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?

  • 2
    $\begingroup$ $U$ should work out to be an exponential random variable and $V$ is the distribution of $\cos \Theta$ for $\Theta \sim U[0,2\pi)$. $\endgroup$ – Dilip Sarwate Jan 13 '13 at 4:43
  • 1
    $\begingroup$ 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. $\endgroup$ – whuber Jan 13 '13 at 17:26
  • $\begingroup$ 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. $\endgroup$ – Taylor Jan 13 '13 at 18:24

Because there's a subtlety here, this question is worth a correct answer. But let's develop it with as little work as possible, in the most straightforward manner.

What subtlety? The variables $(U,V)$ do not determine $(X,Y).$

The change of variables from $(X,Y)$ to $(U,V)$ is two-to-one: because $(U,V)$ gives us information about $Y$ only in the form of $Y^2,$ whenever $(X,Y)$ corresponds to $(U,V),$ so does $(X,-Y).$ Almost surely, $Y\ne -Y$ (the chance of this for a Normal distribution of $Y$ is zero). This means the density of $(U,V)$ will be twice what a mindless application of the routine Calculus formulas indicates.

Those routine formulas are, of course, the computation of the Jacobian. This is just an old-fashioned term for computing the differential element $\mathrm{d}x\,\mathrm{d}y$ in terms of the new variables. One of the easiest ways to work it out starts with the formulas

$$X = V\sqrt{U};\ Y = \sqrt{1-V^2}\sqrt{U}.$$

Taking differentials,

$$\begin{aligned} \mathrm{d}x\,\mathrm{d}y &= \left(\frac{v}{2\sqrt{u}}\mathrm{d}u + \sqrt{u}\mathrm{d}v\right)\,\left(\frac{\sqrt{1-v^2}}{2\sqrt{u}}\mathrm{d}u - \frac{v\sqrt{u}}{\sqrt{1-v^2}}\mathrm{d}v\right) \\ &= \frac{1}{2\sqrt{1-v^2}}\mathrm{d}v\,\mathrm{d}u. \end{aligned}$$

Substitute everything into the original probability element for the bivariate Normal distribution taking care to indicate what the possible values of the variables $u$ and $v$ can be. Omitting this is another pitfall that plagues the uninitiated, so I will be explicit, using $\mathcal{I}$ to represent the indicator function:

$$\begin{aligned} f_{X,Y}(x,y)\,\mathrm{d}x\,\mathrm{d}y &= \frac{1}{2\pi\sigma^2} \exp\left(-\frac{x^2+y^2}{2\sigma^2}\right)\,\mathrm{d}x\,\mathrm{d}y \\ &= \frac{1}{2\pi\sigma^2} \exp\left(-\frac{u}{2\sigma^2}\right)\, \frac{1}{2\sqrt{1-v^2}}\mathrm{d}v\,\mathrm{d}u\ \mathcal{I}(u\ge 0)\,\mathcal{I}(-1\le v\le 1). \end{aligned}$$

(It would be incorrect to write this formula without the indicator functions. Usually readers are expected to notice that $\sqrt{1-v^2}$ is defined only for $-1\le v\le 1,$ so in informal settings we can get away without indicating that explicitly; but it's not quite as noticeable that although $\exp(-u/(2\sigma^2))$ is defined everywhere, its integral diverges unless $u$ is explicitly restricted.)

Introduce the factor of $2$ from the two-to-one transformation and notice the probability element splits into a factor depending only on $u$ and one depending only on $v:$

$$f_{U,V}(u,v)\,\mathrm{d}v\,\mathrm{d}u = \left[\frac{1}{2\sigma^2} \exp\left(-\frac{u}{2\sigma^2}\right)\,\mathrm{d}u\ \mathcal{I}(u\ge 0)\right]\, \left[\frac{1}{\pi\sqrt{1-v^2}}\,\mathrm{d}v\ \mathcal{I}(-1\le v\le 1)\right].$$

In one stroke this answers (a) and (b): because the probability element factors, the variables $U$ and $V$ are independent.

As a check, you may integrate the two factors separately over the set of real numbers: each integrates to $1,$ as it must for any univariate probability distribution.

Question (c) is a routine exercise in a univariate change of variable, so discussing it doesn't add any further interest.


By Box Muller transformation

$X=r\cos(\theta) \hspace{.5cm} Y=r\sin(\theta) \hspace{.5cm} X,Y \sim normal(0,1) \Leftrightarrow \theta \sim Uniform(0,2\pi) \hspace{.5cm} r^2\sim chi(2)$.
$X$ and $Y$ are independent $\Leftrightarrow $ $\theta$ and $r$ are independent.

also $\sin(\theta) \sim \cos(\theta) \sim \sin(2\theta) \sim 2\sin(\theta) \cos(\theta) \sim \cos(2\theta) \sim \cos(2\theta) \sim f $ that $ f(z) =\frac{1}{\pi \sqrt(1-z^2)} I_{[-1,1]}(z) $ since $z=\sin(\theta) \Rightarrow f(z)=|\frac{d}{dz} \sin^{-1}(z)| f_{\theta}(\sin^{-1}(z)) + |\frac{d}{dz} (\pi-\sin^{-1}(z))| f_{\theta}(\pi -\sin^{-1}(z)) =\frac{1}{\sqrt(1-z^2)} \frac{1}{2\pi} +\frac{1}{\sqrt(1-z^2)} \frac{1}{2\pi} =\frac{1}{\pi\sqrt(1-z^2)} $

similar for others.

in hence if $X,Y(i.i.d)\sim N(0,\sigma^2)$ so $X=\sigma r \cos(\theta)$ and $Y=\sigma r \sin(\theta)$.

in hence

$U=\sigma^2 r^2$ and $ V=\frac{\sigma r\cos(\theta)}{r\sigma}=\frac{r\cos(\theta)}{r}=\cos(\theta) \sim f $ and $r$ is independent from any function of $\theta$ like $\sin(\theta)$.

another example $\frac{2XY}{\sqrt(X^2+Y^2)}=\frac{2r^2 \sigma^2\cos(\theta) \sin(\theta)}{r\sigma}=2r \cos(\theta) \sin(\theta) =r \sigma \sin(2\theta) \sim \sigma r \sin(\theta) \sim N(0,\sigma^2)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.