Does the sample variance converge to the true variance?

Question

You estimate the variance \sigma^2 of an i.i.d. sample X1, \dots, Xn (with finite fourth moment) using the sample variance S^2 = \frac{1}{n-1}\sumi (Xi - \bar X)^2. Is S^2 a consistent estimator of \sigma^2?

Accepted Answer

Rewrite as averages. Expand the sum of squares and divide by n: \frac{1}{n}\sumi (Xi - \bar X)^2 = \frac{1}{n}\sumi Xi^2 - \bar X^2. By the law of large numbers, each average converges in probability to its expectation: \frac{1}{n}\sumi Xi^2 \xrightarrow{p} \mathbb{E}[X^2], \qquad \bar X \xrightarrow{p} \mu. Apply the continuous mapping theorem. Squaring is continuous, so \bar X^2 \xrightarrow{p} \mu^2, and subtracting two convergent sequences gives \frac{1}{n}\sumi (Xi - \bar X)^2 \xrightarrow{p} \mathbb{E}[X^2] - \mu^2 = \sigma^2. Handle the divisor. The factor relating S^2 to this average …

Does the sample variance converge to the true variance?

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