A binomial multiplied by a poisson 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.
 A: 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. 
