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Joint expectation of two functions of a random variable

  1. Sep 7, 2010 #1
    Ok I am not sure if I should put this question in the homework category of here but it’s a problem from schaums outline and I know the solution to it but I don’t understand the solution 100% so maybe someone can explain this to me.
    Let [tex]X[/tex] and [tex]Y[/tex] be defined by:
    [tex]\begin{array}{l}
    X = \cos \theta \\
    Y = \sin \theta \\
    \end{array}[/tex]
    Where [tex]\theta [/tex] is a uniform random variable distributed over [tex](0,2\pi )[/tex]
    A) Show that [tex]X[/tex] and [tex]Y[/tex] are uncorrelated
    Attempt at solution:
    Show [tex]{\mathop{\rm cov}} (x,y) = 0[/tex]
    [tex]\begin{array}{l}
    {\mathop{\rm cov}} (x,y) = E[xy] - E[x]E[y] \\
    E[xy] = \int\limits_0^{2\pi } {\int\limits_0^{2\pi } {xy{f_{xy}}(x,y)dxdy} } \\
    \end{array}[/tex]
    Now my question is how do we determine the joint pdf [tex]{{f_{xy}}(x,y)}[/tex] if we only know the marginal pdfs of [tex]\theta [/tex]?
    In the solution to the problem its seems that they assume that
    [tex]{f_{xy}}(x,y) = {f_\theta }(\Theta )[/tex]
    Then the integeral they use becomes
    [tex]E[xy] = \int\limits_0^{2\pi } {xy{f_\theta }(\Theta )d\theta } [/tex]
    But how come it is valid to assume that
    [tex]{f_{xy}}(x,y) = {f_\theta }(\Theta )[/tex]
    Doesn’t the joint (and the marginal) pdf change because of the functions:
    [tex]\begin{array}{l}
    X = \cos \theta \\
    Y = \sin \theta \\
    \end{array}[/tex]
    I f anyone knows what I am trying to ask please give me a little help to what is going on here.
     
  2. jcsd
  3. Sep 7, 2010 #2
    The joint pdf doesn't technically exist, because the random variables (X,Y) have all their mass on a 1-dimensional subset of 2d space, i.e. the circle (cos(theta),sin(theta)). The joint pdf could be written in terms of Dirac delta functions or the expectation could be written as a Stieltjes integral (using the joint cdf) but the theory gets messy and for this example it's much simpler to write the expectation as

    [tex]E[XY] = E[cos(\theta)sin(\theta)][/tex]

    which can be expressed as a single integral because theta has a pdf.
     
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