Why Risk Averages Do Not Show VHM Directly
Observed averages from different risks vary for two reasons. Risks can have different underlying expected outcomes, and the finite observations used to calculate each average can fluctuate around that risk’s expectation. Estimating the variance of hypothetical means (VHM) therefore requires separating genuine between-risk variation from the sampling variation still present in the risk averages.
This distinction matters in the Bühlmann model: VHM describes how much the hypothetical means vary across risks, whereas EPV describes the expected variation of observations around a risk’s own mean. A raw variance of risk averages includes contributions from both. The calculations below are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.
Suppose there are \(R\) risks, with \(m\) observations \(X_{j1},\ldots,X_{jm}\) for risk \(j\). Write that risk’s average as \(\bar X_j\). Under the standard assumptions used here, risks are independent draws from the same risk population; observations within a risk are conditionally independent and identically distributed given that risk’s parameter; and each risk has the same number \(m\) of observations. The conditional mean and variance may differ from risk to risk.
No interest rate or mortality assumption is involved: this is an estimation problem for observations and their conditional moments. No parametric distribution, such as a normal or Poisson distribution, is required. The moment assumptions above are sufficient, provided the relevant variances are finite. If observations are dependent within a risk, or risks do not represent the same target population, the correction below need not apply as stated.
Deriving the Sampling-Variation Correction
For a particular risk, the average \(\bar X_j\) has conditional expectation \(\mu(\Theta_j)\) and conditional variance \(v(\Theta_j)/m\). The law of total variance then separates the across-risk variance of these averages into the desired VHM and the average sampling variance.
Because the \(R\) risk averages are independent and identically distributed under these assumptions, their usual sample variance is an unbiased estimator of \(\operatorname{Var}(\bar X_j)\). Define \(S^2_{\bar X}=\sum_{j=1}^{R}(\bar X_j-\bar{\bar X})^2/(R-1)\), where \(\bar{\bar X}=\sum_{j=1}^{R}\bar X_j/R\). Consequently, if EPV is known, subtracting the sampling contribution gives an unbiased moment estimator of VHM:
The divisor \(m\) is essential: averaging \(m\) conditionally independent observations reduces the conditional variance from \(v(\Theta)\) to \(v(\Theta)/m\). The across-risk sample variance uses \(R-1\), not \(R\), because it estimates the variance of a population of risk averages from a sample.
Worked Examples
Worked Example: Correcting Four Equal-Size Risk Averages
Four hypothetical risks have observed averages of \(30, 40, 50,\) and \(60\) dollars per exposure-year. Each average uses \(m=3\) independent exposure-year observations. Assume the EPV is known to be \(30\) squared (dollars per exposure-year). Estimate VHM. The conditional distribution is not specified; assume only the common-population and conditional moment assumptions stated above. Interest and mortality are not applicable.
Method. Calculate the unbiased sample variance of the four averages, then subtract \(EPV/m\). This avoids treating the within-risk sampling variation as if it were all genuine risk-to-risk variation.
First calculate the mean of the risk averages: \(\bar{\bar X}=(30+40+50+60)/4=180/4=45\) dollars per exposure-year.
The deviations are \(-15,-5,5,15\), and their squared values are \(225,25,25,225\). Their sum is \(500\), so \(S^2_{\bar X}=500/(4-1)=500/3=166.6667\) squared (dollars per exposure-year).
The sampling-variance contribution is \(EPV/m=30/3=10\) squared (dollars per exposure-year). Therefore, \(\widehat{VHM}=166.6667-10=156.6667\), or 156.67 squared (dollars per exposure-year), rounded to two decimal places.
Check. Adding the estimated sampling contribution back gives \(156.6667+10=166.6667\), the observed sample variance. The estimate has the square of the full observation unit: equivalently, dollars squared per exposure-year squared.
Worked Example: Unequal Numbers of Observations
Three hypothetical risks have averages \(10,16,\) and \(22\) dollars per inspection. They use \(m_1=1\), \(m_2=2\), and \(m_3=4\) independent inspections, respectively. Assume common EPV \(=12\) squared (dollars per inspection), a common risk population, and the conditional assumptions already stated. The distribution is otherwise unspecified; interest and mortality are not applicable. Estimate VHM from the unweighted sample variance of the three risk averages.
Method. Since observation counts differ, each average has sampling variance \(EPV/m_j\). The unweighted sample variance across risks therefore contains the average of these three contributions, rather than \(EPV/m\) for one shared \(m\). This avoids using a correction based on a fictitious common count.
The mean of the averages is \((10+16+22)/3=16\). Their deviations are \(-6,0,6\), whose squared values sum to \(36+0+36=72\). Thus \(S^2_{\bar X}=72/(3-1)=36\) squared (dollars per inspection).
The expected sampling contribution to this unweighted sample variance is \[ \frac{EPV}{R}\sum_{j=1}^{R}\frac{1}{m_j} =\frac{12}{3}\left(1+\frac12+\frac14\right) =4(1.75)=7. \] Therefore, \(\widehat{VHM}=36-7=\)29 squared (dollars per inspection).
Check. The three individual sampling variances are \(12/1=12\), \(12/2=6\), and \(12/4=3\); their average is \((12+6+3)/3=7\), matching the correction. Adding \(7\) back to \(29\) recovers the observed variance \(36\).
Worked Example: Estimating EPV from Within-Risk Data
Four hypothetical risks each have two independent observations, measured in dollars per service unit. The observed pairs are \((12,20)\), \((16,24)\), \((20,28)\), and \((24,32)\). Assume a common EPV across risks, conditional independence and identical distribution within each risk, and risks from one common population. Use the pooled within-risk sample variance to estimate EPV, then estimate VHM. No parametric distribution is assumed, and interest and mortality are not applicable.
Method. First estimate EPV from within-risk variation, which measures fluctuations around each risk’s own mean. Then subtract the corresponding sampling contribution from the between-risk sample variance. This avoids using differences between risk means to estimate process variance.
Each pair has a difference of \(8\), so its mean is \(16,20,24,\) or \(28\), respectively. Within any pair, the deviations from its mean are \(-4\) and \(4\); the sum of squared deviations is \(16+16=32\). Each within-risk sample variance is therefore \(32/(2-1)=32\).
Pool the four within-risk sums of squares and divide by their total degrees of freedom: \(\widehat{EPV}=(32+32+32+32)/(4(2-1))=128/4=32\) squared (dollars per service unit).
The risk averages are \(16,20,24,28\), with mean \(22\). Their deviations are \(-6,-2,2,6\), and the squared deviations sum to \(36+4+4+36=80\). Thus \(S^2_{\bar X}=80/(4-1)=80/3=26.6667\). With \(m=2\), the sampling correction is \(\widehat{EPV}/m=32/2=16\). Hence \(\widehat{VHM}=26.6667-16=10.6667\), or 10.67 squared (dollars per service unit), rounded to two decimal places.
Check. The within-risk pooled calculation uses four degrees of freedom, and the correction is \(16\), half the estimated EPV because each average contains two observations. Adding the correction to the VHM estimate gives \(10.6667+16=26.6667\), the observed sample variance of the risk averages.
Unequal Counts and Estimated EPV
For \(R\) unweighted risk averages with different observation counts \(m_j\), assume the risk means still have a common population mean, the risks are independent, and all risks share the same EPV. The expected unweighted sample variance of the averages is \(VHM+\frac{EPV}{R}\sum_{j=1}^{R}1/m_j\). The corresponding moment correction is therefore:
To see why the unequal-count expression averages the individual sampling variances, write \(\bar X_j=\mu(\Theta_j)+\epsilon_j\), where the conditional sampling error has mean zero and variance \(v(\Theta_j)/m_j\). Each risk average has the same unconditional mean, while its unconditional variance is \(VHM+EPV/m_j\). For independent observations with a common mean, the expected unweighted sample variance is the average of their variances. That average is \(VHM+(EPV/R)\sum_j1/m_j\), as claimed.
When EPV is not supplied, it can be estimated from repeated observations within risks. If all risks have \(m\) observations and share EPV, a standard pooled estimator is \(\widehat{EPV}=\sum_{j=1}^{R}\sum_{k=1}^{m}(X_{jk}-\bar X_j)^2/[R(m-1)]\). Each risk contributes \(m-1\) within-risk degrees of freedom. Substituting this estimate into the VHM correction gives a plug-in moment estimate; its exact finite-sample properties can depend on the estimation setup. If EPV is supplied as known, the equal-count estimator derived earlier is unbiased under the stated assumptions.
Common Mistakes and Interpretation
A frequent error is to subtract all of EPV from the observed variance of risk averages. That is wrong when an average contains multiple observations: the average’s conditional variance is EPV divided by its observation count. For the four averages \(30,40,50,60\) in the first example, subtracting the full EPV gives the incorrect calculation \(166.6667-30=136.6667\). The correct calculation is \(166.6667-30/3=156.6667\), or \(156.67\) squared (dollars per exposure-year).
- Using the wrong sample-variance denominator. For \(R\) risk averages, use \(R-1\) in the sample variance, not \(R\).
- Ignoring unequal counts. If \(m_j\) varies, do not use \(EPV/m\) with an arbitrary or average \(m\); use the appropriate average of \(EPV/m_j\) for the unweighted means.
- Mixing units. If \(X\) is dollars per exposure-year, then EPV and VHM are in squared (dollars per exposure-year), or dollars squared per exposure-year squared—not merely dollars squared per exposure-year.
- Changing the risk population. The interpretation depends on comparing risks drawn from a common target population. A shift in the mix of risks can change observed between-risk variation and is not automatically a change in VHM.
- Assuming the correction always produces a positive result. Sampling noise or model mismatch can make a moment estimate negative. The subtraction formula itself does not guarantee a nonnegative estimate; choices for handling boundary estimates require separate consideration.
The correction is useful because it makes the target of the calculation explicit: observed variation among finite risk averages is not identical to variation among the underlying risk means. State the observation unit, the number of observations per average, the EPV assumption, and the variance denominator before calculating; doing so makes both the arithmetic and the interpretation auditable.
Check Your Understanding
For each problem, calculate the requested moment estimate and retain the square of the stated observation unit.
- Four risks have averages \(8,12,16,20\) dollars per policy-month. Each average uses \(m=2\) observations, and known EPV is \(12\) squared (dollars per policy-month). Estimate VHM. (Answer: \(20.67\) squared (dollars per policy-month))
- Five risks have averages \(14,18,22,26,30\) dollars per visit. Each uses \(m=4\) observations, with EPV \(20\) squared (dollars per visit). Estimate VHM. (Answer: \(35\) squared (dollars per visit))
- Three risk averages are \(10,16,22\) dollars per inspection, based on \(m_1=1,m_2=2,m_3=4\), with EPV \(12\) squared (dollars per inspection). Using an unweighted sample variance, estimate VHM. (Answer: \(29\) squared (dollars per inspection))
- Four risks each have two observations. The risk averages are \(3,5,7,9\) claims per unit, and each within-risk sample variance is \(4\) squared (claims per unit). Use the pooled within-risk variance as the EPV estimate, then estimate VHM. (Answer: \(4.67\) squared (claims per unit))
- Three unweighted risk averages are \(40,50,70\) dollars per device-year, based on \(m_1=1,m_2=1,m_3=4\); EPV is \(25\) squared (dollars per device-year). Estimate VHM to two decimal places. (Answer: \(214.58\) squared (dollars per device-year))