Bessel's Correction

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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:

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

By Expectation is Linear:

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

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.