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  • need help with finan problem

    Hi. Can anyone help me with this problem from Finan's book?
    35.1) Let X and Y be independent random variables, both being equally likely to be any of the numbers 1, 2, ..., m. Find E(absolute value(X-Y)).
    How do you deal with the absolute value in this case? The answer is (m+1)(m-1)/(3m). Thanks.
    Last edited by Happychick90; August 12 2011, 01:02 AM.

  • #2
    I can't figure this one out either. Let me know if you find out the solution to solve this one. Thanks!

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    • #3
      Happy Chick

      Consider what the problem would be asking if it were not all the numbers between 1 and m. Try for 1 and 2 at first. Then try 1 and 3. Keep at it until you see what kind of pattern you get in terms of E[|x-y|]

      Thats how I would approach. Hopefully that steers you in the right direction. Looking at that equation doesn't really reveal a whole lot but sometimes just acting out a problem for the first couple values will shed some light on the problem at whole.

      Hope that helps!

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      • #4
        Thanks for the advice! I'll try that out.

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        • #5
          This is how I solved it but it may sounds complicated. Set up the sample space for event (X,Y). You'll see that there m^2 possibilities. In which there are m ways for X=Y (or X-Y=0), (m-1) ways for X=Y+1, ... , 1 way X=Y+(m-1). Similar there are (m-1) ways for Y=X+1, ... , 1 way for Y=X+(m-1). So the E(|X-Y|)=2E(X-Y>0)=2*(1/m^2)*sum[k(m-k)] with k from 1 to (m-1) = 2*(1/m^2)*[sum(km) - sum(k^2)] with k from 1 to m-1.
          Recal the formula of sum(x^2) with x from 1 to n = (n/6)(n+1)(2n+1), hence sum(k^2) with k from 1 to m-1 = (1/6)*(m-1)*(m)*(2m-1). And sum(km) with k from 1 to m-1 = m*(m-1)*m/2. Next you just find the common denominator and simplify and you will get (m+1)(m-1)/(3m). Enjoy.

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