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Let $X_{1}, \dots, X_{n}$ be independent Poisson random variables (parameter $\lambda$), and set $Y=\sum_{i=1}^{n}X_{i}$. Find the conditional probability mass function of the random variable $X_{i}$ given $Y=m$ for an integer $m$.


I have done the first part of the problem, where the distribution of the random variable $Y$ will be given by a Poisson random variable of parameter $n\lambda$ by using the mgf technique to identify the distribution of $Y$. However, I don't know how to compute $P(X_{i}=x | Y=m)$ since I don't know how to compute the joint pmf of $(X_{i},Y)$ (else, this should be straightforward).

How would one proceed with the problem? I initially though that $X$ and $Y$ could be independent, but I don't think that's the case.

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Outline: We want to compute $\Pr(X_i=k\mid Y=m)$. By the definition of conditional probability, this is $$\frac{\Pr((X_i=k)\cap (Y=m))}{\Pr(Y=m)}.\tag{1}$$ Note that $0\le k\le m$. You know how to compute the denominator, so we concentrate on the numerator.

The numerator is $$\Pr(X_i=k)\Pr(W=m-k),$$ where $W$ is the sum of all the $X_j$ except for $X_i$.

You know how to compute $\Pr(W=m-k)$.

Now when you substitute in (1) there will be a pleasant amount of cancellation.

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  • $\begingroup$ Sounds great. However, why would the probability of the intersection of the events be: $P(X_{i}=k)P(W=m-k)$ for $W$ being the sum of all the $X_{j}$ except for $X_{i}$? Do you use independence? I'm not sure on how to arrive at that step. The rest seems doable. Thanks! $\endgroup$ – arcbloom Dec 9 '15 at 4:54
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    $\begingroup$ We have $X_i=k$ and $Y=m$ if $X_i=k$ and the rest of the $X_j$ make up the rest of $m$, meaning they add up to $m-k$. Now use the fact that $X_i$ and the sum of the rest of the $X_j$ are independent random variables, so we multiply. $\endgroup$ – André Nicolas Dec 9 '15 at 5:02
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    $\begingroup$ The expressions will at first look a bit messy. But after the smoke clears, you should find that the conditional distribution of $X_i$ given $Y=m$ is binomial. $\endgroup$ – André Nicolas Dec 9 '15 at 5:08
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    $\begingroup$ You are welcome. By the way the parameters of the $X_k$ need not be all the same, again we get a binomial distribution. And the joint distribution of the $X_i$, given $Y=m$, is multinomial. $\endgroup$ – André Nicolas Dec 9 '15 at 5:28
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    $\begingroup$ Yes, there is a way to prove it, if the $X_i$ are independent, then any function of some of the variables is independent of any function of the rest of the variables. Formal proof is too long for a comment. Pretty easy for the discrete case. $\endgroup$ – André Nicolas Dec 9 '15 at 5:33

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