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Consider the following contingence table:

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$X_1, X_2, X_3 and X_4$ 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!

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$$P(X_1+X_3=n,X_1+X_2+X_3+X_4=N)$$ $$=P(X_1+X_3=n,X_2+X_4=N-n)$$ $$=P(X_1+X_3=n)P(X_2+X_4=N-n)$$ $$=\sum_{i=1}^nP(X_1+i=n)P(X_3=i)+\sum_{i=1}^{N-n}P(X_2+i=N-n)P(X_4=i)$$ $$=\sum_{i=1}^nP(X_1=n-i)P(X_3=i)+\sum_{i=1}^{N-n}P(X_2=N-n-i)P(X_4=i)$$

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: $$=\sum_{i=1}^nP(X_1+i=n)P(X_3=i)\sum_{i=1}^nP(X_2+j=m)P(X_4=j)$$ $$=P(X_1+X_3=n)P(X_2+X_4=m)$$

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  • $\begingroup$ Why is the third step true? $\endgroup$ – Giiovanna Apr 23 '14 at 12:39
  • $\begingroup$ Second to third equation? $\endgroup$ – rlartiga Apr 23 '14 at 12:40
  • $\begingroup$ Yes, the part when you split 1 probability to the product of 2 $\endgroup$ – Giiovanna Apr 23 '14 at 12:41
  • $\begingroup$ $X_1+X_3$ is independent from $X_2+X_4$ $\endgroup$ – rlartiga Apr 23 '14 at 12:42
  • $\begingroup$ Why? I mean, the variables $X_i$ are independent, but why does the sum? $\endgroup$ – Giiovanna Apr 23 '14 at 12:43

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