{ "deckName": "Exam FAM — Credibility", "examCode": "Exam FAM", "cards": [ { "front": "What is the goal of **credibility theory**, and what is the general form of a credibility estimate?", "back": "Credibility blends a risk's **own recent experience** with a broader **prior/manual** estimate to set its premium. The estimate is a weighted average\n$\\text{Estimate}=Z\\,\\bar X+(1-Z)\\,M$,\nwhere $\\bar X$ 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).", "tag": "Partial credibility" }, { "front": "In **limited-fluctuation (classical) credibility**, what does it mean for an estimate to be **fully credible**?", "back": "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\n$\\Pr\\big(|\\bar X-\\theta|\\le k\\,\\theta\\big)\\ge p$.\nIf the data meet this stability standard we use $\\bar X$ alone; if not, we use only partial credibility.", "tag": "Limited fluctuation" }, { "front": "State the **full-credibility standard for claim frequency** (number of expected claims) under limited-fluctuation credibility.", "back": "Assuming Poisson claim counts, full credibility for frequency requires the expected number of claims to be at least\n$\\lambda_F=\\left(\\dfrac{z_p}{k}\\right)^2$,\nwhere $z_p$ 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.", "tag": "Full credibility standard" }, { "front": "Derive the famous **1082-claim standard**: full credibility for frequency at $90\\%$ probability within $\\pm5\\%$.", "back": "Here $p=0.90$, so $z_p=z_{0.95}=1.645$, and $k=0.05$.\n$\\lambda_F=\\left(\\dfrac{z_p}{k}\\right)^2=\\left(\\dfrac{1.645}{0.05}\\right)^2=(32.9)^2\\approx 1082.41$.\nSo about $1{,}082$ expected claims are needed for the frequency estimate to be fully credible at the $90\\%/\\pm5\\%$ standard.", "tag": "Full credibility standard" }, { "front": "Find the full-credibility frequency standard for $90\\%$ probability within $\\pm2.5\\%$, and explain how it compares to the $\\pm5\\%$ standard.", "back": "$z_p=z_{0.95}=1.645$, $k=0.025$.\n$\\lambda_F=\\left(\\dfrac{1.645}{0.025}\\right)^2=(65.8)^2\\approx 4329.64$.\nHalving the tolerance $k$ **quadruples** the required claims (from $\\approx1{,}082$ to $\\approx4{,}330$), because $\\lambda_F$ depends on $1/k^2$.", "tag": "Full credibility standard" }, { "front": "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?", "back": "The claim standard is $\\lambda_F=(1.645/0.05)^2\\approx 1082.41$ expected claims.\nExposures $=\\dfrac{\\lambda_F}{\\lambda}=\\dfrac{1082.41}{0.20}\\approx 5412.05$.\nSo roughly $5{,}412$ policy-years are required to accumulate the $\\approx1{,}082$ expected claims.", "tag": "Full credibility standard" }, { "front": "How is the **full-credibility standard for severity** (claim size) expressed in terms of the number of claims?", "back": "For the mean **severity** to be fully credible, the required number of observed claims is\n$n_F=\\left(\\dfrac{z_p}{k}\\right)^2\\left(\\dfrac{\\sigma_X}{\\mu_X}\\right)^2=\\lambda_{F,0}\\,(\\text{CV}_X)^2$,\nwhere $\\lambda_{F,0}=(z_p/k)^2$ 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.", "tag": "Full credibility standard" }, { "front": "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**?", "back": "Basic standard: $(z_p/k)^2=(1.960/0.05)^2=(39.2)^2\\approx 1536.64$.\nCoefficient of variation squared: $(\\text{CV}_X)^2=\\left(\\dfrac{3000}{1500}\\right)^2=2^2=4$.\n$n_F=1536.64\\times 4\\approx 6146.56$ claims.\nThe heavy-tailed severity ($\\text{CV}=2$) inflates the requirement fourfold over the basic frequency standard.", "tag": "Full credibility standard" }, { "front": "State the **full-credibility standard for aggregate losses** (pure premium) when claim counts are **Poisson**.", "back": "With Poisson frequency, the number of claims needed for the aggregate (pure premium) to be fully credible is\n$n_F=\\left(\\dfrac{z_p}{k}\\right)^2\\left(1+\\dfrac{\\sigma_X^2}{\\mu_X^2}\\right)=\\lambda_{F,0}\\big(1+(\\text{CV}_X)^2\\big)$.\nThe 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 ($\\text{CV}_X=0$).", "tag": "Full credibility standard" }, { "front": "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?", "back": "Basic standard: $\\lambda_{F,0}=(1.645/0.05)^2\\approx 1082.41$.\n$(\\text{CV}_X)^2=\\left(\\dfrac{1000}{2000}\\right)^2=0.25$.\n$n_F=\\lambda_{F,0}\\big(1+0.25\\big)=1082.41(1.25)\\approx 1353.01$ claims.\nIf $\\lambda=0.20$ claims per exposure, the needed exposures are $1353.01/0.20\\approx 6765.06$ policy-years.", "tag": "Full credibility standard" }, { "front": "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.", "back": "$\\lambda_{F,0}=(1.645/0.05)^2\\approx 1082.41$.\n$(\\text{CV}_X)^2=\\dfrac{\\sigma_X^2}{\\mu_X^2}=\\dfrac{750{,}000}{500^2}=\\dfrac{750{,}000}{250{,}000}=3$.\n$n_F=1082.41\\,(1+3)=1082.41\\times 4\\approx 4329.64$ claims.\nThe variance is given directly here, so use $\\sigma_X^2/\\mu_X^2$ without re-squaring.", "tag": "Full credibility standard" }, { "front": "State the **square-root rule** for the partial-credibility factor $Z$ in limited-fluctuation credibility.", "back": "When the data fall short of full credibility, assign\n$Z=\\sqrt{\\dfrac{n}{n_F}}\\quad(\\text{capped at }1),$\nwhere $n$ 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.", "tag": "Partial credibility" }, { "front": "Why does limited-fluctuation partial credibility use $Z=\\sqrt{n/n_F}$ rather than $Z=n/n_F$?", "back": "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 $n$.", "tag": "Partial credibility" }, { "front": "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.", "back": "Partial credibility: $Z=\\sqrt{\\dfrac{500}{1082.41}}=\\sqrt{0.461934}\\approx 0.67966$.\nCredibility estimate $=Z\\bar X+(1-Z)M=0.67966(250)+0.32034(200)$\n$=169.914+64.069\\approx \\$233.98$.", "tag": "Partial credibility" }, { "front": "Poisson claim counts, severity mean $\\$500$ and variance $750{,}000$. At $90\\%/\\pm5\\%$, $1{,}500$ claims are observed (aggregate-loss basis). Find $Z$.", "back": "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.\nPartial credibility: $Z=\\sqrt{\\dfrac{1500}{4329.64}}=\\sqrt{0.346451}\\approx 0.58860$.\nThe $1{,}500$ claims earn about $58.9\\%$ credibility on the aggregate-loss standard.", "tag": "Partial credibility" }, { "front": "What is the central idea of **Bühlmann (greatest-accuracy) credibility**, and how does it differ from limited-fluctuation credibility?", "back": "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.", "tag": "Bühlmann model" }, { "front": "Define the **hypothetical mean** $\\mu(\\theta)$ and the **process variance** $v(\\theta)$ in the Bühlmann model.", "back": "Conditional on a risk's unknown parameter $\\theta$:\n$\\mu(\\theta)=E[X\\mid\\Theta=\\theta]$ — the **hypothetical mean**, the expected outcome for a risk of type $\\theta$.\n$v(\\theta)=\\operatorname{Var}(X\\mid\\Theta=\\theta)$ — the **process variance**, the within-risk variability for that type.\nThe overall (collective) mean is $\\mu=E[\\mu(\\Theta)]$.", "tag": "EPV & VHM" }, { "front": "Define the **Expected Process Variance (EPV)** and the **Variance of the Hypothetical Means (VHM)**.", "back": "$\\text{EPV}=E\\big[v(\\Theta)\\big]=E\\big[\\operatorname{Var}(X\\mid\\Theta)\\big]$ — the average within-risk variability.\n$\\text{VHM}=\\operatorname{Var}\\big(\\mu(\\Theta)\\big)=\\operatorname{Var}\\big(E[X\\mid\\Theta]\\big)$ — the spread of expected outcomes **across** risk types.\nBy the law of total variance the total variance of a single observation is $\\operatorname{Var}(X)=\\text{EPV}+\\text{VHM}$.", "tag": "EPV & VHM" }, { "front": "State the **Bühlmann credibility parameter** $k$ and the **credibility factor** $Z$ for $n$ observations.", "back": "$k=\\dfrac{\\text{EPV}}{\\text{VHM}}=\\dfrac{E[v(\\Theta)]}{\\operatorname{Var}(\\mu(\\Theta))}$,\nand the credibility factor for $n$ identical exposures is\n$Z=\\dfrac{n}{n+k}$.\nThe Bühlmann credibility premium is $Z\\bar X+(1-Z)\\mu$, where $\\mu=E[\\mu(\\Theta)]$ is the collective mean.", "tag": "Bühlmann model" }, { "front": "Interpret the Bühlmann $k=\\text{EPV}/\\text{VHM}$: what does a large versus small $k$ mean?", "back": "$k$ 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.", "tag": "Bühlmann model" }, { "front": "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.", "back": "Each type has probability $0.5$.\nOverall mean: $\\mu=0.5(10)+0.5(20)=15$.\n$\\text{EPV}=E[v(\\Theta)]=0.5(4)+0.5(16)=10$.\n$E[\\mu(\\Theta)^2]=0.5(10^2)+0.5(20^2)=0.5(100)+0.5(400)=250$, so\n$\\text{VHM}=E[\\mu^2]-\\mu^2=250-15^2=250-225=25$.", "tag": "EPV & VHM" }, { "front": "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.", "back": "$k=\\dfrac{\\text{EPV}}{\\text{VHM}}=\\dfrac{10}{25}=0.4$.\n$Z=\\dfrac{n}{n+k}=\\dfrac{3}{3+0.4}=\\dfrac{3}{3.4}\\approx 0.88235$.\nPremium $=Z\\bar X+(1-Z)\\mu=0.88235(18)+0.11765(15)$\n$=15.8824+1.7647\\approx 17.65$.", "tag": "Bühlmann model" }, { "front": "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 $k$.", "back": "For Poisson, $\\mu(\\lambda)=\\lambda$ and $v(\\lambda)=\\lambda$.\n$\\mu=0.6(0.10)+0.4(0.30)=0.06+0.12=0.18$.\n$\\text{EPV}=E[\\Lambda]=0.6(0.10)+0.4(0.30)=0.18$.\n$E[\\Lambda^2]=0.6(0.10^2)+0.4(0.30^2)=0.006+0.036=0.042$, so $\\text{VHM}=0.042-0.18^2=0.042-0.0324=0.0096$.\n$k=\\dfrac{0.18}{0.0096}=18.75$.", "tag": "EPV & VHM" }, { "front": "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.", "back": "$\\bar X=\\dfrac{2}{5}=0.40$ claims/year.\n$Z=\\dfrac{n}{n+k}=\\dfrac{5}{5+18.75}=\\dfrac{5}{23.75}\\approx 0.21053$.\nBühlmann estimate $=Z\\bar X+(1-Z)\\mu=0.21053(0.40)+0.78947(0.18)$\n$=0.084211+0.142105\\approx 0.2263$ claims next year.", "tag": "Bühlmann model" }, { "front": "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 $k$.", "back": "Each type has probability $\\tfrac13$.\n$\\mu=\\tfrac13(1+2+3)=2$.\n$\\text{EPV}=\\tfrac13(4+5+6)=5$.\n$E[\\mu^2]=\\tfrac13(1^2+2^2+3^2)=\\tfrac13(14)\\approx 4.6667$, so $\\text{VHM}=4.6667-2^2=0.6667$.\n$k=\\dfrac{\\text{EPV}}{\\text{VHM}}=\\dfrac{5}{0.6667}=7.5$.", "tag": "EPV & VHM" }, { "front": "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.", "back": "$Z=\\dfrac{n}{n+k}=\\dfrac{4}{4+7.5}=\\dfrac{4}{11.5}\\approx 0.34783$.\nPremium $=Z\\bar X+(1-Z)\\mu=0.34783(2.5)+0.65217(2)$\n$=0.869565+1.304348\\approx 2.1739$.", "tag": "Bühlmann model" }, { "front": "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.", "back": "Poisson: $\\mu(\\lambda)=v(\\lambda)=\\lambda$.\n$\\mu=0.8(0.5)+0.2(1.5)=0.4+0.3=0.70$ and $\\text{EPV}=E[\\Lambda]=0.70$.\n$E[\\Lambda^2]=0.8(0.25)+0.2(2.25)=0.20+0.45=0.65$, so $\\text{VHM}=0.65-0.49=0.16$ and $k=\\dfrac{0.70}{0.16}=4.375$.\n$\\bar X=\\tfrac54=1.25$, $Z=\\dfrac{4}{4+4.375}=\\dfrac{4}{8.375}\\approx 0.47761$.\nEstimate $=0.47761(1.25)+0.52239(0.70)\\approx 0.59701+0.36567=0.9627$ claims/year.", "tag": "Bühlmann model" }, { "front": "A risk's Poisson parameter $\\Lambda$ is **uniform on $(0,1)$**. Claim counts are Poisson$(\\Lambda)$. Compute $\\mu$, EPV, VHM, and $k$.", "back": "For $\\Lambda\\sim U(0,1)$: $E[\\Lambda]=0.5$ and $\\operatorname{Var}(\\Lambda)=\\dfrac{1}{12}\\approx 0.08333$.\nPoisson gives $\\mu(\\lambda)=v(\\lambda)=\\lambda$, so\n$\\mu=E[\\Lambda]=0.5$, $\\text{EPV}=E[\\Lambda]=0.5$, $\\text{VHM}=\\operatorname{Var}(\\Lambda)=\\tfrac{1}{12}$.\n$k=\\dfrac{\\text{EPV}}{\\text{VHM}}=\\dfrac{0.5}{1/12}=6$.", "tag": "EPV & VHM" }, { "front": "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.", "back": "$\\bar X=\\dfrac{1}{2}=0.5$ claims/year.\n$Z=\\dfrac{n}{n+k}=\\dfrac{2}{2+6}=\\dfrac{2}{8}=0.25$.\nEstimate $=Z\\bar X+(1-Z)\\mu=0.25(0.5)+0.75(0.5)=0.5$.\nHere $\\bar X=\\mu$, so the estimate is $0.5$ regardless of $Z$ — credibility only matters when the data disagree with the collective mean.", "tag": "Bühlmann model" }, { "front": "Claim severities are **exponential** with mean $\\Theta$, where $\\Theta=2$ or $4$ each with probability $0.5$. Compute $\\mu$, EPV, VHM, and $k$.", "back": "For an exponential with mean $\\theta$: $\\mu(\\theta)=\\theta$ and $v(\\theta)=\\theta^2$.\n$\\mu=0.5(2)+0.5(4)=3$.\n$\\text{EPV}=E[\\Theta^2]=0.5(2^2)+0.5(4^2)=0.5(4)+0.5(16)=10$.\n$\\text{VHM}=\\operatorname{Var}(\\Theta)=E[\\Theta^2]-\\mu^2=10-9=1$.\n$k=\\dfrac{\\text{EPV}}{\\text{VHM}}=\\dfrac{10}{1}=10$.", "tag": "EPV & VHM" }, { "front": "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.", "back": "$Z=\\dfrac{n}{n+k}=\\dfrac{2}{2+10}=\\dfrac{2}{12}\\approx 0.16667$.\nPremium $=Z\\bar X+(1-Z)\\mu=0.16667(3.5)+0.83333(3)$\n$=0.58333+2.5\\approx 3.0833$.\nWith only $2$ claims against $k=10$, credibility is low and the estimate stays close to the collective mean $3$.", "tag": "Bühlmann model" }, { "front": "What is the **Bayesian premium**, and how does it relate to the Bühlmann credibility premium in general?", "back": "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.\nThe 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.", "tag": "Bühlmann model" }, { "front": "For which conjugate models does the **Bühlmann premium exactly equal the Bayesian premium**?", "back": "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:\n- **Poisson frequency with a gamma prior** on $\\Lambda$;\n- **Exponential / Bernoulli / binomial / negative-binomial (fixed $r$) / normal** likelihoods with their conjugate priors (gamma, beta, normal).\nIn these the credibility factor $Z=n/(n+k)$ matches the posterior-mean weighting exactly.", "tag": "Bühlmann model" }, { "front": "For the **Poisson–gamma** model, derive the Bühlmann $k$ from the gamma prior parameters $\\alpha$ (shape) and $\\theta$ (scale).", "back": "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\n$\\text{EPV}=E[\\Lambda]=\\alpha\\theta$ and $\\text{VHM}=\\operatorname{Var}(\\Lambda)=\\alpha\\theta^2$.\n$k=\\dfrac{\\text{EPV}}{\\text{VHM}}=\\dfrac{\\alpha\\theta}{\\alpha\\theta^2}=\\dfrac{1}{\\theta}$.\nThus $Z=\\dfrac{n}{n+1/\\theta}=\\dfrac{n\\theta}{n\\theta+1}$, matching the gamma posterior weighting exactly.", "tag": "Bühlmann model" }, { "front": "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.", "back": "$\\mu=E[\\Lambda]=\\alpha\\theta=4(0.05)=0.20$. $k=\\dfrac1\\theta=\\dfrac{1}{0.05}=20$.\n$\\bar X=\\dfrac{3}{5}=0.60$, $Z=\\dfrac{5}{5+20}=\\dfrac{5}{25}=0.20$.\nBühlmann $=0.20(0.60)+0.80(0.20)=0.12+0.16=0.28$.\nBayes: 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.**", "tag": "Bühlmann model" }, { "front": "What does the **Bühlmann–Straub** model add beyond the basic Bühlmann model?", "back": "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.", "tag": "Bühlmann-Straub" }, { "front": "State the **Bühlmann–Straub** credibility factor and credibility premium with exposures $m_1,\\dots,m_n$.", "back": "Let total exposure $m=\\sum_{i} m_i$ and the exposure-weighted mean $\\bar X=\\dfrac{\\sum_i m_i X_i}{m}=\\dfrac{\\text{total losses}}{m}$.\nWith $k=\\dfrac{\\text{EPV}}{\\text{VHM}}$ (same definition), the credibility factor is\n$Z=\\dfrac{m}{m+k}$,\nand the credibility premium **per unit of exposure** is $Z\\bar X+(1-Z)\\mu$.", "tag": "Bühlmann-Straub" }, { "front": "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.", "back": "Total exposure $m=100+150+200=450$; total losses $=240+330+500=1{,}070$.\nExposure-weighted mean $\\bar X=\\dfrac{1{,}070}{450}\\approx 2.37778$.\n$k=\\dfrac{\\text{EPV}}{\\text{VHM}}=\\dfrac{8{,}000}{160}=50$, so $Z=\\dfrac{m}{m+k}=\\dfrac{450}{450+50}=\\dfrac{450}{500}=0.90$.\nPure premium per exposure $=0.90(2.37778)+0.10(1.50)=2.14+0.15=\\$2.29$.", "tag": "Bühlmann-Straub" }, { "front": "Continuing the Bühlmann–Straub group ($Z=0.90$, credibility pure premium $\\$2.29$ per exposure): if next year's expected exposure is $250$, what are the expected **total losses**?", "back": "The credibility estimate $\\$2.29$ is a **per-exposure** pure premium. Multiply by next year's exposure:\nExpected total losses $=250\\times 2.29=\\$572.50$.\nThe per-exposure credibility premium scales linearly with the projected exposure base.", "tag": "Bühlmann-Straub" }, { "front": "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$. Find the credibility premium.", "back": "$k=\\dfrac{\\text{EPV}}{\\text{VHM}}=\\dfrac{10{,}000}{200}=50$.\n$Z=\\dfrac{m}{m+k}=\\dfrac{75}{75+50}=\\dfrac{75}{125}=0.60$.\nCredibility premium $=Z\\bar X+(1-Z)\\mu=0.60(1{,}200)+0.40(1{,}000)$\n$=720+400=\\$1{,}120$ per exposure.", "tag": "Bühlmann-Straub" }, { "front": "How does the Bühlmann–Straub factor $Z=\\dfrac{m}{m+k}$ reduce to ordinary Bühlmann, and what does $m$ represent?", "back": "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.", "tag": "Bühlmann-Straub" }, { "front": "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 $v(\\theta)$?", "back": "Treat the type as a discrete random variable $\\Theta$:\n1. $\\mu=E[\\mu(\\Theta)]=\\sum_\\theta p(\\theta)\\,\\mu(\\theta)$.\n2. $\\text{EPV}=E[v(\\Theta)]=\\sum_\\theta p(\\theta)\\,v(\\theta)$.\n3. $\\text{VHM}=\\operatorname{Var}(\\mu(\\Theta))=\\sum_\\theta p(\\theta)\\,\\mu(\\theta)^2-\\mu^2$.\nThen $k=\\text{EPV}/\\text{VHM}$ and $Z=n/(n+k)$ (or $m/(m+k)$ for Bühlmann–Straub).", "tag": "EPV & VHM" }, { "front": "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?", "back": "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.", "tag": "EPV & VHM" }, { "front": "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.", "back": "Limited fluctuation: $Z=\\sqrt{\\dfrac{750}{1082.41}}=\\sqrt{0.692899}\\approx 0.83240$.\nSo about $83.2\\%$ credibility under the square-root rule.\n**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.", "tag": "Limited fluctuation" } ] }