Multiple Poisson r.v.s - conditional probability given the sum of r.v.s is a specific value Suppose $Y_1, Y_2, \dots, Y_n$ are independent Poisson random variables with means $\lambda_1, \lambda_2, \dots, \lambda_n$.
How do you find the conditional probability function of $Y_1$ given that $\displaystyle\sum_{i=1}^n Y_i=m$?
I know that $\displaystyle\sum_{i=1}^n Y_i \sim Poisson(\displaystyle\sum_{i=1}^n \lambda_i)$. 
I've never worked with conditional probabilities, so forgive me if the solution here is trivial...
 A: Ah this is quite nice since the answer that comes out is quite intuitive. Anyway you just apply Bayes' theorem, to get (for $0\leq k \leq m$):
$$
\begin{aligned}
\Pr\left[Y_1 = k \bigg\vert \sum_{i=1}^n Y_i = m\right] &= \frac{\Pr\left[Y_1 = k \cap \sum_{i=1}^n Y_i = m\right]}{\Pr\left[\sum_{i=1}^n Y_i = m\right]} \\
&= \frac{\Pr\left[Y_1 = k \cap \sum_{i=2}^n Y_i = m-k\right]}{\Pr\left[\sum_{i=1}^n Y_i = m\right]} \\
&= \frac{\Pr\left[Y_1 = k \right]\Pr\left[\sum_{i=2}^n Y_i = m-k\right]}{\Pr\left[\sum_{i=1}^n Y_i = m\right]}
\end{aligned} $$
In the final equality we used the independence of $Y_1$ and $\sum_{i=2}^k Y_i $.
Now you know all of the above expressions, since
$Y_1 \sim \mathrm{Pois}(\lambda_1)$, $\sum_{i=2}^n Y_i \sim \mathrm{Pois}(\sum_{i=2}^n \lambda_i)$ and $\sum_{i=1}^n Y_i \sim \mathrm{Pois}(\sum_{i=1}^n \lambda_i)$.
Plugging all of this in will finally yield:
$$ Y_1 \bigg\vert \sum_{i=1}^n Y_i = m \sim Bin\left(m, \frac{\lambda_1}{\sum_{i=1}^n \lambda_i}\right) $$
A: You can assume that $Y_1 = n$ for whatever $n$ you want, and then compute the entire probability that the rest of the variables sum to $m - n$, by exploiting the fact that the sum of the rest of the variables is Poisson with mean given by the sum of means of the individual Poissons.
