Suppose $X\sim \text{Poisson}(\lambda)$, $Y\sim \text{Poisson}(\mu)$. If $X+Y\sim \text{Poisson}(\lambda+\mu),$ can we conclude that $X$ and $Y$ are independent?
I know that, if we assume that the conditional distribution of $X$ given $W=X+Y$ is a Binomial distribution, then $X, Y$ can be shown to be independent. I wonder whether the independence holds in absence of any such assumption.
No. The joint distribution has a lot more degrees of freedom than the marginal distribution of either variable or of the sum, so you can’t fix the former by specifying the latter.
For example, consider how we can change the $k^2$ probabilities $P(X=x\cap Y=y)$ of the independent joint distribution for $0\le x,y\lt k$ without changing any of the given distributions. The marginal distribution of $X$ fixes $k$ sums over these probabilities, the marginal distribution of $Y$ fixes another $k$ sums and the distribution of the sum fixes another $2k-1$ sums, for a total of at most $4k-1$ linear constraints. (These constraints are actually not all linearly independent, but there’s no need to go into that.) For $k\ge4$, we have $k^2\gt4k-1$, so there are fewer constraints than degrees of freedom, and you can solve the corresponding system of linear equations to find increments that you can add to the independent joint distribution without changing any of the given distributions. If you choose the increments small enough, all the probabilities will remain in $[0,1]$.