Exam MAS-II — Credibility Flashcards

Each risk is characterized by a latent risk parameter $\theta$ drawn from a prior. Conditional on $\theta$: **Hypothetical mean** $\mu(\theta)=E[X\mid\theta]$ — the expected claim for a risk of type $\theta$. **Process variance** $v(\theta)=\text{Var}(X\mid\theta)$ — the within-risk variability of that risk's outcomes. The two structural quantities EPV and VHM are built from these by averaging or taking the variance over the prior of $\theta$.

$\text{EPV}=E[v(\theta)]=E[\text{Var}(X\mid\theta)]$ — the average within-risk variance; it measures the noise inside a single risk. $\text{VHM}=\text{Var}[\mu(\theta)]=\text{Var}[E(X\mid\theta)]$ — the spread of the risks' means about the overall mean; it measures how much the risks truly differ. By the variance-decomposition (law of total variance), the total variance of a single observation is $\text{Var}(X)=\text{EPV}+\text{VHM}$.

$k=\dfrac{\text{EPV}}{\text{VHM}}=\dfrac{E[v(\theta)]}{\text{Var}[\mu(\theta)]}$, and with $n$ observations the credibility is $Z=\dfrac{n}{n+k}$. Large 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 $Z\to 1$.

$P_{c}=Z\bar X + (1-Z)\mu$, where $\bar X$ is the risk's own sample mean over $n$ periods and $\mu=E[\mu(\theta)]$ is the overall (collective) mean. It 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.

$Z=\frac{n}{n+k}$ 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. Limits: as $\text{VHM}\to\infty$, $k\to 0$ and $Z\to 1$; as $\text{EPV}\to\infty$, $k\to\infty$ and $Z\to 0$.

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. Both give more weight to larger samples, but Bühlmann's $Z$ is the least-squares-optimal linear weight, not a square-root rule.

$k=\frac{\text{EPV}}{\text{VHM}}=\frac{8000}{500}=16$. $Z=\frac{n}{n+k}=\frac{4}{4+16}=\frac{4}{20}=0.20$. $P_{c}=Z\bar X+(1-Z)\mu=0.20(1500)+0.80(1200)=300+960=\$1{,}260$. The heavy EPV relative to VHM keeps $Z$ low, so the estimate stays close to the collective mean.

Set $Z=\frac{n}{n+16}\ge 0.50$. $\frac{n}{n+16}=0.50 \Rightarrow n = 0.5(n+16) \Rightarrow 0.5n = 8 \Rightarrow n = 16$. So $n=16$ years gives $Z=0.50$ exactly, and any $n>16$ gives $Z>0.50$. In general $Z=0.5$ precisely when $n=k$, a useful check: the credibility equals one-half when the sample size equals the Bühlmann constant.

Here $\mu(\lambda)=\lambda$ and $v(\lambda)=\lambda$ (Poisson mean $=$ variance). Gamma mean $E[\lambda]=\alpha\theta=3(0.1)=0.30$; Gamma variance $\text{Var}[\lambda]=\alpha\theta^{2}=3(0.01)=0.03$. $\mu=E[\mu(\lambda)]=E[\lambda]=0.30$. $\text{EPV}=E[v(\lambda)]=E[\lambda]=0.30$. $\text{VHM}=\text{Var}[\mu(\lambda)]=\text{Var}[\lambda]=0.03$. $k=\frac{\text{EPV}}{\text{VHM}}=\frac{0.30}{0.03}=10$.

Risk's sample mean $\bar X=\frac{4}{5}=0.80$ claims/year; overall mean $\mu=0.30$. $Z=\frac{n}{n+k}=\frac{5}{5+10}=\frac{5}{15}=0.3\overline{3}$. $P_{c}=Z\bar X+(1-Z)\mu=0.3333(0.80)+0.6667(0.30)$ $=0.26667+0.20000=0.46667\approx 0.467$ claims. (For Poisson-Gamma the Bühlmann and exact Bayesian estimates coincide — see the credibility-Bayes link.)

Overall mean $\mu=0.6(100)+0.4(300)=60+120=180$. $\text{EPV}=E[v(\theta)]=0.6(4000)+0.4(9000)=2400+3600=6000$. $\text{VHM}=\text{Var}[\mu(\theta)]=E[\mu^{2}]-\mu^{2}=[0.6(100^{2})+0.4(300^{2})]-180^{2}$ $=[6000+36000]-32400=42000-32400=9600$. $k=\frac{\text{EPV}}{\text{VHM}}=\frac{6000}{9600}=0.625$.

$Z=\frac{n}{n+k}=\frac{3}{3+0.625}=\frac{3}{3.625}\approx 0.8276$. $P_{c}=Z\bar X+(1-Z)\mu=0.8276(250)+0.1724(180)$ $\approx 206.90+31.03=\$237.93$. The large VHM (very different risk types) makes the small constant $k$, so even $3$ years earns high credibility.

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 $\text{Var}(X_{i}\mid\theta)=\frac{v(\theta)}{m_{i}}$. Let $m=\sum_{i} m_{i}$ be total exposure. Then $k=\frac{\text{EPV}}{\text{VHM}}$ as before, but $Z=\frac{m}{m+k}$, and the risk's mean is exposure-weighted: $\bar X=\frac{\sum_i m_i X_i}{m}$.

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. The ordinary Bühlmann model is the special case $m_i=1$ for all $i$, which collapses $\bar X$ to the simple average and $Z$ to $\frac{n}{n+k}$.

Total exposure $m=20+30+50=100$. Weighted mean $\bar X=\frac{20(60)+30(45)+50(58)}{100}=\frac{1200+1350+2900}{100}=\frac{5450}{100}=54.5$. $k=\frac{\text{EPV}}{\text{VHM}}=\frac{10000}{200}=50$. $Z=\frac{m}{m+k}=\frac{100}{100+50}=\frac{100}{150}=0.6\overline{6}$. $P_{c}=Z\bar X+(1-Z)\mu=0.6667(54.5)+0.3333(50)=36.33+16.67=\$53.00$ per exposure.

The credibility premium $\$53.00$ is a rate **per exposure unit**. Multiply by next year's exposure: Expected loss $=P_{c}\times m_{new}=53.00\times 40=\$2{,}120$. Keep the per-exposure rate distinct from the dollar total — credibility is applied to the rate, then scaled to the projected exposure.

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. Classic 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 $\widehat{\text{VHM}}$.

$\hat\mu=\bar X=\frac{1}{rn}\sum_{i=1}^{r}\sum_{j=1}^{n}X_{ij}=\frac{1}{r}\sum_{i=1}^{r}\bar X_i$, the grand mean of all observations (equivalently, the average of the per-risk means when every risk has the same number of years $n$). When exposures differ (Bühlmann-Straub), the grand mean is exposure-weighted instead.

$\widehat{\text{EPV}}=\frac{1}{r}\sum_{i=1}^{r}s_i^{2}$, the average of the per-risk sample variances, where $s_i^{2}=\frac{1}{n-1}\sum_{j=1}^{n}(X_{ij}-\bar X_i)^{2}$. Each $s_i^2$ 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: $\sum_i\sum_j (X_{ij}-\bar X_i)^2 \big/ \sum_i (n_i-1)$.)

$\widehat{\text{VHM}}=\frac{1}{r-1}\sum_{i=1}^{r}(\bar X_i-\bar X)^{2}-\frac{\widehat{\text{EPV}}}{n}$. The 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. Because 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).

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 $\frac{\text{EPV}}{n}$. $E\!\left[\frac{1}{r-1}\sum_i(\bar X_i-\bar X)^2\right]=\text{VHM}+\frac{\text{EPV}}{n}$, so subtracting $\frac{\widehat{\text{EPV}}}{n}$ makes the estimator unbiased for VHM.

Risk A: $\bar X_A=\frac{8+10+9+13}{4}=\frac{40}{4}=10$. Risk B: $\bar X_B=\frac{5+7+6+6}{4}=\frac{24}{4}=6$. Risk C: $\bar X_C=\frac{12+14+11+15}{4}=\frac{52}{4}=13$. Grand mean $\hat\mu=\bar X=\frac{10+6+13}{3}=\frac{29}{3}\approx 9.6667$. (With equal years, averaging the three risk means equals averaging all $12$ observations.)

Use $s_i^2=\frac{1}{n-1}\sum_j(X_{ij}-\bar X_i)^2$ with $n-1=3$. Risk A: deviations $-2,0,-1,3$; squares $4,0,1,9$; sum $=14$; $s_A^2=\frac{14}{3}\approx 4.6667$. Risk B: deviations $-1,1,0,0$; squares $1,1,0,0$; sum $=2$; $s_B^2=\frac{2}{3}\approx 0.6667$. Risk C: deviations $-1,1,-2,2$; squares $1,1,4,4$; sum $=10$; $s_C^2=\frac{10}{3}\approx 3.3333$. $\widehat{\text{EPV}}=\frac{1}{r}\sum_i s_i^2=\frac{1}{3}\left(\frac{14+2+10}{3}\right)=\frac{26}{9}\approx 2.8889$.

Between-risk sum of squares $\sum_i(\bar X_i-\bar X)^2$ with $\bar X=\frac{29}{3}\approx 9.6667$: A: $(10-9.6667)^2=(0.3333)^2\approx 0.1111$; B: $(6-9.6667)^2=(-3.6667)^2\approx 13.4444$; C: $(13-9.6667)^2=(3.3333)^2\approx 11.1111$. Sum $\approx 24.6667$. Divide by $r-1=2$: $\frac{24.6667}{2}=12.3333$. $\widehat{\text{VHM}}=12.3333-\frac{\widehat{\text{EPV}}}{n}=12.3333-\frac{2.8889}{4}=12.3333-0.7222\approx 11.6111$.

$\hat k=\frac{\widehat{\text{EPV}}}{\widehat{\text{VHM}}}=\frac{2.8889}{11.6111}\approx 0.2488$. $\hat Z=\frac{n}{n+\hat k}=\frac{4}{4+0.2488}=\frac{4}{4.2488}\approx 0.9414$. The risks are very different (large VHM) relative to within-risk noise (small EPV), so each risk earns high credibility on only $4$ years.

$P_i=\hat Z\,\bar X_i+(1-\hat Z)\hat\mu$ with $1-\hat Z\approx 0.0586$ and $(1-\hat Z)\hat\mu\approx 0.0586(9.6667)\approx 0.5665$. Risk A: $0.9414(10)+0.5665=9.414+0.5665\approx 9.98$. Risk B: $0.9414(6)+0.5665=5.648+0.5665\approx 6.22$. Risk C: $0.9414(13)+0.5665=12.238+0.5665\approx 12.80$. Balance 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).

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 $\hat\mu$: $\sum_i n_i P_i = Z\sum_i n_i\bar X_i+(1-Z)\hat\mu\sum_i n_i=\sum_i n_i\bar X_i$. Practically, 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.

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. The 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 $\hat\mu=\bar X$.

If counts are conditionally Poisson then $v(\theta)=\mu(\theta)$, so $\text{EPV}=E[v(\theta)]=E[\mu(\theta)]=\mu$. Hence $\widehat{\text{EPV}}=\bar X$ (the overall mean). VHM is still estimated from the spread of risk means with the noise correction: $\widehat{\text{VHM}}=\frac{1}{r-1}\sum_i(\bar X_i-\bar X)^2-\frac{\widehat{\text{EPV}}}{n}$ (set to $0$ if negative). The only difference from nonparametric is that EPV comes from the Poisson identity, not from within-risk sample variances.

Total claims $=0(280)+1(90)+2(24)+3(6)=90+48+18=156$. Total policies $=280+90+24+6=400$. $\bar X=\frac{156}{400}=0.39$ claims/policy. Under the conditional-Poisson assumption $\text{EPV}=\mu$, so $\widehat{\text{EPV}}=\bar X=0.39$.

Mean of squares: $\frac{0^2(280)+1^2(90)+2^2(24)+3^2(6)}{400}=\frac{0+90+96+54}{400}=\frac{240}{400}=0.60$. Sample 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 $n=1$). $\widehat{\text{VHM}}=\hat\sigma^2-\widehat{\text{EPV}}=0.4479-0.39=0.0579$. (Equivalently the between-policy variance of the single-year means minus EPV; with $n=1$ the two coincide.)

$\hat k=\frac{\widehat{\text{EPV}}}{\widehat{\text{VHM}}}=\frac{0.39}{0.0579}\approx 6.736$. $\hat Z=\frac{n}{n+\hat k}=\frac{1}{1+6.736}=\frac{1}{7.736}\approx 0.1293$. For a policy with $\bar X=2$: $P_c=\hat Z(2)+(1-\hat Z)(0.39)=0.1293(2)+0.8707(0.39)$ $\approx 0.2586+0.3396=0.598\approx 0.60$ claims. With just one year and modest VHM, the estimate barely moves off the $0.39$ mean.

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. Thus Bühlmann is the best *linear* estimator; it trades the exact (possibly nonlinear) Bayes rule for a simple, robust line.

When the Bayesian premium is already a **linear** function of the data — the exact-credibility (linear-conjugate) families. Key cases on the syllabus: • **Poisson likelihood with Gamma prior** (claim counts); • **Normal likelihood with Normal prior** (severity/means); • **Bernoulli/Binomial with Beta prior**; • **Exponential/Gamma likelihood with appropriate conjugate prior**. For these the posterior mean is exactly $Z\bar X+(1-Z)\mu$, so Bühlmann and Bayes coincide and no approximation is lost.

Gamma-Poisson conjugacy: posterior is $\text{Gamma}(\alpha+S,\ \beta+n)=\text{Gamma}(2+3,\ 4+3)=\text{Gamma}(5,\ 7)$. Posterior mean $=\frac{\alpha+S}{\beta+n}=\frac{5}{7}\approx 0.7143$ claims. This is the exact Bayesian predictive mean for next year's count.

With rate-$\beta$ Gamma: $\mu=E[\lambda]=\frac{\alpha}{\beta}=\frac{2}{4}=0.5$, $\text{EPV}=E[\lambda]=0.5$, $\text{VHM}=\text{Var}[\lambda]=\frac{\alpha}{\beta^2}=\frac{2}{16}=0.125$. $k=\frac{\text{EPV}}{\text{VHM}}=\frac{0.5}{0.125}=4=\beta$ (indeed $k=\beta$ for Gamma-Poisson). $Z=\frac{n}{n+k}=\frac{3}{3+4}=\frac{3}{7}$; $\bar X=\frac{3}{3}=1$. $P_c=Z\bar X+(1-Z)\mu=\frac{3}{7}(1)+\frac{4}{7}(0.5)=\frac{3}{7}+\frac{2}{7}=\frac{5}{7}\approx 0.7143$. Identical to the posterior mean — exact credibility holds.

Normal-Normal posterior mean is a precision-weighted average: $Z=\frac{n\tau^2}{n\tau^2+\sigma^2}=\frac{4(25)}{4(25)+100}=\frac{100}{200}=0.5$. Posterior mean $=Z\bar X+(1-Z)\mu_0=0.5(60)+0.5(50)=30+25=55$. Equivalently $k=\frac{\sigma^2}{\tau^2}=\frac{100}{25}=4=\text{EPV}/\text{VHM}$, giving $Z=\frac{n}{n+k}=\frac{4}{8}=0.5$ — the Bühlmann $Z$ matches, so Bühlmann equals Bayes here too.

$\mu(\theta)=\theta$ and $v(\theta)=\text{Var}(X\mid\theta)=\sigma^2$ (constant), so $\text{EPV}=E[\sigma^2]=\sigma^2$. $\text{VHM}=\text{Var}[\mu(\theta)]=\text{Var}[\theta]=\tau^2$. Therefore $k=\frac{\text{EPV}}{\text{VHM}}=\frac{\sigma^2}{\tau^2}$ 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.

**Bayesian:** $E[\mu(\theta)\mid \text{data}]$ — the full posterior mean; can be nonlinear in the data; requires the entire model (likelihood + prior). **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. **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.

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 $k=\frac{1.5}{0.75}=2$. $Z=\frac{5}{5+2}=\frac{5}{7}\approx 0.7143$. Bühlmann: $P_c=\frac{5}{7}(2.4)+\frac{2}{7}(1.5)=\frac{12+3}{7}=\frac{15}{7}\approx 2.143$. Bayes: 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.

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. The 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.

Means: $\bar X_1=\frac{15}{3}=5$, $\bar X_2=\frac{15}{3}=5$; grand mean $\hat\mu=5$. Within-risk variances: each set $\{4,5,6\}$ has $s^2=\frac{1+0+1}{2}=1$, so $\widehat{\text{EPV}}=\frac{1+1}{2}=1$. Between-risk: $\frac{1}{r-1}\sum(\bar X_i-\bar X)^2=\frac{(5-5)^2+(5-5)^2}{1}=0$. $\widehat{\text{VHM}}=0-\frac{\widehat{\text{EPV}}}{n}=0-\frac{1}{3}<0\Rightarrow$ set to $0$. Then $\hat Z=0$ and both risks are charged $\hat\mu=5$ — identical means give no evidence of heterogeneity.

Risk 1 total exposure $m_1=10+20=30$; weighted mean $\bar X_1=\frac{10(4.0)+20(5.0)}{30}=\frac{40+100}{30}=\frac{140}{30}\approx 4.667$. Risk 2 total exposure $m_2=30+40=70$; weighted mean $\bar X_2=\frac{30(7.0)+40(6.5)}{70}=\frac{210+260}{70}=\frac{470}{70}\approx 6.714$. Grand (exposure-weighted) mean $\hat\mu=\frac{10(4.0)+20(5.0)+30(7.0)+40(6.5)}{100}=\frac{40+100+210+260}{100}=\frac{610}{100}=6.10$. The grand mean uses total exposure $m=30+70=100$ as the denominator, not a simple average of the two risk means.

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). 2. **Exposures vary across cells**? Use **Bühlmann-Straub**: $Z=\frac{m}{m+k}$, exposure-weighted $\bar X$. 3. **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 $-\widehat{\text{EPV}}/n$. 4. **Counts known to be Poisson but prior unknown**? Use **semiparametric**: $\widehat{\text{EPV}}=\bar X$, then VHM as in step 3.