Calculating Expected Absolute Deviation for Independent Random Variables

[mattclgn]

Homework Statement

Let X and Y be independent random variables, both being equally likely to be any of the numbers 1, 2, ..., m. Show that E[(absolute value(X-Y))] = ((m-l)(m+1)) / 3m.

Homework Equations

None, I guess

The Attempt at a Solution

[/B]
Okay so for sample space, since x and y can be any m; total possible combinations are m*m...and then, wasn't really sure where to go...tried talking through bunch of ideas with friends, but...to no avail.

 

[Dick]

Homework Statement

Let X and Y be independent random variables, both being equally likely to be any of the numbers 1, 2, ..., m. Show that E[(absolute value(X-Y))] = ((m-l)(m+1)) / 3m.

Homework Equations

None, I guess

The Attempt at a Solution

[/B]
Okay so for sample space, since x and y can be any m; total possible combinations are m*m...and then, wasn't really sure where to go...tried talking through bunch of ideas with friends, but...to no avail.

I'd suggest you start by working through the cases explicitly with small numbers. m=1 is no challenge. It's just 0 for the expectation value. m=2 is a little better, you've got the 2^2 cases 1,1 1,2 2,1 2,2. What's the expectation value? Does it match the formula? Now try m=3. Arrange the cases in a square matrix and see if you can think of something to do.

 

[Ray Vickson]

Homework Statement

Let X and Y be independent random variables, both being equally likely to be any of the numbers 1, 2, ..., m. Show that E[(absolute value(X-Y))] = ((m-l)(m+1)) / 3m.

Homework Equations

None, I guess

The Attempt at a Solution

[/B]
Okay so for sample space, since x and y can be any m; total possible combinations are m*m...and then, wasn't really sure where to go...tried talking through bunch of ideas with friends, but...to no avail.

Much of Ross' book emphasizes a "conditioning argument", and this is one case where you can profitably use that approach:
E |X-Y| = \sum_{j=1}^m E\left( |X-Y|\; | Y = j \right) P(Y = j) = \sum_{j=1}^m E |X-j| \, P(Y=j)
The somewhat unfortunate notation $E(|X-Y| |Y=j)$ means $E(g(X,Y)|Y=j)$, where $g(X,Y) = |X-Y|$.

 

[mattclgn]

Okay, cool, I'll give it a shot.