Let $X_1,X_2,..,X_N$ be independent identical samples from a Poission distribution with unknown parameter $\theta$
How would I find the sum of the distribution $X_1+X_2+X_3+..+X_N$ when n is finite. I looked up convolution but that only works for two random variables and I have multiple random variables and they are all poisson.
Best Answer
This what I did not sure if it is right. $W=X_1+X_2+..+X_N$ we have $M_w(t)=E(e^{X_1+X_2+..+X_N})=E(e^{x_1t}*e^{x_2*t}*..e^{x_n*t})$ And this equals
$M_{X_1}(T)*M_{X_2}(T)*..*M_{X_N}(t)=[M_X(t)]^n=[e^{\theta(e^t-1)}]^n$
So now you $W~Poisson(n*\theta))$ which is the same as pmf $(e^{-n*\theta}(n*\theta)^x)/x!$