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Basic proof

  1. Aug 4, 2009 #1
    Hi, I need to show this:

    [tex]b_m \geq\sum_{i=1}^n b_i^2[/tex]

    given these three conditions:

    [tex]b_m \geq b_i[/tex], for [tex]i=1..n[/tex] (in other words [tex]b_m = max(b_i)[/tex]) and

    [tex]0 \leq b_i \leq 1[/tex] for [tex]i=1..n[/tex] and

    [tex]\sum_{i=1}^n b_i=1[/tex]

    I've been working for hours in this without results...Any clue would be really appreciated

    (this is not a homework exercise. I'm just trying to convince myself that the bayes decision error bound is a lower bound for the nearest neighbor rule error bound, and to convince myself of that I've arrived at the conclusion that I have to show the above).

    Thanks,

    Federico
     
  2. jcsd
  3. Aug 4, 2009 #2

    jgens

    User Avatar
    Gold Member

    Maybe this will help . . .

    [tex]0 \leq b_{i}^2 \leq b_i \leq 1[/tex], this implies that [tex]\sum_{i = 1}^n b_{i}^2 \leq \sum_{i = 1}^n b_i = 1[/tex].
     
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