# 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.

• It is even worse than that: $P(x=u)/P(x=v)$ obviously factors into a function of $u$ and a function of $v$ while I absolutely do not understand how you are going to factor $(c+av)^u/(d+bu)^v$ that way unless $a=b=0$, in which case everything is trivial. So, it looks like your system is inconsistent more often than not... – fedja Aug 15 '16 at 19:23

Unless I've made a mistake, this doesn't seem to be possible.

One thing you can do is look at the joint MGF. Conditioning on $X_1$,

\eqalign{M(s,t) &= \mathbb E\left[ \exp(s X_1 + t X_2) \right] = \mathbb E \left[ \exp(s X_1) \mathbb E[\exp(t X_2) | X_1] \right] \cr &= \mathbb E \left[ \exp(s X_1) \exp((\lambda_2 + b X_1)(e^t-1))\right]\cr &= \exp(-\lambda_2 (e^t-1)) M(s+ b (e^t-1),0)}

Similarly, conditioning on $X_2$,

$$M(s,t) =\exp(-\lambda_1 (e^s-1)) M(0, t + a (e^s-1))$$

Combining these and replacing $s$ by $s - b (e^t-1)$:

$$M(s,0) = \exp(\lambda_1 (1 - e^{s + b (1-e^t)})) + \lambda_2(e^t-1)) M(0, a\exp(s + b(1-e^t))-a+t)$$

and similarly

$$M(0,t) = \exp(\lambda_2 (1 - e^{t + a (1-e^s)})) + \lambda_1 (e^s-1)) M(b\exp(t + a(1-e^s))-b+s, 0)$$

But replacing $t$ by $a\exp(s + b(1-e^t))-a+t$ here gives us a rather complicated functional equation for $M(s,0)$:

$$M \left( s,0 \right) =M \left( b{{\rm e}^{a{{\rm e}^{-b{{\rm e}^{t}}+b +s}}-a{{\rm e}^{s}}+t}}-b+s,0 \right) {{\rm e}^{-\lambda_{{2}}{{\rm e} ^{a{{\rm e}^{-b{{\rm e}^{t}}+b+s}}-a{{\rm e}^{s}}+t}}-\lambda_{{1}}{ {\rm e}^{-b{{\rm e}^{t}}+b+s}}+\lambda_{{2}}{{\rm e}^{t}}+{{\rm e}^{s} }\lambda_{{1}}}}$$ Now taking the derivative with respect to $t$ and evaluating at $t=0$ gives us a differential equation for $M(s,0)$:

$$\dfrac{\partial M}{\partial s}(s,0) = -{\frac {M \left( s,0 \right) \left( -\lambda_{{2}} \left( -ab{ {\rm e}^{s}}+1 \right) +\lambda_{{1}}{{\rm e}^{s}}b+\lambda_{{2}} \right) }{b \left( -ab{{\rm e}^{s}}+1 \right) }}$$ which we can solve (with initial condition $M(0,0)=1$): $$M(s,0) = \left( \dfrac{abe^s-1}{ab -1} \right)^{\frac{\lambda_2}{b} + \frac{\lambda_1}{ab}}$$ Unfortunately, this does not satisfy the functional equation. So it seems there is no solution in general.