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$X_1,X_2,$. . . . . . $X_k$ are independent $poisson$ random variables.

We need to show that the conditional distribution of $X_1$ given $X_1+X_2+X_3+....+X_k$ is a binomial distribution.

I started off with finding the $M.G.F$ of $(X_1 | Y=y)$ where $Y= X_1 + X_2 + X_3 +.... X_k$ :

$M_{X_1 | Y=y}(t) = E(e^{t(X_1 | Y=y)})=E(e^{t(X_1 |X_1 + X_2 + X_3 +.... X_k =y)})=E(e^{t(X_1 |X_1 =y-(X_2 + X_3 +.... X_k ))})$

=> $E(e^{t(X_1 |X_1 =y- \sum_{i=2}^{k}X_i)})=E(e^{t(y- \sum_{i=2}^{k}X_i)})$

=> $e^{ty}E(e^{t(- \sum_{i=2}^{k}X_i)})=e^{ty}E(e^{t(-X_2-X_3.....-X_k)})$

=> $e^{ty}E(e^{(-t)X_2})E(e^{(-t)X_3}) ............E(e^{(-t)X_k})$

=> $e^{ty}e^{\lambda_2(e^{-t}-1)}e^{\lambda_3(e^{-t}-1)}..............e^{\lambda_k(e^{-t}-1)}$

It doesn't look like a $M.G.F$ of Binomial distribution(does it ?).

Could anyone help ?

( Assuming , $X_i$ ~ $Pois(\lambda_i)$ )

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  • $\begingroup$ mgf of binomial distribution ${\displaystyle \,\left(1-p+pe^{t}\right)^{n}}$ $\endgroup$ – Nebo Alex Jan 14 '17 at 17:16
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Hint: Firstly, prove that $ S_k = X_1 + \ldots + X_k$ follows a Poisson distribution

with parameter $ \lambda = \lambda_1 + \ldots + \lambda_k$ using MGFs. Then, observe that

$$ P(X_1 = x| S_k = y ) = \frac{P(X_1 = x, S_k = y)}{P(S_k =y)} = \frac{P(X_1 = x,~ X_2 + \ldots + X_k= y -x )}{P(S_k=y)}$$ and use indepedence to proceed.

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  • $\begingroup$ Wouldn't that be $P(X_1=x)$ only ? $X_1 , \sum_{i=2}^k X_i$ being independent ? $\endgroup$ – User9523 Jan 14 '17 at 18:43
  • $\begingroup$ @User9523 Yes you are right. I edited my answer. $\endgroup$ – Greg Jan 14 '17 at 20:52
  • $\begingroup$ Got It. Thanks ! $\endgroup$ – User9523 Jan 15 '17 at 9:32

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