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Homework Help: Prove the sample variance formula.

  1. Sep 8, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that the sample variance of a sample is given by
    S2 = [tex]\frac{\sum^{n}_{j=1}x_j^2 - \frac{1}{n} (\sum^{n}_{j=1}x_j)^2}{n-1}[/tex]

    2. Relevant equations


    3. The attempt at a solution

    For my purposes it is sufficient to show that:

    [tex]\sum^{n}_{j=1}(x_j -x_j^2) = \sum^{n}_{j=1}x_j^2 - \frac{1}{n}\left(\sum^{n}_{j=1}x_j\right)^2[/tex]

    I got as far as this:

    [tex]= \sum^{n}_{j=1}x_j^2 - \sum^{n}_{j=1}2\bar{x}x_j + \sum^{n}_{j=1}\bar{x}^2[/tex]

    I need help getting from there to here:

    [tex] = \sum^{n}_{j=1}x_j^2 - \frac{1}{n}\left(\sum^{n}_{j=1}x_j\right)^2[/tex]

    Thanks in advance and I apologize for any coding error.
    Last edited: Sep 8, 2010
  2. jcsd
  3. Sep 8, 2010 #2


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    Homework Helper

    Hint: remember that [itex] \overline x [/itex] is a constant and can be removed from summations.
  4. Sep 8, 2010 #3
    So if I put the [itex] \overline x [/itex]'s in front

    [tex]= \sum^{n}_{j=1}x_j^2 - 2\bar{x}\sum^{n}_{j=1}x_j + \bar{x}^2\sum^{n}_{j=1}[/tex]

    and rearrange it I get

    [tex]= \sum^{n}_{j=1}x_j^2 +\bar{x}^2\sum^{n}_{j=1} - 2\bar{x}\sum^{n}_{j=1}x_j[/tex]

    My instincts tell me to factor out the [itex] \overline x [/itex] but I don't see how that could help.
  5. Sep 8, 2010 #4


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    Homework Helper


    \sum_{i=1}^n \overline x_i = \overline x_i \sum_{i=1}^N 1

    (you forgot to include the 1: what does the above sum equal?


    \overline x = \frac 1 n \sum_{i=1}^n x_i

    so what must

    \sum_{i=1}^n x_i

  6. Sep 8, 2010 #5
    So, [tex]\sum_{i=1}^n x_i[/tex] must be equal to [tex]n\overline x[/tex] correct?

    The sum from the first sum is n, correct?
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