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Show the two forms of the sample variance are equivalent

  1. Sep 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Showthe two forms of the sample variance are equivalent:
    [itex]\frac{1}{n-1}[/itex][itex]\sum_{i=1}^\n (Yi-Ybar)2[/itex] = [itex]\frac{1}{n(n-1)}[/itex][itex]\sum_{i=1}^\n \sum_{j>i}\n (Yi-Yj)2[/itex]

    The first summation is from i=1 to n, the second is i=1 to n and the third is j>i to n. Sorry, I don't know how to format those.

    2. Relevant equations



    3. The attempt at a solution
    I don't really know where to begin, so I tried just expanding and cancelling where I could. I know the (n-1) in the denominator on both sides cancel, and then after expanding I get

    [itex]\sum (Yi2-2YiYbar + Ybar2[/itex] = [itex]\frac{1}{n}[/itex][itex]\sum \sum (Yi2-2YiYj+Yj2)[/itex]

    Then, I can distribute the 1/n and the summations on the right side. If I do that, and I have a term that does not have j (such as Yi), I can essentially drop the j summation from that term, correct? After I do that, I have the following:

    [itex]\sum Yi2[/itex] -2[itex]\sum YiYbar[/itex] + [itex]\sum Ybar2[/itex] = [itex]\frac{1}{n}[/itex][itex]\sum Yi2 - \frac{2}{n}[/itex][itex]\sum \sum YiYj[/itex]+[itex]\sum \sum Yj2[/itex]

    And here's where I'm stuck (assuming I even did everything right to get here, which I doubt). I don't know how to deal with the Yj, among other things. Any help would be greatly appreciated.

    edit: I'm sorry about the terrible formatting. I tried using the LaTex tag buttons and I've looked at the FAQ; not sure what I'm doing wrong.
     
    Last edited: Sep 10, 2013
  2. jcsd
  3. Sep 11, 2013 #2

    Mark44

    Staff: Mentor

    What you wrote is very hard to read. The problem you're having with formatting comes from mixing HTML tags (e.g., and ) with LaTeX script. Use one or the other, but not both.

    Here's your first equation, cleaned up:
    [itex]\frac{1}{n-1} \sum_{i=1}^n (Y_i - Ybar)^2 = \frac{1}{n(n-1)}\sum_{i=1}^n \sum_{j>i}^n (Y_i-Y_j)^2[/itex]
     
  4. Sep 11, 2013 #3

    Ray Vickson

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    Science Advisor
    Homework Helper



    Even better:
    [tex]\frac{1}{n-1} \sum_{i=1}^n (Y_i - \bar{Y})^2 = \frac{1}{n(n-1)}\sum_{i=1}^{n-1} \sum_{j=i+1}^n (Y_i-Y_j)^2[/tex]
    Note that in the second form, if we have j>i with j going up to n, then i can only go up to (n-1).
     
  5. Sep 11, 2013 #4
    Ah, I see. I was just using the buttons at the top and side. Hopefully I'll get this right this time.

    The original equation.

    [tex]\frac{1}{n-1} \sum_{i=1}^n (Y_i - \bar{Y})^2 = \frac{1}{n(n-1)}\sum_{i=1}^{n-1} \sum_{j=i+1}^n (Y_i-Y_j)^2[/tex]

    Like I said, the (n-1) from both sides cancel, and then I expand both sides

    [tex]\sum_{i=1}^n(Y_i^2 -2Y_i \bar{Y} + \bar{Y}^2 = \frac{1}{n}\sum_{i=1}^{n-1} \sum_{j=i+1}^n(Y_i^2 - 2Y_iY_j + 2Y_j^2)[/tex]

    I then distribute the summations to all the terms

    [itex]\sum_{i=1}^nY_i^2-2 \sum_{i=1}^nY_i\bar{Y} + \sum_{i=1}^n\bar{Y}^2 = \frac{1}{n}\sum_{i=1}^{n-1} \sum_{j=i+1}^nY_i^2 - \frac{2}{n}\sum_{i=1}^{n-1} \sum_{j=i+1}^n Y_iY_j + \frac{1}{n}\sum_{i=1}^{n-1} \sum_{j=i+1}^nY_j^2[/itex]

    I don't really know where to go from here. I know that [itex]\frac{1}{n}\sum_{i=1}^nY_i=\bar{Y}[/itex], but I'm not sure what that does from me. I'll still have the j summations to worry about on the right hand side, as well as the Yj terms.
     
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