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The leave-one-out residual identity for a linear smoother, obtained from the block matrix inverse or the Sherman–Morrison formula, isHence

Compute once the spectral decomposition $K=U\mathrm{diag}(d_1,\dots ,d_n)U^T$ in $O(n^3)$ operations and the vector $U^{T}Y$ in $O(n^2)$. For each ${\lambda }_{\ell }$, setThen computeBoth calculations take $O(n^2)$ operations per tuning parameter, after which the displayed leave-one-out formula costs $O(n)$. All $L$ scores therefore require $O(n^3+L{n}^2)$ operations.

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 $sigma,\tau >0$.