Let $N$ and $X$ distributed Poisson random variables with respect parameter $\lambda$ and $\lambda\cdot p$. Let $X\vert N$ have binomial distribution with number of trials $n$ and success probability $p$. Let $Y=N-X$, find distribution of $Y$.
$f_N(n)=\dfrac{e^{-\lambda}\cdot\lambda^n}{n!},n\in \mathbb{N}$
$f_X(x)=\dfrac{e^{-\lambda p}\cdot(\lambda p)^x}{x!},x\in \mathbb{N}$
To find distribution of $Y$, I tried to find probability mass function of $Y$,
\begin{eqnarray}
f_Y(y)&=&f_{N-X}(y).
\end{eqnarray}
I don't know how to find probability mass function of $Y$. So, someone please help me.
Best Answer
$$P\left(Y=k\mid N=n\right)=P\left(N-X=k\mid N=n\right)=$$$$P\left(X=n-k\mid N=n\right)=\binom{n}{n-k}p^{n-k}\left(1-p\right)^{k}=\binom{n}{k}\left(1-p\right)^{k}p^{n-k}$$
So under condition $N=n$ we are dealing with a binomial distribution with parameters $n$ and $1-p$.
Then $$P\left(Y=k\right)=\sum_{n=k}^{\infty}P\left(Y=k\mid N=n\right)P\left(N=n\right)=$$$$\sum_{n=k}^{\infty}\binom{n}{k}\left(1-p\right)^{k}p^{n-k}e^{-\lambda}\frac{\lambda^{n}}{n!}=e^{-\lambda}\frac{\left[\lambda\left(1-p\right)\right]^{k}}{k!}\sum_{n=k}^{\infty}\frac{\left(\lambda p\right)^{n-k}}{\left(n-k\right)!}=$$$$e^{-\lambda\left(1-p\right)}\frac{\left[\lambda\left(1-p\right)\right]^{k}}{k!}$$