Finding the joint distribution from poisson conditionals Suppose that for two discrete random variables $X_1$ and $X_2$, we know their conditional distributions. Namely
$$X_1~|~X_2 = x_2 \sim \mathrm{Poisson}(\lambda_1 + ax_2),$$
$$X_2~|~X_1 = x_1 \sim \mathrm{Poisson}(\lambda_2 + bx_1).$$
We want to calculate their joint distribution, $p(X_1, X_2)$.
My own idea is to divide the equations above and find $\frac{p_{X_1}(x_1)}{p_{X_2}(x_2)}$ and then sum over all values of $X_1$ to find $1/p_{X_2}(x_2)$. But for this, I have to compute a very bad series. 
Do you have any idea for this problem?
P.S. The series I have to compute is of form
$$\sum_{n=0}^\infty \frac{c^n}{n!(a+nb)^k},$$
which I hardly believe to have a closed form.
 A: Your assumptions quickly leads to contradictions except for certain values of $a$ and $b$.   Keeping in mind that the joint probability mass function and the conditional distributions are proportional, we can derive an expression for $p(x_1,x_2)$ in terms of $p(0,0)$ in two ways.  From the assumed conditional Poisson distributions it follows that
\begin{equation}
\frac{p(0,x_2)}{p(0,0)} = \frac{p_{X_2|X_1=0}(x_2)}{p_{X_2|X_1=0}(0)} = \frac{e^{-\lambda_2}\lambda_2^{x_2}/x_2!}{e^{-\lambda_2}} = \lambda_2^{x_2}/x_2!.
\end{equation} 
Similarly,
\begin{equation}
\frac{p(x_1,x_2)}{p(0,x_2)} = \frac{p_{X_1|X_2=x_2}(x_1)}{p_{X_1|X_2=x_2}(0)} = \frac{e^{-(\lambda_1+a x_2)}(\lambda_1+a x_2)^{x_1}/x_1!}{e^{-(\lambda_1+a x_2)}} = (\lambda_1+a x_2)^{x_1}/x_1!
\end{equation} 
Hence,
\begin{equation}
p(x_1,x_2) = \frac{(\lambda_1+a x_2)^{x_1}\lambda_2^{x_2}}{x_1!x_2!}p(0,0).
\end{equation}
Doing the same argument but going via $p(x_1,0)$ leads to
\begin{equation}
p(x_1,x_2) = \frac{\lambda_1^{x_1}(\lambda_2+b x_1)^{x_2}}{x_1!x_2!}p(0,0).
\end{equation}
The two last equations can both be true only if
\begin{equation}
(\lambda_1+a x_2)^{x_1}\lambda_2^{x_2} = \lambda_1^{x_1}(\lambda_2+b x_1)^{x_2}
\end{equation}
for all $(x_1,x_2)$ which is only possible if $a=b=0$, that is, if $X_1$ and $X_2$ are independent.  Otherwise, the assumption that the conditional distributions are Poisson lead to a contradiction and are thus inconsistent.
