Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 218 / 5

Model1 is the simple linear regressionModel2 is a random-intercept linear mixed modelwith the random intercepts and errors mutually independent.

Repeated measurements from one person can share persistent unobserved spending tendencies, violating model1's independent-error assumption and shifting that person's baseline. Model2 represents this clustered data through a common random intercept $b_i$, which induces within-person covariance $\mathrm{Cov}(Y_{ij},Y_{ik}\mid X)={\sigma }^2$ for $j\ne k$.

The residual deviance of a Gaussian GLM is its residual sum of squares, $154.30$. Thus the empirical training error for squared-error loss isIf training error is defined as the unnormalized total loss, its value is $154.30$.

The first column of the model matrix is the all-ones intercept column. Hence $A_{11}={\sum }_{i,j}1=100$.

The second cross-product entry is $A_{12}={\sum }_{i,j}{X}_{ij}$. Regressing earned on the ten person indicators without an intercept makes $(g_x{)}_i$ the mean of that person's ten earned values. Therefore

Let $q=(\alpha ,\beta {)}^T$, letand recover the sufficient cross-products from the model1 normal equations by setting $c=A{\hat{q}}_0$ and $s=154.30+{\hat{q}}_0^{T}A{\hat{q}}_0$. Thenis the total squared residual about the fixed line. The squared sum of the ten residuals within each person, summed across people, isFor one person's ten observations, the marginal covariance matrix is ${\tau }^2{I}_{10}+{\sigma }^{2}11^T$. The matrix determinant lemma and Sherman–Morrison formula therefore give, up to an additive constant, twice the negative marginal log-likelihoodBecause the model was fitted with REML = FALSE, it minimizes this ordinary marginal maximum-likelihood objective. Thus $(V1,V2,V3,V4)$ belongs to the stated argmin over $alpha,\beta \in \mathbb{R}$ and $sigma,\tau >0$.