Let $X$ be a random variable that follows a Poisson distribution with the mean value $m$.
Let $Y$ be the random variable which conditional probability by $X = n$ follows a binomial distribution with parameters $n,p$.
Prove that:
$$ p(Y = k) = \frac{(pm)^k e^{-mp}}{k!} $$.
$X$ follows a Poisson distribution, which means: $p(X = n) = \frac{m^n}{n!}e^{-m}$.
and $p(Y = k | X = n) = C^k_n p^k (1 – p)^{n-k}$ because it is a binomial distribution.
We have: $$p(Y = k | X = n) = \frac{ p(Y = k \text{ and } X = n) }{p(X = n) }$$
I don't know how to proceed to get the result.
Best Answer
Law of total probability $$P(Y=k)=\sum_{j=0}^{\infty}{P(Y=k|X=j)P(X=j)}$$
$$P(Y=k|X=j)= {j \choose k}p^{k}(1-p)^{j-k} $$
$$P(X=j)=e^{-m}\frac{m^j}{j!}$$
$$P(Y=k)=\sum_{j=0}^{\infty}{{j \choose k}p^{k}(1-p)^{j-k}e^{-m}\frac{m^j}{j!}}=\sum_{j=k}^{\infty}{{j \choose k}p^{k}(1-p)^{j-k}e^{-m}\frac{m^j}{j!}}$$
$$\sum_{j=k}^{\infty}{\frac{j!}{k!(j-k)!}p^{k}(1-p)^{j-k}e^{-m}\frac{m^j}{j!}}=\frac{p^ke^{-m}}{k!}\sum_{j=k}^{\infty}{\frac{1}{(j-k)!}(1-p)^{j-k}m^j}$$
Change of variable $i=j-k$, and use the definition of exponential in terms of series
$$\sum_{j=k}^{\infty}{\frac{1}{(j-k)!}(1-p)^{j-k}m^j}=\sum_{i=0}^{\infty}{\frac{1}{i!}(1-p)^{i}m^{i+k}}=m^ke^{(1-p)m}$$ Finally,
$$P(Y=k)=\frac{p^ke^{-m}}{k!}m^ke^{(1-p)m}=\frac{(pm)^k e^{-pm}}{k!}$$