Bessels Correction

probability It is a correction between sample and the population when measuring their mean and variance.

$$N-1$$

Bessel’s Correction Proof

Goal: Show that the unbiased estimator of population variance ${\sigma }^2$ requires division by $(n-1)$.

Setup

• Population with mean $\mu$ and variance ${\sigma }^2$
• Random sample: $X_1,X_2,...,X_n$ (i.i.d.)
• Sample mean: $\bar{X}=\frac{1}{n}{\sum }_{i=1}^n{X}_i$

The Question

Which estimator is unbiased for ${\sigma }^2$? Biased estimator:

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

Unbiased estimator:

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

Proof

We need to show: $E[S_{n-1}^2]={\sigma }^2$ Step 1: Expand the sum of squared deviations

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

Step 2: Expand the square

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

Step 3: Distribute the summation

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

Step 4: Simplify the middle term Note that:

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

Step 5: Take expectations

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

Step 6: Evaluate each term

For the first term:

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

For the second term, recall that $\text{Var}(\bar{X})=\frac{{\sigma }^2}{n}$:

$$E[(\bar{X}-\mu {)}^2]=\text{Var}(\bar{X})=\frac{{\sigma }^2}{n}$$

Therefore:

$$E[n(\bar{X}-\mu {)}^2]=n\cdot \frac{{\sigma }^2}{n}={\sigma }^2$$

Step 7: Combine results

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

Step 8: Solve for the unbiased estimator

$$E\left[\frac{1}{n-1}\sum _{i=1}^n(X_i-\bar{X}{)}^2\right]=\frac{(n-1){\sigma }^2}{n-1}={\sigma }^2$$ $$\boxed{S^2=\frac{1}{n-1}\sum _{i=1}^n(X_i-\bar{X}{)}^2\text{ is an unbiased estimator of }{\sigma }^2}$$

The division by $(n-1)$ compensates for using the sample mean $\bar{X}$ instead of the true mean $\mu$, which introduces bias by making the deviations artificially smaller.