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The Cov(X,Y) on a square dart board

  1. Jul 28, 2009 #1
    A point (X, Y) is picked at random, uniformly from the square with corners at (0,0), (1,0), (0,1) and (1,1). Compute Cov{X, Y}.

    I think of this as darts thrown at a unit square dart board.

    Cov(X,Y) = E(XY) - E(X)E(Y).

    I compute that E(X)=E(Y)= 0.5 using the integral xf(x).

    But I cannot figure out how to approach computing E(XY). Or is there a better strategy for solving this?

    THANKS!
     
  2. jcsd
  3. Jul 28, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi BookMark440! Welcome to PF! :smile:

    (I'm not sure exactly what you mean by f(x))

    If you got E(X) from ∫x, why can't you get E(XY) from ∫∫xy ?
     
  4. Jul 28, 2009 #3
    Thanks. I did that and ended up getting zero for an answer, which always makes me think I made an error. In this case, I guess it just verifies that X,Y are independent.
     
  5. Jul 29, 2009 #4
    A covariance of zero is very common in textbook examples.
     
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