Let $X, Y$ both be independent normal random variables with mean zero and variance $\sigma^2$. Let $U:= X^2 + Y^2$ and $V:= X/U^{\frac12}$.

Find the joint distribution and find out whether $U$ and $V$ are independent.

For the joint distribution I got: $f_{u,v}(u,v) = \frac{1}{2\pi\sigma} e^{-\frac{v\sqrt{u}}{2\sigma^2}} e^{-\frac{\sqrt{u-v^2u}}{2\sigma^2}}\frac{\sqrt{u}}{2\sqrt{u-v^2u}}$

That does not look like it's independent, is that correct? Also I am not really keen about calculating marginals and therefore integrating that function...

  • $\begingroup$ Are you assuming $X$ and $Y$ are normal? Without some additional information such as that, there's no way to find the joint distribution. $\endgroup$ – Robert Israel Jun 30 at 18:13
  • $\begingroup$ yes, totally forgot that $\endgroup$ – jde Jun 30 at 18:23

I will assume $\sigma=1$ this will not affect the method of proof. Consider an arbitrary positive bounded Borel function $h$. Then $$\eqalign{\mathbb{E} h(U,V)&=\mathbb{E} h\left(X^2+Y^2,\frac{X}{\sqrt{X^2+Y^2}}\right)\cr &=\frac{1}{2\pi}\int_{\mathbb{R}^2} h\left(x^2+y^2,\frac{x}{\sqrt{x^2+y^2}}\right)e^{-(x^2+y^2)/2}dxdy\cr &=\frac{1}{2\pi}\int_0^\infty\int_0^{2\pi} h\left(r^2,\cos\theta\right)e^{-r^2/2}r d\theta dr\cr &=\frac{1}{2\pi}\int_0^\infty\int_0^{\pi} h\left(u,\cos\theta\right)e^{-u/2} d\theta du\cr &=\frac{1}{2\pi}\int_0^\infty\int_{-1}^1 h\left(u,v\right)\frac{e^{-u/2}}{\sqrt{1-v^2}} dv du\cr &=\int_{\mathbb{R}^2}h(u,v)f_{U,V}(u,v)dudv }$$ and because $h$ is arbitrary we conclude that $$\eqalign{f_{U,V}(u,v)&=\mathbb{I}_{(0,+\infty)\times(-1,1)}(u,v) \frac{1}{2\pi}\frac{e^{-u/2}}{\sqrt{1-v^2}}\cr &=\mathbb{I}_{(0,+\infty)}(u) \frac{1}{2}e^{-u/2}\cdot \mathbb{I}_{(-1,1)}(v) \frac{1}{\pi}\frac{1}{\sqrt{1-v^2}}\cr &=f_U(u)\cdot f_V(v) } $$ This proves that $U$ and $V$ are independent and gives their probability density distributions.


Consider the random vector $\vec v=(X,Y)$, which is a normal random vector whose covariance matrix is $\sigma^2 I$. The amazing property of the (multivariate) Gaussian distribution is that it is rotationally symmetric whenever the covariance matrix is isotropic, as it is in this case. Rotational symmetry implies that the direction that the vector $\vec v$ points in is uniformly distributed on the circle, regardless of what the magnitude of $\vec v$ is. More precisely, it means that the conditional distribution of the angle is independent of the magnitude, which is equivalent to saying that they are independent.

At this point, the burning question is: why is the distribution of $\vec v$ rotationally symmetric? One way to see it is by changing to polar coordinates. Indeed, by definition of the normal distribution, we have that for any measurable subset $A$ of the plane, $$ \mathbb P(\vec v\in A)=\frac{1}{2\pi}\int_{A}e^{-x^2/2\sigma^2}e^{-y^2/2\sigma^2}\ dx\ dy $$ $$ =\frac{1}{2\pi}\int_{A}e^{-r^2/2\sigma^2}\ (r\ dr\ d\theta). $$ Here, we see that the integrand does not depend on $\theta$, which implies the rotational symmetry.

Finally, we can answer the question: $U=\|\vec v\|^2$ is a function of the magnitude and $V=\cos\angle \vec v$ is a function of the angle, so by the first paragraph these quantities are independent.


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