# Conditional Distribution of Poisson Variables, given $\sum X_i$

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.

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, $$P\left\{(X_1=x_1, X_2=x_2, \ldots, X_n=x_n)\cap \{Y = N\}\right\}\\[1em] = \begin{cases}e^{-n\lambda}\frac{\lambda^{\sum_i x_i}}{x_1!x_2!\cdots x_n!}, & \text{if}~\sum_i x_i = N,\\0, & \text{if}~\sum_i x_i \neq N,\end{cases}$$ 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\}}\\[1ex] &= \frac{N!}{n^Nx_1!x_2!\cdots x_n!} \quad\text{if}~ \sum_i x_i = N\\[1ex] &= \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.
• 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 Commented Sep 8, 2015 at 14:47
• @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 $y = \sum_{i=1}^n x_i$ Commented Sep 12, 2015 at 21:20