Variance Estimator

If we are to prove that the sample variance is an unbiased estimator of the population variance, can we use the χ² distribution?

 

You don't need to make any distributional assumptions. You know that \[{\mathbb E}\left[\sum{(x_i-\bar{x})}\right]={\mathbb E}\left[\sum{x_i^2}-2\sum{x_i \bar{x}}+\sum{\bar{x}^2}\right]\]and also that\[{\mathbb E}\left[\sum{x_i^2}\right]=n\left[{\mathbb E}(X)^2+\operatorname{var} (X)\right]\]\[{\mathbb E}\left[\sum{x_i \bar{x}}\right]={\mathbb E}\left[\sum{\bar{x}^2}\right]=n{\mathbb E}(X)^2+\operatorname{var}(X)\]Subbing these back into the original equation gives:\[{\mathbb E}\left[\sum{(x_i-\bar{x})}\right]=(n-1)\operatorname{var}(X)\]and so\[{\mathbb E}\left[\frac{\sum{(x_i-\bar{x})}}{n-1}\right]=\operatorname{var}(X)\]