Invariant measure of a reflected random walk Let $(X_n), n \geq 0$ be a Reflected Random Walk defined by:
$X_0 = 0$ and:
$ X_{n+1}=\max( 0 , X_n + \xi )$
$\xi $ is a random variable such that $P(\xi=a)=\theta$ and $P(\xi=-b)=1-\theta$ for a given of strictly positive integers $a$, $b$.
I think that this RRW is positively recurrent iif $\theta \leq \frac{b}{a+b}$. (i computed the drift as if the RW was not reflected).
So if i understood the theory, i know that there exists a stationary distribution in this case.
But when i try to compute it just using the definition $(\pi P = \pi)$ i have a linear equation which links $\pi_n, \pi_{n-a}$ and $\pi_{n+b}$ for large enough $n$. I don't know what i can do to get an explicit expression for $\pi$ from that?
Please let me know if i made a mistake, or if you have any idea for this computation.
Thank you in advance!
 A: Actually if $\theta=b/(a+b)$ the drift is zero on $[b,+\infty)$ hence the chain is null recurrent. Positive recurrence holds for every $\theta\lt b/(a+b)$, which we assume from now on.
As you explain, the stationary distribution solves $\pi=\pi P$. For every $n\geqslant a$, this reads $$\pi(n)=\theta\pi(n-a)+(1-\theta)\pi(n+b).$$ Such linear systems are solved through their characteristic equation $$r^a=\theta+(1-\theta)r^{a+b}.$$ In general, this has $a+b$ solutions $(r_k)$, one of them being $1$, hence $$\pi(n)=\sum_{k=1}^{a+b}c_kr_k^n,$$ for some constants $(c_k)$ determined by the "limit" conditions, that is, by $(\pi(n))_{0\leqslant n\lt a}$, and by  the normalizing condition $$\sum\limits_{n=0}^\infty\pi(n)=1.$$ The normalizing condition imposes that $c_k=0$ for every $|r_k|\geqslant1$ hence the sum over $k$ giving $\pi(n)$ can be restricted to those $k$ such that $|r_k|\lt1$.

Example: Consider the case $(a,b)=(1,2)$ with $\theta=\frac12$, then the chain is indeed positively recurrent and its characteristic equation is $\frac12r^3-r+\frac12=0$, with roots $1$, $p=\frac12(\sqrt5-1)$, and $-\frac1p$, hence $p$ is in $(0,1)$ and $-\frac1p\lt-1$. 
The only root such that $|r|\lt1$ is $p$ hence $\pi(n)=cp^n$ for some constant $c$. The "limit" condition reads $\pi(0)=\frac12\pi(0)+\frac12\pi(1)+\frac12\pi(2)$, which is satisfied by every sequence  $\pi(n)=cp^n$. Finally, the normalizing condition $\sum\limits_{n=0}^\infty\pi(n)=1$ yields $$\pi(n)=(1-p)p^n.$$
