Proof: Sample variance is an unbiased estimator of population variance

Imagine that we are drawing a sample of size (n) from a population which has parameters \(\mu\) and \(\sigma\). The variance of a particular sample (of size n) is given by the following equation. \[s^2 = \frac{\sum_{i=1}^{n} (x_i-\bar X)^2}{n-1}\]

Obviously, the value of individual \(s^2\) is likely to be different from sample to sample. While the sample variance itself is a random variable, the question is “what is the theoretical average or expectation of \(s^2\)?”. That is, if we are to draw a large number of times (say 1 million) -each time a sample of size n, then what can be expected as the average value of \(s^2\)? Taking the expectation both sides,

\[E[s^2] = E[\frac{\sum_{i=1}^{n} (x_i-\bar X)^2}{n-1}]\]

We need some basic results first. \[Var(x_i)= E(x_i^2)-[E(x_i)]^2\] \[\sigma^2 = E(x_i^2)-\mu^2\] \[\therefore E(x_i^2)=\sigma^2 +\mu^2\]

Also, \[Var(\bar X)= E(\bar X^2)-[E(\bar X)]^2\] \[\frac{\sigma^2}{n}=E(\bar X^2)-\mu ^2\] \[\therefore E(\bar X^2)=\frac{\sigma^2}{n}+\mu ^2\]

So, \[E[s^2] = E[\frac{\sum_{i=1}^{n} (x_i-\bar X)^2}{n-1}]\]

\[E[s^2] = \frac{1}{n-1} \sum_{i=1}^{n} E[(x_i-\bar X)^2]== \frac{1}{n-1} \sum_{i=1}^{n} E(x_i^2-2x_i\bar X+\bar X^2)\] \[E[s^2] = \frac{1}{n-1} [\sum_{i=1}^{n} E(x_i^2)-2E(\bar X\sum_{i=1}^{n}x_i)+E(\sum_{i=1}^{n}\bar X^2)]\] \[E[s^2] = \frac{1}{n-1} [\sum_{i=1}^{n} E(x_i^2)-2nE({\bar X}^2)+nE(\bar X^2)]\] \[E[s^2] = \frac{1}{n-1} [\sum_{i=1}^{n} E(x_i^2)-nE({\bar X}^2)\] \[E[s^2] = \frac{1}{n-1} [n E(x_i^2)-nE({\bar X}^2)\] Using earlier results, \[E[s^2] = \frac{1}{n-1} [n (\sigma^2 +\mu^2)-n(\frac{\sigma^2}{n}+\mu ^2))\] \[E[s^2] = \frac{1}{n-1} [n (\sigma^2 +\mu^2)-{\sigma ^2}-n\mu ^2)\] \[\therefore E[s^2] = \sigma ^2\]