Editing Normal distribution (section)

==== Sample variance ====
{{See also|Standard deviation#Estimation|Variance#Estimation}}

The estimator <math style="vertical-align:0">\textstyle\hat\sigma^2</math> is called the ''[[sample variance]]'', since it is the variance of the sample (<math display=inline>(x_1, \ldots, x_n)</math>). In practice, another estimator is often used instead of the <math style="vertical-align:0">\textstyle\hat\sigma^2</math>. This other estimator is denoted  <math display=inline>s^2</math>, and is also called the ''sample variance'', which represents a certain ambiguity in terminology; its square root {{tmath|s}} is called the ''sample standard deviation''. The estimator <math display=inline>s^2</math> differs from <math style="vertical-align:0">\textstyle\hat\sigma^2</math> by having {{math|(''n'' − 1)}} instead of&nbsp;''n'' in the denominator (the so-called [[Bessel's correction]]):
<math display=block>
    s^2 = \frac{n}{n-1} \hat\sigma^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \overline{x})^2.
</math>
The difference between <math display=inline>s^2</math> and <math style="vertical-align:0">\textstyle\hat\sigma^2</math> becomes negligibly small for large ''n''{{'}}s. In finite samples however, the motivation behind the use of <math display=inline>s^2</math> is that it is an [[unbiased estimator]] of the underlying parameter <math display=inline>\sigma^2</math>, whereas <math style="vertical-align:0">\textstyle\hat\sigma^2</math> is biased. Also, by the Lehmann–Scheffé theorem the estimator <math display=inline>s^2</math> is uniformly minimum variance unbiased ([[UMVU]]),<ref name="Krishnamoorthy" /> which makes it the "best" estimator among all unbiased ones. However it can be shown that the biased estimator <math style="vertical-align:0">\textstyle\hat\sigma^2</math> is better than the <math display=inline>s^2</math> in terms of the [[mean squared error]] (MSE) criterion. In finite samples both <math display=inline>s^2</math> and <math style="vertical-align:0">\textstyle\hat\sigma^2</math> have scaled [[chi-squared distribution]] with {{math|(''n'' − 1)}} degrees of freedom:
<math display=block>
    s^2 \sim \frac{\sigma^2}{n-1} \cdot \chi^2_{n-1}, \qquad
    \hat\sigma^2 \sim \frac{\sigma^2}{n} \cdot \chi^2_{n-1}.
</math>
The first of these expressions shows that the variance of <math display=inline>s^2</math> is equal to <math display=inline>2\sigma^4/(n-1)</math>, which is slightly greater than the ''σσ''-element of the inverse Fisher information matrix <math style="vertical-align:0">\textstyle\mathcal{I}^{-1}</math>, which is <math display=inline>2\sigma^4/n</math>. Thus, <math display=inline>s^2</math> is not an efficient estimator for <math display=inline>\sigma^2</math>, and moreover, since <math display=inline>s^2</math> is UMVU, we can conclude that the finite-sample efficient estimator for <math display=inline>\sigma^2</math> does not exist.

Applying the asymptotic theory, both estimators <math display=inline>s^2</math> and <math style="vertical-align:0">\textstyle\hat\sigma^2</math> are consistent, that is they converge in probability to <math display=inline>\sigma^2</math> as the sample size <math display=inline>n\rightarrow\infty</math>. The two estimators are also both asymptotically normal:
<math display=block>
    \sqrt{n}(\hat\sigma^2 - \sigma^2) \simeq
    \sqrt{n}(s^2-\sigma^2) \,\xrightarrow{d}\, \mathcal{N}(0,2\sigma^4).
</math>
In particular, both estimators are asymptotically efficient for  <math display=inline>\sigma^2</math>.