# Bessel's Correction

## Theorem

Let $X_1, X_2, \ldots, X_n$ form a random sample from a population with mean $\mu$ and variance $\sigma^2$.

Let:

$\ds \bar X = \frac 1 n \sum_{i \mathop = 1}^n X_i$

Then:

$\ds \hat {\sigma^2} = \frac 1 {n - 1} \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2$

is an unbiased estimator of $\sigma^2$.

## Proof

If $\hat {\sigma^2}$ is an unbiased estimator of $\sigma^2$, then:

$\expect {\hat {\sigma^2} } = \sigma^2$

In Bias of Sample Variance, it is shown that:

$\ds \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2} = \paren {1 - \frac 1 n} \sigma^2$
 $\ds n \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2}$ $=$ $\ds \expect {\sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2}$ $\ds$ $=$ $\ds \paren {n - 1} \sigma^2$

So:

 $\ds \sigma^2$ $=$ $\ds \frac 1 {n - 1} \expect {\sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2}$ $\ds$ $=$ $\ds \expect {\frac 1 {n - 1} \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2}$ Expectation is Linear $\ds$ $=$ $\ds \expect {\hat {\sigma^2} }$

So $\hat {\sigma^2}$ is an unbiased estimator of $\sigma^2$.

$\blacksquare$

## Source of Name

This entry was named for Friedrich Wilhelm Bessel.