Transform from polar to cartesian

[mathman]

TL;DR

Integrand defined on unit circle. Express in (x,y) coordinates.

Probability distribution - uniform on unit circle. In polar coordinates $dg(r,a)=\frac{1}{2\pi}\delta(r-1)rdrda$. This transforms in $df(x,y)=\frac{1}{2\pi}\delta(\sqrt{x^2+y^2}-1)dxdy$. The problem I ran into was the second integral was 1/2 instead of 1.

 

[pasmith]

This is a univariate distribution. $\Theta$ is uniformly distributed in $[0,2\pi]$; $X = \cos \Theta$ and $Y = \sin\Theta$ can then be determined from that.
$$\begin{align*}<br /> P(X < x) &= \begin{cases}<br /> 0 & x < -1 \\<br /> P(\arccos(x) < \Theta < 2\pi - \arccos(x)) & -1 \leq x \leq 1\\<br /> 1 & x > 1 <br /> \end{cases} \\<br /> P(Y < y) &= \begin{cases}<br /> 0 & y < -1 \\<br /> P(\pi + \arcsin(|y|) < \Theta < 2\pi - \arcsin(|y|)), & -1 \leq y < 0 \\<br /> P(0 < \Theta < \arcsin(y)) + P(\pi - \arcsin(y) < \Theta < 2\pi), & 0 \leq y \leq 1 \\<br /> 1, & y > 1<br /> \end{cases}<br /> \end{align*}$$

$X$ and $Y$ are not independent, but satisfy $X^2 + Y^2 = 1$.

 

[mathman]

I discovered the problem. $\delta(\sqrt{x^2+y^2}-1)$ is satisfied for two values of x, so when integrating to get a function of y, I had to double the single value I had obtained.