How to prove $P(X=a\mid X+Y=r\cap X+Z=n\cap X+Y+Z+W=N)$ has hypergeometric distribution with parameters $(N,n,r)$?

Question

Could anyone help me with this probability problem, please?

Let $X,Y,Z,W$ be independent Poisson random variables with parameters $\lambda_1,\lambda_2,\lambda_3,\lambda_4$ respectively. Prove that $P(X=a\mid X+Y=r\cap X+Z=n\cap X+Y+Z+W=N)$ has hypergeometric distribution with parameters $(N,n,r)$, where $N,n,r$ are fixed integers numbers such as $n\ge0$,$N\ge r$ and $\lambda_1\lambda_4=\lambda_2\lambda_3$. In other words, prove that $P(X=a\mid X+Y=r\cap X+Z=n\cap X+Y+Z+W=N)$ = $\frac{\binom{r}a \binom{N-r}{n-a}}{\binom{N}n}$

So my idea was using Bayes to compute:

$P(X=a\mid X+Y=r\cap X+Z=n\cap X+Y+Z+W=N) = \frac{P(X+Y=r\cap X+Z=n\cap X+Y+Z+W=N\mid X=a) P(X=a)}{P(X+Y=r\cap X+Z=n\cap X+Y+Z+W=N)}$

But then I came to the conclusion that I have no idea how to compute $P(X+Y=r,X+Z=n,X+Y+Z+W=N)$. I tried using the product rule but it just leads nowhere since:

$P(X+Y=r,X+Z=n,X+Y+Z+W=N) = P(X+Y=r) P(X+Z=n\mid X+Y=r) P(X+Y+Z+W=N\mid X+Y=r\cap X+Z=n) $

And I just don't know how to compute $P(X+Z=n\mid X+Y=r)$ nor $P(X+Y+Z+W=N\mid X+Y=r\cap X+Z=n)$ since these don't make sense at all for me.

PS: I'm really sorry for the long post, but this problem has layers which makes difficult to just post the part where I was struggling with. Besides, it's an interesting problem, since the result is used in statistical analysis. Just wanted to say that English is not my native language too, so sorry for any mistakes :)

Answer 1

Could anyone help me with this probability problem, please?

Okay, yes.

So my idea was using Bayes to compute:

Good idea, but you are not quite on the right track.

The mutual independence of $X,Y,Z,$ and $W$ is there to be used.   It is useful.   So use it.

$$\def\e{\operatorname{\mathrm e}}\def\P{\operatorname{\sf P}}\begin{align} &~\P(X=a\mid X+Y=r,X+Z=n,X+Y+Z+W=N) \\[2ex]&=\dfrac{\P(X=a,Y=r-a,Z=n-a,W=N-r-n+a)}{\sum\limits_{k=0}^\infty \P(X=k,Y=r-k,Z=n-k,W=N-r-n+k)} \\[2ex]&=\dfrac{\P_X(a)\P_Y(r-a)\P_Z(n-a)\P_W(N-r-n+a)}{\sum\limits_{k=0}^\infty\P_X(k)\P_Y(r-k)\P_Z(n-k)\P_W(N-r-n+k) } \\[2ex]&=\dfrac{\dfrac{\lambda_1^a\e^{-\lambda_1}}{a!}\dfrac{\lambda_2^{r-a}\e^{-\lambda_2}}{(r-a)!}\dfrac{\lambda_3^{n-a}\e^{-\lambda_3}}{(n-a)!}\dfrac{\lambda_4^{N-r-n+a}\e^{-\lambda_4}}{(N-r-n+a)!}}{\sum\limits_{k=0}^{N-r-n}\dfrac{\lambda_1^k\e^{-\lambda_1}}{k!}\dfrac{\lambda_2^{r-k}\e^{-\lambda_2}}{(r-k)!}\dfrac{\lambda_3^{n-k}\e^{-\lambda_3}}{(n-k)!}\dfrac{\lambda_4^{N-r-n+k}\e^{-\lambda_4}}{(N-r-n+k)!}} \\[2ex]&~~\vdots\\[20ex]\\ \end{align}$$

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PS: Remember $\lambda_1\lambda_4=\lambda_2\lambda_3$.