Consider the following contingence table:
$X_1, X_2, X_3 and X_4$ are r.v with independent Poisson distribution with parameters $(\lambda_i)_{i=1,…,4}$. Show that, (a) if $N\geq 0$ is a given integer, given that $\sum_{i=1}^4 X_i = N$, the conditional distribution of $(X_1, X_2, X_3)$ is multinomial with parameters $N$, $(\alpha_i)_{i=1,…,3} $ and (b) given $N\geq 0$ and $0 \leq n \leq N$, given that $X_1 + X_3 = n$ and $\sum_{i=1}^4 X_i = N$, the conditional distribution of ($X_1, X_2$) is the product of 2 independent binomials.
My attempt
a)
$P(X_1 = n_1 X_2 = n_2, X_3 = n_3 | \sum_{i=1}^4X_i= N) = \\
\\ \frac{P(X_1 = n_1) P(X_2 = n_2) P(X_3 = n_3) P(X_4 = N – n_1 -n_2 – n_3)}{P(\sum_{i=1}^4 X_i = N)}$
Since the $X_i$ are independent poisson, the sum is independent poisson and i just must finish the calculus
b) $P(X_1 = n_1, X_2 = n_2 | X_1 + X_3 = n, \sum_{i=1}^4 X_i = N) = \\
= \frac{P(X_1 = n_1, X=2 = n_2, X_3 = n-n_1, X_4 = N – n -n_2)}{P(X_1 + X_3 = n, \sum_{i=1}^4X_i = N)}$
The numerator we can use the independence , but I have no idea of how to calculate the denominator probability. I also dont know how to use the contingence table to help
Thanks in advance!
Best Answer
$$P(X_1+X_3=n,X_1+X_2+X_3+X_4=N)$$ $$=P(X_1+X_3=n,X_2+X_4=N-n)$$ $$=P(X_1+X_3=n)P(X_2+X_4=N-n)$$ $$=\sum_{i=1}^nP(X_1+i=n)P(X_3=i)+\sum_{i=1}^{N-n}P(X_2+i=N-n)P(X_4=i)$$ $$=\sum_{i=1}^nP(X_1=n-i)P(X_3=i)+\sum_{i=1}^{N-n}P(X_2=N-n-i)P(X_4=i)$$
Edit (proof independence of sum)
$$P(X_1+X_3=n,X_2+X_4=m)=\sum_{i=1}^nP(X_1+i=n,X_2+X_4=m)P(X_3=i)$$ $$=\sum_{i=1}^n\sum_{j=1}^mP(X_1+i=n,X_2+j=m)P(X_4=j)P(X_3=i)$$
How $X_1$ is independent from $X_2$: $$=\sum_{i=1}^n\sum_{j=1}^mP(X_1+i=n)P(X_2+j=m)P(X_4=j)P(X_3=i)$$ Then is possible to separate the double sum in: $$=\sum_{i=1}^nP(X_1+i=n)P(X_3=i)\sum_{i=1}^nP(X_2+j=m)P(X_4=j)$$ $$=P(X_1+X_3=n)P(X_2+X_4=m)$$