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CAS Exam 8 · Bühlmann Credibility Factor · Tutorial 347 of 989

From the Optimal Weight to Z = n/(n+K)

Learn to convert the optimal linear credibility weight into n/(n+K), calculate the resulting estimate, and recognize when the simplification does not apply.

Intermediate 12 min read

What You'll Learn

  • Define the expected process variance and variance of hypothetical means used in the weight formula
  • Derive the credibility factor for n conditionally independent observations with common process variance
  • Simplify the optimal weight to n/(n+K) and calculate K
  • Use the credibility factor to combine an individual’s experience with the collective mean
  • Distinguish an average from a total when applying the formula
  • Identify boundary cases and conditions that invalidate the simple form

From an Optimal Weight to a Simple Formula

In Bühlmann credibility, the estimate for an individual risk combines that risk’s experience with the collective mean. The optimal linear weight on the individual’s experience initially appears to depend on both the number of observations and two variance components. Under a common process variance and conditional independence, it can be written in the compact form \(Z=n/(n+K)\). This form makes it straightforward to calculate the weight and to see how the number of observations enters.

The calculation uses the average of the individual’s \(n\) observations, not their total. The factor \(K\) summarizes the relationship between process variation within a risk and variation between risks. Here the goal is to derive and use the compact weight; the assumptions that make it valid are as important as the algebra.

Definition: Let \(\Theta\) represent the latent risk characteristic, and let \(Y_1,\ldots,Y_n\) be observations for one risk. Write \(M=E[Y_j\mid\Theta]\) for the risk’s conditional mean and \(m=E[M]\) for the collective mean. The expected process variance is \(a=E[\operatorname{Var}(Y_j\mid\Theta)]\), and the variance of hypothetical means is \(v=\operatorname{Var}(M)\). For the common-process-variance case, assume \(\operatorname{Var}(Y_j\mid\Theta)=a\) for each observation. Define \(K=a/v\) when \(v>0\).

The Bühlmann setup assumes that, conditional on \(\Theta\), the observations have the same conditional mean \(M\), are independent, and have the same finite conditional variance \(a\). Consequently, they are identically distributed conditionally in their first two moments. The observations need not be normally distributed: the credibility-weight derivation uses means, variances, and covariances, not a normal likelihood.

Why the Weight Has the Form \(n/(n+K)\)

Let \(\bar Y=(Y_1+\cdots+Y_n)/n\) be the risk’s observed average. A linear credibility estimate of \(M\) has the form \(m+Z(\bar Y-m)\), equivalently \(Z\bar Y+(1-Z)m\). Its optimal linear coefficient under squared-error loss is the covariance of \(\bar Y\) with \(M\), divided by the variance of \(\bar Y\).

1
Find the covariance with the risk mean.
Since \(E[\bar Y\mid\Theta]=M\), the observations’ deviations from \(M\) have conditional mean zero. Thus \(\operatorname{Cov}(\bar Y,M)=\operatorname{Var}(M)=v\).
2
Find the variance of the average.
The variability of \(\bar Y\) has a between-risk part \(v\) and a within-risk part \(a/n\). Hence \(\operatorname{Var}(\bar Y)=v+a/n\).
3
Form and simplify the optimal coefficient.
Divide the covariance by the variance, then divide the numerator and denominator by \(v\), which is positive in this case.
$$ Z=\frac{\operatorname{Cov}(\bar Y,M)}{\operatorname{Var}(\bar Y)} =\frac{v}{v+a/n} =\frac{nv}{nv+a} =\frac{n}{n+a/v} =\frac{n}{n+K}. $$

For the variance calculation, conditional independence gives \(\operatorname{Var}(\bar Y\mid\Theta)=a/n\). The law of total variance then gives \(\operatorname{Var}(\bar Y)=\operatorname{Var}(E[\bar Y\mid\Theta])+E[\operatorname{Var}(\bar Y\mid\Theta)]=v+a/n\). The covariance follows from conditional expectation: \(\operatorname{Cov}(\bar Y,M)=\operatorname{Cov}(E[\bar Y\mid\Theta],M)=\operatorname{Cov}(M,M)=v\). These identities justify the optimal linear coefficient and the simplification.

Theorem: If \(n\geq 1\), \(v>0\), and the observations have a common conditional mean, common process variance \(a\), and are conditionally independent given \(\Theta\), the optimal linear credibility weight on their average is \(Z=n/(n+K)\), where \(K=a/v\). The resulting estimate is \(Z\bar Y+(1-Z)m\).

Both variance components are measured in squared units of the observation, so their ratio \(K\) is dimensionless. The number \(n\) must count observations that each have the assumed process variance. If observations are totals over differing exposure amounts, or if they have unequal conditional variances, simply counting them as identical observations may not satisfy the assumptions.

Worked Calculations

Worked Example: Six Annual Loss-Cost Observations

A hypothetical coverage class uses annual loss cost in dollars as its observation. For each risk, conditional on its risk mean \(\Theta\), annual observations are independent and normally distributed as \(Y_j\mid\Theta\sim N(\Theta,900)\), where the second parameter is the variance. Across risks, \(\Theta\) is 80, 100, or 120 dollars with probabilities 0.375, 0.25, and 0.375, respectively. One risk has six annual observations of 110, 120, 125, 130, 135, and 148 dollars. Calculate its credibility-weighted estimate, using the portfolio mean as the collective mean.

1. Calculate the collective mean. The mean risk level is \(m=0.375(80)+0.25(100)+0.375(120)=30+25+45=100\) dollars.

2. Calculate the variance of hypothetical means. The deviations from 100 are \(-20,0,20\), so \(v=0.375(20^2)+0.25(0^2)+0.375(20^2)=150+0+150=300\) dollars squared. The conditional process variance is \(a=900\) dollars squared.

3. Calculate \(K\) and the credibility factor. \(K=a/v=900/300=3\), so for \(n=6\), \(Z=6/(6+3)=6/9=2/3=0.6667\), rounded to four decimal places.

4. Calculate the risk’s experience average. The total is \(110+120+125+130+135+148=768\) dollars, and \(\bar Y=768/6=128\) dollars.

5. Combine experience and the collective mean. The estimate is \((2/3)(128)+(1/3)(100)=85.3333+33.3333=118.6667\) dollars, or $118.67 rounded to the nearest cent.

Check. Directly, the weight is \(nv/(nv+a)=6(300)/(6(300)+900)=1800/2700=2/3\), matching the simplified result. The method avoids treating six observations as though the individual had only one observation.

Worked Example: Use the Average, Not the Total

A hypothetical fleet’s annual claim-cost observation is conditionally normal: \(Y_j\mid\Theta\sim N(\Theta,16)\), with conditional independence across years. Across risks, \(\Theta\) equals 8 or 12 thousand dollars with probability 0.5 each. For one risk, the total annual claim cost over 12 years is 138 thousand dollars. Calculate its credibility-weighted estimate in thousands of dollars.

1. Find the collective mean and variance of hypothetical means. The collective mean is \(m=0.5(8)+0.5(12)=10\). The variance is \(v=0.5(8-10)^2+0.5(12-10)^2=0.5(4)+0.5(4)=4\), in squared thousands of dollars.

2. Find \(K\) and \(Z\). Here \(a=16\), so \(K=16/4=4\). With \(n=12\), \(Z=12/(12+4)=12/16=0.75\).

3. Convert the total to an average. The observed average is \(\bar Y=138/12=11.5\) thousand dollars.

4. Apply the weight. The estimate is \(0.75(11.5)+0.25(10)=8.625+2.5=11.125\) thousand dollars, or $11,125.00.

Check and trap. The total is 138, but the model’s observation is an annual amount. Treating the total as one observation would give \(Z=1/(1+4)=0.2\) and the nonsensical calculation \(0.2(138)+0.8(10)=35.6\) thousand dollars. The correction is to use \(n=12\) and the average of 11.5. Equivalently, the coefficient on the centered total is \(12(4)/(12^2(4)+12(16))=48/768=0.0625\); applying it to \(138-12(10)=18\) gives \(10+0.0625(18)=11.125\).

Worked Example: Credibility at Current and Target Volumes

A hypothetical insurer models annual normalized claim cost in thousands of dollars. For each risk, observations are conditionally independent with \(Y_j\mid\Theta\sim N(\Theta,720)\). Across risks, \(\Theta\) is 30, 50, or 70 thousand dollars with probabilities 0.225, 0.55, and 0.225. A risk’s five annual observations are 54, 56, 58, 60, and 62 thousand dollars. Calculate its estimate, its credibility at 20 observations, and the smallest number of observations needed for credibility of at least 0.90.

1. Calculate the portfolio mean and \(v\). \(m=0.225(30)+0.55(50)+0.225(70)=6.75+27.5+15.75=50\). The deviations from 50 are \(-20,0,20\), so \(v=0.225(400)+0.55(0)+0.225(400)=90+0+90=180\). Thus \(a=720\) and \(K=720/180=4\).

2. Calculate the observed average and five-observation weight. The sum is \(54+56+58+60+62=290\), so \(\bar Y=290/5=58\). The weight is \(Z_5=5/(5+4)=5/9=0.5556\), rounded to four decimals.

3. Calculate the estimate. \(Z_5(58)+(1-Z_5)(50)=50+(5/9)(58-50)=50+40/9=54.4444\) thousand dollars. Rounded to cents, the estimate is $54,444.44.

4. Calculate credibility at 20 observations. \(Z_{20}=20/(20+4)=20/24=0.8333\), rounded to four decimals.

5. Solve for the minimum volume reaching 0.90. For \(n/(n+4)\geq0.90\), the positive denominator permits cross-multiplication: \(n\geq0.90(n+4)=0.90n+3.6\), so \(0.10n\geq3.6\), or \(n\geq36\). The smallest whole-number count is 36 observations.

Check. At \(n=36\), \(Z=36/40=0.90\); at \(n=35\), \(Z=35/39\approx0.8974\), below the target. This verifies the minimum.

Assumptions, Boundary Cases, and Exam Traps

The short form \(n/(n+K)\) is useful because it separates the count of observations from the variance ratio. But it is not a general substitute for checking the model. The derivation uses the variance \(a/n\) for the average, which follows from equal process variances and conditional independence. When those conditions fail, the variance of the average must be recalculated before finding the optimal linear weight.

  • Using the wrong observation unit. If each observation is annual cost, \(n\) counts years, not the number of claims within those years. Use the same unit for \(a\) that is assumed in the model.
  • Using a total with an average-based formula. The formula \(n/(n+K)\) weights \(\bar Y\). Convert a total to an average before combining it with \(m\).
  • Confusing \(a\) and \(v\). \(a\) describes process variation conditional on the risk; \(v\) describes variation in the risk means across risks. Reversing them reverses the credibility effect.
  • Ignoring dependence. If observations are conditionally correlated, their average generally does not have process variance \(a/n\). The simple factor need not be optimal.

When \(a=0\) and \(v>0\), each observation reveals the risk mean without process noise, so \(Z=1\) for \(n\geq1\); this agrees with \(K=0\). When \(v=0\) and \(a>0\), there is no variation in risk means to distinguish, and experience should receive zero weight for estimating an individual deviation from the collective mean. In that boundary case \(K=a/v\) is undefined, so use the covariance formula or the boundary conclusion rather than substituting into \(n/(n+K)\).

Key takeaway: Under common conditional process variance and conditional independence, the optimal weight on the individual’s experience average is \(Z=n/(n+K)\), with \(K=a/v\). Use the model’s observation unit, calculate the average rather than the total, and verify that the assumptions supporting the variance \(a/n\) hold.

Check Your Understanding

For each hypothetical setting, calculate the requested quantity and check whether the common-process-variance formula applies.

  1. A risk has \(n=3\) independent observations, process variance \(a=600\), and variance of hypothetical means \(v=150\). Find \(K\) and \(Z\). (Answer: \(K=4,\;Z=3/7=0.4286\).)
  2. Annual observations have common process variance \(a=25\) and \(v=5\). What credibility applies to 20 years of experience? (Answer: \(Z=20/(20+5)=0.8000\).)
  3. A risk has eight independent annual observations with total 96. The process variance is 36, \(v=9\), and the collective mean is 10. Find the credibility-weighted estimate. (Answer: \(\bar Y=12,\;K=4,\;Z=8/12=2/3\), estimate \(=10+(2/3)(12-10)=11.3333\).)
  4. With \(K=3\), what is the minimum integer \(n\) for which credibility is at least 0.75? (Answer: \(n=9\), since \(9/(9+3)=0.75\), while \(8/11<0.75\).)
  5. There are 10 observations with conditional variance 200 each, but every pair has conditional covariance 10. Let \(v=50\). Does \(n/(n+K)\), using \(K=200/50\), give the optimal weight? Calculate the weight from the covariance and variance of the average instead. (Answer: No. \(\operatorname{Var}(\bar Y)=50+[200+9(10)]/10=79\) and \(\operatorname{Cov}(\bar Y,M)=50\), so the weight is \(50/79=0.6329\), not \(10/(10+4)=0.7143\).)

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