is, when $a$ is large, it is not so simple to solve the equations –  colinfang Apr 2 '13 at 15:59
    
Now, this should be trivial. –  Did Apr 2 '13 at 16:03
    
Still, it is not trivial. I still need to solve high degree polynomials equation for hitting probability. For mean hitting time, it is trivial. –  colinfang Apr 2 '13 at 16:20
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