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

Question

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.

Answer 1

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.