# Ch 7 problem 4 sheldon ross

1. Mar 6, 2015

### mattclgn

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

None, I guess

3. The attempt at a solution

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.

2. Mar 6, 2015

### Dick

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.

3. Mar 7, 2015

### Ray Vickson

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 $g(X,Y) = |X-Y|$.

Last edited: Mar 7, 2015
4. Mar 10, 2015

### mattclgn

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