1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]
    E[|X-Y|a]=?

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

    jambaugh

    User Avatar
    Science Advisor
    Gold Member

    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

    =[tex]\frac{1}{a+1}[/tex](1-(-1)a+1).

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

    jambaugh

    User Avatar
    Science Advisor
    Gold Member

    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.
    |y
    |___
    |.....|
    |___|___ 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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Expectation of 2 random variable, E(|X-Y|^a)
  1. Reflect e^x at y=2 (Replies: 8)

Loading...