Joint probability

Discussion in 'Math' started by boks, Jan 9, 2009.

  1. boks

    Thread Starter Active Member

    Oct 10, 2008
    218
    0
    If

    f(x,y) = 4xy, for 0<x<1, 0<y<1
    f(x,y) = 0, elsewhere

    What's the probability that X<Y?

    It seems likely that the probability is 0.5, but how can I show it matematically?
     
    Last edited: Jan 9, 2009
  2. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469

    Use can use symmetry here. The problem is symmetrical in the sense that x and y can be swapped and the problem is the same. Hence, the probability that x<y must equal the probalibity that y<x.

    P=0.5, by symmetry.
     
  3. boks

    Thread Starter Active Member

    Oct 10, 2008
    218
    0
    OK, thanks!

    If(x,y)=cxy^2, what would I have to do?
     
  4. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    I can start the approach.

     P=c\; \int^{1}_{0} x \; \int^{x}_{0} y^2 \; dy \;dx

    Simply integrate overt the area of interest. The geometry is simple in this case, so establishing the limits of integration is easy.
     
  5. boks

    Thread Starter Active Member

    Oct 10, 2008
    218
    0
    Makes sense. Great stuff.
     
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