Variance of an Estimator

Why Estimator Variance Matters

An estimator is a statistic used to estimate an unknown population parameter. Because it is calculated from a random sample, its value varies from sample to sample. The variance of an estimator measures the spread of that sampling distribution around the estimator’s own expected value. It is not the variance of the individual observations, though the latter often determines it.

For an actuarial estimate, this distinction matters: the spread of an estimator describes its sampling uncertainty under a specified model. The estimator’s variance alone does not say whether it is centered on the parameter. Bias and unbiasedness address that separate question. Here the focus is on calculating variance accurately, especially for sample means and other linear combinations.

Definition: If $T$ is an estimator with finite second moment, its variance is $$ \operatorname{Var}(T)=E\left[(T-E[T])^2\right] =E[T^2]-(E[T])^2. $$ The expectations are taken over the sampling distribution of $T$, with the model parameter treated as fixed.

Variance From a Sampling Distribution

A sampling distribution gives the possible values of an estimator and their probabilities. For a discrete estimator, multiply each squared deviation from its mean by the probability of that value, then sum. The equivalent second-moment calculation, $E[T^2]-(E[T])^2$, is often shorter. For a continuous estimator with density $f_T(t)$, the corresponding calculation is an integral:

\operatorname{Var}(T)=\int_{-\infty}^{\infty}(t-E[T])^2 f_T(t)\,dt.

Worked Example: Variance From a Mortality Estimator’s Sampling Distribution

A hypothetical study follows two independent lives, each for one year, and estimates the annual mortality probability $q$ with the observed fraction who die. Assume both lives have the same one-year death probability $q=0.25$, independently. There are no insurance cash flows or discounting; the effective annual interest rate is $i=0\%$. The mortality model is Bernoulli for each life, with death indicator $X_j\sim\operatorname{Bernoulli}(0.25)$, and the estimator is $T=(X_1+X_2)/2$.

Method. Find the sampling distribution of $T$, then use the definition of variance. The number of deaths is binomial with $n=2$ and $q=0.25$, so $T$ can equal 0, $0.5$, or 1.

The probability of no deaths is $P(T=0)=(0.75)^2=0.5625$. The probability of exactly one death is $P(T=0.5)=2(0.25)(0.75)=0.3750$. The probability of two deaths is $P(T=1)=(0.25)^2=0.0625$. These probabilities sum to $0.5625+0.3750+0.0625=1.0000$.

First calculate the expectation:

$$ E[T]=(0)(0.5625)+(0.5)(0.3750)+(1)(0.0625) =0+0.1875+0.0625=0.25. $$

Next calculate the second moment:

$$ E[T^2]=(0^2)(0.5625)+(0.5^2)(0.3750)+(1^2)(0.0625) =0+0.09375+0.0625=0.15625. $$

Therefore,

\operatorname{Var}(T)=0.15625-(0.25)^2 =0.15625-0.0625=0.09375.

Check. The sample-mean formula derived below gives $q(1-q)/n=(0.25)(0.75)/2=0.09375$, the same result. A common mistake is to report $q(1-q)=0.1875$, the variance of one death indicator, without dividing by the sample size.

Variances of Sums and Linear Estimators

For random variables $X$ and $Y$ with finite variances, adding them does not generally mean adding their variances. The covariance accounts for how they vary together. For constants $a$ and $b$,

\operatorname{Var}(aX+bY) =a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y) +2ab\operatorname{Cov}(X,Y).

If $X$ and $Y$ are independent, their covariance is zero, so only the first two terms remain. Independence is a sufficient condition for zero covariance; without independence, do not discard the covariance term unless it is known to be zero. For $n$ random variables and fixed coefficients $a_1,\ldots,a_n$, the general expression is

\operatorname{Var}\left(\sum_{j=1}^{n}a_jX_j\right) =\sum_{j=1}^{n}a_j^2\operatorname{Var}(X_j) +2\sum_{j<k}a_ja_k\operatorname{Cov}(X_j,X_k).

Theorem: If $X_1,\ldots,X_n$ are independent with finite variances and $T=\sum_{j=1}^{n}a_jX_j$, then $$ \operatorname{Var}(T)=\sum_{j=1}^{n}a_j^2\operatorname{Var}(X_j). $$ In particular, for independent, identically distributed observations with variance $\sigma^2$, the sample mean $\bar{X}=(X_1+\cdots+X_n)/n$ has variance $\operatorname{Var}(\bar{X})=\sigma^2/n$.

The sample-mean result follows directly from the linear-combination formula. Each observation has coefficient $1/n$, and independence makes every covariance term zero:

\operatorname{Var}(\bar{X}) =\sum_{j=1}^{n}\left(\frac{1}{n}\right)^2\sigma^2 =n\frac{\sigma^2}{n^2} =\frac{\sigma^2}{n}.

This derivation explains both parts of the formula: averaging scales each observation by $1/n$, so its variance contribution is scaled by $1/n^2$; there are $n$ independent contributions, giving a total factor of $1/n$. Identical distributions are not needed for the more general result $\operatorname{Var}(\bar{X})=\sum_j\operatorname{Var}(X_j)/n^2$, provided the observations are independent and have finite variances.

Worked Example: Variance of a Sample Mortality Rate

In a hypothetical one-year study, 200 independent lives share a common death probability $q=0.02$. The estimator is the observed mortality fraction $\hat q=(X_1+\cdots+X_{200})/200$, where each $X_j\sim\operatorname{Bernoulli}(0.02)$. Assume homogeneous, independent one-year mortality and no insurance cash flows or discounting; the effective annual interest rate is $i=0\%$.

Method. A Bernoulli variable with probability $q$ has variance $q(1-q)$. Thus each indicator has variance $(0.02)(0.98)=0.0196$. Independence permits the sample-mean variance formula:

\operatorname{Var}(\hat q)=\frac{0.0196}{200}=0.000098.

The estimator’s standard deviation is $\sqrt{0.000098}=0.009899\ldots$, or approximately $0.0099$ as a proportion. The variance is in squared-proportion units; it is not $0.000098$ deaths or dollars. Check. The equivalent binomial calculation gives $\operatorname{Var}(X_1+\cdots+X_{200})=200(0.0196)=3.92$, and dividing the count by 200 divides its variance by $200^2$: $3.92/40000=0.000098$. The trap is dividing by 200 only once after computing the variance of the count; the sample fraction scales the count by $1/200$, so variance scales by $1/200^2$.

Worked Example: A Weighted Estimate From Two Independent Groups

A hypothetical actuary estimates one-year mortality in two independent groups. Group A has $n_A=100$ independent lives with death probability $q_A=0.10$; Group B has $n_B=50$ independent lives with death probability $q_B=0.20$. Within each group, lives have identical mortality probabilities, and the groups are independent. There are no insurance cash flows or discounting, so take the effective annual interest rate as $i=0\%$. Let $\hat q_A$ and $\hat q_B$ be the observed group death fractions, and define $T=0.4\hat q_A+0.6\hat q_B$.

Method. Calculate each group estimator’s variance, then apply the independent linear-combination formula. In Group A, each death indicator has variance $(0.10)(0.90)=0.09$, so

\operatorname{Var}(\hat q_A)=\frac{0.09}{100}=0.0009.

In Group B, each death indicator has variance $(0.20)(0.80)=0.16$, so

\operatorname{Var}(\hat q_B)=\frac{0.16}{50}=0.0032.

The estimates are independent because they are computed from independent groups. Therefore,

\begin{aligned} \operatorname{Var}(T) &=(0.4)^2(0.0009)+(0.6)^2(0.0032)\\ &=0.16(0.0009)+0.36(0.0032)\\ &=0.000144+0.001152\\ &=0.001296. \end{aligned}

Check. The weights sum to 1, and the expected value is $0.4(0.10)+0.6(0.20)=0.16$. The variance calculation uses the weights squared, not the weights themselves. Using $0.4(0.0009)+0.6(0.0032)$ would give the wrong answer, $0.00228$, because multiplying a random variable by a constant multiplies its variance by the square of that constant.

When the Independence Shortcut Does Not Apply

If observations share a source of variation, or are otherwise dependent, the covariance terms may change the estimator’s variance. For example, repeated measurements on the same person need not be independent. If all $n$ observations have common variance $\sigma^2$ and common pairwise covariance $c$, then

\operatorname{Var}(\bar{X}) =\frac{1}{n^2}\left(n\sigma^2+n(n-1)c\right).

A positive covariance increases the variance relative to the independent case; a negative covariance decreases it. This is why the sampling assumptions must be stated before using $\sigma^2/n$. The shortcut requires independent observations with finite common variance, not merely observations that happen to be listed in a sample.

Identify the estimator.
Write it as a function of the random observations, such as a sample mean or weighted sum.

Record the model assumptions.
Determine the variance of each observation and whether the observations or component estimators are independent.

Apply the appropriate variance rule.
Use the sampling distribution directly, or use the linear-combination formula and retain covariance terms unless they are zero.

Check the scale.
Verify that coefficients are squared and that averaging $n$ independent observations divides their common variance by $n$.

Common Exam Traps

Confusing observation variance with estimator variance. For an independent sample mean, the observation variance is $\sigma^2$, while the estimator variance is $\sigma^2/n$.
Forgetting to square a coefficient. If $T=aX$, then $\operatorname{Var}(T)=a^2\operatorname{Var}(X)$, not $a\operatorname{Var}(X)$.
Assuming all variances add. Covariance terms are present unless independence or another condition establishes that they vanish.
Dividing by $n$ twice or not at all. The variance of a sum of $n$ independent, equal-variance observations is $n\sigma^2$; the variance of their mean is $n\sigma^2/n^2=\sigma^2/n$.

Estimator variance is central to comparing sampling procedures, but it is only one feature of an estimator’s performance. In particular, small variance does not by itself establish that an estimator is close to the target parameter on average. Keep the sampling distribution, the estimator’s center, and its spread conceptually distinct.

The calculations in this tutorial are aligned with the SOA VEE Mathematical Statistics syllabus. The reliable approach is to write the estimator as a random variable, state the sampling assumptions, and then apply the variance definition or an appropriate linear-combination rule.

Check Your Understanding

For each problem, identify the variance rule before calculating. Answers are given at the end of each question.

A hypothetical study observes four independent lives for one year. Each has death probability $q=0.30$, and the estimator is the fraction who die. Assume a common Bernoulli mortality model, no insurance cash flows, and effective annual interest $i=0\%$. Find the estimator’s variance. (Answer: $0.0525$)
Independent observations from a hypothetical claim-severity model have mean $\mu=12$ and variance $9$ (in squared monetary units). For a sample of 10, find the variance of the sample mean. (Answer: $0.9$)
Independent estimators $U$ and $V$ have variances $0.04$ and $0.09$, respectively. Find the variance of $T=2U-\tfrac12V$. (Answer: $0.1825$)
Random variables $X$ and $Y$ have variances $4$ and $9$, and covariance $1.5$. Find $\operatorname{Var}(X+Y)$, and state why the independence shortcut is unavailable. (Answer: $16$; the covariance is nonzero, so include $2\operatorname{Cov}(X,Y)$)
A hypothetical cohort has 25 independent lives with the same one-year death probability $q=0.08$. Estimate $q$ by the observed death fraction. Assume no insurance cash flows and effective annual interest $i=0\%$. Find the variance of the estimate. (Answer: $0.002944$)

What You'll Learn