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Probability- expected value of Z, where z= X/(1+y)^2

  1. Sep 15, 2011 #1
    1. The problem statement, all variables and given/known data

    X & Y are independent r.v.s with uniform distribution between 0 and 1.

    Z= X/(1+Y)^2

    find E[Z].
    2. Relevant equations



    3. The attempt at a solution

    Here is what I did.

    E[Z]= E[X]*E[1/(1+Y)^2]
    E[X]=1/2
    E[1/(1+Y)^2]=?
    I think that once I know the distribution of (1+Y)^(-2), I'll be able to find the answer. Is it 1/(1+Y)^2~ U(1,2) ?
     
  2. jcsd
  3. Sep 15, 2011 #2

    micromass

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    You can also use this theorem:

    [tex]E[g(X,Y)]=\iint{g(x,y)f_{X,Y}(x,y)dxdy}[/tex]

    So, in your case, this comes down to

    [tex]\int_0^1\int_0^1\frac{x}{1+y^2}dxdy[/tex]
     
  4. Sep 15, 2011 #3

    Hey,

    thanks for the answer.

    Is 1+y^2 a typo in your answer?
    so adding a constant to a uniform r.v. doesn't change it's distribution?
     
  5. Sep 15, 2011 #4

    micromass

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    Yes

    Obviously it does change the distribution. But I don't see how that is important here.
     
  6. Sep 15, 2011 #5

    Ray Vickson

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    If W = 1/(1+Y)^2, then for any w > 0 we have P{W <= w} = P{1/(1+Y)^2 <= w} = P{-sqrt(w) <= 1/(1+Y) <= sqrt(w)}. The left-hand inequality -sqrt(w) <= 1/(1+Y) holds automatically because Y >= 0, so P{W <= w} = P{1/(1+Y) <= sqrt(w)} = P{Y >= -1 + 1/sqrt(w)}. Since Y ~ U(0,1), we can easily get P{W <= w} and hence can get the density function of W. Alternatively: just apply the standard formula for the density of a transformed random variable.

    RGV
     
  7. Sep 15, 2011 #6

    Well I was thinking that now its uniformly distributed from 1 to 2... isn't it?

    EDIT:
    Oh right its still uniform from 0 to 1...
    omg... I'm so rusty :\

    Thanks.
     
    Last edited: Sep 15, 2011
  8. Sep 15, 2011 #7

    Ray Vickson

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    You have all the information you need to work out the answer for yourself.

    RGV
     
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