Distribution of $J$ if $K \sim \operatorname{Poisson}(\mu)$ and $J\mid K = k \sim \operatorname{Bn}(k,p)$.

Question

My working so far: $$ p_K(k) = \frac{\mu^k e^{-\mu}}{k!} \quad \text{and} \quad p_{J|K}(j|k) = {k \choose j}p^j(1-p)^{k-j} $$ Then $$ \begin{aligned} p_{J,K}(j,k) &= p_{J|K}(j|K)p_K(k) \\ &= p_{J|K}(j|k) = {k \choose j}p^j(1-p)^{k-j}\frac{\mu^k e^{-\mu}}{k!} \\ &= \frac{p^j(1-p)^{k-j}\mu^ke^{-\mu}}{j!(k-j)!} \end{aligned} $$ Marginalizing this we have $$ \begin{aligned} p_{J}(j) &= \sum_{k=0}^{\infty} \frac{p^j(1-p)^{k-j}\mu^ke^{-\mu}}{j!(k-j)!} \\ &= \frac{p^j}{j!(1-p)^j}\sum_{k=0}^{\infty} \frac{(\mu(1-p))^k}{(k-j)!} \end{aligned} $$ And this is where I get stuck, I can't simplify this any further. Have I done something wrong in my derivations? Thanks.

Answer 1

As $k$ starts from $j$,

$\displaystyle\sum_{k=j}^{\infty}\frac{({\mu}(1-p))^k}{(k-j)!}$ = ${{\mu}^j(1-p)^j}\displaystyle\sum_{n=0}^{\infty}\frac{({\mu}(1-p))^n}{n!}$ = ${{\mu}^j(1-p)^j}e^{{\mu}(1-p)}$

Thus, $p_J(j) = \displaystyle\sum_{k=0}^{\infty}\frac{p^j(1-p)^{k-j}{\mu}^k e^{-{\mu}}}{j!(k-j)!}$ = $\displaystyle\frac{p^j e^{-{\mu}}}{j!(1-p)^j}\sum_{k=j}^{\infty}\frac{({\mu}(1-p))^k}{(k-j)!}$ = $\displaystyle\frac{p^j e^{-{\mu}}}{j!(1-p)^j} * {{\mu}^j(1-p)^j}e^{{\mu}(1-p)}$

=$\displaystyle\frac{({\mu}p)^j e^{-{\mu}p}}{j!}$

which is $Poisson({\mu}p)$.

Sorry for bad LaTeX.