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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?

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  • $\begingroup$ I hope you were taught that the sum of independent Poisson random variables is Poisson. $nX$ is the sum of $n$ Poisson random variables ($X,\; X,\;\ldots X$), but they are not independent. $\endgroup$ – Robert Israel Feb 9 '16 at 23:29
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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 $n^2\lambda$.

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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.

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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 $X$.

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