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Homework Help: Minimize expected value of the absolute difference

  1. Aug 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Let X be a continuous random variable with median m.
    Minimize E[|X - b|] as a function of b. Hint: Show that E[|X - b|] = E[|X - m|] + 2 [tex]\int[/tex] (x - b) f(x) dx , where the integral is from b to m.

    2. Relevant equations



    3. The attempt at a solution
    I wanted to try a solution but I even don't know how to determine whether it is minimum or not. Please help.
     
  2. jcsd
  3. Aug 21, 2010 #2

    lanedance

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    treat is as a function of b
    f(b) = E(|X-b|)

    how would you find the minima of f w.r.t. b?
     
  4. Aug 21, 2010 #3

    statdad

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    Can you show the expression given above? If so, what do you know about the sign of

    [tex]
    \int_b^m (x-b) f(x) \, dx
    [/tex]

    Finally, note that the right-side is a function of the number [itex] b [/itex]. What happens when you evaluate at that function at a (cleverly) chosen value?
     
  5. Sep 16, 2010 #4
    I am having a hard time how to evaluate this integral because the function f(x) is not given.
    [tex]
    \int_b^m (x-b) f(x) \, dx
    [/tex]
     
  6. Sep 16, 2010 #5

    statdad

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    Aaaaah, that's the point: if you were given a specific [tex] f(x) [/tex], any result you obtained would apply only to that particular function , not in general. This exercise is meant to give a general result.

    Hint: In the integral

    [tex]
    \int_a^m (x-b) f(x) \, dx
    [/tex]

    the interval of integration consists of values [tex] x \ge b [/tex], so

    i) What is the sign of [tex] (x-b)f(x) [/tex] over the interval?
    ii) What does this say about the sign of the integral of [tex] (x-b)f(x)[/tex]?
    iii) Using your answers to `i' and `ii'', if you want to choose [tex] b [/tex] to minimize

    [tex]
    E(|X-m|) + 2 \int_b^m (x-b) f(x) \, dx
    [/tex]

    as a function of [tex] b [/tex], what choice does it?
     
  7. Sep 19, 2010 #6

    okay, let me see if I figure this out right.
    i) The sign of [tex] (x-b)f(x) [/tex] over the interval is positive.
    ii) The sign of the integral of [tex] (x-b)f(x)[/tex] is positive.
    iii) Substituting b to m in the expression [tex]
    E(|X-m|) + 2 \int_b^m (x-b) f(x) \, dx
    [/tex]

    results to [tex]
    E(|X-b|) + 2 \int_b^b (x-b) f(x) \, dx
    [/tex]
    which becomes E(|X - b|) + 2*(0)
    and finally to E(|X - b|)
    ...so E[|X - b|] is minimized when b = m.
    I hope this time I finally got it right with your guidance.. =)
     
  8. Sep 20, 2010 #7
    However, I'm wondering what is the use of 2 when the integral is just zero?
     
  9. Sep 20, 2010 #8

    statdad

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    The 2 comes in when the expression on the right is developed.
     
  10. Dec 7, 2010 #9
    Stat Dad -- Thank you for your guidance on this problem -- I was terribly lost and had a very similar question!
     
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