Re: Standard Normal Variable

Think polar coordinates. $\displaystyle R = \sqrt{X^2 + Y^2} $ and $\displaystyle \theta = arctan(Y/X) $

Re: Standard Normal Variable

Well you never replied here is the solution

First notice $\displaystyle g^2 +h^2 = 2y(\cos^2(x) + \sin^2(x)) = 2y $ giving us the inverse transformation $\displaystyle Y=\frac{G^2 +H^2}{2} $.

Similarily $\displaystyle X= \arctan \left(\frac{H}{G}\right) $

$\displaystyle |J| =\left| \begin{array}{cc} g &h \\ \frac{-h}{h^2 + g^2} & \frac{g}{h^2 + g^2} \end{array} \right | = 1 $

Thus

$\displaystyle f_{G,H}\left(x=\arctan\left(\frac{h}{g}\right), y=\frac{g^2 +h^2}{2}\right) |1| = \frac{1}{\sqrt{2 \pi}} e^{-g^2/2} \cdot \frac{1}{\sqrt{2 \pi}} e^{-h^2/2} $

Which are clearly two independent standard normal random variables.