I 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 &lt; x) &amp;= \begin{cases}<br /> 0 &amp; x &lt; -1 \\<br /> P(\arccos(x) &lt; \Theta &lt; 2\pi - \arccos(x)) &amp; -1 \leq x \leq 1\\<br /> 1 &amp; x &gt; 1 <br /> \end{cases} \\<br /> P(Y &lt; y) &amp;= \begin{cases}<br /> 0 &amp; y &lt; -1 \\<br /> P(\pi + \arcsin(|y|) &lt; \Theta &lt; 2\pi - \arcsin(|y|)), &amp; -1 \leq y &lt; 0 \\<br /> P(0 &lt; \Theta &lt; \arcsin(y)) + P(\pi - \arcsin(y) &lt; \Theta &lt; 2\pi), &amp; 0 \leq y \leq 1 \\<br /> 1, &amp; y &gt; 1<br /> \end{cases}<br /> \end{align*}<br />

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.