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.
 A: 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 $X_2$.
A: Suppose $F_{(1)}(x)$ is the cumulative distribution function of $\chi^2_{(1)}$, and $F^{-1}_{(1)}(x)$ its inverse i.e. quantile function, and $F_{(2)}(x)$ is the cumulative distribution function of $\chi^2_{(2)}$.
Then

*

*if $X_1 \sim \chi^2_{(1)}$ so $F_{(1)}(X_1)\sim Unif(0,1)$

*and $X_2 =F^{-1}_{(2)}\left(F_{(1)}(X_1)\right)-X_1$ so $X_2$ is determined by $X_1$ and thus not independent

*then $X_3 =X_1+X_2 \sim \chi^2_{(2)}$
$X_2 \not \sim \chi^2_{(1)}$: for example it has a variance of about $0.409$ while $X_1$ has a variance of $2$ and $X_3$ has a variance of $4$
requested in comments:
This $X_2$ is non-negative because you have $F_{(1)}(x) \ge F_{(2)}(x)$ and then apply a non-decreasing function gives $F^{-1}_{(2)}\left(F_{(1)}(x)\right) \ge x$ and thus $X_2 =F^{-1}_{(2)}\left(F_{(1)}(X_1)\right)-X_1\ge 0$
