{
  "deckName": "Exam MAS-II — Credibility",
  "examCode": "Exam MAS-II",
  "cards": [
    {
      "front": "Define the **hypothetical mean** $\\mu(\\theta)$ and the **process variance** $v(\\theta)$ in the Bühlmann model.",
      "back": "Each risk is characterized by a latent risk parameter $\\theta$ drawn from a prior. Conditional on $\\theta$:\n**Hypothetical mean** $\\mu(\\theta)=E[X\\mid\\theta]$ — the expected claim for a risk of type $\\theta$.\n**Process variance** $v(\\theta)=\\text{Var}(X\\mid\\theta)$ — the within-risk variability of that risk's outcomes.\nThe two structural quantities EPV and VHM are built from these by averaging or taking the variance over the prior of $\\theta$.",
      "tag": "Bühlmann model"
    },
    {
      "front": "Define the **expected process variance (EPV)** and the **variance of the hypothetical means (VHM)**.",
      "back": "$\\text{EPV}=E[v(\\theta)]=E[\\text{Var}(X\\mid\\theta)]$ — the average within-risk variance; it measures the noise inside a single risk.\n$\\text{VHM}=\\text{Var}[\\mu(\\theta)]=\\text{Var}[E(X\\mid\\theta)]$ — the spread of the risks' means about the overall mean; it measures how much the risks truly differ.\nBy the variance-decomposition (law of total variance), the total variance of a single observation is $\\text{Var}(X)=\\text{EPV}+\\text{VHM}$.",
      "tag": "EPV & VHM"
    },
    {
      "front": "State the **Bühlmann credibility factor** $Z$ and the **Bühlmann constant** $k$.",
      "back": "$k=\\dfrac{\\text{EPV}}{\\text{VHM}}=\\dfrac{E[v(\\theta)]}{\\text{Var}[\\mu(\\theta)]}$, and with $n$ observations the credibility is\n$Z=\\dfrac{n}{n+k}$.\nLarge within-risk noise (big EPV) or little true difference between risks (small VHM) makes $k$ large and $Z$ small. More data ($n\\uparrow$) always pushes $Z\\to 1$.",
      "tag": "Bühlmann model"
    },
    {
      "front": "State the **Bühlmann credibility premium** and interpret its two terms.",
      "back": "$P_{c}=Z\\bar X + (1-Z)\\mu$, where $\\bar X$ is the risk's own sample mean over $n$ periods and $\\mu=E[\\mu(\\theta)]$ is the overall (collective) mean.\nIt is a weighted average: the risk's own experience $\\bar X$ gets weight $Z$ and the manual/collective mean $\\mu$ gets weight $1-Z$. As $n\\to\\infty$, $Z\\to 1$ and the premium relies entirely on the risk's own data.",
      "tag": "Bühlmann model"
    },
    {
      "front": "Why does $Z$ increase as VHM grows and decrease as EPV grows?",
      "back": "$Z=\\frac{n}{n+k}$ with $k=\\frac{\\text{EPV}}{\\text{VHM}}$. A **large VHM** means the risks genuinely differ, so a given risk's own data is very informative about which risk it is — $k$ falls, $Z$ rises. A **large EPV** means each risk's data is noisy, so the sample mean is a poor signal — $k$ rises, $Z$ falls.\nLimits: as $\\text{VHM}\\to\\infty$, $k\\to 0$ and $Z\\to 1$; as $\\text{EPV}\\to\\infty$, $k\\to\\infty$ and $Z\\to 0$.",
      "tag": "EPV & VHM"
    },
    {
      "front": "How is the Bühlmann constant $k$ related to the **full-credibility** standard for credibility? (Limited-fluctuation comparison.)",
      "back": "In limited-fluctuation credibility, partial credibility uses $Z=\\sqrt{n/n_{0}}$ with $n_{0}$ the full-credibility standard. In **greatest-accuracy (Bühlmann)** credibility, the analogous formula is $Z=\\frac{n}{n+k}$: the constant $k$ plays the role of the full-credibility threshold but is derived from the variance structure (EPV/VHM) rather than from a confidence/error tolerance.\nBoth give more weight to larger samples, but Bühlmann's $Z$ is the least-squares-optimal linear weight, not a square-root rule.",
      "tag": "Credibility-Bayes link"
    },
    {
      "front": "Worked: a risk has $\\text{EPV}=8000$ and $\\text{VHM}=500$. With $n=4$ years of data and overall mean $\\mu=1200$, the risk's $\\bar X=1500$. Find $k$, $Z$, and the credibility premium.",
      "back": "$k=\\frac{\\text{EPV}}{\\text{VHM}}=\\frac{8000}{500}=16$.\n$Z=\\frac{n}{n+k}=\\frac{4}{4+16}=\\frac{4}{20}=0.20$.\n$P_{c}=Z\\bar X+(1-Z)\\mu=0.20(1500)+0.80(1200)=300+960=\\$1{,}260$.\nThe heavy EPV relative to VHM keeps $Z$ low, so the estimate stays close to the collective mean.",
      "tag": "Bühlmann model"
    },
    {
      "front": "Worked: the same structure has $\\text{EPV}=8000$, $\\text{VHM}=500$ so $k=16$. How many years $n$ of data are needed for $Z\\ge 0.50$?",
      "back": "Set $Z=\\frac{n}{n+16}\\ge 0.50$.\n$\\frac{n}{n+16}=0.50 \\Rightarrow n = 0.5(n+16) \\Rightarrow 0.5n = 8 \\Rightarrow n = 16$.\nSo $n=16$ years gives $Z=0.50$ exactly, and any $n>16$ gives $Z>0.50$.\nIn general $Z=0.5$ precisely when $n=k$, a useful check: the credibility equals one-half when the sample size equals the Bühlmann constant.",
      "tag": "Bühlmann model"
    },
    {
      "front": "Worked (variance decomposition): claim counts are $\\text{Poisson}(\\lambda)$ with $\\lambda\\sim\\text{Gamma}(\\alpha=3,\\theta=0.1)$. Find $\\mu$, EPV, VHM, and $k$.",
      "back": "Here $\\mu(\\lambda)=\\lambda$ and $v(\\lambda)=\\lambda$ (Poisson mean $=$ variance).\nGamma mean $E[\\lambda]=\\alpha\\theta=3(0.1)=0.30$; Gamma variance $\\text{Var}[\\lambda]=\\alpha\\theta^{2}=3(0.01)=0.03$.\n$\\mu=E[\\mu(\\lambda)]=E[\\lambda]=0.30$.\n$\\text{EPV}=E[v(\\lambda)]=E[\\lambda]=0.30$.\n$\\text{VHM}=\\text{Var}[\\mu(\\lambda)]=\\text{Var}[\\lambda]=0.03$.\n$k=\\frac{\\text{EPV}}{\\text{VHM}}=\\frac{0.30}{0.03}=10$.",
      "tag": "EPV & VHM"
    },
    {
      "front": "Continuing the Poisson-Gamma ($\\alpha=3,\\theta=0.1$, so $k=10$): a policyholder is observed for $n=5$ years with a total of $4$ claims. Find the Bühlmann credibility estimate of next year's claim count.",
      "back": "Risk's sample mean $\\bar X=\\frac{4}{5}=0.80$ claims/year; overall mean $\\mu=0.30$.\n$Z=\\frac{n}{n+k}=\\frac{5}{5+10}=\\frac{5}{15}=0.3\\overline{3}$.\n$P_{c}=Z\\bar X+(1-Z)\\mu=0.3333(0.80)+0.6667(0.30)$\n$=0.26667+0.20000=0.46667\\approx 0.467$ claims.\n(For Poisson-Gamma the Bühlmann and exact Bayesian estimates coincide — see the credibility-Bayes link.)",
      "tag": "Credibility-Bayes link"
    },
    {
      "front": "Worked (two-type variance decomposition): a portfolio is $60\\%$ \"good\" risks with $\\mu=100,\\ v=4000$ and $40\\%$ \"bad\" risks with $\\mu=300,\\ v=9000$. Find $\\mu$, EPV, VHM, and $k$.",
      "back": "Overall mean $\\mu=0.6(100)+0.4(300)=60+120=180$.\n$\\text{EPV}=E[v(\\theta)]=0.6(4000)+0.4(9000)=2400+3600=6000$.\n$\\text{VHM}=\\text{Var}[\\mu(\\theta)]=E[\\mu^{2}]-\\mu^{2}=[0.6(100^{2})+0.4(300^{2})]-180^{2}$\n$=[6000+36000]-32400=42000-32400=9600$.\n$k=\\frac{\\text{EPV}}{\\text{VHM}}=\\frac{6000}{9600}=0.625$.",
      "tag": "EPV & VHM"
    },
    {
      "front": "Continuing the two-type portfolio ($\\mu=180$, $k=0.625$): one risk shows $\\bar X=250$ over $n=3$ years. Find $Z$ and the credibility premium.",
      "back": "$Z=\\frac{n}{n+k}=\\frac{3}{3+0.625}=\\frac{3}{3.625}\\approx 0.8276$.\n$P_{c}=Z\\bar X+(1-Z)\\mu=0.8276(250)+0.1724(180)$\n$\\approx 206.90+31.03=\\$237.93$.\nThe large VHM (very different risk types) makes the small constant $k$, so even $3$ years earns high credibility.",
      "tag": "Bühlmann model"
    },
    {
      "front": "State the **Bühlmann-Straub** model and how exposures $m_{i}$ change the credibility formula.",
      "back": "Bühlmann-Straub allows each period (or cell) $i$ to have a different exposure $m_{i}$ with $X_{i}$ the loss per unit exposure: $E[X_{i}\\mid\\theta]=\\mu(\\theta)$ and $\\text{Var}(X_{i}\\mid\\theta)=\\frac{v(\\theta)}{m_{i}}$.\nLet $m=\\sum_{i} m_{i}$ be total exposure. Then $k=\\frac{\\text{EPV}}{\\text{VHM}}$ as before, but\n$Z=\\frac{m}{m+k}$,\nand the risk's mean is exposure-weighted: $\\bar X=\\frac{\\sum_i m_i X_i}{m}$.",
      "tag": "Bühlmann-Straub"
    },
    {
      "front": "In Bühlmann-Straub, why is the credibility-weighted sample mean computed as $\\bar X=\\frac{\\sum_i m_i X_i}{\\sum_i m_i}$ rather than a simple average of the $X_i$?",
      "back": "Periods with more exposure $m_i$ carry more information and less noise (their conditional variance $v(\\theta)/m_i$ is smaller), so they should weight more heavily. The exposure-weighted mean is the minimum-variance unbiased combination of the $X_i$, and it is what the Bühlmann-Straub credibility estimate uses.\nThe ordinary Bühlmann model is the special case $m_i=1$ for all $i$, which collapses $\\bar X$ to the simple average and $Z$ to $\\frac{n}{n+k}$.",
      "tag": "Bühlmann-Straub"
    },
    {
      "front": "Worked Bühlmann-Straub: $\\text{EPV}=10000$, $\\text{VHM}=200$, $\\mu=50$ per exposure. A risk has exposures and pure premiums:\nYear 1: $m=20,\\ X=60$; Year 2: $m=30,\\ X=45$; Year 3: $m=50,\\ X=58$. Find the credibility premium.",
      "back": "Total exposure $m=20+30+50=100$.\nWeighted mean $\\bar X=\\frac{20(60)+30(45)+50(58)}{100}=\\frac{1200+1350+2900}{100}=\\frac{5450}{100}=54.5$.\n$k=\\frac{\\text{EPV}}{\\text{VHM}}=\\frac{10000}{200}=50$.\n$Z=\\frac{m}{m+k}=\\frac{100}{100+50}=\\frac{100}{150}=0.6\\overline{6}$.\n$P_{c}=Z\\bar X+(1-Z)\\mu=0.6667(54.5)+0.3333(50)=36.33+16.67=\\$53.00$ per exposure.",
      "tag": "Bühlmann-Straub"
    },
    {
      "front": "Worked Bühlmann-Straub (expected claims): in the previous example ($Z=0.6\\overline{6}$, $P_c=\\$53.00$ per exposure), if next year's exposure is projected at $m_{new}=40$, what is the expected total credibility loss?",
      "back": "The credibility premium $\\$53.00$ is a rate **per exposure unit**. Multiply by next year's exposure:\nExpected loss $=P_{c}\\times m_{new}=53.00\\times 40=\\$2{,}120$.\nKeep the per-exposure rate distinct from the dollar total — credibility is applied to the rate, then scaled to the projected exposure.",
      "tag": "Bühlmann-Straub"
    },
    {
      "front": "What is the **nonparametric empirical-Bayes** problem, and what data layout does it assume?",
      "back": "When the prior distribution of $\\theta$ is unknown, we estimate the structural parameters $\\mu$, EPV, and VHM **directly from the data** rather than from assumed model forms.\nClassic layout: $r$ risks (policyholders), each observed for $n$ periods, giving a two-way table $X_{ij}$ ($i=1,\\dots,r$ risks; $j=1,\\dots,n$ years). From it we form per-risk means $\\bar X_i$, the grand mean $\\bar X$, and within-risk sample variances $s_i^2$ to estimate $\\hat\\mu$, $\\widehat{\\text{EPV}}$, and $\\widehat{\\text{VHM}}$.",
      "tag": "Empirical Bayes (nonparametric)"
    },
    {
      "front": "Give the nonparametric empirical-Bayes estimator of the **overall mean** $\\hat\\mu$ ($r$ risks, $n$ years each).",
      "back": "$\\hat\\mu=\\bar X=\\frac{1}{rn}\\sum_{i=1}^{r}\\sum_{j=1}^{n}X_{ij}=\\frac{1}{r}\\sum_{i=1}^{r}\\bar X_i$,\nthe grand mean of all observations (equivalently, the average of the per-risk means when every risk has the same number of years $n$).\nWhen exposures differ (Bühlmann-Straub), the grand mean is exposure-weighted instead.",
      "tag": "Empirical Bayes (nonparametric)"
    },
    {
      "front": "Give the nonparametric empirical-Bayes estimator of the **EPV**.",
      "back": "$\\widehat{\\text{EPV}}=\\frac{1}{r}\\sum_{i=1}^{r}s_i^{2}$, the average of the per-risk sample variances, where\n$s_i^{2}=\\frac{1}{n-1}\\sum_{j=1}^{n}(X_{ij}-\\bar X_i)^{2}$.\nEach $s_i^2$ is an unbiased estimate of that risk's process variance $v(\\theta_i)$, so averaging over the $r$ risks estimates $E[v(\\theta)]=\\text{EPV}$. (With unequal years, use a pooled within-risk variance: $\\sum_i\\sum_j (X_{ij}-\\bar X_i)^2 \\big/ \\sum_i (n_i-1)$.)",
      "tag": "Empirical Bayes (nonparametric)"
    },
    {
      "front": "Give the nonparametric empirical-Bayes estimator of the **VHM** and the rule when it comes out negative.",
      "back": "$\\widehat{\\text{VHM}}=\\frac{1}{r-1}\\sum_{i=1}^{r}(\\bar X_i-\\bar X)^{2}-\\frac{\\widehat{\\text{EPV}}}{n}$.\nThe first term is the sample variance of the per-risk means; it overstates VHM because each $\\bar X_i$ carries process noise of size $\\frac{v(\\theta)}{n}$, so we subtract $\\frac{\\widehat{\\text{EPV}}}{n}$ to remove it.\nBecause this is a difference, it can be **negative**; a negative estimate is set to $\\widehat{\\text{VHM}}=0$ (which forces $Z=0$, i.e. give the risk no own-experience weight).",
      "tag": "Empirical Bayes (nonparametric)"
    },
    {
      "front": "Why does the empirical-Bayes VHM estimator subtract $\\widehat{\\text{EPV}}/n$ from the between-risk sample variance?",
      "back": "Each per-risk mean $\\bar X_i$ estimates the true hypothetical mean $\\mu(\\theta_i)$ but with sampling error of variance $\\frac{v(\\theta_i)}{n}$. So the observed spread of the $\\bar X_i$ contains **both** the genuine between-risk variability (VHM) **and** this estimation noise, on average $\\frac{\\text{EPV}}{n}$.\n$E\\!\\left[\\frac{1}{r-1}\\sum_i(\\bar X_i-\\bar X)^2\\right]=\\text{VHM}+\\frac{\\text{EPV}}{n}$, so subtracting $\\frac{\\widehat{\\text{EPV}}}{n}$ makes the estimator unbiased for VHM.",
      "tag": "Empirical Bayes (nonparametric)"
    },
    {
      "front": "Fully worked nonparametric empirical Bayes. Three risks, each observed $n=4$ years:\nRisk A: 8, 10, 9, 13; Risk B: 5, 7, 6, 6; Risk C: 12, 14, 11, 15.\nCompute the per-risk means and the grand mean $\\hat\\mu$.",
      "back": "Risk A: $\\bar X_A=\\frac{8+10+9+13}{4}=\\frac{40}{4}=10$.\nRisk B: $\\bar X_B=\\frac{5+7+6+6}{4}=\\frac{24}{4}=6$.\nRisk C: $\\bar X_C=\\frac{12+14+11+15}{4}=\\frac{52}{4}=13$.\nGrand mean $\\hat\\mu=\\bar X=\\frac{10+6+13}{3}=\\frac{29}{3}\\approx 9.6667$.\n(With equal years, averaging the three risk means equals averaging all $12$ observations.)",
      "tag": "Empirical Bayes (nonparametric)"
    },
    {
      "front": "Same data (A: 8,10,9,13; B: 5,7,6,6; C: 12,14,11,15; $n=4$, $\\bar X_A=10,\\bar X_B=6,\\bar X_C=13$). Compute the within-risk sample variances $s_i^2$ and $\\widehat{\\text{EPV}}$.",
      "back": "Use $s_i^2=\\frac{1}{n-1}\\sum_j(X_{ij}-\\bar X_i)^2$ with $n-1=3$.\nRisk A: deviations $-2,0,-1,3$; squares $4,0,1,9$; sum $=14$; $s_A^2=\\frac{14}{3}\\approx 4.6667$.\nRisk B: deviations $-1,1,0,0$; squares $1,1,0,0$; sum $=2$; $s_B^2=\\frac{2}{3}\\approx 0.6667$.\nRisk C: deviations $-1,1,-2,2$; squares $1,1,4,4$; sum $=10$; $s_C^2=\\frac{10}{3}\\approx 3.3333$.\n$\\widehat{\\text{EPV}}=\\frac{1}{r}\\sum_i s_i^2=\\frac{1}{3}\\left(\\frac{14+2+10}{3}\\right)=\\frac{26}{9}\\approx 2.8889$.",
      "tag": "Empirical Bayes (nonparametric)"
    },
    {
      "front": "Same data ($\\bar X_A=10,\\bar X_B=6,\\bar X_C=13$; $\\hat\\mu=\\frac{29}{3}$; $\\widehat{\\text{EPV}}=\\frac{26}{9}$; $r=3$, $n=4$). Compute $\\widehat{\\text{VHM}}$.",
      "back": "Between-risk sum of squares $\\sum_i(\\bar X_i-\\bar X)^2$ with $\\bar X=\\frac{29}{3}\\approx 9.6667$:\nA: $(10-9.6667)^2=(0.3333)^2\\approx 0.1111$;\nB: $(6-9.6667)^2=(-3.6667)^2\\approx 13.4444$;\nC: $(13-9.6667)^2=(3.3333)^2\\approx 11.1111$.\nSum $\\approx 24.6667$. Divide by $r-1=2$: $\\frac{24.6667}{2}=12.3333$.\n$\\widehat{\\text{VHM}}=12.3333-\\frac{\\widehat{\\text{EPV}}}{n}=12.3333-\\frac{2.8889}{4}=12.3333-0.7222\\approx 11.6111$.",
      "tag": "Empirical Bayes (nonparametric)"
    },
    {
      "front": "Same data ($\\widehat{\\text{EPV}}=\\frac{26}{9}\\approx 2.8889$, $\\widehat{\\text{VHM}}\\approx 11.6111$, $n=4$). Find $\\hat k$ and $\\hat Z$.",
      "back": "$\\hat k=\\frac{\\widehat{\\text{EPV}}}{\\widehat{\\text{VHM}}}=\\frac{2.8889}{11.6111}\\approx 0.2488$.\n$\\hat Z=\\frac{n}{n+\\hat k}=\\frac{4}{4+0.2488}=\\frac{4}{4.2488}\\approx 0.9414$.\nThe risks are very different (large VHM) relative to within-risk noise (small EPV), so each risk earns high credibility on only $4$ years.",
      "tag": "Empirical Bayes (nonparametric)"
    },
    {
      "front": "Same data ($\\hat\\mu=\\frac{29}{3}\\approx 9.6667$, $\\hat Z\\approx 0.9414$). Find the empirical-Bayes credibility premiums for Risks A, B, and C, and verify they sum back to the total.",
      "back": "$P_i=\\hat Z\\,\\bar X_i+(1-\\hat Z)\\hat\\mu$ with $1-\\hat Z\\approx 0.0586$ and $(1-\\hat Z)\\hat\\mu\\approx 0.0586(9.6667)\\approx 0.5665$.\nRisk A: $0.9414(10)+0.5665=9.414+0.5665\\approx 9.98$.\nRisk B: $0.9414(6)+0.5665=5.648+0.5665\\approx 6.22$.\nRisk C: $0.9414(13)+0.5665=12.238+0.5665\\approx 12.80$.\nBalance check: $9.98+6.22+12.80=29.00=3\\hat\\mu=3(9.6667)$ — the credibility premiums preserve the grand total (the estimator is balanced).",
      "tag": "Empirical Bayes (nonparametric)"
    },
    {
      "front": "Why must the empirical-Bayes credibility premiums satisfy the **balance** condition $\\sum_i n_i P_i=\\sum_i n_i\\bar X_i$?",
      "back": "Because $P_i=Z\\bar X_i+(1-Z)\\hat\\mu$ and $\\hat\\mu=\\bar X$ is the (exposure-weighted) grand mean, averaging the $P_i$ over the same weights returns $\\hat\\mu$: $\\sum_i n_i P_i = Z\\sum_i n_i\\bar X_i+(1-Z)\\hat\\mu\\sum_i n_i=\\sum_i n_i\\bar X_i$.\nPractically, this means crediting individual risks toward the mean does **not** change the total premium collected — credibility redistributes premium among risks but is revenue-neutral in aggregate.",
      "tag": "Empirical Bayes (nonparametric)"
    },
    {
      "front": "What is **semiparametric** estimation in credibility, and when is it used?",
      "back": "Semiparametric estimation assumes a **parametric form for the conditional distribution** of the data (so the process variance is a known function of the mean) but leaves the **prior** of $\\theta$ unspecified, estimating its features empirically.\nThe headline case is claim counts modeled as conditionally **Poisson**, where $v(\\theta)=\\mu(\\theta)$, so the EPV equals the overall mean and need not be estimated from within-risk variances — it is read straight off $\\hat\\mu=\\bar X$.",
      "tag": "Semiparametric"
    },
    {
      "front": "Give the **semiparametric (Poisson)** estimators of EPV and VHM.",
      "back": "If counts are conditionally Poisson then $v(\\theta)=\\mu(\\theta)$, so $\\text{EPV}=E[v(\\theta)]=E[\\mu(\\theta)]=\\mu$. Hence\n$\\widehat{\\text{EPV}}=\\bar X$ (the overall mean).\nVHM is still estimated from the spread of risk means with the noise correction:\n$\\widehat{\\text{VHM}}=\\frac{1}{r-1}\\sum_i(\\bar X_i-\\bar X)^2-\\frac{\\widehat{\\text{EPV}}}{n}$ (set to $0$ if negative).\nThe only difference from nonparametric is that EPV comes from the Poisson identity, not from within-risk sample variances.",
      "tag": "Semiparametric"
    },
    {
      "front": "Fully worked semiparametric (Poisson). $400$ policyholders each observed $1$ year. Claim-count distribution:\n0 claims: 280 policies; 1 claim: 90; 2 claims: 24; 3 claims: 6. Estimate $\\widehat{\\text{EPV}}$ via the Poisson identity.",
      "back": "Total claims $=0(280)+1(90)+2(24)+3(6)=90+48+18=156$.\nTotal policies $=280+90+24+6=400$.\n$\\bar X=\\frac{156}{400}=0.39$ claims/policy.\nUnder the conditional-Poisson assumption $\\text{EPV}=\\mu$, so $\\widehat{\\text{EPV}}=\\bar X=0.39$.",
      "tag": "Semiparametric"
    },
    {
      "front": "Continuing the Poisson semiparametric example ($n=1$ year per policy, $\\bar X=0.39$, $\\widehat{\\text{EPV}}=0.39$): estimate the **total sample variance** of claim counts, then $\\widehat{\\text{VHM}}$.",
      "back": "Mean of squares: $\\frac{0^2(280)+1^2(90)+2^2(24)+3^2(6)}{400}=\\frac{0+90+96+54}{400}=\\frac{240}{400}=0.60$.\nSample variance $\\hat\\sigma^2=E[X^2]-\\bar X^2=0.60-0.39^2=0.60-0.1521=0.4479$ (this estimates total variance $=\\text{EPV}+\\text{VHM}$ when $n=1$).\n$\\widehat{\\text{VHM}}=\\hat\\sigma^2-\\widehat{\\text{EPV}}=0.4479-0.39=0.0579$.\n(Equivalently the between-policy variance of the single-year means minus EPV; with $n=1$ the two coincide.)",
      "tag": "Semiparametric"
    },
    {
      "front": "Finishing the Poisson semiparametric example ($\\widehat{\\text{EPV}}=0.39$, $\\widehat{\\text{VHM}}=0.0579$, $n=1$): find $\\hat k$ and the credibility estimate for a policyholder who had $2$ claims in the year.",
      "back": "$\\hat k=\\frac{\\widehat{\\text{EPV}}}{\\widehat{\\text{VHM}}}=\\frac{0.39}{0.0579}\\approx 6.736$.\n$\\hat Z=\\frac{n}{n+\\hat k}=\\frac{1}{1+6.736}=\\frac{1}{7.736}\\approx 0.1293$.\nFor a policy with $\\bar X=2$: $P_c=\\hat Z(2)+(1-\\hat Z)(0.39)=0.1293(2)+0.8707(0.39)$\n$\\approx 0.2586+0.3396=0.598\\approx 0.60$ claims.\nWith just one year and modest VHM, the estimate barely moves off the $0.39$ mean.",
      "tag": "Semiparametric"
    },
    {
      "front": "State the **credibility-Bayes link**: in what sense is the Bühlmann estimate \"the best\" approximation to the Bayesian premium?",
      "back": "The exact Bayesian premium is $E[\\mu(\\theta)\\mid X_1,\\dots,X_n]$, the posterior mean of the hypothetical mean. The **Bühlmann credibility premium** $Z\\bar X+(1-Z)\\mu$ is the **least-squares linear approximation** to that Bayesian estimate — it minimizes $E\\big[(\\,\\text{Bayesian premium}-(a+b\\bar X)\\,)^2\\big]$ over all linear functions of the data.\nThus Bühlmann is the best *linear* estimator; it trades the exact (possibly nonlinear) Bayes rule for a simple, robust line.",
      "tag": "Credibility-Bayes link"
    },
    {
      "front": "When does the Bühlmann credibility estimate equal the **exact Bayesian** premium?",
      "back": "When the Bayesian premium is already a **linear** function of the data — the exact-credibility (linear-conjugate) families. Key cases on the syllabus:\n• **Poisson likelihood with Gamma prior** (claim counts);\n• **Normal likelihood with Normal prior** (severity/means);\n• **Bernoulli/Binomial with Beta prior**;\n• **Exponential/Gamma likelihood with appropriate conjugate prior**.\nFor these the posterior mean is exactly $Z\\bar X+(1-Z)\\mu$, so Bühlmann and Bayes coincide and no approximation is lost.",
      "tag": "Credibility-Bayes link"
    },
    {
      "front": "Worked credibility-Bayes (Poisson-Gamma exact). Counts are $\\text{Poisson}(\\lambda)$, prior $\\lambda\\sim\\text{Gamma}(\\alpha=2,\\ \\beta=4)$ (rate $\\beta$, so $E[\\lambda]=\\alpha/\\beta=0.5$). Observe $n=3$ years totaling $S=3$ claims. Find the Bayesian (posterior-mean) estimate.",
      "back": "Gamma-Poisson conjugacy: posterior is $\\text{Gamma}(\\alpha+S,\\ \\beta+n)=\\text{Gamma}(2+3,\\ 4+3)=\\text{Gamma}(5,\\ 7)$.\nPosterior mean $=\\frac{\\alpha+S}{\\beta+n}=\\frac{5}{7}\\approx 0.7143$ claims.\nThis is the exact Bayesian predictive mean for next year's count.",
      "tag": "Credibility-Bayes link"
    },
    {
      "front": "Verify the previous Poisson-Gamma result with the **Bühlmann** formula, showing the two agree. (Prior $\\text{Gamma}(\\alpha=2,\\beta=4)$ rate-form; $n=3$, $\\bar X=1$.)",
      "back": "With rate-$\\beta$ Gamma: $\\mu=E[\\lambda]=\\frac{\\alpha}{\\beta}=\\frac{2}{4}=0.5$, $\\text{EPV}=E[\\lambda]=0.5$, $\\text{VHM}=\\text{Var}[\\lambda]=\\frac{\\alpha}{\\beta^2}=\\frac{2}{16}=0.125$.\n$k=\\frac{\\text{EPV}}{\\text{VHM}}=\\frac{0.5}{0.125}=4=\\beta$ (indeed $k=\\beta$ for Gamma-Poisson).\n$Z=\\frac{n}{n+k}=\\frac{3}{3+4}=\\frac{3}{7}$; $\\bar X=\\frac{3}{3}=1$.\n$P_c=Z\\bar X+(1-Z)\\mu=\\frac{3}{7}(1)+\\frac{4}{7}(0.5)=\\frac{3}{7}+\\frac{2}{7}=\\frac{5}{7}\\approx 0.7143$.\nIdentical to the posterior mean — exact credibility holds.",
      "tag": "Credibility-Bayes link"
    },
    {
      "front": "Worked credibility-Bayes (Normal-Normal exact). Severity $X\\mid\\theta\\sim N(\\theta,\\ \\sigma^2=100)$ with prior $\\theta\\sim N(\\mu_0=50,\\ \\tau^2=25)$. Observe $n=4$ values averaging $\\bar X=60$. Find the posterior-mean (Bayesian) estimate.",
      "back": "Normal-Normal posterior mean is a precision-weighted average:\n$Z=\\frac{n\\tau^2}{n\\tau^2+\\sigma^2}=\\frac{4(25)}{4(25)+100}=\\frac{100}{200}=0.5$.\nPosterior mean $=Z\\bar X+(1-Z)\\mu_0=0.5(60)+0.5(50)=30+25=55$.\nEquivalently $k=\\frac{\\sigma^2}{\\tau^2}=\\frac{100}{25}=4=\\text{EPV}/\\text{VHM}$, giving $Z=\\frac{n}{n+k}=\\frac{4}{8}=0.5$ — the Bühlmann $Z$ matches, so Bühlmann equals Bayes here too.",
      "tag": "Credibility-Bayes link"
    },
    {
      "front": "For the Normal-Normal model, show that the Bühlmann constant $k$ equals $\\sigma^2/\\tau^2$.",
      "back": "$\\mu(\\theta)=\\theta$ and $v(\\theta)=\\text{Var}(X\\mid\\theta)=\\sigma^2$ (constant), so $\\text{EPV}=E[\\sigma^2]=\\sigma^2$.\n$\\text{VHM}=\\text{Var}[\\mu(\\theta)]=\\text{Var}[\\theta]=\\tau^2$.\nTherefore $k=\\frac{\\text{EPV}}{\\text{VHM}}=\\frac{\\sigma^2}{\\tau^2}$ and $Z=\\frac{n}{n+\\sigma^2/\\tau^2}=\\frac{n\\tau^2}{n\\tau^2+\\sigma^2}$, exactly the Bayesian posterior-mean weight — Normal-Normal is an exact-credibility model.",
      "tag": "Credibility-Bayes link"
    },
    {
      "front": "Contrast the **Bühlmann** estimate, the **Bayesian** estimate, and the **limited-fluctuation** estimate of a future premium.",
      "back": "**Bayesian:** $E[\\mu(\\theta)\\mid \\text{data}]$ — the full posterior mean; can be nonlinear in the data; requires the entire model (likelihood + prior).\n**Bühlmann (greatest-accuracy):** $Z\\bar X+(1-Z)\\mu$ with $Z=\\frac{n}{n+k}$ — the least-squares linear approximation to Bayes; needs only $\\mu$, EPV, VHM; equals Bayes for conjugate-linear models.\n**Limited-fluctuation (classical):** $Z\\bar X+(1-Z)M$ with $Z=\\min\\!\\big(\\sqrt{n/n_0},1\\big)$ — a square-root partial-credibility rule from a full-credibility standard $n_0$; not derived from the variance structure and generally not the optimal linear estimator.",
      "tag": "Credibility-Bayes link"
    },
    {
      "front": "Worked comparison: Poisson-Gamma with prior $\\text{Gamma}(\\alpha=3,\\beta=2)$ (rate form, $E[\\lambda]=1.5$); a risk shows total claims $S=12$ over $n=5$ years ($\\bar X=2.4$). Compute both the Bühlmann and the exact Bayesian estimate and confirm equality.",
      "back": "Structure: $\\mu=\\frac{\\alpha}{\\beta}=\\frac{3}{2}=1.5$, $\\text{EPV}=1.5$, $\\text{VHM}=\\frac{\\alpha}{\\beta^2}=\\frac{3}{4}=0.75$, so $k=\\frac{1.5}{0.75}=2$.\n$Z=\\frac{5}{5+2}=\\frac{5}{7}\\approx 0.7143$.\nBühlmann: $P_c=\\frac{5}{7}(2.4)+\\frac{2}{7}(1.5)=\\frac{12+3}{7}=\\frac{15}{7}\\approx 2.143$.\nBayes: with total claims $S=12$ (an integer Poisson count), posterior $\\text{Gamma}(3+12,\\ 2+5)=\\text{Gamma}(15,7)$, mean $\\frac{15}{7}\\approx 2.143$. They agree exactly.",
      "tag": "Credibility-Bayes link"
    },
    {
      "front": "In empirical Bayes, what does it mean — and what do you do — when $\\widehat{\\text{VHM}}\\le 0$?",
      "back": "A non-positive $\\widehat{\\text{VHM}}$ means the observed spread among the risk means is no larger than what the within-risk process noise $\\frac{\\widehat{\\text{EPV}}}{n}$ alone would produce — there is no statistical evidence the risks truly differ.\nThe convention is to set $\\widehat{\\text{VHM}}=0$, which makes $\\hat k=\\infty$ and therefore $\\hat Z=0$. Every risk is then charged the grand mean $\\hat\\mu$: with no detectable heterogeneity, individual experience earns zero credibility.",
      "tag": "Empirical Bayes (nonparametric)"
    },
    {
      "front": "Worked empirical Bayes producing $\\widehat{\\text{VHM}}=0$. Two risks, $n=3$ years each — Risk 1: 4, 6, 5; Risk 2: 5, 4, 6. Show the credibility collapses to the mean.",
      "back": "Means: $\\bar X_1=\\frac{15}{3}=5$, $\\bar X_2=\\frac{15}{3}=5$; grand mean $\\hat\\mu=5$.\nWithin-risk variances: each set $\\{4,5,6\\}$ has $s^2=\\frac{1+0+1}{2}=1$, so $\\widehat{\\text{EPV}}=\\frac{1+1}{2}=1$.\nBetween-risk: $\\frac{1}{r-1}\\sum(\\bar X_i-\\bar X)^2=\\frac{(5-5)^2+(5-5)^2}{1}=0$.\n$\\widehat{\\text{VHM}}=0-\\frac{\\widehat{\\text{EPV}}}{n}=0-\\frac{1}{3}<0\\Rightarrow$ set to $0$.\nThen $\\hat Z=0$ and both risks are charged $\\hat\\mu=5$ — identical means give no evidence of heterogeneity.",
      "tag": "Empirical Bayes (nonparametric)"
    },
    {
      "front": "Worked Bühlmann-Straub empirical Bayes with unequal exposures. Two risks:\nRisk 1: year exposures/losses $(m,X)=(10,4.0),(20,5.0)$; Risk 2: $(30,7.0),(40,6.5)$. Find the exposure-weighted risk means and the grand mean.",
      "back": "Risk 1 total exposure $m_1=10+20=30$; weighted mean $\\bar X_1=\\frac{10(4.0)+20(5.0)}{30}=\\frac{40+100}{30}=\\frac{140}{30}\\approx 4.667$.\nRisk 2 total exposure $m_2=30+40=70$; weighted mean $\\bar X_2=\\frac{30(7.0)+40(6.5)}{70}=\\frac{210+260}{70}=\\frac{470}{70}\\approx 6.714$.\nGrand (exposure-weighted) mean $\\hat\\mu=\\frac{10(4.0)+20(5.0)+30(7.0)+40(6.5)}{100}=\\frac{40+100+210+260}{100}=\\frac{610}{100}=6.10$.\nThe grand mean uses total exposure $m=30+70=100$ as the denominator, not a simple average of the two risk means.",
      "tag": "Bühlmann-Straub"
    },
    {
      "front": "Summarize the **decision flow** for choosing a credibility method on an exam problem.",
      "back": "1. **Prior fully specified** (parametric likelihood + prior given)? Compute $\\mu$, EPV, VHM analytically; use Bühlmann $Z=\\frac{n}{n+k}$, or get the exact Bayesian posterior mean (they match for conjugate-linear models).\n2. **Exposures vary across cells**? Use **Bühlmann-Straub**: $Z=\\frac{m}{m+k}$, exposure-weighted $\\bar X$.\n3. **Prior unknown, raw two-way data given**? Use **nonparametric empirical Bayes**: $\\hat\\mu=\\bar X$, $\\widehat{\\text{EPV}}=\\overline{s_i^2}$, $\\widehat{\\text{VHM}}=$ between-variance $-\\widehat{\\text{EPV}}/n$.\n4. **Counts known to be Poisson but prior unknown**? Use **semiparametric**: $\\widehat{\\text{EPV}}=\\bar X$, then VHM as in step 3.",
      "tag": "Credibility-Bayes link"
    }
  ]
}