10
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ I have used Latex to alter the presentation of your question. If this is not what you intended, let me know $\endgroup$
    – Henry
    Jun 1 '11 at 9:34
17
$\begingroup$

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

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

$\endgroup$
5
  • 2
    $\begingroup$ 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. $\endgroup$
    – caracal
    Jun 1 '11 at 13:48
  • $\begingroup$ Could you clarify which one is called "empirical variance"? $\endgroup$ Jan 29 '18 at 15:18
  • 1
    $\begingroup$ @GuillaumeChérel - I did not use the word "empirical" because the point I was trying to make is the key question is distinguishing between estimating the variance (or standard deviation) of the population and estimating the distribution of the error in estimating the mean. Others might call any estimation of parameters from observations as empirical $\endgroup$
    – Henry
    Jan 29 '18 at 18:28
  • 1
    $\begingroup$ I see. Indeed, the word empirical is vague, and that is precisely my problem: I came upon the term empirical variance when reading an article presenting an algorithm I want to implement. I can't find any clue as to whether I should divide the sum of squared differences by $n$ or $n - 1$. So I supposed people using the term empirical were implicitly refering to one or the other. Any idea? $\endgroup$ Jan 31 '18 at 9:03
  • $\begingroup$ Just a guess, but they may mean "sample variance" and they may be more likely to divide by $n-1$ when calculating that $\endgroup$
    – Henry
    Jan 31 '18 at 9:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.