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Random Variables Transformation

  1. Apr 22, 2010 #1
    1. The problem statement, all variables and given/known data

    X, Y, Z random variables, independent of each other, with uniform distribution in (0,1). C = XY and R = Z2. Without using the joint probability function, find P(C>R).

    3. The attempt at a solution

    So far:

    P(C > R) = P(C - R > 0) = P(XY - Z2 > 0) = P(g(X,Y,Z) > 0).

    Now, I don't know how to find g-1(-infinity, 0) = {(x,y,z) in (0,1)3 : xy - z2 > 0} to continue to operate. Or any other way to solve this.

    Any suggestions?
     
  2. jcsd
  3. Apr 22, 2010 #2
    Try [itex]-\sqrt{XY} < Z < \sqrt{XY}[/itex] instead, and remember that all variables are uniform over (0,1).
     
  4. Apr 22, 2010 #3
    Let me see.

    P(C > R) = P(Z2<XY) = P([tex] - \sqrt {XY} < Z < \sqrt {XY} [/tex]) = FZ([tex]\sqrt {XY} [/tex]) - FZ(-[tex]\sqrt {XY} [/tex]), where in the second equality I used that all values of XY are positive and in the last one that FZ is absolutely continuous.

    But now, since Z ~ U(0,1), its FZ(z) = z 1{0<z<1} + 1{z>1}. So FZ(-[tex]\sqrt {XY} [/tex]) = 0, because FZ(z) is 0 for all z<0.

    So, I would get that P(C > R) = FZ([tex]\sqrt {XY} [/tex]) = sqrt(xy) 1{0<sqrt(xy)<1} + 1{sqrt(xy)>1}. XY is bounded by 1, so sqrt(xy) can't be >1. Then this is sqrt(xy) 1{0<sqrt(xy)<1}.

    I'm sorry, but I don't know how to continue...
     
  5. Apr 23, 2010 #4
    And now what do I do? How can I use that X and Y have uniform distributions?
     
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