# Homework Help: Change of variable - joint p.d.f

1. Aug 11, 2010

### Gekko

1. The problem statement, all variables and given/known data

X and Y uniformly distributed on the disc:

1/pi, x^2+y^2 <=1

X=Rcos(theta), Y =Rsin(theta)

0<=R<=1, 0<=theta<=2pi

Find the joint PDF of R,theta

3. The attempt at a solution

g(R,theta)=f(x,y)*abs(jacobian)

The Jacobian is R

The joint PDF is then simply R/pi = 1/pi (since area is R^2 pi = 1 and R = 1)

Is it that simple?

Last edited: Aug 11, 2010
2. Aug 11, 2010

### lanedance

as X & Y are unifromly distrubted, the probability of ocurring in any area A is proportinal to A

in cartesian coordinates an area element is dx.dy so the joint pdf is
$$p_{X,Y}(x,y).dx.dy \propto dx.dy$$

in polar coordinates an area element is $r.dr.d\theta$ so the joint pdf is
$$p_{R, \Theta}(r,\theta).dr.d\theta \propto r.dr.d\theta$$

Last edited: Aug 11, 2010
3. Aug 11, 2010

### Gekko

Thanks. So in the polar space, the area is R.pi where R is 1. Since this should equal 1 then each point and hence the joint PDF is 1/pi
This makes sense because we are still looking at the same area just in polar coordinates.
I'd just like to validate that the joint PDF is 1/pi. Is that the way to define it?

4. Aug 11, 2010

### lanedance

not quite - see above I was a little mroe explicit in writng out the form of the pdfs

in cartesian coords, infinitesiaml area elelment is $dA = dx.dy$ , and the normalised joint pdf is $$p_{X,Y}(x,y)dxdy = \frac{1}{\pi}dx.dy$$

in polar coordinates you need to account for the jacobian you found. The infinitesimal area elelment is $dA = r.dr.d\theta$

5. Aug 11, 2010

### Gekko

Thanks a lot. Understand now. Appreciate your help

Last edited: Aug 11, 2010
6. Aug 11, 2010

### lanedance

no - the area of a circle is $\pi r^2$

an ifinitesiaml area element in polar coords $dA = r.dr.d\theta$

try integrating your joint pdf over $r \in [0,1), \ \ \theta \in [0, 2 \pi)$ and see if it sums to 1