Suppose that $X \sim \mathcal{X}^2_n$ and $Y \sim \mathcal{X}^2_m$ are independent. Can we say that $\frac{X}{Y}$ is independent of $X+Y$? For example, can we show that
$$p(X/Y|X+Y) = p(X/Y)?$$
We know that the sum of two independent chi-squared random variables is chi-squared, so $X+Y \sim \mathcal{X}^2_{n+m}$ and $\frac{X}{Y}$ has an F distribution.
Best Answer
Fix some function $\phi : \mathbb{R}^2 \to \mathbb{R}$. By a change of variables \begin{align} &E[\phi(X+Y, X/Y)] \\ &=\int_0^\infty \int_0^\infty \phi(x+y,x/y) f(x,y) \, dx \, dy \\ &= \int_0^\infty \int_0^\infty \phi(x+y, x/y) c_n x^{n/2 - 1} e^{-x/2} \cdot c_m y^{m/2 - 1} e^{-y/2} \, dx \, dy \\ &= c_n c_m \int_0^\infty \int_y^\infty \phi(u, (u-y)/y) (u-y)^{n/2 - 1} y^{m/2 - 1} e^{-u/2} \, du \, dy & u = x+y \\ &= c_n c_m \int_0^\infty \int_0^u \phi(u, (u-y)/y) (u-y)^{n/2 - 1} y^{m/2 - 1} e^{-u/2} \, dy \, du \\ &= c_n c_m \int_0^\infty \int_0^\infty \phi(u, v) \left(\frac{uv}{v+1}\right)^{n/2-1} \left(\frac{u}{v+1}\right)^{m/2-1} e^{-u/2} \frac{u}{(v+1)^2} \, dv \, du & v = \frac{u}{y} - 1 \end{align} This holds for any $\phi$ for which the expectation exists. This implies the joint PDF of $(U,V) := (X+Y, X/Y)$ is $$f(u,v) \propto u^{(n+m)/2-1} e^{-u/2} \cdot \frac{v^{n/2-1}}{(v+1)^{(n+m)/2}}.$$ Since the joint PDF is separable (can be written as $f(u,v) = g(u)h(v)$), we see that $U$ and $V$ are independent.