Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm reviewing for a test, and I am not sure if I am getting the right solution.

Let $X$ and $Y$ be iid $\mathcal{N}(0, \sigma^2)$ random variables.

a. Find the distribution of $U = X^2 + Y^2$, $V = \frac{X}{\sqrt{X^2 + Y^2}}$,

b. are $U,V$ independent?

c. Suppose $\sin(\theta) = V$. Find distribution of $\theta$ when $0 \le \theta \le \pi/2$.

(tentative answers):

I get

  1. $f_{U,V}(u,v) = \frac{1}{4\pi} \sigma^{-2} \exp \left[ -u/(2\sigma^2) \right]| (1-v^2)^{1/2} + (1-v^2)v^2|$,

  2. yes (density factors and supports dont rely on each other)and

  3. $g(\theta) = \left[\cos^2(\theta) + \cos^3(\theta)\sin^2(\theta)\right]\frac{1}{8 \pi \sigma^4}$.

Anybody recognize any of these distributions?

share|improve this question
$U$ should work out to be an exponential random variable and $V$ is the distribution of $\cos \Theta$ for $\Theta \sim U[0,2\pi)$. – Dilip Sarwate Jan 13 '13 at 4:43
1. Your answer to (3) is puzzling, because the total probability (the integral of $g$) will depend on $\sigma$, whereas it cannot: it must always equal $1$. 2. For some insight into this question, read about the Box Muller transform. – whuber Jan 13 '13 at 17:26
Thanks guys. (1) simplifies to $\left[ \frac{1}{2\pi}\frac{1}{\sqrt{1-v^2}}\right] \left[\frac{1}{2\sigma^2}\exp\left[ \frac{-u}{2\sigma^2}\right] \right]$, and that makes it easier to find the density of $\theta$, as well. – Taylor Jan 13 '13 at 18:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.