Probability – Are X and Y Independent if X, Y ~ P(?) and X + Y ~ P(2?)?

Question

Let $X$ and $Y$ be two random variables that follow the Poisson distribution $\mathscr P(\lambda)$ (with $\lambda > 0$).

It is well-known that if $X$ and $Y$ are independent then the random variable $X+Y$ follows the Poisson distribution $\mathscr P(2\lambda)$ (by direct computations).

Is the converse true ? If $X+Y$ follows the Poisson distribution $\mathscr P(2\lambda)$, are $X$ and $Y$ necessarily independent?

Answer 1

No. Consider $r(01)=-r(02)=r(12)=-r(21)=r(20)=-r(10)=\epsilon$ and $r(ij)=0$ outside. If $p_i=e^{-\alpha}\alpha^i/i!$ then $(p_ip_j+r(ij))$ is a counter example if $\epsilon$ is small enough.