Estimate the mean and variance given a sample.

A pricing actuary pulls 30 claim amounts from last quarter and needs two numbers: a best guess for the true average claim and a best guess for the spread. The sample mean and sample variance answer both, but only if you apply the right denominator and respect the standard error.

The estimators. Given an iid sample $X_1,\dots ,X_n$ from a distribution with mean $\mu$ and variance ${\sigma }^2$, define:

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

KEY: Both estimators are unbiased: $E[\bar{X}]=\mu$ and $E[S^2]={\sigma }^2$. The $n-1$ (called Bessel's correction) is not cosmetic. It corrects for the fact that $\bar{X}$, not $\mu$, sits in the squared term and absorbs one degree of freedom.

Why $n-1$? The deviations $(X_i-\bar{X})$ sum to zero by construction, so only $n-1$ of them are free. Algebraically,

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

Dividing by $n-1$ recovers ${\sigma }^2$. Dividing by $n$ gives a biased low estimator with expected value $(n-1){\sigma }^2/n$.

TRAP: Some calculators offer two keys: ${\sigma }_n$ (population, divides by $n$) and ${\sigma }_{n-1}$ (sample, divides by $n-1$). On MAS-I, "sample variance" always means the $n-1$ version.

Common mistakes

• Dividing by $n$ instead of $n-1$. The textbook "population variance" formula gives a value that is too small by a factor of $(n-1)/n$. At $n=5$ the shortfall is 20%. Always convert when the source uses ${\sigma }_n^2$.
• Using $S$ directly as the standard error. The standard error of the mean is $S/\sqrt{n}$, not $S$. Skipping the $\sqrt{n}$ inflates the interval width by a factor of $\sqrt{n}$.
• Treating $S$ as unbiased for $\sigma$. $S^2$ is unbiased; $S$ is slightly biased downward. The exam rarely tests the correction, but never assert $E[S]=\sigma$.

Bottom line

• Sample mean $\bar{X}=\frac{1}{n}\sum X_i$ is unbiased: $E[\bar{X}]=\mu$, with variance ${\sigma }^2/n$.
• Sample variance $S^2=\frac{1}{n-1}\sum (X_i-\bar{X}{)}^2$. The $n-1$ denominator is what makes $E[S^2]={\sigma }^2$.
• Computational shortcut: $S^2=\frac{1}{n-1}\left(\sum X_i^2-n{\bar{X}}^2\right)$ avoids recomputing each deviation.
• Standard error of the mean: $\text{SE}(\bar{X})=S/\sqrt{n}$. Use this, not $S$, in confidence intervals for $\mu$.

Exam shortcut

When a problem reports both $\sum X_i$ and $\sum X_i^2$, skip the deviation table entirely. Compute $\bar{X}$, then $\sum X_i^2-n{\bar{X}}^2$, then divide by $n-1$. When the stem gives a "variance" with no qualifier, check the denominator: if it matches $n$, multiply by $n/(n-1)$ before using it as $S^2$. For confidence-interval questions, write the SE first ($S/\sqrt{n}$); plugging $S$ directly into the margin is the single most common arithmetic trap.

The full lesson (about 1,598 words, 11 min read) adds 2 worked examples, all 6 common mistakes, a self-check, free in the app.

Learning objectives

• B1

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