Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
2  
$(X,Y)=(R\cos\Theta,R\sin\Theta)$. –  André Nicolas Dec 19 '12 at 19:12

1 Answer 1

$(X,Y)=(R\cos\Theta,R\sin\Theta)$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.