Exam MAS-II · practice
Exam MAS-II — Linear Mixed Models Practice Flashcards
Thirty exam-realistic multiple-choice problems on CAS Exam MAS-II linear mixed models — fixed vs random effects, random-intercept and random-slope structures, variance components, the intraclass correlation, ML vs REML estimation, and the BLUP/shrinkage estimator and its exact equivalence to Buhlmann credibility — each with a fully worked solution.
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Bayesian Analysis Bayesian Analysis Practice Credibility Credibility Practice Generalized Linear Models Generalized Linear Models Practice
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- Fixed vs random effectsA factory runs an experiment comparing exactly two machines it owns, $M_1$ and $M_2$, with no intent to generalize beyond them. In a separate study, $40$ policyholders are randomly sampled from a large book and the analyst wants to generalize to the whole book. How should the machine factor and the policyholder factor be modeled? (A) Both as fixed effects (B) Both as random effects (C) Machine fixed; policyholder random (D) Machine random; policyholder fixed (E) Machine fixed; policyholder fixed only if $n>30$**Answer: (C).** The rule is whether the observed levels are the *only* levels of interest or a *sample* from a larger population. The two machines are the entire population of interest — inference is conditional on exactly those two levels, and we want a separate parameter for each. That is a **fixed** effect. The $40$ policyholders are a random sample drawn from a large book; we want to generalize to the population of policyholders and borrow strength across them, estimating the *variance* $\sigma_u^{2}$ rather than $40$ individual parameters. That is a **random** effect. Choice (E) is the common error of using a sample-size threshold to decide fixed vs random — the decision depends on the inferential target, not on $n$.
- Random intercept & slopeIn the model $y_{ij}=\beta_0+\beta_1 x_{ij}+u_i+\epsilon_{ij}$ with $u_i\sim N(0,\sigma_u^{2})$, which statement is correct? (A) Each group has its own slope and its own intercept (B) Each group has the same slope $\beta_1$ but a group-specific intercept $\beta_0+u_i$ (C) Each group has the same intercept but a group-specific slope (D) The fitted lines for different groups all cross at $x=0$ (E) $u_i$ is a fixed parameter estimated separately for each group**Answer: (B).** This is a **random-intercept** model. The covariate enters only through the common fixed slope $\beta_1$, while the random effect $u_i$ shifts the whole line up or down. Group $i$'s line is $(\beta_0+u_i)+\beta_1 x$ — a family of **parallel** lines, one per group. Choice (A)/(C) describe a random-slope model, which would need a $u_{1i}x_{ij}$ term. Choice (E) is wrong because $u_i$ is a *random variable* we predict (its variance is estimated), not a fixed parameter. The lines are parallel, so they never cross (D is false).
- ICC & credibility linkA random-intercept claims model is fit and reports a between-group (policyholder) variance of $\hat\sigma_u^{2}=24$ and a within-group residual variance of $\hat\sigma_e^{2}=56$. Calculate the intraclass correlation $\rho$. (A) $0.30$ (B) $0.43$ (C) $0.57$ (D) $0.70$ (E) $2.33$**Answer: (A).** The ICC is the between-group share of total variance: $\rho=\dfrac{\sigma_u^{2}}{\sigma_u^{2}+\sigma_e^{2}}=\dfrac{24}{24+56}=\dfrac{24}{80}=0.30$. So $30\%$ of the total variation is between policyholders and any two observations from the same policyholder correlate $0.30$. Choice (D) $0.70$ is the *within*-group share $\sigma_e^{2}/(\sigma_u^{2}+\sigma_e^{2})$ (computing the complement). Choice (E) $2.33$ is $k=\sigma_e^{2}/\sigma_u^{2}$, the credibility constant, not the ICC.
- Variance componentsA study of a random-intercept model reports total variance $\sigma_u^{2}+\sigma_e^{2}=80$ and an intraclass correlation $\rho=0.40$. Recover the between-group variance $\sigma_u^{2}$ and the within-group variance $\sigma_e^{2}$. (A) $\sigma_u^{2}=20,\ \sigma_e^{2}=60$ (B) $\sigma_u^{2}=32,\ \sigma_e^{2}=48$ (C) $\sigma_u^{2}=40,\ \sigma_e^{2}=40$ (D) $\sigma_u^{2}=48,\ \sigma_e^{2}=32$ (E) $\sigma_u^{2}=60,\ \sigma_e^{2}=20$**Answer: (B).** From $\rho=\dfrac{\sigma_u^{2}}{\sigma_u^{2}+\sigma_e^{2}}$, the between-group variance is $\sigma_u^{2}=\rho\,(\sigma_u^{2}+\sigma_e^{2})=0.40(80)=32$. Then $\sigma_e^{2}=80-32=48$. Check: $\dfrac{32}{32+48}=\dfrac{32}{80}=0.40$. ✓ Choice (D) swaps the two components (it applies $\rho$ to get $\sigma_e^{2}$ instead of $\sigma_u^{2}$).
- Variance componentsFor the random-intercept model $y_{ij}=\beta_0+u_i+\epsilon_{ij}$ with $u_i\sim N(0,\sigma_u^{2})$ and $\epsilon_{ij}\sim N(0,\sigma_e^{2})$, what is the covariance between two distinct observations $y_{ij}$ and $y_{ik}$ ($j\neq k$) in the same group? (A) $0$ (B) $\sigma_e^{2}$ (C) $\sigma_u^{2}$ (D) $\sigma_u^{2}+\sigma_e^{2}$ (E) $\sigma_u^{2}\sigma_e^{2}$**Answer: (C).** Two observations in the same group share the same random intercept $u_i$ but have independent residuals. Writing $y_{ij}=\beta_0+u_i+\epsilon_{ij}$ and $y_{ik}=\beta_0+u_i+\epsilon_{ik}$: $\text{Cov}(y_{ij},y_{ik})=\text{Var}(u_i)+\text{Cov}(\epsilon_{ij},\epsilon_{ik})=\sigma_u^{2}+0=\sigma_u^{2}$. Choice (D) $\sigma_u^{2}+\sigma_e^{2}$ is the *variance* of a single observation $\text{Var}(y_{ij})$, which mistakenly adds the residual variance. Choice (A) $0$ is the covariance across *different* groups, where the $u$'s are independent.
- ICC & credibility linkIn a random-intercept model the within-group (residual) variance is $\sigma_e^{2}=270$ and the between-group variance is $\sigma_u^{2}=30$. Calculate the intraclass correlation, then state the credibility constant $k=\sigma_e^{2}/\sigma_u^{2}$. (A) $\rho=0.10,\ k=9$ (B) $\rho=0.10,\ k=0.111$ (C) $\rho=0.90,\ k=9$ (D) $\rho=0.11,\ k=9$ (E) $\rho=0.90,\ k=0.111$**Answer: (A).** ICC: $\rho=\dfrac{\sigma_u^{2}}{\sigma_u^{2}+\sigma_e^{2}}=\dfrac{30}{30+270}=\dfrac{30}{300}=0.10$. Credibility constant: $k=\dfrac{\sigma_e^{2}}{\sigma_u^{2}}=\dfrac{270}{30}=9$. Note the consistency check $k=\dfrac{1-\rho}{\rho}=\dfrac{0.90}{0.10}=9$. ✓ Choice (C) reports $\rho=0.90$, the within-group share. Choice (B) inverts $k$ as $\sigma_u^{2}/\sigma_e^{2}$.
- BLUP & shrinkageA random-intercept model has $\hat\beta_0=300$, $\sigma_u^{2}=400$, and $\sigma_e^{2}=1600$. Group A has $n_A=4$ observations with sample mean $\bar y_A=420$. Calculate the BLUP-predicted mean for group A. (A) $300$ (B) $330$ (C) $360$ (D) $384$ (E) $420$**Answer: (C).** The BLUP shrinkage / credibility factor is $Z=\dfrac{n_A\sigma_u^{2}}{n_A\sigma_u^{2}+\sigma_e^{2}}=\dfrac{4(400)}{4(400)+1600}=\dfrac{1600}{1600+1600}=\dfrac{1600}{3200}=0.50$. The predicted random intercept is $\hat u_A=Z(\bar y_A-\hat\beta_0)=0.50(420-300)=60$, so the predicted group mean is $\hat\beta_0+\hat u_A=300+60=360$. Equivalently $Z\bar y_A+(1-Z)\hat\beta_0=0.50(420)+0.50(300)=360$. The raw $420$ is pulled halfway to the grand mean. Choice (E) is the unshrunk $\bar y_A$ (no pooling); choice (A) is the grand mean (complete pooling).
- BLUP & shrinkageA random-intercept model has $\hat\beta_0=1000$, $\sigma_u^{2}=200$, and $\sigma_e^{2}=5400$. Group B has $n_B=3$ observations averaging $\bar y_B=1300$. Calculate the BLUP-predicted mean for group B. (A) $1000$ (B) $1030$ (C) $1090$ (D) $1150$ (E) $1300$**Answer: (B).** First the credibility constant: $k=\dfrac{\sigma_e^{2}}{\sigma_u^{2}}=\dfrac{5400}{200}=27$. Shrinkage factor: $Z=\dfrac{n_B}{n_B+k}=\dfrac{3}{3+27}=\dfrac{3}{30}=0.10$. Predicted random intercept: $\hat u_B=Z(\bar y_B-\hat\beta_0)=0.10(1300-1000)=30$. Predicted mean: $1000+30=1030$. With only $3$ observations and large $k$, the group earns just $10\%$ credibility, so its own $\bar y_B=1300$ is **heavily shrunk** toward the grand mean. Choice (E) ignores the shrinkage; choice (A) over-shrinks to the grand mean.