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Help with proof that E[x_i] is X bar

  1. Apr 5, 2014 #1
    I'm having a little trouble with the proof that the expected value of [tex] x_i [/tex] is [tex] \bar{X} [/tex]

    What I have is

    [tex] E[x_i]=\sum_{j=1}^{N}X_j Pr(x_i=X_j) [/tex]

    Then

    [tex] Pr(x_i=X_j) = 1/N [/tex]

    This is the bit I can't understand, how does that probability evaluate to that value.

    I know the denominator is how many ways you can choose n out N. I think that the numerator should be how many ways you can choose (n-1) out of (N-1). But I seem to have an extra n.

    Any advice ?
     
  2. jcsd
  3. Apr 5, 2014 #2

    FactChecker

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    I am used to [tex]\bar{X}[/tex]being defined as E(X) or as a sample mean. That would make the identity either true by definition or false. Can you define that notation, please?
     
  4. Apr 5, 2014 #3
    I'm trying to show that [tex] E[x_i]=\bar{X} [/tex]
    where [tex]\bar{x}[/tex] is the sample mean and [tex]\bar{X} [/tex] is the population mean
     
  5. Apr 5, 2014 #4

    mathman

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    By definition E(xi) is the population mean, assuming all xi have the same mean.
     
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