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:


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

Then:


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

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

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


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

In Bias of Sample Variance, it is shown that:


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

By Linearity of Expectation Function:

So:

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