Bayesian Experience Rating For Loss Costs

Combining Collective Expectations With Risk Experience

An experience-rated loss cost should reflect both the risk's own observations and what is known about comparable risks. An observation based on little or volatile experience should not overwhelm a well-supported collective expectation. Conversely, repeated, precise experience should move the estimate toward the risk's observed level. A Bayesian normal-normal model makes this balance explicit: it combines the collective prior and observed loss costs according to their relative precision.

The quantity estimated here is the risk's underlying expected loss cost, denoted by $\theta$. It is not the realized loss cost in a future period. The model treats observed risk loss costs as noisy measurements of $\theta$; the posterior mean is the estimate after conditioning on the observations. This distinction matters when a later calculation concerns a future realized loss amount rather than the underlying expected cost.

Definition: Let $\theta$ be a risk's underlying expected loss cost. A normal-normal experience-rating model specifies a collective prior $\theta\sim N(\mu,\tau^2)$, where $\mu$ is the collective mean and $\tau^2$ is the variance of underlying risk means around that collective. Conditional on $\theta$, observed loss costs $Y_j$ are independent and satisfy $Y_j\mid\theta\sim N(\theta,\sigma_j^2)$. The posterior distribution is the conditional distribution of $\theta$ given the observed values.

The variances have different roles. The prior variance $\tau^2$ measures variation among underlying risk means in the collective. Each $\sigma_j^2$ measures the process or sampling variation in an observed loss cost around its risk mean. All values and standard deviations must use consistent loss-cost units; variances consequently use squared units. The model assumes the collective parameters and observation variances are known for the calculation.

The observed loss cost may be based on a period or exposure unit, but its variance must match that same observation. If exposures differ, do not treat the loss costs as equally precise without justification. The exposure-specific variance can be supplied directly or derived under an appropriate process model, as in the third example.

The Posterior Estimate and Its Precision Weights

For one observation $y$ with conditional variance $\sigma^2$, the posterior variance and mean are

V=\left(\frac{1}{\tau^2}+\frac{1}{\sigma^2}\right)^{-1} =\frac{\tau^2\sigma^2}{\tau^2+\sigma^2}, \qquad E[\theta\mid Y=y] =V\left(\frac{\mu}{\tau^2}+\frac{y}{\sigma^2}\right).

Equivalently, define the experience weight $Z=\tau^2/(\tau^2+\sigma^2)$. Then

E[\theta\mid Y=y]=Zy+(1-Z)\mu.

A larger prior variance means the collective is less informative relative to the observation, so experience receives more weight. A larger observation variance means the observed loss cost is less precise, so experience receives less weight. These directions are useful checks on a calculation.

For multiple conditionally independent observations with values $y_j$ and variances $\sigma_j^2$, the formula becomes

V=\left(\frac{1}{\tau^2}+\sum_{j=1}^{k}\frac{1}{\sigma_j^2}\right)^{-1}, \qquad m=E[\theta\mid y_1,\ldots,y_k] =V\left(\frac{\mu}{\tau^2}+\sum_{j=1}^{k}\frac{y_j}{\sigma_j^2}\right).

Each component's normalized weight is its precision, the reciprocal of its variance, divided by total precision. Thus the prior weight is $V/\tau^2$, and observation $j$'s weight is $V/\sigma_j^2$. These weights are nonnegative and sum to one. The posterior distribution is $N(m,V)$.

Why Precision Weighting Holds

The prior density contributes a term proportional to $\exp[-(\theta-\mu)^2/(2\tau^2)]$. Conditional independence means the likelihood is the product of the observation densities, each contributing $\exp[-(y_j-\theta)^2/(2\sigma_j^2)]$. Multiplying prior and likelihood adds their quadratic exponents. The terms involving $\theta$ are

-\frac{1}{2}\left[ \theta^2\left(\frac{1}{\tau^2}+\sum_j\frac{1}{\sigma_j^2}\right) -2\theta\left(\frac{\mu}{\tau^2}+\sum_j\frac{y_j}{\sigma_j^2}\right) \right].

Let $A=1/\tau^2+\sum_j1/\sigma_j^2$ and $B=\mu/\tau^2+\sum_jy_j/\sigma_j^2$. Completing the square gives $A(\theta-B/A)^2$, plus a term that does not depend on $\theta$. The resulting normal density has variance $1/A=V$ and mean $B/A=m$, which gives the stated formulas. The derivation depends on a normal prior, normal conditional observations, known variances, and conditional independence; it is not the general posterior formula for arbitrary distributions.

Worked Example: One Loss-Cost Observation

A hypothetical risk has a collective prior mean of $1,200 and prior variance of 90,000 dollars squared. Its observed loss cost is $1,500, with conditional variance 160,000 dollars squared. Calculate the posterior mean and variance. Assume no discounting or mortality adjustment is used. The distributional assumptions are $\theta\sim N(1200,90000)$ and $Y\mid\theta\sim N(\theta,160000)$.

Method. Use the single-observation normal-normal update. First calculate the experience weight:

Z=\frac{90{,}000}{90{,}000+160{,}000} =\frac{90{,}000}{250{,}000}=0.36.

The collective weight is $1-Z=0.64$. Therefore the posterior mean is

m=0.36(1{,}500)+0.64(1{,}200) =540+768=\$1{,}308.

The posterior variance is

V=\frac{90{,}000(160{,}000)}{250{,}000} =57{,}600\text{ dollars}^2.

As a check, the experience weight is below one-half because its variance exceeds the prior variance, and the estimate of $1,308 lies between $1,200 and $1,500. The posterior standard deviation is $\sqrt{57{,}600}=\$240$. No rounding was needed. The common mistake this avoids is assigning more weight to the larger variance: that would incorrectly give the noisier observation greater influence.

Worked Example: Two Observations With Unequal Variances

A hypothetical specialty contractor has a collective prior $\theta\sim N(800,40000)$. Two independent historical loss-cost observations are $950 with variance 90,000 dollars squared and $1,100 with variance 40,000 dollars squared. Find the posterior mean and variance. Assume no discounting or mortality adjustment. Conditional on $\theta$, the observations are normal with the stated variances and mean $\theta$.

Method. Add prior and observation precisions. To keep the arithmetic clear, multiply every precision by 360,000:

\frac{360{,}000}{40{,}000}=9,\qquad \frac{360{,}000}{90{,}000}=4,\qquad \frac{360{,}000}{40{,}000}=9.

The total scaled precision is $9+4+9=22$. Consequently the normalized weights are $9/22$ for the collective prior, $4/22$ for the first observation, and $9/22$ for the second. The posterior mean is

m=\frac{9(800)+4(950)+9(1{,}100)}{22} =\frac{7{,}200+3{,}800+9{,}900}{22} =\frac{20{,}900}{22} =\$950.

The posterior variance is the reciprocal of total precision:

V=\frac{360{,}000}{22} =16{,}363.636\ldots\text{ dollars}^2.

Rounded only at the final step, the variance is 16,363.64 dollars squared. The result is checked by noting that the second observation, which has the smaller variance, receives more weight than the first. The posterior mean of $950 is within the range of the prior mean and observations. The method avoids taking an unweighted average, which would ignore the specified difference in observation precision.

Accounting for Exposure

A loss cost calculated from more exposure is often more precise than one calculated from less exposure. Under a model in which the loss-cost variance is inversely proportional to exposure $e$, write $\operatorname{Var}(Y\mid\theta)=\sigma_0^2/e$, where $\sigma_0^2$ is the variance for one exposure unit. This relationship is an assumption, not a universal property of every loss-cost measure; use it only when supported by the process model. Once the variances are established, the same posterior formulas apply.

Worked Example: Exposure-Adjusted Loss-Cost Variances

For a hypothetical pool, the collective prior is $\theta\sim N(1000,40000)$. The variance of a loss cost at one exposure unit is 160,000 dollars squared. Two independent experience periods have exposures 2 and 4, and observed loss costs of $1,300 and $900, respectively. Assume the loss-cost variance is the one-unit variance divided by exposure. Calculate the posterior estimate and variance. No discounting or mortality adjustment applies.

Method. Convert the one-unit process variance into each observation's variance:

\sigma_1^2=\frac{160{,}000}{2}=80{,}000, \qquad \sigma_2^2=\frac{160{,}000}{4}=40{,}000.

Scale the three precisions by 80,000. The prior precision becomes $80{,}000/40{,}000=2$, the first observation precision becomes $80{,}000/80{,}000=1$, and the second becomes $80{,}000/40{,}000=2$. Total scaled precision is $2+1+2=5$. The posterior mean is

m=\frac{2(1{,}000)+1(1{,}300)+2(900)}{5} =\frac{2{,}000+1{,}300+1{,}800}{5} =\$1{,}020.

The posterior variance is $80{,}000/5=16{,}000$ dollars squared. The four-exposure observation gets twice the precision of the two-exposure observation, despite both being reported as loss costs. The estimate lies between the collective mean and the two observations. The common mistake avoided is multiplying the loss cost itself by exposure or treating the raw observations as equally precise; exposure changes the assumed variance here, not the observed loss-cost value.

Assumptions, Interpretation, and Exam Traps

The estimate is appropriate when the collective prior represents a relevant population, the observation model reasonably describes the risk's loss-cost variation, and the specified variances are on compatible bases. A poor collective complement can pull the estimate in the wrong direction even when the arithmetic is flawless. The normal model also assigns some probability to negative loss costs; for strongly skewed or strictly positive costs, that approximation may be unsuitable. Do not apply the normal-normal formula mechanically when the data-generating model or prior is materially different.

The posterior mean is a credibility-weighted estimate of $\theta$, and under squared-error loss it minimizes posterior expected squared error. That interpretation does not turn the posterior variance into a future loss variance: it measures remaining uncertainty about $\theta$ under this model. A future observed loss cost would also reflect its own process variation. Keep those quantities separate.

A frequent exam trap is reversing the variance ratio in the single-observation formula. If $\tau^2=90{,}000$ and $\sigma^2=160{,}000$, the incorrect calculation $Z=160{,}000/(90{,}000+160{,}000)=0.64$ would produce $0.64(1{,}500)+0.36(1{,}200)=\$1{,}392$. The correction is $Z=\tau^2/(\tau^2+\sigma^2)=0.36$, giving $1,308. The larger observation variance means less, not more, weight for the observed loss cost.

Before calculating, check that each variance describes the observation as entered, not a total-loss variance or a variance based on a different exposure basis. Then check that posterior weights sum to one, the estimate is between the prior mean and observed values, and increased observation variance would reduce that observation's weight. These checks often reveal a reversed ratio or a unit mismatch. This treatment is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Check Your Understanding

For each hypothetical case, calculate the requested posterior estimate. Assume the stated normal model, known parameters, conditional independence where there are multiple observations, and no discounting or mortality adjustment.

A prior is $N(500,10000)$, and one loss-cost observation is $650 with conditional variance 15,000 dollars squared. Find the posterior mean. (Answer: $560)
A prior is $N(2000,90000)$. Two independent observations are $2,300 with variance 90,000 dollars squared and $1,700 with variance 30,000 dollars squared. Find the posterior mean. (Answer: $1,880)
A prior is $N(1200,64000)$, and an observation is $1,500 with variance 36,000 dollars squared. Find both posterior mean and variance. (Answer: Mean $1,392; variance 23,040 dollars squared)
A prior is $N(700,25000)$. One-unit loss-cost variance is 100,000 dollars squared; observations are $900 at exposure 2 and $600 at exposure 5, with variance inversely proportional to exposure. Find the posterior mean to the nearest cent. (Answer: $690.91)

What You'll Learn