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Ch 7 problem 4 sheldon ross

  1. Mar 6, 2015 #1
    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. jcsd
  3. Mar 6, 2015 #2

    Dick

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    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.
     
  4. Mar 7, 2015 #3

    Ray Vickson

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    Much of Ross' book emphasizes a "conditioning argument", and this is one case where you can profitably use that approach:
    [tex] 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) [/tex]
    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
  5. Mar 10, 2015 #4
    Okay, cool, I'll give it a shot.
     
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