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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.

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    $\begingroup$ 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. $\endgroup$ – Aaron Montgomery Nov 7 '19 at 15:03
  • $\begingroup$ @AaronMontgomery Thanks for pointing that out. Yes, they are independent. I updated the question. $\endgroup$ – KRL Nov 10 '19 at 23:12
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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.

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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.)

I guess all that remains from here to prove the result is some gnarly integration, which a good table of integrals of good CAS should be help with. There also is very likely some much more straightforward to show this result that I'm just not seeing right this second. I also imagine it's a theorem that's been proved in some book somewhere.

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