So I have been taught that the sum of Poisson random variables has a Poisson distribution. However, I have a problem with this.
Suppose you have a Poisson random variable $X$ with $E(X) = a$.
Then a sum of $n$ Poisson random variables gives a mean of $E(nX) = n(a).$
However the variance will be $\text{Var}(nX) = n^2(a)$
Here lies the problem. If $nX$ has a Poisson distribution then the mean and the variance has to be the same; this is a property of Poisson distribution. But since they are different, the distribution can't be Poisson. So how do I resolve this discrepancy?
Best Answer
The sum of independent Poisson has Poisson distribution. Let $X_i$, $i=1$ to $n$, be independent Poisson and have parameters $\lambda_i$ ($i=1$ to $n$). Then the variance of $X_1+\cdots +X_n$ is $\lambda_1+\cdots +\lambda_n$. In particular, if all the $\lambda_i$ are equal to $\lambda$, the variance is $n\lambda$, not $n^2\lambda$.