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Minimum Value of Absolute Deviation

  1. Apr 11, 2008 #1
    Hi,

    How can I rigorously prove that the quantity

    [tex]S = \sum_{i=1}^{n}|X_{i} - a|[/tex]

    (where [itex]X_{1},\ldots,X_{n}[/itex] is a random sample and a is some real number) is minimum when a is the median of the [itex]X_{i}[/itex]'s?

    Thanks.
     
    Last edited: Apr 11, 2008
  2. jcsd
  3. Apr 11, 2008 #2
    Ok I think I got it. If a is the median, then there are just as many numbers less than it than there are greater than it...but how do I write the median in terms of the random sample?
     
  4. Apr 11, 2008 #3
    There's no simple way to write it like you can with, say, the mean. What you can do is re-order the sample in increasing order, such that [itex]X_1 \leq \ldots \leq X_{n/2} \leq a \leq X_{n/2 + 1} \leq \ldots \leq X_n[/itex]. For the actual proof, you might try working by contradiction: assume some other value of a results in the lowest value for the sum, and then show that you can construct an even lower value by moving a towards the median.
     
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