statistic

**thereddevils** · July 21st 2009, 04:19 AM

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

$<br /> \frac{1}{n}\sum^{n}_{i=1}y_i^2- \overline{y}^2<br />$

**mr fantastic** · July 21st 2009, 07:31 PM

Originally Posted by thereddevils

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

$<br /> \frac{1}{n}\sum^{n}_{i=1}y_i^2- \overline{y}^2<br />$

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

Then the variance is equal to:

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

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

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

The completion is left for you.

**thereddevils** · July 22nd 2009, 04:46 AM

Originally Posted by mr fantastic

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

Then the variance is equal to:

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

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

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

The completion is left for you.

ok thanks i got it .

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