Transform from polar to cartesian

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SUMMARY

The discussion focuses on transforming a uniform probability distribution on a unit circle from polar coordinates to Cartesian coordinates. The initial polar distribution is represented as ##dg(r,a)=\frac{1}{2\pi}\delta(r-1)rdrda##, which transforms to ##df(x,y)=\frac{1}{2\pi}\delta(\sqrt{x^2+y^2}-1)dxdy##. A key issue identified was that the second integral yielded a value of 1/2 instead of 1, indicating a misunderstanding in the integration process. The relationship between the variables X and Y is established through the equations X = cos(Θ) and Y = sin(Θ), revealing that they are not independent but constrained by the equation X² + Y² = 1.

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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.
 
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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.
 
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.
 
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