Imagine we have a biased random walk on an interval $[0, L]$, where the probability of taking a $+1$ step is $p$ and the probability of taking a $-1$ step is $(1-p)$. At the reflecting boundary $0$, the walker will take a $+1$ step with probability $p$ or remain in place, and at the reflecting boundary $L$, the walker will take a $-1$ step with probability $(1-p)$ or remain in place.
My guess is that having some $+1$ step probability of $p$ and $-1$ step probability of $q$, where $(p + q) < 1$, would make no difference, that what matters is the ratio $\frac{p}{q}$.
Is it possible to find an exact stationary distribution for this Markov process by solving the following recurrence relations?:
$P(X=0) = (1-p)P(X=0) + (1-p)P(X=1)$
$P(X=L) = pP(X=L-1) + p(X=L)$
$P(X=n) = pP(X=n-1) + (1-p)P(X=n+1)$
Where: $0 < n < L$
If anyone is able to find a solution, please do let me know what techniques you used! I haven't made much traction here myself, and I'd actually like to learn how to solve these sorts of constant coefficient recurrence relation problems.
For the user Henry's exact solution for the above problem without the reflecting boundary at $L$:
$\pi(X=n) = (\frac{p}{q})^n(1-\frac{p}{q})$
Please see: Probability distribution for the position of a biased random walker on the positive integers