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Homework Help: Expectation of 2 random variable, E(|X-Y|^a)

  1. Feb 25, 2009 #1
    Hi, anyone help please.

    Let X and Y are independent uniform random variables over the interval [0,1]

    where, a>0
  2. jcsd
  3. Feb 25, 2009 #2


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    Do you have any restrictions on a? Is it integer or any real number?

    I suggest you start with the definition of the expectation value... you sum or integrate over values times the probability distribution:

    [tex] E(g(X,Y)) = \int_{x,y\in[0,1]} g(x,y)\cdot f(x,y) dxdy[/tex]
    where f(x,y) is the joint probability distribution. Since X and Y are both uniform the joint distribution will be f(x,y) = constant. You can determine this constant by normalization:
    [tex] \int_{x,y\in [0,1]} f(x,y) dx dy = 1[/tex]

    As far as doing the integral involving the absolute value you should break the square over which you are integrating into the region where (x-y)>0 and (x-y)<0.
  4. Feb 25, 2009 #3
    The exercise doesn't tell that a is interger or real number, then I just consider it as real number.

    As your suggestion, I do following,
    Let Z = X-Y, where |Z|>0 and Z [tex]\in[/tex][-1,1]

    E[|Z|a] = [tex]\int[/tex][tex]^{1}_{-1}[/tex]Zadz
    = [tex]\int[/tex][tex]^{1}_{0}[/tex]Zadz - [tex]\int[/tex][tex]^{0}_{-1}[/tex]Zadz
    = [tex]\frac{1}{a+1}[/tex] - [tex]\frac{1}{a+1}[/tex](-1)a+1


    Sorry for my poorness, this is corrected or not?
  5. Feb 25, 2009 #4


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    Not quite. You still have to integrate over an area so you need two variables. I would stick to x and y but use the substitution for for finding indefinite integral.

    Consider the area over which you must integrate is the square 0<x<1,0<y<1.
    |___|___ x

    The boundary at which the absolute value changes from equal, to negative of is the line y=x which is the diagonal of this square.

    You integrate over each of the two triangles separately.
    | / /|
    |/ / | (hope the spacing comes out right)
    In the upper triangle y>x so |x-y| = (y-x).
    In the lower triangle x>y so |x-y| = (x-y).

    So integrate |x-y|^a over each triangle making the above substitutions.

    Something like:
    [tex]\int_{y=0}^1\int_{x=0}^y(y-x)^a dx dy[/tex]
    for the lower triangle.
    You may use (z=x-y) or (z=y-x) as a variable substitution to do the indefinite integrals but convert your answer back to x's and y's to evaluate at the limits.
  6. Feb 27, 2009 #5
    Thanks for your advice, let me try!
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