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Expected value and joint probability density function

  1. Jan 10, 2009 #1
    Assume that two random variables (X,Y) are uniformly distributed on a circle with radius a. Then the joint probability density function is

    [tex]f(x,y) = \frac{1}{\pi a^2}, x^2 + y^2 <= a^2[/tex]
    [tex]f(x,y) = 0, otherwise[/tex]

    Find the expected value of X.

    E(X) = [tex]\int^{\infty}_{- \infty}\int^{\infty}_{- \infty}\frac{x}{\pi a^2} dxdy[/tex]

    Is this correct so far? What are the limits of the integral supposed to be?
  2. jcsd
  3. Jan 10, 2009 #2


    Staff: Mentor

    Since your pdf is nonzero only on the circle, I think this is what you want:
    [tex]\int^{a}_{-a}\int^{\sqrt{a^2 - y^2}}_{- \sqrt{a^2 - y^2}}\frac{x}{\pi a^2} dxdy[/tex]
  4. Jan 11, 2009 #3
    I think this integral would be easier in polar coordinates.
  5. Jan 11, 2009 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    No that is not correct because the limits of integration cannot be from [itex]-\infty[/itex] to [itex]\infty[/itex]. The probability outside the circle of radius a is 0, not [itex]1/(\pi a^2)[/itex].

    In polar coordinates the integral would be
    [tex]\frac{1}{\pi a^2}\int_r\int_\theta r cos(\theta) (r drd\theta)[/tex]
    and you want to cover the circle of radius a. What are the limits of integration for that?
  6. Jan 21, 2009 #5
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