# Standard Normal Variable

Printable View

• Dec 2nd 2012, 05:28 AM
Solaris1972
Standard Normal Variable
Attachment 26019

I started by trying to solve for G for x and H for y, so that we could proceed to the Jacobian. However

it lead to this mess X=cot^-1(2/H) but then I'm left with Y=(H/sin(X))^2/2 (here clearly I don't want an X in the RHS). So I'm stuck. Is there a better way to go about this?
• Dec 2nd 2012, 12:16 PM
Scopur
Re: Standard Normal Variable
Think polar coordinates. $R = \sqrt{X^2 + Y^2}$ and $\theta = arctan(Y/X)$
• Dec 2nd 2012, 06:39 PM
Scopur
Re: Standard Normal Variable
Well you never replied here is the solution
First notice $g^2 +h^2 = 2y(\cos^2(x) + \sin^2(x)) = 2y$ giving us the inverse transformation $Y=\frac{G^2 +H^2}{2}$.
Similarily $X= \arctan \left(\frac{H}{G}\right)$

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

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

Which are clearly two independent standard normal random variables.