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.
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.
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