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Help on continous probability question!

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data

    The r.v (X,Y) is distributed uniformly in the triangle with vertices (0,0), (1,0), (1,-1).

    claim (a): The variable (X^2, Y^2) is distributed uniformly in the region {(x,y): 0 <= x <= 1, 0 <= y <= x^2}.

    2. Relevant equations

    -

    3. The attempt at a solution

    I have the solution but I do not understand it.

    "If (X^2, Y^2) would be distributed uniformly in {(x,y): 0 <= x <= 1,
    0 <= y <= x^2}, then in particular,

    P(Y^2 > (X^2)^2) = 0"

    This makes sense, I doubt I'd try proving the new r.v wasn't distributed uniformly in the new interval in this way, but OK.

    Continuing..

    "...However: P(Y^2 > (X^2)^2) >= P(X <= 0.1, Y <= -0.9) = 2*(0.1^2)/2 = 0.01"

    How in heaven's name the second inequality was derived? I do not understand this and the choice of the numbers ...

    Please help. :)
     
  2. jcsd
  3. Jan 30, 2012 #2

    Ray Vickson

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    Something is wrong: you say that (X,Y) ranges over the triangle with vertices (0,0), (1,0), (1,-1) which lies in the region Y < 0. Therefore, we cannot have positive probability on the region 0 <= x <= 1, 0 <= y <= x^2.

    RGV
     
  4. Jan 30, 2012 #3
    For the original variable, but the new r.v is x squared, y squared: (X^2, Y^2)

    It would make some sense that it would be distributed in that region...
     
  5. Jan 30, 2012 #4
    I don't believe the claim, either for the area over which it is distributed, or for uniformity.

    edit to add: For example, the zone near (0.7,-0.7) for the original rv will be transformed to a zone near (0.5,0.5) - won't it? - but that seems to be excluded from the claimed area.
     
    Last edited: Jan 30, 2012
  6. Jan 30, 2012 #5

    LCKurtz

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    I don't think the OP has clearly stated what the problem is. Perhaps it is to show ##(X^2,Y^2)## is distributed over ##0\le y \le x^2,\ 0\le x\le 1)## but is not uniform. That I could agree with.
     
  7. Jan 30, 2012 #6

    Ray Vickson

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    I guess I am still confused. The random pair (U=X^2,V=Y^2) ranges over the set {0 <= u <= 1, 0 <= v <= u} because the boundary y=-x of the original set becomes u=v in the new variables.

    RGV
     
  8. Jan 30, 2012 #7

    LCKurtz

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    That makes sense to me.
     
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