# What is the difference between empirical variance and variance?

As far as I know variance is calculated as

$$\text{variance} = \frac{(x-\text{mean})^2}{n}$$

while

$$\text{Empirical Variance} = \frac{(x-\text{mean})^2}{n(n-1)}$$

Is it correct? Or is there some other definition? Kindly explain with example or any refence for reading on this topic

-
 I have used Latex to alter the presentation of your question. If this is not what you intended, let me know – Henry Jun 1 '11 at 9:34

In your expression for the variance, you need to take a sum (or integral) across the population

$$\text{variance} = \frac{\sum_i(x_i-\text{mean})^2}{n}$$

If your data is a sample from the population then this expression will give you a biased estimate of the population variance. An unbiased estimate would be as follows (note the change in the denominator from your expression), often called the sample variance

$$\text{Sample variance} = \frac{\sum_i(x_i-\text{mean})^2}{n-1}$$

If on the other hand you were trying to estimate the variance of the sample mean, then you vould have a smaller number, closer to your expression. The square root of this is called the standard error of the mean and a reasonable estimate is

$$\text{Standard error} = \sqrt{\frac{\sum_i (x_i-\text{mean})^2}{n(n-1)}}$$

-
See en.wikipedia.org/wiki/Bias_of_an_estimator#Sample_variance for an explanation why the variance $1/n \sum_{i} (x_{i} - \bar{x})^2$ is a biased estimator, and vdov.net/~acosta/content/mle-normal for an explanation why it is the maximum-likelihood estimator for normal variables. – caracal Jun 1 '11 at 13:48