Why Averages Stabilize With More Observations

Why More Observations Usually Produce a More Stable Average

A group’s observed claim average can move noticeably when only a few observations are available. With more observations, unusually high and unusually low results have more opportunity to offset one another. The mathematical version of this idea is a statement about variance: under specified conditions, the variance of an average is inversely proportional to the number of observations.

For credibility work, the result explains one important way that volume affects the stability of observed experience. It does not say that every added observation must move the actual average closer to its expected value. It describes the average’s variation across repeated samples under a stated model.

Definition: Let $X_1,\ldots,X_n$ be observations with finite variances. Their arithmetic average is $\bar{X}=(X_1+\cdots+X_n)/n$. If the observations have a common expected value $\mu$, then $E[\bar{X}]=\mu$. The variance of $\bar{X}$ measures how much this average varies across repeated samples.

In the examples, claim amounts are hypothetical. No interest rate or mortality assumption applies; each example states the probability model used for its observations. Variances are in squared dollars when the observations are dollar amounts, while standard deviations are in dollars.

The Variance Formula and Its Conditions

The general formula makes the role of dependence explicit. Variance of a sum includes both the individual variances and the covariances between distinct observations.

\operatorname{Var}(\bar{X}) = \frac{1}{n^2} \left( \sum_{i=1}^{n}\operatorname{Var}(X_i) + 2\sum_{1\leq i<j\leq n}\operatorname{Cov}(X_i,X_j) \right).

This follows because multiplying a random variable by $1/n$ multiplies its variance by $1/n^2$, and the variance of a sum is the sum of the variances plus twice the sum of the pairwise covariances. The covariance terms cannot be discarded merely because the observations have equal variances.

If all observations have variance $\sigma^2$ and are independent, every covariance between distinct observations is zero. More generally, pairwise zero covariance is enough for this variance calculation; independence is sufficient but not necessary. Under either condition, the variance simplifies to:

\operatorname{Var}(\bar{X}) = \frac{n\sigma^2}{n^2} = \frac{\sigma^2}{n}.

The expectation of the average is the common mean $\mu$, whether or not the observations are independent, provided each has that expectation. Independence or zero covariance is relevant to the variance calculation, not to this expectation calculation. Identical distributions are not required for the variance result if all observations have the same variance and are pairwise uncorrelated.

Write the average.
Express it as the sum of the $n$ observations multiplied by $1/n$.

Apply the variance-of-a-sum identity.
Include a covariance term for every distinct pair of observations.

Apply the model assumptions.
Equal variances and zero pairwise covariances reduce the variance to $\sigma^2/n$.

Interpret the result.
The standard deviation is $\sigma/\sqrt{n}$, so quadrupling the sample size halves the standard deviation.

The distinction between variance and standard deviation matters. Variance falls in proportion to $1/n$; standard deviation falls in proportion to $1/\sqrt{n}$. A tenfold increase in sample size therefore divides the variance by ten, but divides the standard deviation only by $\sqrt{10}$.

Worked Example: Independent Claim Amounts With a Common Distribution

A hypothetical portfolio produces independent observations $X_i$ of claim amount, each with the following distribution: $P(X_i=\$600)=0.25$, $P(X_i=\$1{,}000)=0.50$, and $P(X_i=\$1{,}400)=0.25$. Find the variance and standard deviation of the average of 16 observations, then compare them with those for 64 observations.

Method and model. The observations are independent and identically distributed with the stated discrete distribution. No interest or mortality assumption applies.

First calculate the mean:

E[X_i]=0.25(600)+0.50(1{,}000)+0.25(1{,}400) =150+500+350 =\$1{,}000.

The deviations from the mean are $-\$400$, $0$, and $\$400$. Thus the variance of one observation is:

\operatorname{Var}(X_i) =0.25(400^2)+0.50(0^2)+0.25(400^2) =40{,}000+0+40{,}000 =80{,}000\text{ dollars}^2.

For $n=16$, independence and the common variance give:

E[\bar{X}]=\$1{,}000,\qquad \operatorname{Var}(\bar{X})=\frac{80{,}000}{16}=5{,}000\text{ dollars}^2, \qquad \operatorname{SD}(\bar{X})=\sqrt{5{,}000}=\$70.71.

For $n=64$:

\operatorname{Var}(\bar{X})=\frac{80{,}000}{64}=1{,}250\text{ dollars}^2, \qquad \operatorname{SD}(\bar{X})=\sqrt{1{,}250}=\$35.36.

The sample size increased by a factor of four. The variance fell to one-quarter, and the standard deviation fell to one-half. The mean remained $\$1{,}000$. Standard deviations are rounded to the nearest cent. The result checks against the square-root relationship: $\$70.71/2$ is approximately $\$35.36$.

Common mistake avoided: Dividing the standard deviation by four when the sample size quadruples. It is the variance that is divided by four; the standard deviation is divided by two.

Independent Observations With Unequal Variances

The compact formula $\sigma^2/n$ requires equal individual variances as well as zero pairwise covariances. If independent observations have different variances, the variance of the average is still computable, but each variance must be included in the sum:

\operatorname{Var}(\bar{X}) = \frac{\sum_{i=1}^{n}\operatorname{Var}(X_i)}{n^2} \quad\text{when the observations are independent.}

Worked Example: Four Independent Observations With Different Variances

Suppose four independent hypothetical claim-cost estimates are modeled as $X_1\sim N(800,100)$, $X_2\sim N(900,400)$, $X_3\sim N(1{,}000,900)$, and $X_4\sim N(1{,}100,1{,}600)$, with dollar means and dollar-squared variances. Find the expected average, its variance, and its standard deviation.

Method and model. The observations are independent normal variables, with the means and variances specified by the parameters above. They do not have equal variances. No interest or mortality assumption applies.

By linearity of expectation:

E[\bar{X}]=\frac{800+900+1{,}000+1{,}100}{4} =\frac{3{,}800}{4} =\$950.

Independence means the covariance terms are zero. The sum of the four variances is $100+400+900+1{,}600=3{,}000$ dollars squared, so:

\operatorname{Var}(\bar{X}) = \frac{3{,}000}{4^2} = \frac{3{,}000}{16} = 187.5\text{ dollars}^2, \qquad \operatorname{SD}(\bar{X})=\sqrt{187.5}=\$13.69.

The expected average is $\$950$, and the standard deviation is rounded to the nearest cent. As a check, each observation contributes its variance divided by $16$; these contributions sum to $6.25+25+56.25+100=187.5$ dollars squared.

Common mistake avoided: Using one observation’s variance as though it applied to all four. For example, using $100/4=25$ dollars squared incorrectly treats all four variances as $100$. The correct calculation sums the four distinct variances and divides by $16$.

Zero Covariance Is More General Than Independence

Independence is a convenient assumption because it guarantees zero covariance between observations with finite variances. It is not necessary for the variance-of-a-sum calculation. If every distinct pair has covariance zero, the same sum-of-variances result follows even when some observations are dependent.

Worked Example: Dependent Variables With Zero Covariance

Let $X$ have probabilities $1/3$ each of being $-1$, $0$, or $1$. Define $Y=1$ when $X=0$, and $Y=0$ otherwise. Calculate the covariance and the variance of $(X+Y)/2$. Explain why this example does not justify assuming independence.

Method and model. This is a finite discrete model. $Y$ is determined by $X$, so the variables are dependent. No interest or mortality assumption applies.

First, $E[X]=(-1+0+1)/3=0$, and $E[Y]=1/3$. Since $X^2$ is one when $X=-1$ or $X=1$, and zero otherwise:

\operatorname{Var}(X)=E[X^2]-E[X]^2=\frac{2}{3}, \qquad \operatorname{Var}(Y)=\frac{1}{3}\left(1-\frac{1}{3}\right)=\frac{2}{9}.

The product $XY$ is always zero: when $Y=1$, $X=0$, and when $X$ is nonzero, $Y=0$. Therefore:

\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=0-(0)(1/3)=0.

The average has expectation $(0+1/3)/2=1/6$. Its variance is:

\operatorname{Var}\left(\frac{X+Y}{2}\right) = \frac{1}{4}\left(\frac{2}{3}+\frac{2}{9}+2(0)\right) = \frac{1}{4}\left(\frac{8}{9}\right) = \frac{2}{9}.

The variables are dependent because knowing $X$ determines $Y$, yet their covariance is zero. Thus zero covariance is enough for the variance addition here; independence is not. The example also shows why zero covariance should not be described as equivalent to independence.

When Dependence Prevents the Usual Stabilization

If observations share a common source of variation, their covariances can remain positive as the sample grows. Then averaging does not remove that shared component. The following example calculates the effect directly.

Worked Example: A Shared Component Across Four Observations

Let $U$ take values $-3$ and $3$ with equal probability. Let $E_1,\ldots,E_4$ each take values $-1$ and $1$ with equal probability, independently of one another and of $U$. Define $X_i=U+E_i$. Find the variance of the average of the four observations, and compare it with the value obtained by incorrectly treating the $X_i$ as independent.

Method and model. This is a shared-component model with independent, equally likely two-point variables as specified. No interest or mortality assumption applies. Each $X_i$ has mean zero and variance $\operatorname{Var}(U)+\operatorname{Var}(E_i)=9+1=10$.

For distinct $i$ and $j$, the errors $E_i$ and $E_j$ are independent, and are independent of $U$. Therefore:

\operatorname{Cov}(X_i,X_j) = \operatorname{Cov}(U+E_i,U+E_j) = \operatorname{Var}(U) = 9.

There are four individual variances and six distinct pairs. Applying the full covariance formula:

\operatorname{Var}(\bar{X}) = \frac{4(10)+2(6)(9)}{4^2} = \frac{40+108}{16} = 9.25.

If the covariance terms were incorrectly omitted, the result would be $4(10)/16=2.5$. The correct variance is $9.25$, because averaging reduces the separate error variation but does not eliminate the common $U$ component. The calculation checks by writing $\bar{X}=U+(E_1+E_2+E_3+E_4)/4$: its variance is $9+4/16=9.25$.

Interpretation and Exam Traps

Under the equal-variance, zero-covariance conditions, $\operatorname{Var}(\bar{X})=\sigma^2/n$ tends to zero as $n$ grows. If all observations have common mean $\mu$, this means the average’s variation around $\mu$ decreases. For example, Chebyshev’s inequality gives, for any $\epsilon>0$:

P\left(|\bar{X}-\mu|\geq\epsilon\right) \leq \frac{\sigma^2}{n\epsilon^2}.

As $n$ increases, this upper bound decreases toward zero. This formalizes stabilization in probability under the stated conditions; it does not promise that every realized average gets closer to $\mu$ each time an observation is added.

Do not confuse variance with standard deviation. The variance changes by a factor of $1/n$; the standard deviation changes by a factor of $1/\sqrt{n}$.
Do not omit covariance terms without justification. Independence is sufficient, and pairwise zero covariance is also sufficient for the variance sum.
Do not use $\sigma^2/n$ for unequal variances. For independent observations with different variances, sum those variances and divide by $n^2$.
Do not assume more observations always reduce every realized average’s error. The result concerns variation across repeated samples under a model, not the direction of each individual change.

Key takeaway: An average of $n$ equal-variance observations has variance $\sigma^2/n$ when distinct observations have zero covariance. Independence is one way to ensure this condition, but it is not the only way. Shared positive covariance can make the average much less stable than the independent-observation formula suggests.

Understanding how sample size and dependence affect the variance of experience is aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus.

Check Your Understanding

For each problem, identify the assumptions needed and calculate the requested quantity.

For each exposure, claim occurrence $X_i$ is Bernoulli with $P(X_i=1)=0.10$, and the 25 exposures are independent. Find the expected claim count per exposure, the variance of the average claim count per exposure, and its standard deviation. (Answer: mean $0.10$, variance $0.0036$, standard deviation $0.0600$)
Independent claim amounts each take values $\$500$ and $\$1{,}500$ with equal probability. For an average of 100 such observations, calculate the individual variance, average variance, and average standard deviation. (Answer: individual variance $\$250{,}000\text{ dollars}^2$, average variance $\$2{,}500\text{ dollars}^2$, standard deviation $\$50.00$)
Three independent estimates have distributions $X_1\sim N(120,16)$, $X_2\sim N(150,25)$, and $X_3\sim N(180,64)$, where the second parameter is variance. Find the mean and standard deviation of their arithmetic average. (Answer: mean $150$, standard deviation $\sqrt{105/9}\approx 3.416$)
Five observations each have variance $36$, and each distinct pair has covariance $9$. Find the variance of their average and compare it with the result from incorrectly assuming independence. (Answer: correct variance $14.4$; independence-based variance $7.2$)
Independent observations have common variance $144$. If an average based on 16 observations has standard deviation $3$, how many observations are needed to reduce its standard deviation to $1.5$? (Answer: $64$ observations)

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