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Expected Value: E(X^Y)

  1. Dec 6, 2008 #1
    1. The problem statement, all variables and given/known data
    X and Y are joointly distributed discrete random variables with probability mass function
    pX,Y(x,y)=(c/ec)(1-c)xcy/y!, x,y=0,1,2,..., 0<c<1
    Find E(XY)


    2. Relevant equations
    3. The attempt at a solution
    Code (Text):
    By definition,
    E(X[SUP]Y[/SUP])
      ∞    ∞
    = ∑    ∑  x[SUP]y[/SUP] (c/e[SUP]c[/SUP])(1-c)[SUP]x[/SUP]c[SUP]y[/SUP]/y!
     x=0  y=0
     
    How can we calculate this double sum? We can pull out the constant c/ec, but what's next?

    Thanks for any help!
     
  2. jcsd
  3. Dec 6, 2008 #2
    Do the y-sum first and use
    [tex]
    \sum_{y=0}^\infty{\frac{Z^y}{y!}}=e^Z
    [/tex]
    What is Z here?
    Next do the x-sum using
    [tex]
    \sum_{x=0}^\infty{a^x}=\frac{1}{1-a}
    [/tex]
    What is a here? Is |a|<1?
     
  4. Dec 6, 2008 #3
    The xy is giving me some trouble doing the y-sum, it depends on both x and y, and I cannot separate it into a product, what should I do?
     
  5. Dec 7, 2008 #4
    When doing the y-sum treat x as a constant.
     
  6. Dec 9, 2008 #5
    OK, so
    E(XY)

    = ∑ [ec (1-c)]x (c/ec)
    x=0
    ={1/[1-ec (1-c)]} (c/ec)

    Did I get it right?


    Also, why is |ec (1-c)|<1 ? How can we see this? I can only show 0<c<1 => 0<1-c<1 => 0<(1-c)ec<ec < e, but we need (1-c)ec<ec < 1, how can we prove this? This is giving me so much headache...please help...


    Thank you!
     
    Last edited: Dec 9, 2008
  7. Dec 9, 2008 #6
    Just out of curiousity, in this case, is it possible to sum over x FIRST, and then sum over y?

    Code (Text):
    ∞    ∞
    ∑    ∑  x[SUP]y[/SUP] (c/e[SUP]c[/SUP])(1-c)[SUP]x[/SUP]c[SUP]y[/SUP]/y!
    y=0 x=0
     
  8. Dec 9, 2008 #7
    Yes.
    Try and find the maximum of ec (1-c) on the interval [0,1]. You surely know how to find the maximum of a differentiable function:smile: You will find that the function has its maximum when c=0 and this maximum value is 1.

    For your other question: Sure you can but it's a lot more complicated.
     
  9. Dec 9, 2008 #8
    "Try and find the maximum of ec (1-c) on the interval [0,1]. You surely know how to find the maximum of a differentiable function You will find that the function has its maximum when c=0 and this maximum value is 1."

    OK, I got it! Thanks!
    Is it possible to prove it without using calculus?
     
  10. Dec 11, 2008 #9
    Still wondering...
     
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