Does the sum of Poisson random variables have a Poisson distribution? 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?
 A: Let $X, Y$ be independent and both distributed $\mathcal{Poiss}(\lambda)$.
$$\begin{align}
\mathsf{Var}(X+Y) & = \mathsf E((X+Y)^2)-(\mathsf E(X+Y))^2
\\ & = \mathsf E(X^2+2XY+Y^2)-(\mathsf E(X)^2+2\mathsf E(X)\mathsf E(Y)+\mathsf E(Y)^2)
\\ & = \mathsf E(X^2)-\mathsf E(X)^2+\mathsf E(Y^2)-\mathsf E(Y)^2 \color{gainsboro}{+ 2(\mathsf E(XY)-\mathsf E(X)\mathsf E(Y))} 
\\ & = 2\lambda
\\[4ex]
\mathsf{Var}(2X) & = \mathsf E((2X)^2)-(\mathsf E(2X))^2
\\ & = 4\mathsf E(X^2)-4\mathsf E(X)^2
\\ & = 4\lambda
\\[4ex]
\therefore \mathsf {Var}(\sum_{k=1}^n X_k) & \neq \mathsf {Var}(nX_1) & n>1
\end{align}$$
That is the difference.
$\Box$

tl;dr: In general $\mathsf {Var}(X+Y) = \mathsf {Var}(X)+\mathsf {Var}(Y)+2\mathsf {Cov}(X,Y)$.   When $X,Y$ are independent the covariance term vanishes, but when assuredly $X=Y$ then the covariance is equal to the variance of $X$.
A: 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 $n^2\lambda$.
