Willys Flashcards Download
Become an Actuary › Exams › Flashcards › Exam FAM › Credibility
Exam FAM · ~7-9%

Exam FAM — Credibility Flashcards

Credibility theory for SOA Exam FAM: classical limited-fluctuation credibility (the full-credibility standard for frequency, severity, and aggregate losses, plus the square-root partial-credibility rule), and greatest-accuracy Bühlmann/Bühlmann–Straub credibility (hypothetical means, process variance, EPV, VHM, the Bühlmann $k$ and credibility factor $Z$, and varying exposures), with the Bayesian connection through the Poisson–gamma and exponential conjugate models — all carried through with fully worked EPV/VHM/$k$/$Z$ computations from explicit risk structures.

44 cards6 topicsFree · fact-checked · LaTeX math
Tap card or press Space to flip
Answer

Import this deck

Download all 44 cards and import them into your flashcard app (JSON or CSV — works with Anki). Using the Willys app? No import needed — this deck is already built in (Settings → Library → Browse).

Download JSON Download CSV

Every deck is built into the Willys app

All of these decks — including the full practice problem banks — come built into Willys AI Flashcards & Quizzes for iPhone & iPad (Mac version coming soon), with FSRS + SM-2 spaced repetition, streaks, and exam-date cram mode. 14-day free trial, then $14.99. To load a deck in the app: Settings → Library → Browse, then pick your exam and deck.

Download on the App Store

More Exam FAM decks:

Aggregate Loss Models Aggregate Loss Models Practice Coverage Modifications Coverage Modifications Practice Credibility Practice Life Annuities

← All Exam FAM decks

Browse all 44 cards as a list
  1. Partial credibility
    What is the goal of **credibility theory**, and what is the general form of a credibility estimate?
    Credibility blends a risk's **own recent experience** with a broader **prior/manual** estimate to set its premium. The estimate is a weighted average $\text{Estimate}=Z\,\bar X+(1-Z)\,M$, where $\bar X$ is the observed mean from the risk's data, $M$ is the prior (manual/collective) mean, and $Z\in[0,1]$ is the **credibility factor**. More/less-variable data pushes $Z$ toward $1$ (trust the data) or $0$ (trust the manual).
  2. Limited fluctuation
    In **limited-fluctuation (classical) credibility**, what does it mean for an estimate to be **fully credible**?
    The observed mean is fully credible ($Z=1$) when there is **enough data** that the observed mean is, with high probability $p$, within a relative margin $\pm k$ of its true expected value. Formally we require $\Pr\big(|\bar X-\theta|\le k\,\theta\big)\ge p$. If the data meet this stability standard we use $\bar X$ alone; if not, we use only partial credibility.
  3. Full credibility standard
    State the **full-credibility standard for claim frequency** (number of expected claims) under limited-fluctuation credibility.
    Assuming Poisson claim counts, full credibility for frequency requires the expected number of claims to be at least $\lambda_F=\left(\dfrac{z_p}{k}\right)^2$, where $z_p$ is the standard-normal percentile for probability $p$ (so $\Pr(-z_p\le Z\le z_p)=p$) and $k$ is the relative tolerance. This $\lambda_F$ is the **standard for full credibility** expressed as expected claims.
  4. Full credibility standard
    Derive the famous **1082-claim standard**: full credibility for frequency at $90\%$ probability within $\pm5\%$.
    Here $p=0.90$, so $z_p=z_{0.95}=1.645$, and $k=0.05$. $\lambda_F=\left(\dfrac{z_p}{k}\right)^2=\left(\dfrac{1.645}{0.05}\right)^2=(32.9)^2\approx 1082.41$. So about $1{,}082$ expected claims are needed for the frequency estimate to be fully credible at the $90\%/\pm5\%$ standard.
  5. Full credibility standard
    Find the full-credibility frequency standard for $90\%$ probability within $\pm2.5\%$, and explain how it compares to the $\pm5\%$ standard.
    $z_p=z_{0.95}=1.645$, $k=0.025$. $\lambda_F=\left(\dfrac{1.645}{0.025}\right)^2=(65.8)^2\approx 4329.64$. Halving the tolerance $k$ **quadruples** the required claims (from $\approx1{,}082$ to $\approx4{,}330$), because $\lambda_F$ depends on $1/k^2$.
  6. Full credibility standard
    An insurer expects $\lambda=0.20$ claims per policy per year and wants frequency to be fully credible at $90\%/\pm5\%$. How many **policy-years of exposure** are needed?
    The claim standard is $\lambda_F=(1.645/0.05)^2\approx 1082.41$ expected claims. Exposures $=\dfrac{\lambda_F}{\lambda}=\dfrac{1082.41}{0.20}\approx 5412.05$. So roughly $5{,}412$ policy-years are required to accumulate the $\approx1{,}082$ expected claims.
  7. Full credibility standard
    How is the **full-credibility standard for severity** (claim size) expressed in terms of the number of claims?
    For the mean **severity** to be fully credible, the required number of observed claims is $n_F=\left(\dfrac{z_p}{k}\right)^2\left(\dfrac{\sigma_X}{\mu_X}\right)^2=\lambda_{F,0}\,(\text{CV}_X)^2$, where $\lambda_{F,0}=(z_p/k)^2$ is the basic frequency standard and $\text{CV}_X=\sigma_X/\mu_X$ is the coefficient of variation of individual claim sizes. Severity needs more claims when claim sizes are more dispersed.
  8. Full credibility standard
    Claim sizes have mean $\mu_X=\$1{,}500$ and standard deviation $\sigma_X=\$3{,}000$. Using a $95\%/\pm5\%$ standard, how many claims are needed for **full credibility of severity**?
    Basic standard: $(z_p/k)^2=(1.960/0.05)^2=(39.2)^2\approx 1536.64$. Coefficient of variation squared: $(\text{CV}_X)^2=\left(\dfrac{3000}{1500}\right)^2=2^2=4$. $n_F=1536.64\times 4\approx 6146.56$ claims. The heavy-tailed severity ($\text{CV}=2$) inflates the requirement fourfold over the basic frequency standard.
  9. Full credibility standard
    State the **full-credibility standard for aggregate losses** (pure premium) when claim counts are **Poisson**.
    With Poisson frequency, the number of claims needed for the aggregate (pure premium) to be fully credible is $n_F=\left(\dfrac{z_p}{k}\right)^2\left(1+\dfrac{\sigma_X^2}{\mu_X^2}\right)=\lambda_{F,0}\big(1+(\text{CV}_X)^2\big)$. The extra term $1+(\text{CV}_X)^2$ combines the Poisson frequency variability (the $1$) with the severity variability (the $(\text{CV}_X)^2$). It collapses to the pure frequency standard when severity is constant ($\text{CV}_X=0$).
  10. Full credibility standard
    Claim counts are Poisson; claim sizes have mean $\mu_X=\$2{,}000$ and standard deviation $\sigma_X=\$1{,}000$. At a $90\%/\pm5\%$ standard, how many claims make the **aggregate losses** fully credible?
    Basic standard: $\lambda_{F,0}=(1.645/0.05)^2\approx 1082.41$. $(\text{CV}_X)^2=\left(\dfrac{1000}{2000}\right)^2=0.25$. $n_F=\lambda_{F,0}\big(1+0.25\big)=1082.41(1.25)\approx 1353.01$ claims. If $\lambda=0.20$ claims per exposure, the needed exposures are $1353.01/0.20\approx 6765.06$ policy-years.
  11. Full credibility standard
    Claim counts are Poisson; severity has mean $\$500$ and **variance** $750{,}000$. At a $90\%/\pm5\%$ standard, find the full-credibility standard for **aggregate losses** in claims.
    $\lambda_{F,0}=(1.645/0.05)^2\approx 1082.41$. $(\text{CV}_X)^2=\dfrac{\sigma_X^2}{\mu_X^2}=\dfrac{750{,}000}{500^2}=\dfrac{750{,}000}{250{,}000}=3$. $n_F=1082.41\,(1+3)=1082.41\times 4\approx 4329.64$ claims. The variance is given directly here, so use $\sigma_X^2/\mu_X^2$ without re-squaring.
  12. Partial credibility
    State the **square-root rule** for the partial-credibility factor $Z$ in limited-fluctuation credibility.
    When the data fall short of full credibility, assign $Z=\sqrt{\dfrac{n}{n_F}}\quad(\text{capped at }1),$ where $n$ is the actual number of observed claims (or exposures) and $n_F$ is the full-credibility standard in the same units. The square root makes $Z$ scale with the *standard deviation* of the estimator rather than its variance.
  13. Partial credibility
    Why does limited-fluctuation partial credibility use $Z=\sqrt{n/n_F}$ rather than $Z=n/n_F$?
    Full credibility requires the **standard deviation** of $\bar X$ to be small enough. The standard error of the mean shrinks like $1/\sqrt{n}$, so doubling precision needs four times the data. Setting $Z=\sqrt{n/n_F}$ makes the credibility-weighted estimator's added fluctuation match the same tolerance: at $n=n_F$, $Z=1$, and below that $Z$ scales with $\sqrt n$, not $n$.
  14. Partial credibility
    A risk has $500$ observed claims; the full-credibility standard is $n_F\approx 1{,}082$ claims. The observed mean pure premium is $\$250$ and the manual rate is $\$200$. Find the credibility estimate.
    Partial credibility: $Z=\sqrt{\dfrac{500}{1082.41}}=\sqrt{0.461934}\approx 0.67966$. Credibility estimate $=Z\bar X+(1-Z)M=0.67966(250)+0.32034(200)$ $=169.914+64.069\approx \$233.98$.
  15. Partial credibility
    Poisson claim counts, severity mean $\$500$ and variance $750{,}000$. At $90\%/\pm5\%$, $1{,}500$ claims are observed (aggregate-loss basis). Find $Z$.
    Aggregate standard: $n_F=(1.645/0.05)^2(1+\tfrac{750{,}000}{500^2})=1082.41(1+3)=1082.41\times4\approx 4329.64$ claims. Partial credibility: $Z=\sqrt{\dfrac{1500}{4329.64}}=\sqrt{0.346451}\approx 0.58860$. The $1{,}500$ claims earn about $58.9\%$ credibility on the aggregate-loss standard.
  16. Bühlmann model
    What is the central idea of **Bühlmann (greatest-accuracy) credibility**, and how does it differ from limited-fluctuation credibility?
    Bühlmann credibility is a **least-squares Bayesian** approach: among all linear estimators $\hat\mu=a+b\bar X$ it picks the $a,b$ minimizing the expected squared error against the true Bayesian premium. It produces $Z=\dfrac{n}{n+k}$ with $k$ driven by the **variance structure of the risks**, rather than a fluctuation tolerance. Unlike limited-fluctuation credibility, $Z$ is never exactly $1$ for finite $n$ and uses no arbitrary $z_p,k$ inputs.
  17. EPV & VHM
    Define the **hypothetical mean** $\mu(\theta)$ and the **process variance** $v(\theta)$ in the Bühlmann model.
    Conditional on a risk's unknown parameter $\theta$: $\mu(\theta)=E[X\mid\Theta=\theta]$ — the **hypothetical mean**, the expected outcome for a risk of type $\theta$. $v(\theta)=\operatorname{Var}(X\mid\Theta=\theta)$ — the **process variance**, the within-risk variability for that type. The overall (collective) mean is $\mu=E[\mu(\Theta)]$.
  18. EPV & VHM
    Define the **Expected Process Variance (EPV)** and the **Variance of the Hypothetical Means (VHM)**.
    $\text{EPV}=E\big[v(\Theta)\big]=E\big[\operatorname{Var}(X\mid\Theta)\big]$ — the average within-risk variability. $\text{VHM}=\operatorname{Var}\big(\mu(\Theta)\big)=\operatorname{Var}\big(E[X\mid\Theta]\big)$ — the spread of expected outcomes **across** risk types. By the law of total variance the total variance of a single observation is $\operatorname{Var}(X)=\text{EPV}+\text{VHM}$.
  19. Bühlmann model
    State the **Bühlmann credibility parameter** $k$ and the **credibility factor** $Z$ for $n$ observations.
    $k=\dfrac{\text{EPV}}{\text{VHM}}=\dfrac{E[v(\Theta)]}{\operatorname{Var}(\mu(\Theta))}$, and the credibility factor for $n$ identical exposures is $Z=\dfrac{n}{n+k}$. The Bühlmann credibility premium is $Z\bar X+(1-Z)\mu$, where $\mu=E[\mu(\Theta)]$ is the collective mean.
  20. Bühlmann model
    Interpret the Bühlmann $k=\text{EPV}/\text{VHM}$: what does a large versus small $k$ mean?
    $k$ measures noise relative to signal. **Large $k$** (EPV $\gg$ VHM): risks are similar to one another but each risk's data is noisy, so individual experience is barely informative → $Z$ small, lean on the manual. **Small $k$** (VHM $\gg$ EPV): risks differ a lot but each is measured precisely, so own experience is informative → $Z$ near $1$. Since $Z=n/(n+k)$, $Z$ rises toward $1$ as $n$ grows and as $k$ shrinks.
  21. EPV & VHM
    Two equally-likely risk types have constant per-period means $\mu_A=10$ (variance $4$) and $\mu_B=20$ (variance $16$). Compute $\mu$, EPV, and VHM.
    Each type has probability $0.5$. Overall mean: $\mu=0.5(10)+0.5(20)=15$. $\text{EPV}=E[v(\Theta)]=0.5(4)+0.5(16)=10$. $E[\mu(\Theta)^2]=0.5(10^2)+0.5(20^2)=0.5(100)+0.5(400)=250$, so $\text{VHM}=E[\mu^2]-\mu^2=250-15^2=250-225=25$.
  22. Bühlmann model
    For the two-type risk above ($\text{EPV}=10$, $\text{VHM}=25$, $\mu=15$), a risk is observed for $n=3$ periods with mean $\bar X=18$. Find the Bühlmann credibility premium.
    $k=\dfrac{\text{EPV}}{\text{VHM}}=\dfrac{10}{25}=0.4$. $Z=\dfrac{n}{n+k}=\dfrac{3}{3+0.4}=\dfrac{3}{3.4}\approx 0.88235$. Premium $=Z\bar X+(1-Z)\mu=0.88235(18)+0.11765(15)$ $=15.8824+1.7647\approx 17.65$.
  23. EPV & VHM
    Annual claim counts are Poisson$(\Lambda)$, with $\Lambda=0.10$ for $60\%$ of risks and $\Lambda=0.30$ for $40\%$. Compute $\mu$, EPV, VHM, and $k$.
    For Poisson, $\mu(\lambda)=\lambda$ and $v(\lambda)=\lambda$. $\mu=0.6(0.10)+0.4(0.30)=0.06+0.12=0.18$. $\text{EPV}=E[\Lambda]=0.6(0.10)+0.4(0.30)=0.18$. $E[\Lambda^2]=0.6(0.10^2)+0.4(0.30^2)=0.006+0.036=0.042$, so $\text{VHM}=0.042-0.18^2=0.042-0.0324=0.0096$. $k=\dfrac{0.18}{0.0096}=18.75$.
  24. Bühlmann model
    For the Poisson mixture above ($k=18.75$, $\mu=0.18$), a policyholder is observed $n=5$ years with $2$ total claims. Find the Bühlmann estimate of next year's claim count.
    $\bar X=\dfrac{2}{5}=0.40$ claims/year. $Z=\dfrac{n}{n+k}=\dfrac{5}{5+18.75}=\dfrac{5}{23.75}\approx 0.21053$. Bühlmann estimate $=Z\bar X+(1-Z)\mu=0.21053(0.40)+0.78947(0.18)$ $=0.084211+0.142105\approx 0.2263$ claims next year.
  25. EPV & VHM
    Three equally-likely risk types have hypothetical means $\mu(\theta)=1,2,3$ and process variances $v(\theta)=4,5,6$. Compute $\mu$, EPV, VHM, and $k$.
    Each type has probability $\tfrac13$. $\mu=\tfrac13(1+2+3)=2$. $\text{EPV}=\tfrac13(4+5+6)=5$. $E[\mu^2]=\tfrac13(1^2+2^2+3^2)=\tfrac13(14)\approx 4.6667$, so $\text{VHM}=4.6667-2^2=0.6667$. $k=\dfrac{\text{EPV}}{\text{VHM}}=\dfrac{5}{0.6667}=7.5$.
  26. Bühlmann model
    For the three-type structure above ($k=7.5$, $\mu=2$), a risk is observed for $n=4$ periods with sample mean $\bar X=2.5$. Find the Bühlmann premium.
    $Z=\dfrac{n}{n+k}=\dfrac{4}{4+7.5}=\dfrac{4}{11.5}\approx 0.34783$. Premium $=Z\bar X+(1-Z)\mu=0.34783(2.5)+0.65217(2)$ $=0.869565+1.304348\approx 2.1739$.
  27. Bühlmann model
    Claim counts are Poisson$(\Lambda)$ with $\Lambda=0.5$ for $80\%$ of risks and $\Lambda=1.5$ for $20\%$. A policyholder has $5$ claims over $4$ years. Find the Bühlmann credibility estimate of the annual frequency.
    Poisson: $\mu(\lambda)=v(\lambda)=\lambda$. $\mu=0.8(0.5)+0.2(1.5)=0.4+0.3=0.70$ and $\text{EPV}=E[\Lambda]=0.70$. $E[\Lambda^2]=0.8(0.25)+0.2(2.25)=0.20+0.45=0.65$, so $\text{VHM}=0.65-0.49=0.16$ and $k=\dfrac{0.70}{0.16}=4.375$. $\bar X=\tfrac54=1.25$, $Z=\dfrac{4}{4+4.375}=\dfrac{4}{8.375}\approx 0.47761$. Estimate $=0.47761(1.25)+0.52239(0.70)\approx 0.59701+0.36567=0.9627$ claims/year.
  28. EPV & VHM
    A risk's Poisson parameter $\Lambda$ is **uniform on $(0,1)$**. Claim counts are Poisson$(\Lambda)$. Compute $\mu$, EPV, VHM, and $k$.
    For $\Lambda\sim U(0,1)$: $E[\Lambda]=0.5$ and $\operatorname{Var}(\Lambda)=\dfrac{1}{12}\approx 0.08333$. Poisson gives $\mu(\lambda)=v(\lambda)=\lambda$, so $\mu=E[\Lambda]=0.5$, $\text{EPV}=E[\Lambda]=0.5$, $\text{VHM}=\operatorname{Var}(\Lambda)=\tfrac{1}{12}$. $k=\dfrac{\text{EPV}}{\text{VHM}}=\dfrac{0.5}{1/12}=6$.
  29. Bühlmann model
    For the $U(0,1)$ Poisson model above ($k=6$, $\mu=0.5$), a policyholder records $1$ claim over $2$ years. Find the Bühlmann estimate.
    $\bar X=\dfrac{1}{2}=0.5$ claims/year. $Z=\dfrac{n}{n+k}=\dfrac{2}{2+6}=\dfrac{2}{8}=0.25$. Estimate $=Z\bar X+(1-Z)\mu=0.25(0.5)+0.75(0.5)=0.5$. Here $\bar X=\mu$, so the estimate is $0.5$ regardless of $Z$ — credibility only matters when the data disagree with the collective mean.
  30. EPV & VHM
    Claim severities are **exponential** with mean $\Theta$, where $\Theta=2$ or $4$ each with probability $0.5$. Compute $\mu$, EPV, VHM, and $k$.
    For an exponential with mean $\theta$: $\mu(\theta)=\theta$ and $v(\theta)=\theta^2$. $\mu=0.5(2)+0.5(4)=3$. $\text{EPV}=E[\Theta^2]=0.5(2^2)+0.5(4^2)=0.5(4)+0.5(16)=10$. $\text{VHM}=\operatorname{Var}(\Theta)=E[\Theta^2]-\mu^2=10-9=1$. $k=\dfrac{\text{EPV}}{\text{VHM}}=\dfrac{10}{1}=10$.
  31. Bühlmann model
    For the exponential-severity model above ($k=10$, $\mu=3$), a risk's $2$ observed claims average $\bar X=3.5$. Find the Bühlmann credibility premium for severity.
    $Z=\dfrac{n}{n+k}=\dfrac{2}{2+10}=\dfrac{2}{12}\approx 0.16667$. Premium $=Z\bar X+(1-Z)\mu=0.16667(3.5)+0.83333(3)$ $=0.58333+2.5\approx 3.0833$. With only $2$ claims against $k=10$, credibility is low and the estimate stays close to the collective mean $3$.
  32. Bühlmann model
    What is the **Bayesian premium**, and how does it relate to the Bühlmann credibility premium in general?
    The Bayesian premium is the exact posterior expectation $E[X_{n+1}\mid \text{data}]$ of the next observation given the observed data — the best (least-squares) predictor with no linearity restriction. The Bühlmann premium is the **best linear approximation** to the Bayesian premium. In general they differ, but the Bühlmann estimate always lies on the least-squares regression line through the Bayesian estimates, so it is sometimes called the *credibility/linearized Bayes* estimate.
  33. Bühlmann model
    For which conjugate models does the **Bühlmann premium exactly equal the Bayesian premium**?
    When the model/prior is a **linear-exponential-family conjugate pair**, the Bayesian premium is already linear in $\bar X$, so Bühlmann reproduces it exactly. The exam-standard cases are: - **Poisson frequency with a gamma prior** on $\Lambda$; - **Exponential / Bernoulli / binomial / negative-binomial (fixed $r$) / normal** likelihoods with their conjugate priors (gamma, beta, normal). In these the credibility factor $Z=n/(n+k)$ matches the posterior-mean weighting exactly.
  34. Bühlmann model
    For the **Poisson–gamma** model, derive the Bühlmann $k$ from the gamma prior parameters $\alpha$ (shape) and $\theta$ (scale).
    Prior $\Lambda\sim\text{Gamma}(\alpha,\theta)$ has $E[\Lambda]=\alpha\theta$ and $\operatorname{Var}(\Lambda)=\alpha\theta^2$. Poisson gives $\mu(\lambda)=v(\lambda)=\lambda$, so $\text{EPV}=E[\Lambda]=\alpha\theta$ and $\text{VHM}=\operatorname{Var}(\Lambda)=\alpha\theta^2$. $k=\dfrac{\text{EPV}}{\text{VHM}}=\dfrac{\alpha\theta}{\alpha\theta^2}=\dfrac{1}{\theta}$. Thus $Z=\dfrac{n}{n+1/\theta}=\dfrac{n\theta}{n\theta+1}$, matching the gamma posterior weighting exactly.
  35. Bühlmann model
    Poisson counts with a Gamma$(\alpha=4,\ \theta=0.05)$ prior on $\Lambda$. A policyholder has $3$ claims in $5$ years. Find the Bühlmann premium and confirm it equals the Bayesian premium.
    $\mu=E[\Lambda]=\alpha\theta=4(0.05)=0.20$. $k=\dfrac1\theta=\dfrac{1}{0.05}=20$. $\bar X=\dfrac{3}{5}=0.60$, $Z=\dfrac{5}{5+20}=\dfrac{5}{25}=0.20$. Bühlmann $=0.20(0.60)+0.80(0.20)=0.12+0.16=0.28$. Bayes: posterior is Gamma$(\alpha+\sum x,\ \theta/(1+n\theta))$, mean $=\dfrac{(\alpha+\sum x)\theta}{1+n\theta}=\dfrac{(4+3)(0.05)}{1+5(0.05)}=\dfrac{0.35}{1.25}=0.28$. **Equal.**
  36. Bühlmann-Straub
    What does the **Bühlmann–Straub** model add beyond the basic Bühlmann model?
    Bühlmann–Straub handles risks observed with **different exposure volumes** $m_i$ each period (e.g. varying numbers of insureds, policy-years, or claim counts), rather than a fixed unit exposure. The per-exposure outcomes $X_i$ have process variance inversely proportional to $m_i$. It is the standard model for group/aggregate experience rating where exposure differs year to year.
  37. Bühlmann-Straub
    State the **Bühlmann–Straub** credibility factor and credibility premium with exposures $m_1,\dots,m_n$.
    Let total exposure $m=\sum_{i} m_i$ and the exposure-weighted mean $\bar X=\dfrac{\sum_i m_i X_i}{m}=\dfrac{\text{total losses}}{m}$. With $k=\dfrac{\text{EPV}}{\text{VHM}}$ (same definition), the credibility factor is $Z=\dfrac{m}{m+k}$, and the credibility premium **per unit of exposure** is $Z\bar X+(1-Z)\mu$.
  38. Bühlmann-Straub
    A group's losses by year are: $m_1=100$ exposures / $\$240$ losses, $m_2=150$ / $\$330$, $m_3=200$ / $\$500$. Given $\text{EPV}=8{,}000$, $\text{VHM}=160$, and collective mean $\mu=\$1.50$ per exposure, find the Bühlmann–Straub pure premium per exposure.
    Total exposure $m=100+150+200=450$; total losses $=240+330+500=1{,}070$. Exposure-weighted mean $\bar X=\dfrac{1{,}070}{450}\approx 2.37778$. $k=\dfrac{\text{EPV}}{\text{VHM}}=\dfrac{8{,}000}{160}=50$, so $Z=\dfrac{m}{m+k}=\dfrac{450}{450+50}=\dfrac{450}{500}=0.90$. Pure premium per exposure $=0.90(2.37778)+0.10(1.50)=2.14+0.15=\$2.29$.
  39. Bühlmann-Straub
    Continuing the Bühlmann–Straub group ($Z=0.90$, credibility pure premium $\$2.29$ per exposure): if next year's expected exposure is $250$, what are the expected **total losses**?
    The credibility estimate $\$2.29$ is a **per-exposure** pure premium. Multiply by next year's exposure: Expected total losses $=250\times 2.29=\$572.50$. The per-exposure credibility premium scales linearly with the projected exposure base.
  40. Bühlmann-Straub
    In Bühlmann–Straub, EPV $=10{,}000$ and VHM $=200$. A policyholder has total exposure $m=75$ with weighted-average pure premium $\bar X=\$1{,}200$; the manual rate is $\mu=\$1{,}000$. Find the credibility premium.
    $k=\dfrac{\text{EPV}}{\text{VHM}}=\dfrac{10{,}000}{200}=50$. $Z=\dfrac{m}{m+k}=\dfrac{75}{75+50}=\dfrac{75}{125}=0.60$. Credibility premium $=Z\bar X+(1-Z)\mu=0.60(1{,}200)+0.40(1{,}000)$ $=720+400=\$1{,}120$ per exposure.
  41. Bühlmann-Straub
    How does the Bühlmann–Straub factor $Z=\dfrac{m}{m+k}$ reduce to ordinary Bühlmann, and what does $m$ represent?
    When every period carries the **same unit exposure** ($m_i=1$), the total exposure $m=\sum m_i$ equals the number of periods $n$, and $Z=\dfrac{m}{m+k}=\dfrac{n}{n+k}$ — the ordinary Bühlmann factor. In general $m$ is the **total exposure** (sum of the $m_i$), not a count of periods, so a few high-exposure years can earn as much credibility as many small ones.
  42. EPV & VHM
    How do you compute **EPV** and **VHM** from a discrete risk-structure table giving, for each type $\theta$, its probability, mean $\mu(\theta)$, and process variance $v(\theta)$?
    Treat the type as a discrete random variable $\Theta$: 1. $\mu=E[\mu(\Theta)]=\sum_\theta p(\theta)\,\mu(\theta)$. 2. $\text{EPV}=E[v(\Theta)]=\sum_\theta p(\theta)\,v(\theta)$. 3. $\text{VHM}=\operatorname{Var}(\mu(\Theta))=\sum_\theta p(\theta)\,\mu(\theta)^2-\mu^2$. Then $k=\text{EPV}/\text{VHM}$ and $Z=n/(n+k)$ (or $m/(m+k)$ for Bühlmann–Straub).
  43. EPV & VHM
    When the EPV and VHM are **estimated from data** (nonparametric empirical Bayes), why might the estimated $\hat Z$ behave oddly, and what is the practical fix?
    With unbiased empirical estimators $\widehat{\text{VHM}}$ can come out **negative** (sampling noise can make the between-risk variation estimate fall below zero). Then $\hat k=\widehat{\text{EPV}}/\widehat{\text{VHM}}$ is negative or undefined and $\hat Z$ is nonsensical. The practical fix is to **truncate** $\widehat{\text{VHM}}$ at $0$ (treat all risks as identical, $Z=0$) and lean entirely on the collective mean.
  44. Limited fluctuation
    A risk has $750$ observed claims with a full-credibility frequency standard of $\approx 1{,}082$ claims. Compute its limited-fluctuation credibility $Z$, and contrast with how Bühlmann would assign credibility.
    Limited fluctuation: $Z=\sqrt{\dfrac{750}{1082.41}}=\sqrt{0.692899}\approx 0.83240$. So about $83.2\%$ credibility under the square-root rule. **Contrast:** Bühlmann would instead set $Z=\dfrac{n}{n+k}$ using the risk-structure $k=\text{EPV}/\text{VHM}$ — it never reaches exactly $1$, needs no $z_p$/$k$ tolerance, and is grounded in the actual variance components rather than a normal-approximation stability target.