My question is a variant on this one. It concerns a random walk which is asymmetric in three ways. First, the probabilities of moving left or right are not necessarily equal. Second, the distance moved when moving one direction is greater than the other. Third, the random walk does not start at zero (I know this last one doesn't make a massive difference as it is just a case of labelling points, but I find it easier to think about in this way.)
Consider a random walk with starting point $S_0 = n$. The next steps are given by $$ S_{n+1} = \begin{cases} S_n + m \text{ with probability } p \\ S_n - 1 \text{ with probability } 1-p =q \\ \end{cases} $$ where $m>1$ is an integer.
What is the probability that this random walk eventually reaches the origin (ie. $S_n =0$ for some $n$)?
When $m=2$, this is the same as the earlier question, but can we say anything about larger values of $m$. I tried solving this using the characteristic equation method found here but had no luck.
What I've tried so far (copying from the this answer):
Let $f_n$ be the probability we are looking for:
$$f_n = {\bf Pr}\big[ S_0 =n \land \exists t > 0, S_t = 0 \big]$$
$f_n$ satisfies the recurrance relation:
$$f_n = \begin{cases} 1, & n = 0,\\ p f_{n+m} + q f_{n-1}, & n > 0 \end{cases}\tag{*1}$$
Assuming a polynomial solution $f_n = \lambda^n$, this leads to a characteristic equation:
$$ \lambda^{m+1} + \frac{q}{p} -\frac{1}{p}\lambda = 0 $$ But at this point I get stuck, since there isn't a general way to solve degree $m$ polynomials. When $m=2$, the polynomial factorises, but I don't think this is true in general. I suspect I will need a different method...
Any help, including just references is appreciated!
