Sample Variance

Study guide: Exam MAS-I

The Sample Variance $S^2$ measures the spread of a random sample around its Sample Mean. Dividing the sum of squared deviations by $n - 1$ rather than $n$ makes it an unbiased estimator of the population variance $\sigma^2$ .

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

$E[S^2] = \sigma^2, \qquad \frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1} \ \text{ (Normal data)}$

The divisor $n - 1$ is the degrees of freedom: one is spent estimating $\mu$ by $\bar{X}$ , and the deviations $X_i - \bar{X}$ sum to zero, so only $n-1$ of them are free
Dividing by $n$ instead gives $\hat\sigma^2 = \frac{n-1}{n}S^2$ , which is biased low by $\sigma^2/n$ for any population. For a Normal population it is the MLE and has smaller Mean Square Error than $S^2$ — the classic bias-variance trade-off in estimation. The MSE comparison needs normality: for Bernoulli $(1/2)$ data, $S^2$ has the smaller MSE once $n \ge 5$
The computing form $S^2 = \frac{1}{n-1}\left(\sum X_i^2 - n\bar{X}^2\right)$ is what exam problems usually need
For Normal data, $(n-1)S^2/\sigma^2 \sim \chi^2_{n-1}$ independent of $\bar{X}$ ; this is what makes the $t$ -statistic $\frac{\bar{X}-\mu}{S/\sqrt{n}} \sim t_{n-1}$ work, and it gives $\text{Var}(S^2) = 2\sigma^4/(n-1)$
Tests and intervals for $\sigma^2$ come from the same $\chi^2$ Sampling Distribution
$S^2$ is even less resistant than $\bar{X}$ — deviations are squared, so a single outlier can dominate it entirely

Computing the Sample Variance Five claim amounts (in thousands) are $4, 6, 9, 11, 20$ . Find $\bar{x}$ and $s^2$ .

Answer $\bar{x} = \frac{4 + 6 + 9 + 11 + 20}{5} = 10$ $\sum x_i^2 = 16 + 36 + 81 + 121 + 400 = 654$ $s^2 = \frac{654 - 5(10)^2}{5 - 1} = \frac{154}{4} = 38.5$ So $s = 6.20$ (thousand). Dividing by $5$ instead — the MLE, if the claims are a Normal sample — gives $\hat\sigma^2 = 154/5 = 30.8$ .

Testing a Variance A sample of $n = 25$ losses from a Normal population gives $s^2 = 150$ . Test $H_0: \sigma^2 = 100$ against $H_1: \sigma^2 > 100$ at $\alpha = 0.05$ , given $\chi^2_{0.05, 24} = 36.42$ .

Answer $\chi^2 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{24 \times 150}{100} = 36.0$ Since $36.0 < 36.42$ , fail to reject $H_0$ at the $5\%$ level — the observed spread is not quite strong enough evidence that the true variance exceeds $100$ .