Transforming two normal random variables

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 $0 \le \theta \le \pi/2$.

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?

• $U$ should work out to be an exponential random variable and $V$ is the distribution of $\cos \Theta$ for $\Theta \sim U[0,2\pi)$. 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. 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 $$\frac{2XY}{\sqrt(X^2+Y^2)}=\frac{2r^2 \sigma^2\cos(\theta) \sin(\theta)}{r\sigma}=2r \cos(\theta) \sin(\theta) =r \sigma \sin(2\theta) \sim \sigma r \sin(\theta) \sim N(0,\sigma^2)$$