Prove the sample variance formula.

[The_Iceflash]

Homework Statement

Prove that the sample variance of a sample is given by
S² = \frac{\sum^{n}_{j=1}x_j^2 - \frac{1}{n} (\sum^{n}_{j=1}x_j)^2}{n-1}

Homework Equations

N/A

The Attempt at a Solution

For my purposes it is sufficient to show that:

\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



I got as far as this:

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

I need help getting from there to here:

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

Thanks in advance and I apologize for any coding error.

 

[statdad]

Hint: remember that \overline x is a constant and can be removed from summations.

 

[The_Iceflash]

Hint: remember that \overline x is a constant and can be removed from summations.

So if I put the \overline x's in front

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

and rearrange it I get

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

My instincts tell me to factor out the \overline x but I don't see how that could help.

 

[statdad]

So if I put the \overline x's in front

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

and rearrange it I get

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

My instincts tell me to factor out the \overline x but I don't see how that could help.

Hint:

<br /> \sum_{i=1}^n \overline x_i = \overline x_i \sum_{i=1}^N 1<br />

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

Hint:

<br /> \overline x = \frac 1 n \sum_{i=1}^n x_i<br />

so what must

<br /> \sum_{i=1}^n x_i<br />

equal?

 

[The_Iceflash]

Hint:

<br /> \sum_{i=1}^n \overline x_i = \overline x_i \sum_{i=1}^N 1<br />

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

Hint:

<br /> \overline x = \frac 1 n \sum_{i=1}^n x_i<br />

so what must

<br /> \sum_{i=1}^n x_i<br />

equal?

So, \sum_{i=1}^n x_i must be equal to n\overline x correct?

The sum from the first sum is n, correct?