Asymmetric random walk with unequal step size other than 1. 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?
 A: 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 $t_1$?
A: It can be calculated by simply raising the transition matrix that encodes the random walk by powers of n
So for the two steps forward, one step back process, with an absorbing barrier at 0 and another at 5, and p=1/2, you would have:
$ \begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 \\
\frac{1}{2} & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 \\
0 & \frac{1}{2} & 0 & 0 & \frac{1}{2} & 0 & 0 \\
0 & 0 & \frac{1}{2} & 0 & 0 & \frac{1}{2} & 0 \\
0 & 0 & 0 & \frac{1}{2} & 0 & 0 & \frac{1}{2} \\
0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 
\end{bmatrix}  $
The right most entries give the probabilities of being absorbed at the upper barrier at step 'n' assuming it has not been absorbed before. The left entry gives the probability of being absorbed at 0.
Nice, closed-form solutions do not exist though for uneven steps for arbitrarily high powers, but generating functions can be derived.
A closed form solution for a 2 step forward, 1-step back process, with a barrier at $m$ and zero, assuming the particle starts at $m-1$ is
$\frac{b+b^2}{z}$  where b is the series solution of the smallest root of $x^3-2x+z=0$ as a power series in terms of $z$, setting z=1 and summing the series gives the exact probability of absorption for the upper barrier at up to $m$ steps.
