# Product of three Poisson distributions

Product of two Poisson distributions is a Bessel function:

$$\sum_{r=0}^\infty \frac{e^{-f} f^r }{\Gamma(r+1)} \frac{e^{-g} g^r }{\Gamma(r+1)} = e^{-f-g} I_0\left(2 \sqrt{f g} \right)$$

What I need is the product of three:

$$\sum_{r=0}^\infty \frac{e^{-f} f^r }{\Gamma(r+1)} \frac{e^{-g} g^r }{\Gamma(r+1)} \frac{e^{-h} h^r }{\Gamma(r+1)} = \quad\mbox{?}$$

Is there a known special function? Any idea?

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It's doubtful that the answer is simpler than a hypergeometric function. Curiously though, are you really interested in the probability that 3 Poisson's are equal, or are you trying to find the distribution of the product of three Poisson random variables? – Alex R. Feb 24 '13 at 2:04
@Alex I'm optimizing a Bayesian cost function involving a Bayes least squares solution. The integral equation I need to solve happens to contain such expression. – Memming Feb 24 '13 at 17:08

Well, I guess according to the definition $$_pF_q(a_1,\ldots,a_p;b_1,\ldots,b_q;z) = \sum_{n=0}^\infty \frac{(a_1)_n\dots(a_p)_n}{(b_1)_n\dots(b_q)_n} \, \frac {z^n} {n!}$$ you have a generalized hypergeometric function: $$\sum _{r=0}^{\infty } \frac{f^r g^r h^r}{(r!)^3}= \ _0F_2(\cdot \ ;1,1;f g h)$$ And there is no way you can simplify it without using some approximations.