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Consider the random point $(X,Y)$ in $\mathbb{R}^2$. The ratio $X/Y$ tells us what angle the segment from $(0,0)$ to $(X,Y)$ makes with the $x$-axis, while $X^2+Y^2$ tells us how far $(X,Y)$ is from $(0,0)$.

The distribution of $(X,Y)$ is symmetric under rotations, so the distribution of the angle $\Theta$ is uniform and independent of the radius $R=\sqrt{X^2+Y^2}$.

This intuitive explanation can be made more rigorous by converting to polar coordinates.

The joint density of two independent standard normals $(X,Y)$ is $$f(x,y)={1\over 2\pi} \exp(-(x^2+y^2)/2).$$ Converting to polar coordinates we get the joint density of $(R,\Theta)$ as $$g(r,\theta)={r\over 2\pi}\exp(-r^2/2)={1\over 2\pi}\cdot r\exp(-r^2/2).$$ This product form of $g(r,\theta)$ shows that $\Theta$ and $R$ are independent.

I don't understand how to map it backwards, that is get the gaussian random variable from the radium and angle pdf ?

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$(X,Y)=(R\cos\Theta,R\sin\Theta)$. –  André Nicolas Dec 19 '12 at 19:12
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Can you please show me how I can substitute it into the independent pdf found ? –  user1675999 Dec 19 '12 at 21:57
Go through the process of converting that you already went through, this time in the opposite direction. So again you need to find a Jacobian. –  André Nicolas Dec 19 '12 at 22:06
thank you, i found the answer, I just didn't know what a jacobian was till you mentioned it. –  user1675999 Dec 20 '12 at 0:44
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