# Are sum and ratio of two independent chi-squared random variables are independent?

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.

• Minor corrections: I think you mean that $X, Y$ are independent, right? Also, $X/Y$ does not have an $F$-distribution, but $(X/m)/(Y/n)$ does. – Aaron Montgomery Nov 7 '19 at 15:03
• @AaronMontgomery Thanks for pointing that out. Yes, they are independent. I updated the question. – KRL Nov 10 '19 at 23:12

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 $$f(u,v) = f(u)f(v)$$), we see that $$U$$ and $$V$$ are independent.
As in my comment, I'll assume you meant for $$X, Y$$ to be independent. I originally thought the answer to your question was no, and I set out to write that down, and now I'm pretty sure the answer is yes. Note that what follows is not a proof, but rather intended to be a convincing heuristic with a possible sketch of a proof. If that's not what you need, my apologies.
You're looking to verify the equality $$\mathbb P \left( \frac X Y \leq a, X + Y \leq b \right) = \mathbb P \left( \frac X Y \leq a\right) \mathbb P \left(X + Y \leq b \right).$$ You can evaluate these probabilities with integrals. For the left side of the inequality, the region on quadrant I of the $$xy$$-plane described by $$x/y \leq a$$ and $$x + y \leq b$$ is the triangle bounded by $$(0,0)$$, $$(\frac{ab}{a+1}, \frac{b}{a+1})$$, and $$(0,b)$$. The left side of that equation is therefore $$\int_0^{\frac{ab}{a+1}} \int_{x/a}^{b-x} f_X(x) f_Y(y) \, \textrm{d}y \, \textrm{d} x$$ where $$f_X$$ and $$f_Y$$ are the respective density functions of $$X, Y$$. Similarly, by analyzing the regions described on the right side of the original equation, we see that the right side can be evaluated as $$\left(\int_0^{\infty} \int_{x/a}^{\infty} f_X(x) f_Y(y) \, \textrm{d} y \, \textrm{d} x\right) \left(\int_0^b \int_0^{b-x} f_X(x) f_Y(y) \, \textrm{d} y \, \textrm{d} x \right).$$ I was fairly sure that those two expressions wouldn't be equal to one another, so I set up a Desmos calculator to play around with them, and... well, they seem to be. (Note that I had to use $$10,000$$ as a stand-in for $$\infty$$ in the integral limits because Desmos doesn't like infinity there.)