Are $U=\frac{2X_1^2}{(X_2+X_3)^2}$ and $V=\frac{2(X_2-X_3)^2}{2X_1^2+(X_2+X_3)^2}$ independent?

Question

Consider i.i.d standard normal variables $X_1,X_2,X_3$. How can I determine whether $U=\frac{2X_1^2}{(X_2+X_3)^2}$ and $V=\frac{2(X_2-X_3)^2}{2X_1^2+(X_2+X_3)^2}$ are independently distributed?

This was part of a multiple choice question, so I am wondering if there is a short argument. I can show that $Y_1=X_1,Y_2=X_2-X_3,Y_3=X_2+X_3$ are all independent of each other. And $U$ is a function of $(Y_1,Y_3)$ while $V$ is a function of $(Y_1,Y_2,Y_3)$. Also, $U$ and $V$ are functionally dependent since $V=\frac{2Y_2^2/Y_3^2}{U+1}$. But that doesn't help me answer the question. It can be seen that the marginals of $U,V$ are $F$ distributions. Do I have to find the joint density of $(U,V)$ through a change of variables?

Another idea was to try applying Basu's theorem. So I introduced a parameter $\sigma^2$ as the variance of the $X_i$'s. But then both $U$ and $V$ seem to be an ancillary statistic for $\sigma^2$.

Answer 1

Let $X=X_1,$ $Y=(X_2+X_3)/\sqrt2,$ and $Z=(X_2-X_3)/\sqrt2.$ As in your question, it is apparent that these are iid standard Normal. Moreover,

$$U = \frac{X^2}{Y^2}\ \text{ and }\ V = 2\frac{Z^2}{X^2 + Y^2}.$$

Consider, then, how $X/Y$ and $X^2+Y^2$ are related. The latter is the squared distance from the origin while the former is the cotangent of the angle $\Theta$ made by $(X,Y)$ to the $X$-axis. Because iid Normal variables are spherical, the angle and the distance are independent.

Recall that any (measurable) functions of independent variables are independent. Call these functions $f$ and $g.$ Beginning with the variables $\Theta$ and $(R, Z) = (\sqrt{X^2+Y^2}, Z),$ we have just seen $(\Theta,R,Z)$ are independent. Observing that we may express $V = 2Z^2/R^2 = g(R,Z)$ and $U = \cot^2(\Theta) = f(\Theta),$ we immediately see $(U,V)$ are independent.

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In retrospect, it is evident $(\Theta, R, Z)$ is a cylindrical coordinate system for the original Cartesian coordinates $(X_1,X_2,X_3).$ Thus, if you prefer an explicitly rigorous, Calculus-based derivation, consider computing the joint distribution function in these cylindrical coordinates: it should separate into a term for $\theta$ and a term for $(r,z).$

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BTW, this is a challenge for those of us who like to draw inspiration from simulations: both $U$ and $V$ have infinite means and, in even fairly large simulations, the $(U,V)$ scatterplot can look decidedly dependent. Here, for instance, is a log-log plot for 4,000 random $(U,V)$ pairs.

The lack of many extreme values of $U$ makes it look like $V$ tends to be high when $U$ is extreme.

Here's the R simulation code.

n <- 4e3
x <- rnorm(n)
y <- rnorm(n)
z <- rnorm(n)
u <- 2x^2 / (y+z)^2
v <- 2(y-z)^2 / (2*z^2 + (y+z)^2) 
plot(u,v, log="xy", col=gray(0, alpha=0.1), asp=1)