# 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.

• Please add the [self-study] tag & read its wiki. – gung - Reinstate Monica Aug 15 '16 at 13:01
• @gung Tag added. – M.R.Karimi Aug 15 '16 at 14:12
• What is "$k$"? I rather suspect your two equations are inconsistent except for special values of $a, b, \lambda_1, \lambda_2$ (such as $a=b=0$). From what information have you deduced them? – whuber Aug 15 '16 at 14:44
• @whuber I just explained the general form. – M.R.Karimi Aug 15 '16 at 18:21
• I see no additional explanation since I posted my comment. One reason for asking is that there are closed forms for your sum provided $k$ is a positive integer, but they get rapidly more complicated as $k$ increases. Since you haven't provided any information about $k$, it's unclear whether such solutions would be applicable. – whuber Aug 15 '16 at 19:52

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.