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Cross-validation estimates predictive performance by repeatedly fitting on one part of the data and evaluating on held-out observations. In K-fold cross-validation, divide training data into disjoint folds $I_1,\dots ,I_K$; $k(i)$ identifies observation $i$'s fold, and ${\hat{f}}^{(-k)}$ is fitted without fold $I_k$. The displayed average scores each observation using a rule that did not train on it. Weight by observation count when fold sizes differ. Leave-one-out cross-validation uses one observation per fold. ^[ref]

For the training error example, three folds contain indices $i$ with equal remainder modulo 3. Each fit uses eight observations. The squared-error loss averages for degrees 1, 2, and 7 are $0.724132$, $0.027192$, and $1.532790$ respectively. Select degree 2 and refit on all twelve training data observations. The folds reuse scarce data while challenging each candidate to predict unseen responses; they do not touch the final test set.

Any learned preprocessing, including variable selection, must also be fitted within each training fold to avoid data leakage. ^[ref] Repeated tuning can itself overfit the cross-validation score; nested cross-validation uses inner folds for model selection and outer folds for assessment of that entire procedure. ^[ref]

Folds use fewer training observations than the final refit, and their scores are dependent because training observations overlap. ^[ref] Splitting must match deployment: for time series forecasting, rolling-origin cross-validation trains on the past and evaluates later observations, preventing future information from entering a forecast. Ordinary random folds need not answer that forecasting question. ^[ref]

At fraction $0.8$, the Lasso regularization path lies between the vertical lines marked steps 4 and 5. In that interval the first five predictors have nonzero slopes: $X_1,X_2,X_4,X_5$ are positive and $X_3$ is negative. The sixth predictor remains at zero until the later part of the path, beyond this fraction. Therefore the chosen Lasso model includesThe regression intercept is retained. In particular the negative $X_3$ trace has already left zero at fraction $0.8$; a small slope is not the same as a zero slope. The selected fraction came from ten-fold K-fold cross-validation, and is not a numerical value of the ridge penalty in the preceding parts.

One suitable criterion among the three unpenalized normal linear models is the Akaike information criterion. With $d$ estimated mean coefficients and the variance also estimated, minimizing AIC is equivalent to minimizingThe values of $d$ are $2,2,3$ for model1, model2 and model3. This compares fit and complexity rather than choosing solely by individual coefficient significance. The Bayesian information criterion or a common-fold predictive comparison would also be reasonable, depending on the selection objective.

For the ridge regression tuning parameter, use K-fold cross-validation with squared prediction loss:Use the same folds for every candidate, estimate any centering or scaling within each training fold, choose a minimizing $\lambda$, and refit on all observations. This estimates mean squared prediction error while avoiding data leakage. The printed summaries do not supply all residual sums of squares or validation losses, so they do not determine which model or penalty wins.

Because models 1 and 3 are full likelihood models for the same response, one can compare their Akaike information criterion values; that favors model 1. One can also compare held-out count prediction by K-fold cross-validation, using a common loss such as Poisson deviance or negative log predictive density.

The AIC comparison cannot include model 2 because a Quasi-Poisson fit specifies only mean and variance and has no full likelihood. Cross-validation can compare model 2 with model 3 if all predictions are scored by the same proper out-of-sample loss.

Choose $({\lambda }_1,{\lambda }_2)$ on a grid by K-fold cross-validation, comparing the same held-out prediction loss and optionally applying the one-standard-error rule for a simpler model. A genuinely untouched test set can then estimate final prediction error.

Ordinary model-based intervals after selecting nonzero coefficients ignore selection and are generally invalid. Valid approaches include Debiased Lasso or a selective-inference procedure under its assumptions, sample splitting followed by an unpenalized refit and inference on the independent half, or a bootstrap that repeats both tuning and fitting and is interpreted with care near the nonsmooth zero threshold.

Choose $\lambda$ to minimize an estimate of out-of-sample mean squared prediction error, commonly K-fold cross-validation or a separate validation set. As $\lambda$ increases, the coefficient vector is shrunk toward zero. This generally increases bias of an estimator but decreases variance of an estimator; the minimizing value balances the two contributions in the bias-variance tradeoff. The independent test set in the question can assess the final choice, but repeatedly selecting $\lambda$ on that same set would cause data leakage.

The code computes Leave-one-out cross-validation. Ifthen observation $i$ is not a support vector. Removing it leaves the optimum unchanged, and the resulting classifier still classifies it correctly. A leave-one-out error can therefore occur only for an observation on or inside the margin, which provesThus the fraction of training observations on or inside the margin is an upper bound on leave-one-out error. One can refit only after deleting support vectors, reusing the full fit for every other observation. Alternatively, K-fold cross-validation needs only $K$ fits and is often preferable for larger data sets.

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]