Results 1 to 5 of 5

Math Help - Showing equality of two expressions for calculating variance.

  1. #1
    Newbie
    Joined
    Jul 2010
    Posts
    11

    Showing equality of two expressions for calculating variance.

    Showing equality of two expressions for calculating variance.-ghg.png
    Last edited by mr fantastic; July 5th 2010 at 05:42 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by gaidd View Post
    Click image for larger version. 

Name:	ghg.PNG 
Views:	24 
Size:	16.6 KB 
ID:	18101
    \displaystyle \frac{1}{n-1} \sum_{i=1}^n (x_i - \overline{x})^2 = \frac{1}{n-1} \sum_{i=1}^n x_i^2 - \frac{2 \overline{x}}{n-1} \sum_{i=1}^n x_i + \frac{1}{n-1} \sum_{i=1}^n \overline{x}^2


    \displaystyle = \frac{1}{n-1} \sum_{i=1}^n x_i^2 - \frac{2 n \overline{x}}{(n-1)} \cdot \frac{1}{n} \sum_{i=1}^n x_i + \frac{n}{n-1} \overline{x}^2


    \displaystyle \frac{1}{n-1} \sum_{i=1}^n x_i^2 - \frac{2 n \overline{x}^2}{n-1} +  \frac{n}{n-1} \overline{x}^2


    \displaystyle \frac{1}{n-1} \sum_{i=1}^n x_i^2 - \frac{n \overline{x}^2}{n-1}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,824
    Thanks
    713
    Hello, gaidd!

    Show that: . \displaystyle{\tfrac{1}{n-1}\sum^n_{i=1}(x_i - \overline x)^2 \;=\;\tfrac{1}{n-1}\sum^n_{i=1} x_i^2 - \tfrac{n}{n-1}\,\overline{x}^2 }

    All sums are from i = 1\text{ to }n.

    Recall that \overline{x} is a constant.


    \frac{1}{n-1}\sum(x_i - \overline x)^2

    . . = \;\frac{1}{n-1}\sum\left(x_i^2 - 2x_i\overline{x} + \overline{x}^2\right)

    . . =\; \frac{1}{n-1}\bigg[\sum x_i^2 - \sum 2x_i\overline{x} + \sum \overline{x}^2\bigg]

    . . =\; \frac{1}{n-1}\bigg[\sum x_i^2 - 2\overline{x}\sum x_i + \overline{x}^2\sum 1 \bigg]

    Multiply the middle term by \frac{n}{n}
    Note that: . \sum 1 \,=\,n

    . . . . \frac{1}{n-1}\left[\sum x_i^2 - 2\overline{x}\,n\underbrace{\left(\frac{\sum x_i}{n}\right)} + \;\overline{x}^2\!\cdot\!n \right]
    . . . . . . . . . . . . . . . . This is \overline{x}

    . . =\; \frac{1}{n-1}\bigg[\sum x_i^2 - 2\overline{x}n\!\cdot\!\overline{x} + n\overline{x}^2\bigg]

    . . =\;\frac{1}{n-1}\bigg[\sum x_i^2 - 2n\overline{x}^2 + n\overline{x}^2\bigg]

    . . =\;\frac{1}{n-1}\bigg[\sum x_i^2 - n\overline{x}^2\bigg]

    . . =\;\frac{1}{n-1}\sum x_i^2 - \frac{n}{n-1}\,\overline{x}^2



    Edit: Too slow . . . again!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2010
    Posts
    11
    Quote Originally Posted by mr fantastic View Post
    \displaystyle \frac{1}{n-1} \sum_{i=1}^n (x_i - \overline{x})^2 = \frac{1}{n-1} \sum_{i=1}^n x_i^2 - \frac{2 \overline{x}}{n-1} \sum_{i=1}^n x_i + \frac{1}{n-1} \sum_{i=1}^n \overline{x}^2


    \displaystyle = \frac{1}{n-1} \sum_{i=1}^n x_i^2 - \frac{2 n \overline{x}}{(n-1)} \cdot \frac{1}{n} \sum_{i=1}^n x_i + \frac{n}{n-1} \overline{x}^2


    \displaystyle \frac{1}{n-1} \sum_{i=1}^n x_i^2 - \frac{2 n \overline{x}^2}{n-1} +  \frac{n}{n-1} \overline{x}^2


    \displaystyle \frac{1}{n-1} \sum_{i=1}^n x_i^2 - \frac{n \overline{x}^2}{n-1}
    thank you mr fantastic,,, and i m sorry for the mistake i v done,,,"_*
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jul 2010
    Posts
    11
    Quote Originally Posted by Soroban View Post
    Hello, gaidd!


    All sums are from i = 1\text{ to }n.

    Recall that \overline{x} is a constant.


    \frac{1}{n-1}\sum(x_i - \overline x)^2

    . . = \;\frac{1}{n-1}\sum\left(x_i^2 - 2x_i\overline{x} + \overline{x}^2\right)

    . . =\; \frac{1}{n-1}\bigg[\sum x_i^2 - \sum 2x_i\overline{x} + \sum \overline{x}^2\bigg]

    . . =\; \frac{1}{n-1}\bigg[\sum x_i^2 - 2\overline{x}\sum x_i + \overline{x}^2\sum 1 \bigg]

    Multiply the middle term by \frac{n}{n}
    Note that: . \sum 1 \,=\,n

    . . . . \frac{1}{n-1}\left[\sum x_i^2 - 2\overline{x}\,n\underbrace{\left(\frac{\sum x_i}{n}\right)} + \;\overline{x}^2\!\cdot\!n \right]
    . . . . . . . . . . . . . . . . This is \overline{x}

    . . =\; \frac{1}{n-1}\bigg[\sum x_i^2 - 2\overline{x}n\!\cdot\!\overline{x} + n\overline{x}^2\bigg]

    . . =\;\frac{1}{n-1}\bigg[\sum x_i^2 - 2n\overline{x}^2 + n\overline{x}^2\bigg]

    . . =\;\frac{1}{n-1}\bigg[\sum x_i^2 - n\overline{x}^2\bigg]

    . . =\;\frac{1}{n-1}\sum x_i^2 - \frac{n}{n-1}\,\overline{x}^2



    Edit: Too slow . . . again!
    thank you Sir/ Soroban.....
    you are really amazing,,,"_"
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Showing equality via a simple calculation
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: October 2nd 2011, 01:33 AM
  2. Replies: 1
    Last Post: July 6th 2010, 07:19 AM
  3. Testing equality of variance
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 22nd 2008, 11:46 PM
  4. Calculating Variance
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 6th 2008, 06:35 PM
  5. Calculating variance of u_n - u_2n
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: July 8th 2008, 08:37 AM

Search Tags


/mathhelpforum @mathhelpforum