Poisson Distribution - sum of RVs Question:
$X$ balls are thrown to $n$ bins (each ball has an equal chance to get to each bin). Let $X_1,\dots, X_n$ be the amount of balls in each cell.
a. Show that if $X \sim \text{Poisson}(\lambda)$ then $\displaystyle X_i \sim \text{Poisson} \left(\frac \lambda n\right)$ and that the $X_i$'s are independent.
My thought:
I know how to prove it the other way around, but not from the sum to the parts of it...
Here's a chance to reverse the steps (I'm not sure how I can justify the last move, and how I should figure indepndence:
$P(X=k)=e^{-\lambda} \frac {\lambda^k}{k!}=
e^{-(\frac \lambda n)n} \frac {((\frac \lambda n)n)^k}{k!}=
\frac {e^{-(\frac \lambda n)n}}{k!} \sum_{m=0}^k \frac {k!}{m!(k-m!)}(\frac \lambda n)^m{((\frac \lambda n)(n-1))^{k-m}}=
 {e^{-(\frac \lambda n)n}} \sum_{m=0}^k \frac 1{m!(k-m)!}(\frac \lambda n)^m{((\frac \lambda n)(n-1))^{k-m}}=$
$ {e^{-(\frac \lambda n)n}} \sum_{m=0}^k \frac {(\frac \lambda n)^m}{m!}\frac {((\frac \lambda n)(n-1))^{k-m}}{(k-m)!}=$
$ \sum_{m=0}^k {e^{-(\frac \lambda n)}} \frac {(\frac \lambda n)^m}{m!} {e^{-(\frac \lambda n(n-1))}}\frac {((\frac \lambda n)(n-1))^{k-m}}{(k-m)!}=$
$\sum_{m=0}^k P(X_1=m)P(\sum_{i=2}^nX_i=k-m)$
 A: *

*First we calculate the probability of : $P(X_i=k | X=m)=\binom mk (\frac 1n)^k(1-\frac 1n)^{m-k}$ . This is true because we pick k out of m balls to put in the $i$'th bin, so k balls fall into that bin with probability $(\frac 1n)$ and the rest of the balls are are not in it with probability $(1-\frac 1n)^{m-k}$.

*Now using the law of total probability:
$ P(X_i=k)=
\sum _{m=k}^\infty P(X_i=k | X=m) P(X=m)=
\sum _{m=k}^\infty \binom mk (\frac 1n)^k(1-\frac 1n)^{m-k} e^{-\lambda} \frac {\lambda ^m}{m!}=
\sum _{m=k}^\infty \frac 1{k!(m-k)!} (\frac 1n)^k(1-\frac 1n)^{m-k} e^{-\lambda} {\lambda ^m}=$
$\frac {e^{-\lambda}}{k!} (\frac {\lambda}n)^k\sum _{m=k}^\infty \frac 1{(m-k)!} (1-\frac 1n)^{m-k} \lambda ^{m-k}=$
$\frac {e^{-\lambda}}{k!} (\frac {\lambda}n)^k\sum _{l=0}^\infty \frac {(\lambda -\frac {\lambda }n)^l} {l!} =$
$ \frac {e^{-\lambda}}{k!} (\frac {\lambda}n)^k e^{(\lambda -\frac {\lambda }n)} =
 \frac {e^{\frac {\lambda }n}}{k!} (\frac {\lambda}n)^k \Rightarrow X_i \sim Poisson(\frac np) $


Now I still need to find out how to show independence...
A: MGF's are easy and by far the most convenient way to solve your problem.
$$
\begin{align}
\psi_X(u) &= \mathbb{E} e^{uX}\\ 
&=\sum_{k=0}^{\infty} e^{uk} \frac{\lambda^k}{k !} e^{-\lambda}\\
&=e^{-\lambda} \sum_{k=0}^{\infty} \frac{(\lambda e^{u})^k}{k !}\\
&=\exp\{ \lambda(e^u -1) \}
\end{align}
$$
Now, since you know that $X_i$'s are independent, you know that $\psi_X(u) = \prod_{i=1}^n \psi_{X_i}(u) $. All you need to prove that the only way this is going to happen is that $X_i \sim  Poisson(\frac{\lambda}{n}) $. 
You can start by computing an MGF of $X_i$ following the computation above.
A: You need to try and reverse the steps in your previous answer.
The important point is that each ball is goes independently and uniformly (equally as likely to go in any of the buckets)
So for a sequence $X_1,\dots, X_n$, with $\sum_{i=1}^n X_n = X$ you can write down the probability that I get $X$ from a Poisson distribution  and the probability that when I throw $X$ balls randomly into $n$ buckets I get the sequence $X_1,\dots X_n$.
So you can write down the probability that you get the sequence $X_1,\dots,X_n$, but you need to manipulate it so it looks like a product of Poisson probabilities.
You've done all the equations, you just need to write them down in the reverse order and keep your head clear to make sure each step is justified. 
