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The question is:

Let $Y_1, Y_2, ..., Y_k$ be $k$ independent Poisson random variables with parameters $\lambda_1, \lambda_2, ..., \lambda_k$ respectively. Find the conditional distribution of $Y_1, Y_2, ..., Y_k,$ given $\sum_{i=1}^k Y_i=y$.

I know that $f(y_1, y_2, ..., y_k|y)=f(y_1, y_2, ..., y_k, y)/f(y)$. Does this mean I need to find $f(y_1,y_2,...,y_k,y)$? How do I do that?

I've done transformations with the same number of variables, but not with adding in another variable.

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  • $\begingroup$ Are you interested in the joint distribution of $(Y_1,Y_2,\dots,Y_k)$ or in the marginal distributions, i.e in the distribution of each $Y_i$? $\endgroup$ – zhoraster Jan 19 '16 at 15:37
  • $\begingroup$ I want to find the conditional distribution, $f(y_1, y_2, ..., y_k|y)$. The only way I know how to find that is by using the equation I gave below the block quote, but I don't know how to find the joint distribution $f(y_1, y_2, ..., y_k,y)$. $\endgroup$ – Stephanie Jan 19 '16 at 15:42
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I understand that by $f(y_1,y_2,\dots,y_k,y)$ you mean $P(Y_1 = y_1,Y_2=y_2\dots,Y_k=y_k,Y = y)$.

Hint Note that by independence $$ f(y_1,y_2,\dots,y_k,y) = \prod_{i=1}^k \frac{\lambda_i^{y_i}}{y_i!}e^{-\lambda_i} $$ whenever $y_1 + y_2 + \dots + y_k = y$ and $0$ otherwise.

Now since $Y$ has Poisson distribution with parameter $\lambda = \lambda_1 + \dots + \lambda_k$, it is a matter of simple computation to conclude that, given $Y = y$, $(Y_1,\dots,Y_k)$ has the multinomial distribution with parameters $y$ (the number of trials) and $\lambda_1/\lambda,\dots,\lambda_k/\lambda$ (the probabilities of different outcomes).

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  • $\begingroup$ Thank you for your response! I asked another grad student about this and he agreed with you. Is that good enough for me to click the check mark? I don't have a solution to check it against. $\endgroup$ – Stephanie Jan 19 '16 at 18:08

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