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

  1. Aug 11, 2010 #1
    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. jcsd
  3. Aug 11, 2010 #2

    lanedance

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    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
    [tex] p_{X,Y}(x,y).dx.dy \propto dx.dy [/tex]

    in polar coordinates an area element is [itex]r.dr.d\theta [/itex] so the joint pdf is
    [tex] p_{R, \Theta}(r,\theta).dr.d\theta \propto r.dr.d\theta [/tex]
     
    Last edited: Aug 11, 2010
  4. Aug 11, 2010 #3
    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?
     
  5. Aug 11, 2010 #4

    lanedance

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    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 [itex] dA = dx.dy [/itex] , and the normalised joint pdf is [tex] p_{X,Y}(x,y)dxdy = \frac{1}{\pi}dx.dy [/tex]

    in polar coordinates you need to account for the jacobian you found. The infinitesimal area elelment is [itex]dA = r.dr.d\theta [/itex]
     
  6. Aug 11, 2010 #5
    Thanks a lot. Understand now. Appreciate your help
     
    Last edited: Aug 11, 2010
  7. Aug 11, 2010 #6

    lanedance

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    no - the area of a circle is [itex] \pi r^2 [/itex]

    an ifinitesiaml area element in polar coords [itex]dA = r.dr.d\theta [/itex]

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