Bessel’s Correction

Bessel’s correction is an approach to reduce the Bias due to finite sample size.

In statistics, it’s the use of $n-1$ instead of $n$ in the formula for the sample variance and sample standard deviation.

Proof of Bessel’s Correction

From ChatGPT.

Let $X_1,X_2,\dots ,X_n$ be i.i.d. samples from a population with mean $\mu$ and variance ${\sigma }^2$.

Define:

• The sample mean:

  $$\bar{X}=\frac{1}{n}\sum _{i=1}^n{X}_i$$

• The sample variance:

  $$S^2=\frac{1}{n-1}\sum _{i=1}^n(X_i-\bar{X}{)}^2$$

We now prove that $\mathbb{E}[S^2]={\sigma }^2$, i.e., this estimator is unbiased.

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Step 1: Expand the squared term

Note that:

$$\sum _{i=1}^n(X_i-\bar{X}{)}^2=\sum _{i=1}^n{X}_i^2-n{\bar{X}}^2$$

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Step 2: Take expectation

Take the expectation of both terms:

$$\mathbb{E}\left[\sum _{i=1}^n(X_i-\bar{X}{)}^2\right]=\sum _{i=1}^n\mathbb{E}[X_i^2]-n\mathbb{E}[{\bar{X}}^2]$$

• Since $\mathbb{E}[X_i^2]={\sigma }^2+{\mu }^2$, we have:

  $$\sum _{i=1}^n\mathbb{E}[X_i^2]=n({\sigma }^2+{\mu }^2)$$

• For $\mathbb{E}[{\bar{X}}^2]$, we use:

  $$\mathbb{E}[{\bar{X}}^2]=\mathrm{Var}(\bar{X})+\mathbb{E}[\bar{X}{]}^2=\frac{{\sigma }^2}{n}+{\mu }^2$$

So:

$$n\mathbb{E}[{\bar{X}}^2]={\sigma }^2+n{\mu }^2$$

Putting it all together:

$$\mathbb{E}\left[\sum _{i=1}^n(X_i-\bar{X}{)}^2\right]=n({\sigma }^2+{\mu }^2)-({\sigma }^2+n{\mu }^2)=(n-1){\sigma }^2$$

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Step 3: Divide by $n-1$

Since

$$S^2=\frac{1}{n-1}\sum _{i=1}^n(X_i-\bar{X}{)}^2$$

Taking expectation:

$$\mathbb{E}[S^2]=\frac{1}{n-1}\mathbb{E}\left[\sum _{i=1}^n(X_i-\bar{X}{)}^2\right]=\frac{(n-1){\sigma }^2}{n-1}={\sigma }^2$$

✅ Hence, the sample variance with denominator $n-1$ is an unbiased estimator of the population variance.