When sum of two Poisson variables is not Poisson

Question

If $X$ and $Y$ are independent Poisson random variables, then $X+Y$ is Poisson. We also know that sum of two dependent Poisson variables might result in another Poisson variable.

My question is: what would be the counterexample (not the rigorous proof which might exist) that negates the claim that sum of two Poissons, irrespective of their dependency, would always result in Poisson.

Answer 1

The expected value of a Poisson random variable is equal to its variance.

Now just take two Poisson random variables with parameters $\lambda_1$ and $\lambda_2$ with non-vanishing covariance. Then their sum can’t be Poisson as the sum’s expected value is $\lambda_1+\lambda_2$ but the sum’s variance doesn’t equal $\lambda_1+\lambda_2$.

Answer 2

If $X\sim \mathcal{P}(\lambda)$, then $$\Bbb P(X+X=1)=0$$ Therefore $X+X$ does not follow a Poisson distribution.