# Prove the sample variance formula.

• Sep 7th 2010, 11:51 AM
Iceflash234
Prove the sample variance formula.
Basically I have to prove that the sample variance of a sample is given by:

$s^2 =$ ${\sum_{j=1}^nx_j^2-{1\over n}}(\sum_{j=1}^nx_j)^2\over n-1}$

For my purposes it's sufficient to show that ${\sum_{j=1}^n(x_j-x_j^2) = {\sum_{j=1}^nx_j^2-{1\over n}}(\sum_{j=1}^nx_j)^2$

Help is appreciated. Thanks in advance.

If I made any mistakes inputting the sums please let me know.
• Sep 7th 2010, 12:51 PM
mr fantastic
Quote:

Originally Posted by Iceflash234
Basically I have to prove that the sample variance of a sample is given by:

$s^2 =$ ${\sum_{j=1}^nx_j^2-{1\over n}}(\sum_{j=1}^nx_j)^2\over n-1}$

For my purposes it's sufficient to show that ${\sum_{j=1}^n(x_j-x_j^2) = {\sum_{j=1}^nx_j^2-{1\over n}}(\sum_{j=1}^nx_j)^2$

Help is appreciated. Thanks in advance.

If I made any mistakes inputting the sums please let me know.

This is a standard proof found in most textbooks on mathematical statistics. Have you consulted any?
• Sep 7th 2010, 01:52 PM
Iceflash234
Quote:

Originally Posted by mr fantastic
This is a standard proof found in most textbooks on mathematical statistics. Have you consulted any?

I checked all the ones I have access to. I even tried googling it.
• Sep 9th 2010, 03:06 AM
mr fantastic
Quote:

Originally Posted by Iceflash234
Basically I have to prove that the sample variance of a sample is given by:

$s^2 =$ ${\sum_{j=1}^nx_j^2-{1\over n}}(\sum_{j=1}^nx_j)^2\over n-1}$

For my purposes it's sufficient to show that ${\sum_{j=1}^n(x_j-x_j^2) = {\sum_{j=1}^nx_j^2-{1\over n}}(\sum_{j=1}^nx_j)^2$

Help is appreciated. Thanks in advance.

If I made any mistakes inputting the sums please let me know.

Your starting point should be the definition of sample variance: Sample Variance -- from Wolfram MathWorld

What progress have you made from this?