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Pruning balances a tree's empirical loss $R(T)$ against the number of leaves. Increasing $\alpha$ favors a nested sequence of smaller subtrees, obtained by removing splits with the smallest effective cost per removed leaf. Choose complexity using cross-validation or a validation set, and assess the selected tree on a separate test set. The loss can be deviance or misclassification loss; the chosen convention must be specified.

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]

The misclassification rate is the expected zero-one loss of a classification in statistical learning rule. Its empirical version is the fraction of incorrectly labeled observations. Evaluation on the fitting sample gives a training error; evaluation on an independent test set estimates prediction error. For a majority-class leaf of a classification tree, the training error count is $n_t-{\max }_k{n}_{tk}$, unlike classification-tree deviance, which also uses the fitted class probabilities.

Overfitting occurs when further adaptation improves training error while degrading performance on new data, for example by fitting incidental disturbances instead of a stable regression function. A gap between training error and test error alone does not prove that a simpler rule would generalize better. Compare candidates using cross-validation or a validation set; reserve the test set for assessment. Regularization and early stopping can limit adaptation. ^[ref]

Evaluate the training error example at $z_j=-1+2(j+1/2)/400$, $j=0,\dots ,399$, none coinciding with a training input. The mean squared discrepancy $400^{-1}{\sum }_j(\hat{f}(z_j)-f(z_j){)}^2$ is $0.373259$, $0.000255$, and $0.006319$ for degrees 1, 2, and 7. Thus degree 7 improves training error but worsens recovery of the target relative to degree 2. Degree 1 misses the target's curvature, illustrating underfitting.

These are errors against a known noiseless target, not test errors against fresh noisy responses. In a separate noiseless control, degrees 2 and 7 both recover this target exactly; their numerical held-out errors are approximately $3.77\times 10^{-31}$ and $3.72\times 10^{-31}$. The fixed-disturbance simulation illustrates a mechanism, not a universal numerical benchmark or a claim that higher polynomial degree always harms generalization.

The formula makes species the categorical response and the dot denotes all four other columns as candidate predictors. The classification tree is grown by recursive binary partitioning, not by fitting a linear regression to numerical species codes. At a node $t$, count observations of each class, $n_{tk}$, and set ${\hat{p}}_{tk}=n_{tk}/n_t$. These are the multinomial maximum-likelihood probabilities. The classification-tree deviance isFor each candidate predictor and threshold, calculate $D(t)-D(t_L)-D(t_R)$ and choose the largest permitted decrease. Continue recursively until a node is pure or the stopping controls prevent another split, for example because it has too few observations or insufficient deviance to warrant further growth. Only petal length, petal width and sepal length appear in the selected splits; sepal width was available but did not win a split. This greedy construction need not find a globally optimal tree. The default deviance criterion and recursive partitioning are documented in the tree package manual.

Each node number encodes its position: node $j$ has left and right children $2j$ and $2j+1$. The reported sample size is the number reaching the node; its deviance is calculated from that node's class counts. The fitted class is the class with largest fitted probability, with a convention resolving ties. The root's three counts are equal, so its displayed class is only a tie choice. An asterisk marks a terminal node, where prediction stops.

The resulting rule is displayed below. A left edge satisfies the stated strict inequality and a right edge its complement. None of the recorded training values lies exactly at these midpoint thresholds; the diagram adopts the right branch at equality for prediction.

Its six terminal regions have the following derived class counts, in the order setosa, versicolor, virginica:The root split identifies all 50 setosa observations. Among the remaining specimens, the petal-width split is the principal separation of versicolor from virginica; further petal-length and sepal-length splits refine it. The leaf errors add to $4/150\simeq 2.67\%$. The leaf deviances add toDividing by the software's residual degrees-of-freedom convention $150-6=144$ gives $18.0489/144\simeq 0.1253$, as reported. The root deviance is $-2(150)\log (1/3)=300\log 3\simeq 329.6$. The deviance is a likelihood measure, whereas the error rate counts wrong majority-class predictions; these are different fit summaries.

Both quantities are in-sample, so the low error does not establish prediction accuracy on new specimens. Cost-complexity tree pruning, with tree size chosen by cross-validation, can remove weak splits and reduce overfitting. An independent test set is needed for a final estimate of the selected tree's prediction error.

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.

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]

Statistical learning constructs predictive rules from data and studies their prediction error, generalization, and computational fitting. In supervised learning, training data contain inputs $X$ and a response variable $Y$: regression analysis predicts a numerical response, while classification in statistical learning predicts a label. Unsupervised learning instead seeks structure without supplied responses, as in cluster analysis and principal component analysis. These tasks connect learning to probability and mathematical optimization. ^[ref]

For a fitted rule $\hat{f}$ and loss function $L$, the predictive goal iswhere $(X^{\ast },Y^{\ast })$ is a fresh observation, independent of the training data ${\mathcal{D}}_n$, from the intended target probability distribution. This prediction error fixes the fitted rule and averages over future observations; averaging again over training data gives the learning procedure's average performance. Its practical beauty is that a rule compresses past observations into useful future predictions. Its central difficulty is that excellent training error need not imply good generalization: cross-validation supports model selection, and a reserved test set assesses the resulting choice.

A test set is held out from both parameter fitting and model selection and is used to estimate the final selected procedure's prediction error. A validation set influences the choice; a test set assesses that choice after it is fixed. The distinction concerns how observations are used, not their file names.

For a fixed rule and an independent test set of size $m$ from the target probability distribution,This test error is an estimate, with sampling uncertainty; it is not a guarantee for each future prediction. Dataset shift changes what performance the estimate represents.

In the overfitting example, all degree choices are fixed in advance, cross-validation uses only training data, and the selected degree is locked before evaluating 400 separate points. Displaying the other curves afterward explains their behavior; it does not authorize selecting by those scores. Repeatedly inspecting a test set and changing the model makes it part of model selection, introducing data leakage and potentially optimistic assessment. ^[ref]

Underfitting occurs when a prediction rule fails to capture useful structure in the target relation. A straight-line polynomial regression fitted to a genuinely quadratic noiseless target illustrates a representational limitation. Poor prediction alone does not identify the cause: compare candidate rules using cross-validation or a validation set, while keeping the final test set reserved for assessment.