Independence of $\chi^2$ distribution Suppose we are given that $X_1$ and $X_2$ are two non-negative random variable such that $$X_1+X_2\sim\chi^2_{(2)}$$. Also we are given that $X_1\sim\chi^2_{(1)}$. Can we say the following

*

*$X_2$ is independent of $X_1$?

*$X_2\sim\chi^2_{(1)}$?

If not, then please provide a counterexample.
This is just a fun question question. I'm asking it just for curiosity. Any kind of help appreciated.

Another way to construct a counterexample: let $Z\sim\chi^2_2$ be independent of $X_1$, and define $X_2=Z-X_1$. Since $X_2$ is not even strictly non-negative, it's not $\chi^2_1$.
However, if $X_2$ is independent of $X_1$, then $X_2\sim \chi^2_1$: the characteristic function of $X_1+X_2$ is the product of the two individual characteristic functions, which is enough to determine the distribution of $X_2$.
