Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
add comment

1 Answer 1

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.