# Conditional distribution of random sample from Poisson distribution

I'm dealing with this problem for a while, and although I have some intuition, I think I'm missing something, please take a look:

We are given a random sample $X_1, X_2,...,X_n$ from Poisson$(\theta)$ distribution. I'm asked to prove that the conditional distribution of: $X_1, X_2,...,X_{n-1}$ given $Y=\sum_{i=1}^n X_i$ is multinomial.

What I have tried so far:

$P(X_1, X_2,...,X_{n-1} \mid Y = y) = \frac{P(X_1, X_2,...,X_{n-1},Y = y)}{P(Y=y)}$

Then, although I know the $n-1$ samples of $X_i$ are independent among each other, and that I could get their joint pdf easily, the problem comes up when I need to related such joint pdf to the pdf of $Y$. Since, I'm not sure whether they are independent, otherwise, the product of the pdfs would suffice.

I know that $Y \sim \text{Poisson}(n\theta)$.

Any clue? Thanks!

We have $$P(X_1=x_1,\ldots, X_{n-1}=x_{n-1}, Y=y) = P(X_1=x_1,\ldots, X_{n-1}=x_{n-1}, X_n=y-\sum_{i=1}^{n-1}x_i) \\=f(x_1)f(x_2)\ldots f(x_{n-1}) f\left(y-\sum_{i=1}^{n-1}x_i \right)$$ where $$f(x) = \frac{\theta^xe^{-\theta}}{x!}$$
• OK. I worked out the conditional distribution: $\frac{P(X_1=x_1,X_2=x_2,...,X_{n-1}=x_{n-1},X_n = y - \sum_{i=1}^{n-1} x_i)}{P(Y=y)}$ And obtained: $\frac{y!}{\prod_{i=1}^{n-1} x_i!} \cdot \frac{1}{n^y (y- \sum_{i=1}^{n-1} x_i) !}$ Do you guys recognize the multinomial distribution somehow? – pkenneth81 Apr 16 '18 at 2:39
• @htennek2k That doesn't look wrong math-wise, but can't say I do. Are you sure you aren't actually supposed to compute $$P(X_1=x_1,\ldots, X_n=x_n\mid Y=y)$$ where $Y=\sum_{i=1}^n X_i$? – spaceisdarkgreen Apr 16 '18 at 3:02
• Well, literally, the problem says: Show that the conditional pdf of $X_1,X_2,...,X_{n-1}$, given $Y=y$, is multinomial. This is from Introduction to Mathematical Statistics of Hogg, Problem 7.6.6. – pkenneth81 Apr 16 '18 at 3:23