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Exam MAS-I — Credibility Flashcards

Credibility theory for CAS Exam MAS-I: classical limited-fluctuation full and partial credibility standards (the 1082-claim frequency benchmark and the severity/aggregate $(1+\mathrm{CV}^2)$ extensions), Bühlmann greatest-accuracy credibility built from the hypothetical mean, process variance, EPV and VHM, the Bühlmann-Straub model with exposures, and Bayesian credibility under the Poisson-gamma, Bernoulli-beta, and normal-normal conjugate pairs — with fully worked EPV/VHM/$k$/$Z$ and posterior-mean computations.

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  1. Limited fluctuation
    What problem does **credibility theory** solve, and what is the general form of a credibility estimate?
    Credibility blends a risk's own experience with broader (collateral) information to estimate its expected losses. The estimate is the 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/collective (manual) mean, and $Z\in[0,1]$ is the **credibility factor**. Large, stable data sets earn $Z$ near $1$; thin data leans on $M$ with $Z$ near $0$.
  2. Limited fluctuation
    State the **limited-fluctuation (classical)** standard for full credibility of claim frequency.
    Full credibility requires the observed number of claims to be within $\pm k$ of the mean with probability $P$. For Poisson frequency the expected number of claims needed is $\lambda_F = \left(\frac{z_{(1+P)/2}}{k}\right)^{2}$, where $z_{(1+P)/2}$ is the standard-normal percentile. At $P=90\%$ within $\pm5\%$: $z_{0.95}=1.645$, $k=0.05$, giving $\lambda_F=\left(\frac{1.645}{0.05}\right)^{2}=1082.41\approx 1082$ claims.
  3. Limited fluctuation
    Derive the **1082-claim** full-credibility standard for frequency at $90\%$ within $\pm5\%$.
    We want $P\!\left(|N - \lambda| \le k\lambda\right)=0.90$ with $N$ Poisson, so $\sigma_N=\sqrt{\lambda}$. Using the normal approximation, $k\lambda = z_{0.95}\sqrt{\lambda}\;\Rightarrow\; \sqrt{\lambda}=\frac{z_{0.95}}{k}$. Thus $\lambda_F=\left(\frac{z_{0.95}}{k}\right)^{2}=\left(\frac{1.645}{0.05}\right)^{2}=\frac{2.706}{0.0025}=1082.41$. So about $1082$ expected claims are required for full credibility of frequency.
  4. Limited fluctuation
    Find the full-credibility standard (in expected claims) for frequency at $95\%$ within $\pm10\%$.
    Here $P=0.95\Rightarrow z_{0.975}=1.960$ and $k=0.10$. $\lambda_F=\left(\frac{1.960}{0.10}\right)^{2}=(19.6)^{2}=384.16$. So about $384$ expected claims are needed. Tightening $k$ or raising $P$ would increase the requirement; loosening either reduces it.
  5. Limited fluctuation
    How does the full-credibility standard for **claim severity** differ from the frequency standard?
    For severity (average claim size) the number of *claims* needed for full credibility is $n_F = \lambda_F \,\mathrm{CV}_X^{2} = \left(\frac{z_{(1+P)/2}}{k}\right)^{2}\frac{\sigma_X^{2}}{\mu_X^{2}}$, where $\mathrm{CV}_X=\sigma_X/\mu_X$ is the coefficient of variation of individual claim sizes. The frequency standard $\lambda_F$ is scaled by $\mathrm{CV}_X^{2}$ because the only randomness now comes from the size distribution.
  6. Limited fluctuation
    State the full-credibility standard (in expected claims) for **aggregate losses** and explain the $(1+\mathrm{CV}_X^{2})$ factor.
    For aggregate losses $S=\sum X_i$ with Poisson frequency, the expected number of claims for full credibility is $n_F = \lambda_F\,(1+\mathrm{CV}_X^{2}) = \left(\frac{z_{(1+P)/2}}{k}\right)^{2}\!\left(1+\frac{\sigma_X^{2}}{\mu_X^{2}}\right)$. The $1$ captures frequency (Poisson) variability and the $\mathrm{CV}_X^{2}$ captures severity variability; together they reflect that aggregate losses are riskier than either component alone.
  7. Limited fluctuation
    Claim sizes have mean $\mu_X=2{,}000$ and standard deviation $\sigma_X=3{,}000$. Using the $90\%/\pm5\%$ standard ($\lambda_F=1082$), find the expected claims needed for full credibility of **aggregate losses**.
    Coefficient of variation: $\mathrm{CV}_X=\frac{3000}{2000}=1.5$, so $\mathrm{CV}_X^{2}=2.25$. Aggregate standard: $n_F=\lambda_F(1+\mathrm{CV}_X^{2})=1082(1+2.25)=1082(3.25)\approx 3517$ expected claims. For comparison, full credibility of **severity** alone would need $\lambda_F\,\mathrm{CV}_X^{2}=1082(2.25)\approx 2435$ claims.
  8. Full & partial credibility
    State the **partial credibility** formula and the resulting credibility estimate.
    When the data falls short of the full-credibility standard, assign the square-root rule $Z=\sqrt{\frac{n}{n_F}}$ (capped at $1$), where $n$ is the observed number of claims (or exposures) and $n_F$ is the full-credibility standard. The credibility estimate is $Z\,\bar X + (1-Z)\,M$, blending observed mean $\bar X$ with the manual/collective mean $M$.
  9. Full & partial credibility
    Why does limited-fluctuation credibility use the **square root** $Z=\sqrt{n/n_F}$ rather than $n/n_F$?
    Credibility is chosen to standardize the fluctuation of the *partially credible* estimator to the same tolerance as the full-credibility case. The variance of the sample mean scales like $1/n$, so its standard deviation scales like $1/\sqrt{n}$. Setting the weighted estimator's relative standard deviation equal to the full-credibility tolerance gives $Z\propto\sqrt{n}$, hence $Z=\sqrt{n/n_F}$.
  10. Full & partial credibility
    A risk has generated $400$ claims. The full-credibility standard is $1082$ claims. The observed average pure premium is $\bar X=520$ and the manual premium is $M=600$. Find the credibility premium.
    Partial credibility: $Z=\sqrt{\frac{400}{1082}}=\sqrt{0.36969}\approx 0.6080$. Credibility premium: $Z\bar X+(1-Z)M = 0.6080(520)+0.3920(600)$ $=316.16 + 235.20 = \$551.36$. The estimate sits between the observed $520$ and the manual $600$, closer to the observation because $Z>0.5$.
  11. Full & partial credibility
    A class generated $300$ claims with observed mean severity $\bar X=4{,}200$. The full-credibility standard for severity is $1{,}200$ claims and the manual mean is $M=4{,}000$. Find the credibility estimate of severity.
    Partial credibility: $Z=\sqrt{\frac{300}{1200}}=\sqrt{0.25}=0.5$. Estimate: $Z\bar X+(1-Z)M = 0.5(4200)+0.5(4000)$ $=2100 + 2000 = \$4{,}100$. The estimate sits exactly halfway between observation and manual because $Z=0.5$.
  12. Full & partial credibility
    How many claims would a risk need to reach a credibility of $Z=0.80$ when the full-credibility standard is $n_F=1082$ claims?
    Solve $Z=\sqrt{n/n_F}$ for $n$: $n = Z^{2}\,n_F$. $n = (0.80)^{2}(1082) = 0.64(1082) = 692.48 \approx 693$ claims. Intuitively, since $Z$ scales with $\sqrt{n}$, getting $80\%$ credibility needs $(0.8)^2=64\%$ of the full-credibility claim count.
  13. Limited fluctuation
    What are the key **limitations** of limited-fluctuation (classical) credibility?
    1. The full-credibility standard depends on arbitrary choices of $P$ and $k$. 2. It assigns $Z=1$ once the standard is met, ignoring further data (no improvement beyond full credibility). 3. The complement of credibility $M$ is chosen heuristically, not derived. 4. It relies on a normal approximation and a Poisson frequency assumption. Greatest-accuracy (Bühlmann/Bayesian) credibility addresses these by minimizing mean squared error.
  14. Bühlmann model
    Define the **hypothetical mean** $\mu(\theta)$ and the **process variance** $v(\theta)$ in the Bühlmann model.
    Each risk is characterized by an unknown risk parameter $\theta$ drawn from a prior distribution. Given $\theta$: **Hypothetical mean** $\mu(\theta)=E[X\mid\theta]$ — the expected outcome for a risk with parameter $\theta$. **Process variance** $v(\theta)=\mathrm{Var}(X\mid\theta)$ — the within-risk variability of outcomes for that fixed $\theta$. Both are functions of the random $\theta$, so they are themselves random variables across the population of risks.
  15. EPV & VHM
    Define **EPV** (expected process variance) and **VHM** (variance of hypothetical means).
    $\text{EPV} = E[v(\theta)] = E\big[\mathrm{Var}(X\mid\theta)\big]$ — the average within-risk variance, measuring noise inside a risk. $\text{VHM} = \mathrm{Var}[\mu(\theta)] = \mathrm{Var}\big(E[X\mid\theta]\big)$ — the spread of risk means across the population, measuring how different the risks are. By the law of total variance, the total (unconditional) variance is $\mathrm{Var}(X)=\text{EPV}+\text{VHM}$.
  16. Bühlmann model
    State the Bühlmann credibility factor $Z$ and the Bühlmann credibility constant $k$.
    $k=\frac{\text{EPV}}{\text{VHM}}=\frac{E[v(\theta)]}{\mathrm{Var}[\mu(\theta)]}$, and with $n$ observations $Z=\frac{n}{n+k}$. The Bühlmann credibility premium is $Z\,\bar X + (1-Z)\,\mu$, where $\mu=E[\mu(\theta)]$ is the overall (collective) mean. Smaller $k$ (risks differ a lot relative to internal noise) gives higher credibility.
  17. Bühlmann model
    Interpret the Bühlmann constant $k=\text{EPV}/\text{VHM}$. When is credibility high?
    $k$ compares within-risk noise (EPV) to between-risk spread (VHM). - Small $k$ (low EPV, high VHM): risks are very different from each other and each risk's data is stable, so own experience is informative → $Z=\frac{n}{n+k}$ is large. - Large $k$ (high EPV, low VHM): risks look alike and individual data is noisy, so lean on the collective mean → $Z$ is small. As $n\to\infty$, $Z\to 1$ for any finite $k$.
  18. Bühlmann model
    How is the Bühlmann premium the **least-squares** approximation to the Bayesian premium?
    The exact Bayesian premium is $E[\mu(\theta)\mid\mathbf X]$, often nonlinear in the data. The Bühlmann premium $Z\bar X+(1-Z)\mu$ is the **linear** function of the observations $\bar X$ that minimizes the expected squared error against the Bayesian premium (equivalently against the next observation). So Bühlmann is the best *linear* (least-squares) approximation to Bayes, exact when the Bayesian estimate is itself linear in the data (e.g. the conjugate exponential-family cases).
  19. EPV & VHM
    A risk's outcomes given $\theta$ are Poisson with mean $\theta$, and $\theta$ has $E[\theta]=0.20$, $\mathrm{Var}(\theta)=0.05$. Find EPV, VHM, and $k$.
    For Poisson, $\mu(\theta)=\theta$ and $v(\theta)=\theta$. $\text{EPV}=E[v(\theta)]=E[\theta]=0.20$. $\text{VHM}=\mathrm{Var}[\mu(\theta)]=\mathrm{Var}(\theta)=0.05$. $k=\frac{\text{EPV}}{\text{VHM}}=\frac{0.20}{0.05}=4$. With, say, $n=3$ years of data, $Z=\frac{3}{3+4}=\frac{3}{7}\approx 0.4286$.
  20. EPV & VHM
    Two risk types are equally likely. Type A has mean $10$ and variance $20$; Type B has mean $30$ and variance $60$. Find EPV, VHM, $\mu$, and $k$.
    Overall mean $\mu = 0.5(10)+0.5(30)=20$. $\text{EPV}=E[v(\theta)] = 0.5(20)+0.5(60)=40$. $\text{VHM}=\mathrm{Var}[\mu(\theta)] = E[\mu^2]-\mu^2 = \big(0.5(10^2)+0.5(30^2)\big)-20^2$ $=(50+450)-400=100$. $k=\frac{\text{EPV}}{\text{VHM}}=\frac{40}{100}=0.4$.
  21. Bühlmann model
    Using the two-type setup ($\mu=20$, $k=0.4$), a policyholder is observed for $n=2$ years with mean $\bar X=25$. Find the Bühlmann credibility premium.
    $Z=\frac{n}{n+k}=\frac{2}{2+0.4}=\frac{2}{2.4}\approx 0.8333$. Bühlmann premium: $Z\bar X+(1-Z)\mu = 0.8333(25)+0.1667(20)$ $=20.833 + 3.333 = \$24.17$. The estimate is pulled most of the way toward the observed $25$ because $Z$ is high (risks differ a lot relative to their internal noise).
  22. Bühlmann model
    A class is modeled with $\mu(\theta)=\theta$, process variance $v(\theta)=\theta$ ; $\theta\sim\text{Gamma}$ with $E[\theta]=0.10$ and $\mathrm{Var}(\theta)=0.01$. Over $n=5$ years the insured had $\bar X=0.30$ claims/year. Find the Bühlmann estimate.
    $\text{EPV}=E[\theta]=0.10$, $\text{VHM}=\mathrm{Var}(\theta)=0.01$, $\mu=E[\theta]=0.10$. $k=\frac{0.10}{0.01}=10$, so $Z=\frac{5}{5+10}=\frac{5}{15}=\frac{1}{3}\approx 0.3333$. Estimate: $Z\bar X+(1-Z)\mu = \tfrac{1}{3}(0.30)+\tfrac{2}{3}(0.10)$ $=0.10 + 0.0667 = 0.1667$ claims/year.
  23. EPV & VHM
    Outcomes given $\theta$ are uniform on $(0,\theta)$, and $\theta\sim\text{Uniform}(0,6)$. Find $\mu(\theta)$, $v(\theta)$, EPV, VHM, and $k$.
    For $X\mid\theta\sim U(0,\theta)$: $\mu(\theta)=\frac{\theta}{2}$, $v(\theta)=\frac{\theta^{2}}{12}$. With $\theta\sim U(0,6)$: $E[\theta]=3$, $E[\theta^{2}]=\frac{6^{2}}{3}=12$, $\mathrm{Var}(\theta)=12-9=3$. $\text{EPV}=E\!\left[\frac{\theta^{2}}{12}\right]=\frac{12}{12}=1$. $\text{VHM}=\mathrm{Var}\!\left(\frac{\theta}{2}\right)=\frac{1}{4}\mathrm{Var}(\theta)=\frac{3}{4}=0.75$. $k=\frac{1}{0.75}=1.333$.
  24. EPV & VHM
    Why does the law of total variance, $\mathrm{Var}(X)=\text{EPV}+\text{VHM}$, motivate the Bühlmann weight?
    Total variability of an outcome splits into the part inside risks (EPV) and the part between risks (VHM). Credibility should reward the *between-risk* signal and discount the *within-risk* noise. The Bühlmann weight $Z=\frac{n\,\text{VHM}}{n\,\text{VHM}+\text{EPV}}=\frac{n}{n+k}$ is exactly the fraction of the observed-mean's variance explained by genuine differences between risks, so $Z$ rises as VHM grows or EPV shrinks.
  25. Bühlmann-Straub
    State the **Bühlmann-Straub** model and its credibility factor when exposures differ across periods.
    Bühlmann-Straub generalizes Bühlmann to varying exposures $m_i$ (e.g. policy counts, earned car-years). Each period's loss ratio $X_i$ has $\mathrm{Var}(X_i\mid\theta)=\frac{v(\theta)}{m_i}$. With total exposure $m=\sum_i m_i$, $Z=\frac{m}{m+k}$, where $k=\frac{\text{EPV}}{\text{VHM}}$, and the credibility premium is $Z\,\bar X+(1-Z)\mu$ with the **exposure-weighted** mean $\bar X=\frac{\sum_i m_i X_i}{\sum_i m_i}$.
  26. Bühlmann-Straub
    How does the credibility-weighted mean $\bar X$ in Bühlmann-Straub differ from a plain average?
    Bühlmann-Straub weights each period by its **exposure** $m_i$, not equally: $\bar X=\frac{\sum_i m_i X_i}{\sum_i m_i}$. A year with more exposure (more policies/claims) contributes proportionally more to the estimate because its loss ratio is a more reliable signal. Ordinary Bühlmann is the special case where every $m_i=1$ (equal exposures), reducing $\bar X$ to the simple mean and $m=n$.
  27. Bühlmann-Straub
    A group has $\text{EPV}=8{,}000$ and $\text{VHM}=200$. Over three years the exposures and pure premiums are $(m_1,X_1)=(10,90)$, $(m_2,X_2)=(20,110)$, $(m_3,X_3)=(30,130)$. The manual premium is $\mu=100$. Find the Bühlmann-Straub credibility premium.
    $k=\frac{\text{EPV}}{\text{VHM}}=\frac{8000}{200}=40$. Total exposure $m=10+20+30=60$. $Z=\frac{m}{m+k}=\frac{60}{60+40}=\frac{60}{100}=0.60$. Exposure-weighted mean: $\bar X=\frac{10(90)+20(110)+30(130)}{60}=\frac{900+2200+3900}{60}=\frac{7000}{60}=116.67$. Premium: $Z\bar X+(1-Z)\mu = 0.60(116.67)+0.40(100)=70.00+40=\$110.00$.
  28. Bühlmann-Straub
    A risk has exposures $m_1=50$, $m_2=70$ and claim counts $12$ then $20$. The Poisson model gives $\text{EPV}=0.30$ and $\text{VHM}=0.02$; collective mean $\mu=0.25$. Find the Bühlmann-Straub estimate of the claim frequency.
    Total exposure $m=50+70=120$; total claims $=12+20=32$, so $\bar X=\frac{32}{120}=0.2667$. $k=\frac{\text{EPV}}{\text{VHM}}=\frac{0.30}{0.02}=15$. $Z=\frac{m}{m+k}=\frac{120}{120+15}=\frac{120}{135}\approx 0.8889$. Estimate: $Z\bar X+(1-Z)\mu = 0.8889(0.2667)+0.1111(0.25)$ $=0.2370 + 0.0278 = 0.2648$ claims per exposure.
  29. Bühlmann-Straub
    In Bühlmann-Straub, how do you estimate the **predicted total losses** for next period given a future exposure $m^{*}$?
    First compute the credibility-weighted *rate* (pure premium per exposure): $\hat P = Z\bar X+(1-Z)\mu$, with $Z=\frac{m}{m+k}$ and exposure-weighted $\bar X$. Then multiply by the projected exposure: predicted losses $=m^{*}\,\hat P$. For example, if $\hat P=\$110$ per exposure and next year's exposure is $m^{*}=40$, the predicted total losses are $40(110)=\$4{,}400$.
  30. Bayesian credibility
    What makes a prior **conjugate** to a likelihood in Bayesian credibility, and why is it convenient?
    A prior is **conjugate** when the posterior distribution belongs to the same family as the prior. The data simply updates the prior's parameters, so the posterior is available in closed form without integration. The three exam workhorses are the conjugate pairs: **Poisson–gamma**, **Bernoulli/binomial–beta**, and **normal–normal**. In each, the Bayesian predictive mean turns out to be a credibility-weighted average, matching the Bühlmann form exactly.
  31. Bayesian credibility
    For the **Poisson–gamma** model, state the posterior distribution of $\lambda$ and the predictive mean.
    Prior $\lambda\sim\text{Gamma}(\alpha,\theta)$ (mean $\alpha\theta$); data $X_1,\dots,X_n\mid\lambda\sim\text{Poisson}(\lambda)$ with total claims $\sum X_i$. The posterior is $\lambda\mid\text{data}\sim\text{Gamma}\!\left(\alpha+\textstyle\sum X_i,\ \frac{\theta}{1+n\theta}\right)$. Posterior (predictive) mean: $E[\lambda\mid\text{data}]=\big(\alpha+\textstyle\sum X_i\big)\frac{\theta}{1+n\theta}$.
  32. Bayesian credibility
    Poisson–gamma: prior $\lambda\sim\text{Gamma}(\alpha=2,\theta=0.10)$. An insured is observed $n=4$ years with claim counts $0,1,0,2$ (total $3$). Find the posterior mean number of claims.
    Total claims $\sum X_i = 0+1+0+2 = 3$. Posterior shape $=\alpha+\sum X_i = 2+3 = 5$. Posterior scale $=\frac{\theta}{1+n\theta}=\frac{0.10}{1+4(0.10)}=\frac{0.10}{1.4}=0.071429$. Posterior mean $=5(0.071429)=0.3571$ claims/year. (Prior mean was $\alpha\theta=2(0.10)=0.20$; the higher observed $\bar X=3/4=0.75$ pulled the estimate up.)
  33. Bayesian credibility
    Show that the Poisson–gamma posterior mean is a credibility weighting, and identify $Z$.
    The posterior mean rearranges as $E[\lambda\mid\text{data}]=Z\bar X+(1-Z)\mu$, with $\bar X=\frac{\sum X_i}{n}$, $\mu=\alpha\theta$ (the prior mean), and $Z=\frac{n\theta}{1+n\theta}=\frac{n}{n+1/\theta}$. Thus the Bühlmann constant is $k=\frac{1}{\theta}$ — and indeed $\text{EPV}=E[\lambda]=\alpha\theta$ while $\text{VHM}=\mathrm{Var}(\lambda)=\alpha\theta^{2}$, so $k=\frac{\alpha\theta}{\alpha\theta^{2}}=\frac{1}{\theta}$. Bühlmann is **exact** here.
  34. Bayesian credibility
    Verify Bühlmann equals Bayes for the Poisson–gamma case with $\alpha=2$, $\theta=0.10$, $n=4$, $\bar X=0.75$.
    $\text{EPV}=\alpha\theta=0.20$, $\text{VHM}=\alpha\theta^{2}=2(0.01)=0.02$, so $k=\frac{0.20}{0.02}=10=\frac{1}{\theta}$. ✓ $Z=\frac{n}{n+k}=\frac{4}{4+10}=\frac{4}{14}=0.2857$. Bühlmann: $Z\bar X+(1-Z)\mu=0.2857(0.75)+0.7143(0.20)$ $=0.2143 + 0.1429 = 0.3571$ claims/year — identical to the posterior mean computed directly.
  35. Bayesian credibility
    For the **Bernoulli–beta** model, state the posterior distribution and predictive mean of $p$.
    Prior $p\sim\text{Beta}(\alpha,\beta)$; data are $n$ Bernoulli trials with $s=\sum X_i$ successes. The posterior is $p\mid\text{data}\sim\text{Beta}(\alpha+s,\ \beta+n-s)$. Predictive (posterior) mean: $E[p\mid\text{data}]=\frac{\alpha+s}{\alpha+\beta+n}$. This is a credibility blend of the sample proportion $\frac{s}{n}$ and the prior mean $\frac{\alpha}{\alpha+\beta}$ with $Z=\frac{n}{n+\alpha+\beta}$.
  36. Bayesian credibility
    Bernoulli–beta: prior $p\sim\text{Beta}(\alpha=3,\beta=7)$. In $n=20$ trials there are $s=8$ successes. Find the posterior mean of $p$ and the implied $Z$.
    Posterior $\text{Beta}(\alpha+s,\beta+n-s)=\text{Beta}(3+8,\,7+12)=\text{Beta}(11,19)$. Posterior mean $=\frac{11}{11+19}=\frac{11}{30}\approx 0.3667$. As a credibility blend: sample proportion $\frac{8}{20}=0.40$, prior mean $\frac{3}{10}=0.30$, $Z=\frac{n}{n+\alpha+\beta}=\frac{20}{20+10}=\frac{2}{3}$. Check: $\frac{2}{3}(0.40)+\frac{1}{3}(0.30)=0.2667+0.10=0.3667$. ✓
  37. Bayesian credibility
    For the **normal–normal** model (known variance), state the posterior of the mean $\theta$.
    Prior $\theta\sim N(\mu,a)$ (variance $a$); data $X_1,\dots,X_n\mid\theta\sim N(\theta,v)$ with known process variance $v$ and sample mean $\bar X$. The posterior is normal with mean $E[\theta\mid\text{data}]=Z\bar X+(1-Z)\mu$, where $Z=\frac{n}{n+v/a}=\frac{n}{n+k}$, $k=\frac{v}{a}=\frac{\text{EPV}}{\text{VHM}}$. Here $\text{EPV}=v$ and $\text{VHM}=a$, so Bühlmann is again **exact**.
  38. Bayesian credibility
    Normal–normal: prior $\theta\sim N(\mu=1000,\,a=40{,}000)$, process variance $v=160{,}000$. A risk shows $\bar X=1{,}300$ over $n=4$ observations. Find the posterior (credibility) mean.
    $k=\frac{v}{a}=\frac{160000}{40000}=4$. $Z=\frac{n}{n+k}=\frac{4}{4+4}=0.5$. Posterior mean: $Z\bar X+(1-Z)\mu=0.5(1300)+0.5(1000)$ $=650 + 500 = \$1{,}150$. The estimate lands halfway between the observed $1{,}300$ and the prior $1{,}000$ because $Z=0.5$.
  39. Bayesian credibility
    How does the exact **Bayesian premium** relate to the Bühlmann premium in general (non-conjugate) settings?
    The Bayesian premium $E[X_{n+1}\mid\mathbf X]$ is the minimum-mean-squared-error predictor among *all* functions of the data. The Bühlmann premium is the best predictor restricted to *linear* functions of $\bar X$. So in general the Bühlmann premium only approximates Bayes (it is the projection of the Bayesian estimator onto linear functions). In the linear-exponential conjugate cases (Poisson–gamma, Bernoulli–beta, normal–normal) the Bayesian premium is already linear in $\bar X$, so the two coincide exactly.
  40. EPV & VHM
    Three urns are equally likely. Urn 1 has $E[X]=1$, $\mathrm{Var}(X)=1$; Urn 2 has $E[X]=2$, $\mathrm{Var}(X)=4$; Urn 3 has $E[X]=3$, $\mathrm{Var}(X)=9$. Find $k$ and the credibility $Z$ for $n=3$ draws.
    Overall mean $\mu=\frac{1+2+3}{3}=2$. $\text{EPV}=E[v(\theta)]=\frac{1+4+9}{3}=\frac{14}{3}\approx 4.6667$. $\text{VHM}=\mathrm{Var}[\mu(\theta)]=\frac{1^2+2^2+3^2}{3}-2^2=\frac{14}{3}-4=\frac{2}{3}\approx 0.6667$. $k=\frac{14/3}{2/3}=7$, so $Z=\frac{3}{3+7}=0.30$.
  41. Bühlmann model
    Continuing the three-urn problem ($\mu=2$, $k=7$): a player draws $n=3$ values averaging $\bar X=2.4$. Give the Bühlmann credibility estimate of the next draw.
    $Z=\frac{n}{n+k}=\frac{3}{3+7}=0.30$. Estimate: $Z\bar X+(1-Z)\mu = 0.30(2.4)+0.70(2)$ $=0.72 + 1.40 = 2.12$. Because $k$ is large (urns are fairly similar relative to their internal noise), only $30\%$ weight goes to the observed average and the estimate stays close to the collective mean $2$.
  42. EPV & VHM
    An exponential model has $X\mid\theta\sim\text{Exponential}(\text{mean }\theta)$, with $\theta\sim\text{Inverse Gamma}$ such that $E[\theta]=500$ and $E[\theta^{2}]=375{,}000$. Find EPV, VHM, and $k$.
    For $X\mid\theta\sim\text{Exp}(\text{mean }\theta)$: $\mu(\theta)=\theta$ and $v(\theta)=\theta^{2}$. $\text{EPV}=E[v(\theta)]=E[\theta^{2}]=375{,}000$. $\text{VHM}=\mathrm{Var}[\mu(\theta)]=\mathrm{Var}(\theta)=E[\theta^{2}]-(E[\theta])^{2}=375{,}000-500^{2}=375{,}000-250{,}000=125{,}000$. $k=\frac{375{,}000}{125{,}000}=3$.
  43. Bühlmann model
    Using the exponential model above ($\mu=500$, $k=3$): a policyholder reports $n=6$ claims averaging $\bar X=620$. Find the Bühlmann credibility estimate of severity.
    $Z=\frac{n}{n+k}=\frac{6}{6+3}=\frac{6}{9}=0.6667$. Estimate: $Z\bar X+(1-Z)\mu = 0.6667(620)+0.3333(500)$ $=413.33 + 166.67 = \$580.00$. With $\tfrac{2}{3}$ credibility, the estimate moves two-thirds of the way from the prior $500$ toward the observed $620$.
  44. Bühlmann model
    Compare the credibility assigned by **limited fluctuation** versus **Bühlmann** for the same data, conceptually.
    Limited fluctuation gives $Z=\sqrt{n/n_F}$ and jumps to $Z=1$ once $n\ge n_F$, with the standard $n_F$ set by tolerance choices $P,k$. Bühlmann gives $Z=\frac{n}{n+k}$, which **never** reaches $1$ for finite $n$ and is derived to minimize squared error using the population's EPV and VHM. Bühlmann uses the data's actual variance structure; classical credibility relies on a normal/Poisson approximation and arbitrary tolerances. For small $n$, Bühlmann is generally preferred when EPV and VHM can be estimated.