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An exercise asks me to show that if $X_1,X_2,\ldots,X_n$ is a random sample from a Poisson distribution with parameter $\theta$, the conditional distribution of $X_1,X_2,\ldots, X_{n-1}$, given $Y=\sum_{i=1}^n X_i$, is multinomial.

Normally, this would have been an easy one but unfortunately the joint distribution here throws me off a bit. This is where I would appreciate some help.

Thank you in advance.

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The joint probability mass function of the $X_i$is $$p_{\mathbf X}(\mathbf x) = \prod_{i=1}^n e^{-\lambda}\frac{\lambda^{x_i}}{x_i!} = e^{-n\lambda}\frac{\lambda^{\sum_i x_i}}{x_1!x_2!\cdots x_n!}.$$ $Y = \sum_i X_i$ is a Poisson random variable with parameter $n\lambda$ and so $P\{Y = N\} = e^{-n\lambda}\frac{(n\lambda)^{N}}{N!}$. Now, $$P\left\{(X_1=x_1, X_2=x_2, \ldots, X_n=x_n)\cap \{Y = N\}\right\} = \begin{cases}e^{-n\lambda}\frac{\lambda^{\sum_i x_i}}{x_1!x_2!\cdots x_n!}, & \text{if}~\sum_i x_i = N,\\0, & \text{if}~\sum_i x_i \neq N,\end{cases}$$ and so $$\begin{align} p_{\mathbf X}(\mathbf x \mid Y=N) &= \frac{P\{(X_1=x_1, X_2=x_2, \ldots, X_n=x_n)\cap (Y = N)\}}{P\{Y=N\}}\\ &= \frac{N!}{n^Nx_1!x_2!\cdots x_n!} ~\text{if}~ \sum_i x_i = N\\ &= \frac{N!}{x_1!x_2!\cdots x_n!}\left(\frac{1}{n}\right)^{x_1} \left(\frac{1}{n}\right)^{x_2}\cdots\left(\frac{1}{n}\right)^{x_n} ~\text{where}~ \sum_i x_i = N \end{align}$$ which is a multinomial distribution.

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  • $\begingroup$ I am wondering if there are differences between $P(X_1=x_1, X_2,=x_2, ...,X_{n-1}=x_{n-1}\mid Y) $and $P(X_1=x_1, X_2,=x_2, ...,X_n=x_n \mid Y)$, would you like to explain, thanks $\endgroup$ – Deep North Sep 8 '15 at 14:47
  • $\begingroup$ @DeepNorth $$P\left(X_1=x_1, X_2,=x_2, ...,X_{n-1}=x_{n-1}\mid Y=y\right) =P \left(X_1=x_1, X_2,=x_2, ...,X_{n}=x_{n}\mid Y=y\right) $$ since both sides are $0$ if $y \neq \sum_{i=1}^n x_i$; and have value as given in my answer if $y = \sum_{i=1}^n x_i$ $\endgroup$ – Dilip Sarwate Sep 12 '15 at 21:20
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    $\begingroup$ What is the intuition for this result? $\endgroup$ – jII Nov 1 '15 at 17:59

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