Let $f(.)$ be a probability mass function on the non-negative integers such that $0<f(n)<1$ and $f(0)+f(1)+...=1$. Then does there exist an irreducible birth-death process with stationary distribution $f(.)$? If it does not exist then what is a counter example?
So far, I fixed the following:
Let $$ a_k=P(X_{n+1}=k-1|X_n=k) \\ b_k=P(X_{n+1}=k+1|X_n=k) \\ c_k=P(X_{n+1}=k|X_n=k) $$ where $a_k>0$, $b_k>0$ to make the birth-death process irreducible. Then if $f(.)$ is a stationary distribution to this irreducible birth-death process then it must be the case that
$$ f(0)=\frac{1}{1+\sum_{n=1}^{\infty}\prod_{j=0}^{n-1}\frac{b_j}{a_{j+1}}} $$ and $$ f(n+1)=\frac{b_n ...b_0}{a_{n+1}...a_1} f(0) $$
But I am not sure how to use these results to conclude.
We need to find values for $\{a_n\},\{b_n\},\{c_n\}.$ From the conditions for stationarity we see that
$$\frac{f(n+1)}{f(n)}=\frac{b_n}{a_{n+1}},n\ge0.$$
This will be satisfied if we define $b_n=f(n+1)/2, n\ge0$ and $a_n=f(n-1)/2, n\ge1.$
Then $c_n=1-(a_n+b_n)=1-(f(n-1)+f(n+1))/2.$ The divisor of $2$ guarantees $c_n\ge0.$
The fact that we have $0\lt a_n\lt 1$ and $0<b_n\lt 1$ makes the chain irreducible.
The final issue is to assure the stationary distribution exists. For this we need to show:
$$ \sum_{n=1}^{\infty}\frac{b_{n-1}b_{n-2}...b_0}{a_na_{n-1}...a_1}\lt\infty $$ This term is $$\sum_{n=1}^{\infty}\frac{f(n)}{f(0)}=\frac{1-f(0)}{f(0)}\lt\infty$$