What Is an Unbiased Estimator of the Population Mean and Variance?

An unbiased estimator is a sample statistic whose expected value equals the population parameter being estimated. The sample mean $\bar{X}$ is unbiased for the population mean $\mu$, while the corrected sample variance $S^2$, using denominator $n-1$, is unbiased for the population variance $\sigma^2$.

The Estimators

For a random sample $X_1,X_2,\ldots,X_n$:

Unbiasedness is a long-run property of an estimator, not a claim that one sample produces the true parameter. A particular value of $\bar{X}$ or $S^2$ may be above or below the population value. However, if the sampling process were repeated many times, the mean of all resulting estimates would equal the parameter.

The sample mean is unbiased because

$$E(\bar{X})=\mu.$$

Therefore, across many random samples, the average of the sample means equals the true population mean.

For the uncorrected variance $s_n^2$, the expectation is $E(s_n^2)=\frac{n-1}{n}\sigma^2$. Multiplying by the correction factor therefore removes the systematic downward bias.

For variance, dividing by $n$ tends to underestimate $\sigma^2$. This happens because the deviations are measured from the sample mean $\bar{X}$, which has already been estimated from the same data. Once $\bar{X}$ is known, only deviations can vary independently, giving .

For example, consider the sample $2,4,6$. Its mean is $4$, and the sum of squared deviations is

$$(2-4)^2+(4-4)^2+(6-4)^2=8.$$

Dividing by $n=3$ gives $\frac{8}{3}$, whereas the unbiased sample variance is

$$S^2=\frac{8}{3-1}=4.$$

In IB notation, this correction may also be written as $s_{n-1}^2=\frac{n}{n-1}s_n^2$.

Exam Technique

This is AHL 4.14, so it is HL-only and may appear in GDC-active Paper 1, Paper 2, or Paper 3. On a GDC, distinguish variance using divisor $n$ from corrected sample variance using $n-1$; calculator labels vary, so check the denominator.

A common misconception is that the sample standard deviation $S$ is unbiased for $\sigma$; generally, it is not. State clearly that $\bar{X}$ estimates $\mu$ unbiasedly and $S^2$, not , estimates unbiasedly.