Exam MAS-II — Generalized Linear Models Practice Flashcards

**Answer: (C).** With the offset, the linear predictor is $\eta = \ln E + \beta_0 + \beta_1 x = \ln 3 + (-2.3) + 0.5(1)$. $\ln 3 \approx 1.098612$, so $\eta \approx 1.098612 - 2.3 + 0.5 = -0.701388$. Expected claims $= e^{\eta} = e^{-0.701388} \approx 0.4959 \approx 0.496$. Equivalently $\mu = E\,e^{\beta_0}e^{\beta_1} = 3\,e^{-2.3}e^{0.5} = 3(0.100259)(1.648721) \approx 3(0.165299) = 0.4959$. (Dropping the offset gives the per-exposure rate $e^{-1.8} \approx 0.165$ — distractor A; using $E=1$ but keeping the $0.5$ urban term also lands near there.)

**Answer: (B).** Linear predictor: $\eta = 6.0 + 0.25(2) - 0.10(1) = 6.0 + 0.5 - 0.1 = 6.4$. Predicted mean: $\mu = e^{6.4} \approx \$601.85$. (Forgetting the $x_2$ term, $\eta = 6.5$, gives $e^{6.5} \approx \$665.14$ — distractor C; using $x_1 = 1$ instead of $2$, $\eta = 6.15$, gives $\approx \$469$.)

**Answer: (A).** A log link produces a multiplicative model, so the relativities multiply: $\mu = e^{\beta_0}\,e^{\beta_E}\,e^{\beta_3}\,e^{\beta_D} = 400(1.15)(0.85)(0.70)$. $400 \times 1.15 = 460$; $460 \times 0.85 = 391$; $391 \times 0.70 = 273.70$. Predicted pure premium $= \$273.70$. (Adding the relativities additively, $400(1 + 0.15 - 0.15 - 0.30) = \$280$, or stopping after two factors at $\$391$, are the common errors behind distractors D and the near-A values.)

**Answer: (C).** Under a log link, a one-unit change in the covariate multiplies the mean by $e^{\hat\beta}$. For an indicator turning from $0$ to $1$: Relativity $= e^{0.336} \approx 1.3993$. Percent change $= (e^{0.336} - 1)\times 100\% \approx 39.9\%$ increase. (Reading $\hat\beta$ itself as the percent change gives relativity $1.336$ — distractor B; rounding $e^{0.336}$ to a tidy $1.40$ is distractor D; $e^{-0.336} \approx 0.715$ would be the relativity if the sign were flipped — distractor A.)

**Answer: (D).** With the offset, $\mu = E\,e^{-1.6 + 0.3x}$. Holding the covariate $x$ fixed, the non-exposure factor $e^{-1.6+0.3x}$ is identical for both risks, so $\dfrac{\mu_B}{\mu_A} = \dfrac{E_B}{E_A} = \dfrac{4}{1} = 4.00$. An offset enters with a fixed coefficient of $1$, so expected counts scale **linearly** with exposure. (Treating exposure as an ordinary covariate with the $0.3$ slope, $e^{0.3(4-1)} \approx 2.46$, or mixing in $e^{-1.6}$, produces the other distractors.)

**Answer: (B).** Linear predictor: $\eta = -1.2 + 0.9(1) = -0.3$. Invert the logit: $\mu = \dfrac{1}{1 + e^{-\eta}} = \dfrac{1}{1 + e^{0.3}} = \dfrac{1}{1 + 1.349859} = \dfrac{1}{2.349859} \approx 0.4256$. So the predicted lapse probability is about $0.426$. (Using $\mu = \dfrac{1}{1+e^{-0.3}}$ — i.e. the wrong sign on $\eta$ — gives $0.574$, distractor D; treating $\eta$ as the probability directly is another trap.)

**Answer: (D).** In logistic regression $e^{\hat\beta}$ is the odds ratio for a one-unit change in the covariate: Odds ratio $= e^{0.7} \approx 2.0138$. So having a prior claim multiplies the **odds** of the event by about $2.01$ — the odds roughly double. The odds ratio acts on the odds, not directly on the probability. (Reading $\hat\beta$ as the odds ratio gives $0.7$ — distractor B; $e^{-0.7} \approx 0.497$ flips the sign — distractor A.)

**Answer: (C).** In the exponential dispersion family $\text{Var}(Y) = \phi\,V(\mu)$ where $V(\mu) = b''(\theta)$. The variance functions are: normal $V=1$, Poisson $V=\mu$, gamma $V=\mu^{2}$, inverse Gaussian $V=\mu^{3}$, binomial proportion $V=\mu(1-\mu)$. For the **gamma**, $V(\mu) = \mu^{2}$, so $\text{Var}(Y) = \phi\mu^{2}$ — the standard deviation is proportional to the mean (constant coefficient of variation), which is why the gamma suits right-skewed claim severities. Distractor B is the Poisson, D the inverse Gaussian, E the binomial.