A binomial multiplied by a poisson

Question

What distribution do you obtain when you multiply a poisson distribution and a binomial distribution, and why? I'm assuming you obtain a poisson distribution.

Answer 1

Here is a physical example:

If we interpret $\mu$ and the probability that a photon produces an electron, then for a given number of photons entering the photo detector (say $n$) the probability distribution of electrons coming out is a binomial distribution with n trials and a probability of success of $\mu$.

$$ P(m)=\frac{n!}{m!(n-m)!}\mu^m (1-\mu)^{n-m}_{} =\binom{n}{m}\mu^m (1-\mu)^{n-m}_{} $$

The number of photons that go into the binomial distribution is the output of a Poisson distribution. We cannot get more electrons out than photons that went into the photo detector. We can sum up all the possible binomial distributions with a Poisson distribution weighting factor.

$$ P(m,\lambda,\mu)=\sum_{j=m}^{\infty} \frac{j!}{(j-m)!m!}\mu^m(1-\mu)^{j-m} \frac{\lambda^j e^{-\lambda}}{j!} $$ Simplify and bring terms that do not depend on j outside of the sum.

$$ P(m,\lambda,\mu)= \frac{\mu^m e^{-\lambda}}{m!} \sum_{j=m}^{\infty} \frac{\lambda^j(1-\mu)^{j-m}}{(j-m)!} $$ Let $n=j-m$

$$ P(m,\lambda,\mu)= \frac{\mu^m e^{-\lambda}}{m!} \sum_{n=0}^{\infty} \frac{\lambda^{n+m}(1-\mu)^{n}}{n!} $$

$$ P(m,\lambda,\mu)= \frac{(\lambda \mu)^m e^{-\lambda}}{m!} \sum_{n=0}^{\infty} \frac{(\lambda (1-\mu))^{n}}{n!} $$

$$ P(m,\lambda,\mu)= \frac{(\lambda \mu)^m e^{-\lambda \mu}}{m!} $$

This is a Poisson distribution with a mean of $\lambda \mu$!

Note: the "!" in the last line of text should be understood as excitement and a not a factorial.