# Probability - Expectation and conditional distribution

I'm finding it hard to solve a specific question... Tried solving it for the past hour without success. I'll state the problem and show my attempt of solving it so maybe someone could help me. The question:

"Let $n$ random variables be given, $X[1],X[2],...,X[n];$ (The number in the brackets is the index and $n$ is natural) such that all of them are independent with the same probability distribution; Poisson with parameter $Lambda=3$. The random variable $Sn$ is defined as follows : $S[n]=X[1]+X[2]+...+X[n]$.

For $m>=1$ find: $E(Exp(S[n+m])|S[n])$; in other words the expectation of $Exp(S[n+m])$ given $S[n]$."

My attempt of solving it: Since I can neither upload any images showing the solution (low rating) nor type in the required mathematics language for this website, I'll summarize what I did==>

1) Using the convolution theorem, the probability distribution of $S[n]$ is Poisson with parameter $3n$.

2) $S[n+m]= S[n]+ X[n+1]+...+X[n+m]$, so if I define a R.V $M=X[n+1]+...+X[n+m]$ and call $U=S[n+m]$ we get that $U=S[n]+M$ such that M has a Poisson probability distribution with parameter $3m$.

3) $P(U=u|S[n]=s)=P(U=u\;AND\;S[n]=s)/P(S[n]=s)=P(M=u-s)/P(S[n]=s)$ and by substituting the relative probabilities one can find the probability distribution of $P(U|S[n])$ for $u>=s+1$.

4) Hence the answer is the sum over $Exp(U)*P(U|S[n])$ for all $u>=s+1$ which unfortunately doesn't give me the correct answer which is: $Exp(S[n]+3m(e-1))$.

Sorry for the long post, and thanks in advance for any help.

## 1 Answer

The mistake is where you write

$$\frac{P(U=u\land S[n]=s)}{P(S[n]=s)}=\frac{P(M=u-s)}{P(S[n]=s)}\;.$$

The probability in the numerator of the left-hand side is the probability of the intersection of two independent events, and on the right-hand side you only have the probability of one of those two events in the numerator. Since the events are independent, you don't need to go through the detour of setting up this fraction and then cancelling the denominator; you can simply write

$$P(U=u\mid S[n]=s)=P(M=u-s\mid S[n]=s)=P(M=u-s)\;.$$