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The generalized cross-validation curve reaches its minimum at $\lambda =5$ among the supplied grid values. This balances improved conditioning against shrinkage bias using an estimate of prediction error, rather than choosing the penalty with the smallest training residual sum of squares. Nearby values have similar errors, so the plot does not establish a highly precise optimal penalty.

Reading the PDF table at $\lambda =5$ gives the fitted regression intercept and slopes:These table entries are necessary because the TeX stores the table only inside a figure. The ridge regression slopes are shrunk relative to the zero-penalty fit; their small nonzero values do not represent variable exclusion.

There is a minor source inconsistency: the prose says the predictors are centered, but the table's intercept varies with $\lambda$. With exactly centered predictor columns and an unpenalized intercept it would remain $\overline{Y}$. The numbers above faithfully report the printed table, while part (a) gives the centered formula and the general uncentered conversion. The table therefore reflects a different or incompletely described preprocessing convention.

Let $Q={\sum }_i{x}_i^2>0$. Under the usual nondegenerate regression assumption $n>2$, maximizing the normal likelihood is equivalent to minimizing the residual sum of squares. The normal equations and ${\sum }_i{x}_i=0$ give the maximum-likelihood estimatorsWrite $A=\overline{\epsilon }$ and $B={\sum }_i{x}_i{\epsilon }_i/Q$. Then $\hat{\alpha }=\alpha +A$, $\hat{\beta }=\beta +B$, and the residual is $r_i={\epsilon }_i-A-x_{i}B$. Independence of the errors givesFor each $i$,Thus the three requested variables are pairwise uncorrelated.

They are linear functions of the jointly normal error vector. More strongly, the whole vector $(A,B,r_1,\dots ,r_n)$ has a multivariate normal distribution, and its block covariances show that $A$, $B$ and the residual vector are independent, by uncorrelated jointly normal variables are independent. Independence is preserved when applying a measurable function to the residual block. HenceThe residual projection is $I-11^{\mathsf{T}}/n-x{x}^{\mathsf{T}}/Q$, of rank $n-2$, so additionally ${\sum }_i{r}_i^2/{\sigma }^2\sim {\chi }_{n-2}^2$. This is the orthogonal projection of a Gaussian vector interpretation of the same result. The printed design conditions alone allow $n=2$; then the fit is exact, the positive-variance likelihood has no finite maximizer, and the residual test in part (b) is undefined. The standard testing question therefore needs $n>2$.

Use a common intercept and laboratory-specific slopes in the normal linear model. With $t_i$ denoting days and $I_{Bi},I_{Ci}$ the indicator variables for laboratories B and C, the design matrix has row $x_i^T=(1,t_i,t_i{I}_{Bi},t_i{I}_{Ci})$, andThe interaction terms change slopes, without introducing laboratory-specific intercepts. Thus the estimated daily growth rates arein millimetres per day, with fitted common intercept $0.6464$ mm. The residual degrees of freedom are $30-4=26$. The usual unbiased error variance estimate is the square of the residual standard error:The small difference from $5921.7/26\approx 227.758$ arises from rounding the standard error to two decimal places; the latter uses the more precise residual sum of squares supplied later. A maximum-likelihood estimate would instead divide that residual sum of squares by $30$.

Write the one-column design as $x\in {\mathbb{R}}^n\setminus \{0\}$ and $S=x^{T}x$. Maximizing the normal linear model likelihood first gives $\hat{\beta }=x^{T}Y/S$. Its minimized residual sum of squares is $R=\Vert Y-x\hat{\beta }{\Vert }^2$. For variance $v>0$, the remaining log likelihood is $-\frac{n}{2}\log v-R/(2v)$ plus a constant, whose maximum is atHere $R>0$ almost surely because $n>1$ and the true variance is positive. In the exceptional data case $R=0$, the likelihood is unbounded as $v\downarrow 0$, so there is no positive-variance maximum.

Let $P=x{x}^T/S$, the orthogonal projection matrix onto the design column. Then $\hat{\beta }-\beta =x^T\epsilon /S$ and $Y-x\hat{\beta }=(I-P)\epsilon$. These are orthogonal linear projections of an isotropic multivariate normal distribution, hence independent. An orthonormal coordinate system adapted to $x$ makes the residual squared length a sum of $n-1$ independent squared standard normals. Thus the joint distribution of least-squares and variance estimators is the product lawIn particular the maximum likelihood variance estimator is biased: $\mathbb{E}{\hat{\sigma }}^2=(n-1){\sigma }^2/n$; the unbiased residual estimator uses denominator $n-1$.

The command performs a nested-model F-test for $H_0:{\beta }_1=0$ against $H_1:{\beta }_1\ne 0$, retaining the intercept and $x_2$ coefficient in both models. Write $RS{S}_2$ and $RS{S}_3$ for the residual sum of squares in the reduced and full fits. Under $H_0$ and the independent normal-error assumptions,By Cochran's theorem, the numerator and denominator are independent scaled chi-squares with one and $n-3$ degrees of freedom. Equivalently, for this single-coefficient restriction, $F=t_{{\beta }_1}^2$, so its p-value equals that of the two-sided coefficient Student's t-test in model3.

Using the printed $t$ value gives $F=0.800^2=0.640$ and $p\approx 0.43$. The printed estimate and standard error instead giveThis discrepancy is present in the PDF and is too large to be explained by rounding at the shown precision; the underlying ANOVA output is omitted, so an exact unique statistic cannot be recovered. The univariate correlation and $t$ value imply $n\approx 2+2.575^2(1-0.2517297^2)/0.2517297^2\approx 100.006$, consistent with $n=100$ after rounding. With denominator degrees of freedom 97, the two versions give p-values approximately $0.4257$ and $0.4298$, respectively. Either way the partial effect is not significant at 5%.

Apart from terms independent of $\beta$, the Gaussian log-likelihood isDifferentiating with respect to $\beta$ givesThe unique zero, and hence the maximum-likelihood estimate, isThe same value minimizes the residual sum of squares ${\sum }_i(Y_i-\beta x_i{)}^2$, since maximizing the Gaussian likelihood for fixed ${\sigma }^2$ is equivalent to minimizing this sum. It is therefore also the least-squares estimator.

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 $154.30$.

With repeated responses at an identical predictor configuration, pure error is their within-configuration variability. Its residual sum of squares can estimate measurement variance independently of a candidate regression mean function. Without replication, a fitted residual can combine pure error and lack of fit.

For a correctly specified full-rank fixed-design Gaussian linear regression with $p<n$, dividing the residual sum of squares by the residual degrees of freedom $n-p$ gives an unbiased estimator of the noise variance. The projection removes the whole deterministic mean, so the numerator's expectation is $(n-p){\sigma }^2$. A misspecified subset model need not have this property.

The residual sum of squares measures how far the observed responses lie from their fitted values. If $r_i=Y_i-{\hat{Y}}_i$ is the $i$th regression residual, thenIt is the squared Euclidean norm of the regression residual vector. Squaring prevents positive and negative deviations from cancelling and gives large deviations more weight. In the constructed five-point example in linear regression, the regression residuals are $(1/5,1/5,-4/5,1/5,1/5)$, so the residual sum of squares is $4/5$.

Ordinary least squares minimizes the residual sum of squares over the chosen linear regression family; ridge regression minimizes the residual sum of squares plus a penalty. Consequently the ridge regression fit can have a larger residual sum of squares than the ordinary least squares fit on exactly the same design matrix. For the one-predictor construction in ridge regression, the residual sum of squares rises from $4/5$ to $33/10$ when the penalty changes from $0$ to $10$; that increase is the cost of shrinking the fitted slope.

Dividing the residual sum of squares by $n$ gives the training error for squared-error loss, not an estimate based on fresh responses. Its units are squared response units, and it adds across observations, so raw totals need not be comparable across different response scales or sample sizes. Comparing fits to the same responses is useful, but choosing a ridge regression penalty by minimum training residual sum of squares favors the unpenalized fit. Assess prediction error using a validation set or cross-validation instead.^[ref]

Ridge regression fits a linear regression model while penalizing the size of its slope regression coefficients. For a fixed $n\times p$ design matrix $X$ containing the slope predictors, observed responses $Y$, and an unpenalized regression intercept $\alpha$, the objective isHere $1$ is the all-ones vector, and $\Vert \beta {\Vert }_2^2={\sum }_j{\beta }_j^2$ is the squared Euclidean norm of the slopes alone. The first term is the residual sum of squares; the second implements regularization. No random-error assumption is needed to define this fit. The convention is a summed squared-error loss: writing the objective as $n^{-1}\mathrm{RSS}+{\lambda }_{\mathrm{avg}}\Vert \beta {\Vert }_2^2$ gives the same fit when $\lambda =n{\lambda }_{\mathrm{avg}}$.^[ref]

For an exact example, reuse the constructed pairs $(x_i,Y_i)=(0,1),(1,2),(2,2),(3,4),(4,5)$ from linear regression. Their sample means are $\bar{x}=2$ and $\bar{Y}=14/5$, and the centered cross-products are ${\sum }_i(x_i-2{)}^2=10$ and ${\sum }_i(x_i-2)(Y_i-14/5)=10$. ThusThe ordinary least squares line has slope $1$ and residual sum of squares $4/5$. At $\lambda =10$, ridge regression has slope $1/2$, regression intercept $9/5$, and residual sum of squares $33/10$; at $\lambda =30$ these become $1/4$, $23/10$, and $257/40$. Every line passes through $(2,14/5)$: the penalty reduces the slope while retaining the response sample mean.

The computational benefit is visible after predictor centering. Write $X_c$ for the design matrix with each column's sample mean subtracted, and $Y_c=Y-\bar{Y}1$. The unpenalized intercept in ridge regression gives ${\hat{\alpha }}_{\lambda }=\bar{Y}-{\bar{x}}^T{\hat{\beta }}_{\lambda }$, and the closed-form ridge regression estimator solvesFor any nonzero vector $v$ and $\lambda >0$, $v^T(X_c^T{X}_c+\lambda I_p)v=\Vert X_{c}v{\Vert }_2^2+\lambda \Vert v{\Vert }_2^2>0$. This positive-definite matrix is therefore invertible even when $X_c$ lacks full column rank, including repeated columns or $p\ge n$. The unique penalized slopes resolve the ambiguity of rank-deficient ordinary least squares; they do not supply evidence about slope directions absent from the data. This makes ridge regression useful for multicollinearity.^[ref]

Predictor centering and scaling serve different purposes. Centering handles the free regression intercept. Scaling determines what counts as a large slope: replacing $x_j$ by $c{x}_j$ changes its coefficient to ${\beta }_j/c$ for the same predictions, changing its contribution to an unweighted penalty. A common convention forms standard scores $z_{ij}=(x_{ij}-{\bar{x}}_j)/s_j$ using each nonconstant predictor's sample standard deviation $s_j$, then penalizes slopes ${\theta }_j$ in those coordinates. Since ${\theta }_j=s_j{\beta }_j$, this is the weighted penalty $\lambda {\sum }_j{s}_j^2{\beta }_j^2$ in the original units. Constant columns have $s_j=0$ and must be removed before this scaling. Scaling is a modelling choice; equal penalties in original units and in standardized units define different fits.^[ref]

The singular value decomposition explains the shrinkage. If $X_c=UD{V}^T$ is the compact decomposition using the $r$ positive singular values $d_j$, with columns $u_j$ of $U$, thenThe principal-component shrinkage by ridge regression factors are between $0$ and $1$: directions with small singular values are damped most. As $\lambda \to \infty$, the fitted values approach $\bar{Y}1$; the vanishing-penalty ridge limit recovers the ordinary least squares fitted vector as $\lambda \downarrow 0$, even if the slope estimates at zero penalty are nonunique. The covariance and bias of a ridge regression estimator describes the resulting bias-variance tradeoff. A penalty can reduce variance of an estimator while introducing bias of an estimator; the unbounded directional risk of fixed ridge shrinkage rules out a uniform improvement for a fixed positive penalty over all possible signals.^[ref]

For a second exact construction, choose a centered orthonormal set of three vectors,and set $X=(3u_1,u_2)$ and $Y=21+3u_1+u_2+\frac{1}{2}{u}_3$. The squared singular values are $9$ and $1$. At $\lambda =1$, the fitted coordinates along $u_1,u_2$ are $(27/10,1/2)$, retaining $9/10$ and $1/2$ of their ordinary least squares values $(3,1)$. At $\lambda =9$ they become $(3/2,1/10)$, retaining $1/2$ and $1/10$. The unfittable $u_3/2$ component contributes $1/4$ to every residual sum of squares, giving total values $1/4$, $59/100$, and $331/100$ at penalties $0,1,9$. These are constructed fits, not measured prediction results.

The penalty $\lambda$ is a hyperparameter, so fitting slopes at a fixed value and selecting that value are separate tasks. Use a validation set or K-fold cross-validation to compare candidate penalties by held-out squared-error loss. In each split, compute predictor sample means and scales from that split's training data, apply the same transformation to its validation set, and refit the slopes; using held-out information in preprocessing introduces data leakage. Minimum training residual sum of squares cannot select useful shrinkage: ordinary least squares already attains the smallest possible value.^[ref] After model selection, refit on all training data with the selected penalty. A separate test set or nested cross-validation assesses the entire selection procedure; a low selected validation score alone is no performance guarantee.^[ref]