# statistic

• Jul 21st 2009, 04:19 AM
thereddevils
statistic
Show that ( from basic definition) the variances for a set of observations $\displaystyle y_1,y_2,y_3,......,y_n$ with the mean $\displaystyle \overline{y}$ can be determined by using

$\displaystyle \frac{1}{n}\sum^{n}_{i=1}y_i^2- \overline{y}^2$
• Jul 21st 2009, 07:31 PM
mr fantastic
Quote:

Originally Posted by thereddevils
Show that ( from basic definition) the variances for a set of observations $\displaystyle y_1,y_2,y_3,......,y_n$ with the mean $\displaystyle \overline{y}$ can be determined by using

$\displaystyle \frac{1}{n}\sum^{n}_{i=1}y_i^2- \overline{y}^2$

You should know the basic definition: variance $\displaystyle =\frac{1}{n} \sum^{n}_{i=1}(y_i - \overline{y})^2$.

Then the variance is equal to:

$\displaystyle \frac{1}{n} \sum^{n}_{i=1}(y_i^2 - 2 y_i \overline{y} + \overline{y}^2)$

$\displaystyle = \frac{1}{n} \left(\sum^{n}_{i=1}y_i^2\right) - 2 \overline{y} \frac{1}{n}\left(\sum^{n}_{i=1}y_i \right) + \frac{1}{n} \left(\sum^{n}_{i=1}\overline{y}^2\right)$

$\displaystyle = \frac{1}{n} \left(\sum^{n}_{i=1}y_i^2\right) - 2 \overline{y} \frac{1}{n}\left(\sum^{n}_{i=1}y_i \right) + \overline{y}^2$

The completion is left for you.
• Jul 22nd 2009, 04:46 AM
thereddevils
Quote:

Originally Posted by mr fantastic
You should know the basic definition: variance $\displaystyle =\frac{1}{n} \sum^{n}_{i=1}(y_i - \overline{y})^2$.

Then the variance is equal to:

$\displaystyle \frac{1}{n} \sum^{n}_{i=1}(y_i^2 - 2 y_i \overline{y} + \overline{y}^2)$

$\displaystyle = \frac{1}{n} \left(\sum^{n}_{i=1}y_i^2\right) - 2 \overline{y} \frac{1}{n}\left(\sum^{n}_{i=1}y_i \right) + \frac{1}{n} \left(\sum^{n}_{i=1}\overline{y}^2\right)$

$\displaystyle = \frac{1}{n} \left(\sum^{n}_{i=1}y_i^2\right) - 2 \overline{y} \frac{1}{n}\left(\sum^{n}_{i=1}y_i \right) + \overline{y}^2$

The completion is left for you.

ok thanks i got it .