Let $N$~Pois$(\lambda)$, $X|(N=n)$~Bin$(N,p)$, $Y=N-X$. Show $X$, $Y$ are independent and Poisson with parameters $\lambda p$ and $\lambda (1-p)$.

Any direction on this problem would be much appreciated.

So far I know the joint distribution of $X$ and $Y$ is

$\begin{align} \mathsf P(X=x, Y=y) & = \mathsf P(X=x, N-X=y) \\ & = \mathsf P(X=x, N=y+x) \\ & = \mathsf P(X=x\mid N=y+x)\cdot \mathsf P(N=y+x) \\ & = \binom{y+x}{x}p^{x}(1-p)^{y}\cdot\frac{\lambda^{y+x}e^{-\lambda}}{(y+x)!} \end{align}$.

From here, how do I show $X$ and $Y$ independent with $X\sim\operatorname{Pois}(\lambda p)$ and $Y\sim\operatorname{Pois}(\lambda(1-p))$?

Thanks in advance!

Solution 1:

Conditioning on $N$ you get

$P(X=x)=\sum_{n=x}^{\infty}\binom{n}{x}p^x(1-p)^{(n-x)}e^{-\lambda}\frac{\lambda^n}{n!} $

$=e^{-\lambda}\frac{(\lambda p)^x}{x!}\sum_{n=x}^{\infty}\frac{(\lambda (1-p))^{n-x}}{(n-x)!}$ $=e^{-\lambda}\frac{(\lambda p)^x}{x!}e^{\lambda (1-p)}=e^{-\lambda p} \frac{(\lambda p)^x}{x!}$

Similarly you can get the distribution of Y.

Now to check independence multiply these two and check if you get the joint distribution you calculated.

Solution 2:

\begin{align} \binom{y+x}{x} p^{x}(1-p)^{y}\cdot\frac{\lambda^{y+x}e^{-\lambda}}{(y+x)!} \\ =\frac{\ (y+x)!}{y!x!} p^{x}(1-p)^{y}\cdot\frac{\lambda^{y+x}e^{-\lambda}}{(y+x)!} \\ =\frac{\lambda^{x} p^{x} e^{-\lambda p}}{x!} \cdot\frac{\lambda^{y}(1-p)^{y}e^{-\lambda (1-p)}}{y!} \end{align}

Hence X and Y are independent Poisson with λ and λ(1-p) respectively.

It is not necessary to sum the above from 0 to infinity wrt to y to get pdf of x, and to sum the above from 0 to infinity wrt to x to get pdf of y, since it is clear that the first part does not contain any y variable and the second any x variable.