Show that the conditional pmf of $X_i$, given $T = t$, is $Binomial(t, \lambda_i/\lambda).$ The Statement of the Problem:
The random variables $X_1, ..., X_n$ are independent and $X_i \sim  Poisson(\lambda _i), i = 1, ..., n$. Set
$$ T = \sum_{i=1}^n X_i \qquad \text{and} \qquad \lambda = \sum_{i=1}^n \lambda_i$$
and show that
(a) The conditional density ( I assume that this is actually referring to the conditional mass) of $X_i$, given $T = t$, is $Binomial(t, \lambda_i/\lambda), i = 1, ..., n.$
(b) What does the distribution in part (a) become for $\lambda_1 = \cdots=\lambda_n = 0.5$?
Where I Am:
I'm not really sure what to do with this problem. I assume that, for part (a), the first step is to simply convert the problem into its formal state:
$$ P(X_i=x_i|T=t) = \frac{P(X_i = x_i, T = t)}{P(T=t)} $$
aaaannnnndddd that's about as far as I can get. I know that a Poisson r.v. with parameter $\lambda = np$ provides a good approximation to a binomial r.v. with parameters $n,p$ when $n$ is large and $p$ is small, but since there's not information about the size of these parameters, I assume that I can't use that fact. Anyway, any help here would be appreciated because I'm pretty lost. Thanks.
 A: First we can assume without loss of generality that $n = 2$ since a sum independent Poisson r.v.'s is itself Poisson.  For $0 \leq k \leq t, k \in \mathbb{N}$ we have
\begin{align}
P(X_1 = k \mid X_1 + X_2 = t) &= \frac{P(X_1 = k \cap X_1 + X_2 = t)}{P(X_1 + X_2 = t)} \\
&= \frac{P(X_1 = k) P(X_2 = t - k)}{P(X_1 + X_2 = t)} \\
&= \frac{e^{-\lambda_1} \frac{\lambda_1^k}{k!} e^{-\lambda_2} \frac{\lambda_2^{t - k}}{(t - k)!}}{e^{-(\lambda_1 + \lambda_2)}\frac{(\lambda_1 + \lambda_2)^t}{t!}} \\
&= \frac{t!}{k! (t - k)!} \left ( \frac{\lambda_1}{\lambda_2} \right )^k \left ( \frac{\lambda_2}{\lambda_1 + \lambda_2} \right )^{t - k} \left ( \frac{\lambda_2}{\lambda_1 + \lambda_2} \right )^k \\
&= \binom{t}{k} \left ( \frac{\lambda_1}{\lambda_1 + \lambda_2} \right )^{k} \left ( \frac{\lambda_2}{\lambda_1 + \lambda_2} \right )^{t - k} .
\end{align}
A: Hint: $T=\sum_{k=1}^nX_k \sim Poisson(\lambda)$
$T_1=\sum_{k=2}^nX_k\sim Poisson(\lambda-\lambda_1)$
and $P(X_1=x_1,T=t)=P(X_1=x_1, \sum_{k=2}^nX_k=t-k)=P(X_1=x_1)P( \sum_{k=2}^nX_k=t-k)$
now just use Poisson formula and you will get the result.
