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In the Bühlmann model, one policy has a random risk parameter $\Theta$. Conditional on $\Theta$, its annual observations $X_i$ are independent and identically distributed random variables, with mean $m(\Theta )$ and variance $v(\Theta )$. Put $m_0=\mathbb{E}[m(\Theta )]$, let $v=\mathbb{E}[v(\Theta )]$ be the expected process variance, and let $a=\mathrm{Var}(m(\Theta ))$ be the variance of hypothetical means. Assume the required second moments are finite.

The Bühlmann credibility premium is the best affine estimate of $m(\Theta )$ under mean squared error. After choosing the intercept to make the estimate unbiased, write it as $m_0+{\sum }_{i=1}^n{b}_i(X_i-m_0)$. If $B={\sum }_i{b}_i$, the law of total variance and conditional independence give its errorFor fixed $B$, the Cauchy-Schwarz inequality minimizes the last sum at $b_i=B/n$. Differentiating $a(1-B{)}^2+v{B}^2/n$ gives $B=na/(na+v)$. Thus the credibility factor and credibility premium areThe ratio form assumes $a>0$; when $a=0$ and $v>0$, the conditional mean is constant and $Z=0$. Predicting $X_{n+1}$ rather than its conditional mean adds the constant $v$ to the error and leaves the optimal affine estimate unchanged.

Here the latent intensity has a Pareto distribution with shape three and lower bound one. For $r<3$, its moments are $\mathbb{E}[{\Theta }^r]=3{\int }_1^{\infty }{\theta }^{r-4}d\theta =3/(3-r)$. ConsequentlyThe equality for $v$ uses the Poisson distribution conditional variance $v(\Theta )=\Theta$. There are two years of experience, so $Z=1/2$ and the observed sample mean is $k/2$. The Poisson credibility with a shape-three Pareto intensity is thereforeIn particular it equals $2$ when $k=5$.

For the exact Bayesian credibility estimate, the total count conditional on $\Theta =\theta$ has Poisson distribution of mean $2\theta$. Multiplication of its likelihood by the prior distribution density gives a Bayesian posterior density proportional to ${\theta }^{k-4}{e}^{-2\theta }$ on $\theta >1$. At $k=5$, its normalizing integral and first-moment integral areHence the next-year predictive expected value isThe Bühlmann credibility premium optimizes over affine estimates, whereas the Bayesian posterior mean optimizes over all square-integrable estimates. This explains why the two credibility estimates need not agree for this nonconjugate prior.

A statistical functional is a map $T$ defined on a class of probability distributions. Its associated functional statistic is $T(F_n)$, where $F_n=n^{-1}{\sum }_i{\delta }_{X_i}$ is the empirical distribution and each ${\delta }_{X_i}$ is a Dirac measure. For example, $T(F)=\int xdF(x)$ yields the sample mean. The influence function records the derivative in the direction of point contamination:It is a local, infinitesimal robustness diagnostic, not a description of arbitrary large contamination.

Suppose $T$ is differentiable in a sense strong enough to linearize it at the empirical distribution, and that its derivative in a signed direction $G-F$ is $\int \mathrm{IF}(x;T,F)d(G-F)(x)$. Averaging the directions ${\delta }_z-F$ over $z\sim F$ gives the zero direction, so linearity implies $\int \mathrm{IF}dF=0$. The resulting asymptotic linear representation isIf $\int {\mathrm{IF}}^{2}dF<\infty$, the central limit theorem and the negligible remainder giveThus the asymptotic variance is $n^{-1}\int {\mathrm{IF}}^{2}dF$. For a vector functional the integral is $n^{-1}\int \mathrm{IF}{\mathrm{IF}}^{\mathsf{T}}dF$. Convergence of the actual finite-sample variance to this leading expression additionally requires control of second moments of the remainder, or an appropriate uniform integrability condition; convergence in distribution alone does not provide that. Existence of a contamination derivative by itself also does not guarantee the required empirical linearization.

Several complementary robustness measures follow. Gross-error sensitivity is ${\gamma }^{\ast }={\sup }_z|\mathrm{IF}(z)|$. Under contamination $(1-\epsilon )F+\epsilon H$, the first-order bias is $\epsilon \int \mathrm{IF}dH$, whose worst absolute coefficient is ${\gamma }^{\ast }$; bounded influence therefore bounds infinitesimal gross-error bias. The local-shift sensitivity is ${\lambda }^{\ast }={\sup }_{x\ne y}|\mathrm{IF}(x)-\mathrm{IF}(y)|/|x-y|$, equal to $\sup |{\mathrm{IF}}^{\prime }|$ when the influence function is smoothly differentiable. It measures the effect of moving a contaminant a small distance. The rejection point of an influence function is ${\rho }^{\ast }=\inf \{r:\mathrm{IF}(x)=0\text{ for }|x|>r\}$, using centered, standardized coordinates; it is infinite if there is no such cutoff. A finite cutoff completely rejects sufficiently extreme observations to first order. These measures address different aspects of robustness. The integral of squared influence controls asymptotic precision and efficiency, while breakdown point concerns finite amounts of contamination and is not determined by bounded influence alone.

For the quantile, assume $0<p<1$, a unique solution $q=q_p(F)$, and a density continuous and positive at $q$. Define quantiles of the contaminated distribution by generalized inversion, since a distribution with a point mass need not attain $p$ as an equality. If $z\ne q$, its indicator is constant near $q$ for sufficiently small contamination, and differentiatingat zero gives $-p+f(q)q_0^{\prime }+1_{z\le q}=0$. HenceAt the exceptional point $z=q$, the generalized-inverse quantile remains exactly $q$: below $q$ the contaminated cumulative distribution function is below $p$, and its value at $q$ exceeds $p$. Its actual derivative there is zero. The usual step-function expression is an almost-everywhere version, which is all that the asymptotic variance needs. Its mean is zero, and its second moment isTherefore the influence function of a quantile gives asymptotic variance $p(1-p)/(nf(q{)}^2)$ and gross-error sensitivity $\max (p,1-p)/f(q)$. Its jump makes local-shift sensitivity infinite, and its nonzero constant tails make its rejection point infinite. If $f(q)=0$ or the quantile is not unique, this regular derivation does not apply.

Saddlepoint approximation methods retain the nonquadratic shape of a distribution through its cumulant-generating function. Suppose $X_1,\dots ,X_n$ are independent and identically distributed, $K(t)=\log E(e^{t{X}_1})$ is finite on an open interval containing zero, and $K^{\prime \prime }>0$. For an interior target mean $x$, solve the saddlepoint equation $K^{\prime }(t)=x$. Introduce exponential tilting byUnder the tilted distribution the mean and variance are $K^{\prime }(t)=x$ and $K^{\prime \prime }(t)$. For the product sample the Radon-Nikodym derivative is $e^{t{\sum }_i{X}_i-nK(t)}$. Consequently the exact density relation for the sample mean isA local central limit approximation to the tilted density at its own mean gives $f_{\overline{X},t}(x)\simeq \sqrt{n/(2\pi K^{\prime \prime }(t))}$. HenceFor the sum $S_n$, the equivalent expression is $f_{S_n}(s)\simeq e^{nK(t)-ts}/\sqrt{2\pi n{K}^{\prime \prime }(t)}$, with $K^{\prime }(t)=s/n$. Under the usual smooth nonlattice density conditions this approximation has relative error of order $n^{-1}$ for a fixed interior target. In the tilted Edgeworth expansion the order $n^{-1/2}$ odd term vanishes at the mean. The exponent retains the Legendre-transform rate $I(x)=tx-K(t)$, rather than a quadratic Taylor approximation to that rate, explaining the useful tail and skewness behavior. Normalization is not automatic, and accuracy statements need their regularity and range restrictions.

For a normal distribution, $K(t)=\mu t+{\sigma }^2{t}^2/2$ and $t=(x-\mu )/{\sigma }^2$. Substitution recovers exactly the normal density of the sample mean. For an exponential distribution of rate $\lambda$, $K(t)=-\log (1-t/\lambda )$, $t=\lambda -1/x$ and $K^{\prime \prime }(t)=x^2$. The approximation becomesIt has exactly the gamma density shape. Replacing its leading constant by the exact normalizing constant makes it exact, illustrating why exactness of shape and exactness of the unnormalized approximation are separate questions.

For a vector sample mean, tilt by $e^{t^{\mathsf{T}}y-K(t)}$, solve $\nabla K(t)=x$, and use the local multivariate normal density at the tilted mean. This givesJoint and marginal saddlepoint densities can also be divided to approximate conditional densities, for example after conditioning on a nuisance sufficient statistic. The p* approximation uses likelihood and fitted-information coordinates to obtain a closely related approximation to a maximum-likelihood estimator's density, usually conditional on an ancillary statistic.

Tail probabilities need a corresponding integration approximation. The Lugannani-Rice saddlepoint tail approximation usesand givesHere $\Phi ,\varphi$ are the standard normal distribution functions. The upper-tail correction has the opposite sign: $1-\Phi (w)+\varphi (w)(1/u-1/w)$. Inverting a cumulative distribution function introduces a pole in addition to the density's saddlepoint; treating both supplies the correction term, which is why simply integrating a normal approximation at the untilted mean is less accurate. At $x=K^{\prime }(0)$ the apparent singularities cancel. With $v=K^{\prime \prime }(0)$ and $k=K^{\prime \prime \prime }(0)$, expansions give $w=t\sqrt{nv}(1+kt/(3v)+O(t^2))$ and $u=t\sqrt{nv}(1+kt/(2v)+O(t^2))$, so $1/w-1/u\to k/(6\sqrt{n}{v}^{3/2})$. For the normal distribution the correction is zero everywhere.

These methods require care for lattice distributions, where summation and lattice corrections replace the density inversion, and near boundaries or singular covariance matrices. A distribution with no moment-generating function in a neighborhood of zero does not satisfy the starting assumptions. A successful saddlepoint approximation is therefore a structured refinement of regular asymptotic inference, not a formula guaranteed uniformly in every tail or for every model.

Use the divisor-$n$ sample covariance matrix, since this is the normalization that enters the multivariate normal likelihood function:For a covariance matrix that is a positive-definite matrix $V$, the product of the multivariate normal densities givesTo separate the sample mean from the sample covariance matrix, expand $x_j-\mu =(x_j-\bar{x})+(\bar{x}-\mu )$. The cross terms sum to zero because ${\sum }_j(x_j-\bar{x})=0$. Also $u^T{V}^{-1}u=\mathrm{tr}(V^{-1}u{u}^T)$ by the trace identity. ThereforeThe printed expression suppresses the parameter-independent constant $np\log (2\pi )$; dropping this constant has no effect on maximum likelihood estimation or the likelihood-ratio test.

Assume independent samples $X_1,\dots ,X_n$ and $Y_1,\dots ,Y_m$, each consisting of independent normal random variables, with respective means ${\mu }_1,{\mu }_2$ and a common unknown variance ${\sigma }^2>0$. Take $n,m\ge 2$. Normality, independence and equal variances are the assumptions giving the exact pooled Student's t-test; unequal variances would require a different test. We test $H_0:{\mu }_1={\mu }_2$ against an unrestricted difference.

Put $N=n+m$ andThe normal likelihood is proportional to $({\sigma }^2{)}^{-N/2}\exp [-S/(2{\sigma }^2)]$, where $S$ is the residual sum of squares. Under the unrestricted model, the maximizing means are $\bar{X},\bar{Y}$ and $S=W$. Under $H_0$, the maximizing common mean is $(n\bar{X}+m\bar{Y})/N$ and decomposition about the sample means gives $S=W+B$. For either model, maximizing over variance gives ${\hat{\sigma }}^2=S/N$. Thus the generalized likelihood-ratio test statistic isSmall $\Lambda$ is therefore equivalent to large $|T|$.

Under the null, $Z=(\bar{X}-\bar{Y})/[\sigma \sqrt{1/n+1/m}]$ is standard normal. In each sample an orthogonal change of Gaussian coordinates separates the sample-mean coordinate from the centered residual coordinates; those coordinates are independent. Consequently $V=W/{\sigma }^2$ has a chi-squared distribution with $N-2$ degrees of freedom and is independent of $Z$. Therefore $T=Z/\sqrt{V/(N-2)}$ has the Student's t-distribution with $N-2$ degrees of freedom. At significance level $\alpha$, reject exactly when $\boxed{|T|>t_{N-2,1-\alpha /2}}$, the two-sample pooled t-test.

The first output uses the equal-variance, pooled Student's t-test, as shown by its seven statistical degrees of freedom. Its exact model assumes two independent samples from normal distributions, with means ${\mu }_X,{\mu }_Y$ and a common unknown variance ${\sigma }^2$. The hypotheses are $H_0:{\mu }_X={\mu }_Y$ against $H_1:{\mu }_X<{\mu }_Y$. The sample means are $\bar{x}=4.35$ and $\bar{y}=9.88$, and the unbiased within-sample sample variances areThe pooled sample variance and its resulting standard error areHenceTo justify its reference distribution, under the null $(\bar{X}-\bar{Y})/(\sigma \sqrt{1/4+1/5})$ is standard normal, while $(3s_X^2+4s_Y^2)/{\sigma }^2$ is ${\chi }_7^2$, independently of the two sample means. Dividing the first variable by the square root of the second divided by seven gives the exact pooled two-sample t statistic, with Student t-distribution $t_7$. Its one-sided p-value is the lower tailThe associated one-sided 95% confidence interval for ${\mu }_X-{\mu }_Y$ isThe printed missing lower bound means that this is a one-sided upper confidence limit, not that a finite lower endpoint should be estimated. With such small samples the normality and common-variance assumptions matter; a Welch analysis would use a different variance estimate and generally different degrees of freedom.

The second output is the Wilcoxon rank sum test for two independent continuous samples. Under a common-distribution null hypothesis, labels are exchangeable conditional on the pooled values, so the four ranks assigned to the first sample form a uniformly chosen four-element subset of $\{1,\dots ,9\}$. The first sample receives ranks $2,1,3,5$, giving raw rank sumEquivalently the Mann–Whitney U test statistic is $U_X=W_X-4\cdot 5/2=1$, the number of cross-sample pairs with an $X$ observation larger than a $Y$ observation. This output reports the raw rank sum, rather than the shifted statistic used by some software.

A first sample tending to be smaller has a small rank sum. Under the null, all $\left(\genfrac{}{}{0ex}{}{9}{4}\right)=126$ rank allocations have equal probability. Only $\{1,2,3,4\}$ and $\{1,2,3,5\}$ have rank sum at most 11. By the exact rank-sum tail one above the minimum, the exact lower-tail p-value isNo large-sample normal approximation or continuity correction is involved; the absence of ties makes this simple enumeration valid. For a common-shape location-shift family, the test's alternative is a negative location shift of $X$ relative to $Y$. Without that model, the null is equality of distributions and the rank comparison is not simply a test of equality of means or medians.

Both tests reject at the 5% level in the direction of smaller $X$ observations; only the pooled t-test rejects at 1%. The t-test uses the numerical spacing and targets a mean difference under its normal model. The rank-sum test uses ordering and an exact exchangeability calculation, avoiding the normality and common-normal-variance assumptions. Their different p-values therefore reflect both different statistics and different modelling assumptions, not a numerical contradiction.

The within-group and between-group scatter decomposition separates noise from group separation: $W$ measures variation about each group sample mean, and $B$ measures separation of those means about the overall sample mean. In linear discriminant analysis, seek a projection $a^{T}x$ maximizing between-group relative to within-group variation,Normalize by $a^{T}Wa=1$; a Lagrange multiplier calculation gives the generalized eigenvalue problem $Ba=\lambda Wa$. The leading canonical discriminant directions have the largest eigenvalues, and subsequent directions are orthogonal in the $W$ inner product. There are at most $\min (p,g-1)$ nonzero separating directions. Projecting observations onto these directions gives a low-dimensional display of group separation.

For classification, estimate the common covariance matrix by $S_p=W/(n-g)$ and use the linear discriminant analysis scoreswhere ${\pi }_{\nu }$ are class prior probabilities. Assign to the largest score. The same within-group scatter sets the noise scale, while between-group scatter selects the directions in which the group means separate most strongly.

Use the usual sample mean and the divisor-$n-1$ sample covariance matrix,The residual sum is zero, so expanding $x_j-\mu =(x_j-\bar{x})+(\bar{x}-\mu )$ givesMultiplying the independent multivariate normal densities and taking logarithms yields the log-likelihoodHere $|V|$ is the determinant and $\mathrm{tr}$ denotes the trace. The maximum-likelihood estimators areIf $S$ instead denotes the divisor-$n$ sample covariance matrix, replace $(n-1)S$ by $nS$ and then $\hat{V}=S$. These formulas describe a nonsingular maximum-likelihood estimator when the empirical scatter is positive-definite; if it is singular, there is no maximizer over positive-definite covariance matrices.

princomp(): principal component analysis. Supply observations in rows and numeric variables in columns. After centering, let $Z$ denote the resulting matrix and $C=Z^{T}Z/n$ its sample covariance matrix, with an explicitly stated normalization. WriteThe first direction maximizes the Rayleigh quotient $q^{T}Cq$ over unit vectors; successive directions maximize it subject to orthogonality to the earlier directions. The principal component scores are $z_{ij}=q_j^T(x_i-\bar{x})$, with variances ${\lambda }_j$ and zero pairwise sample covariance. Zero covariance alone does not imply independence. Each principal component loading describes a direction in variable space; its sign is arbitrary, and repeated eigenvalues leave the basis inside the corresponding eigenspace arbitrary.

The function returns principal component loadings, component standard deviations sdev and, when requested, principal component scores. By default it uses a covariance matrix; cor=T instead performs principal component analysis on a correlation matrix, standardizing nonconstant variables to variance one. This choice matters when variables have different measurement units. These interface details are documented in the S-Plus help for princomp.

A scree plot shows the ordered eigenvalues; a plot of the first two principal component scores can reveal groups, unusual observations or a dominant gradient. The first $k$ components explain ${\sum }_{j\le k}{\lambda }_j/{\sum }_j{\lambda }_j$ of the total variance. Reconstructing each centered observation by its orthogonal projection onto these directions minimizes average squared reconstruction error among $k$-dimensional linear subspaces. Thus PCA gives an optimal linear compression for squared reconstruction error, with scaling and the retained dimension chosen explicitly. It is an unsupervised learning method, rather than a classification rule fitted to supplied labels.

tree(): classification trees and regression trees. A call such as tree(response ~ predictors, data=dat) fits a recursive partitioning model. At a current node, test numerical predictors against candidate thresholds, or divide the levels of a categorical predictor into two sets. Select a split that most improves the fitting criterion, then repeat within its children. Numerical splits give axis-aligned regions. A categorical response produces a classification tree; a numeric response produces a regression tree, as specified in the S-Plus tree documentation.

For a regression tree, the constant prediction in a terminal region is its sample mean, minimizing the sum of squared residuals there. For a classification tree, estimate class probabilities by the observed proportions in a leaf and predict a most probable class under equal misclassification costs. The classification-tree deviance of a node iswith $0\log 0=0$; splitting maximizes the decrease from parent to children. This fitting criterion differs from the number of wrong classifications. The middle sketches show the same three-region partition as a binary tree: first split on variable one, then split the right-hand branch on variable two.

Growing to small leaves can overfit. Use cost-complexity tree pruning, balancing fitted loss $R(T)$ against leaf count,and select complexity by cross-validation. Plotting the fitted tree and annotating nodes displays the rules; prediction routes each new observation through the fitted splits. A tree gives an interpretable, piecewise-constant prediction rule; training fit alone does not measure prediction accuracy.

The rpart() alternative. This also fits classification and regression trees by recursive partitioning, with squared-error splitting for ordinary numeric responses and, by default for classification, Gini impurity $1-{\sum }_r{p}_r^2$. Classification settings can instead use information-based splitting and specify prior probabilities and a loss matrix. It supports surrogate splits to approximate a node's primary left/right assignment when the primary predictor is missing. Its complexity-parameter path and cross-validated errors support cost-complexity tree pruning; for example, select a complexity parameter and apply prune(fit, cp=...). The maintainer's rpart documentation describes the criteria and the distinction in handling surrogate predictors. The partition and tree sketches apply to either implementation; the fitted tree need not be the same because splitting criteria and controls can differ.

cmdscale(): classical multidimensional scaling. Its input is a dissimilarity matrix or stored distance object. It outputs coordinates for the observations in a chosen dimension, usually two for visualization. Let $D^{(2)}$ contain the squared entries of $D$, and putFor centered Euclidean coordinates $X$, the distance identity $d_{ij}^2=\Vert x_i{\Vert }^2+\Vert x_j{\Vert }^2-2x_i^T{x}_j$ gives $G=X{X}^T$: multiplying by $H$ removes the row and column norm terms. Diagonalize this Gram matrix and retain its $k$ largest positive eigenvalues. The coordinates areThus classical scaling exactly recovers Euclidean distances when all positive eigenvalues are retained. Reducing dimension approximates the centered Gram matrix; it does not in general minimize squared discrepancies in the original distances. Negative eigenvalues identify a failure of the Euclidean distance matrix criterion, so their size matters when interpreting a display. The historical S-Plus cmdscale help describes the dimension argument, eigenvalue output and optional additive adjustment to dissimilarities.

Plot the two coordinate columns on equally scaled axes and label observations. Nearby plotted points suggest small dissimilarities only insofar as the dimension reduction fits the input well. The bottom sketches give an exact Euclidean example: a configuration and its independently recovered two-dimensional coordinates have identical pairwise distances, despite different coordinate axes. Classical scaling converts distances into geometry; translations, rotations and reflections do not change the represented distances. For Euclidean distances between observations, its coordinates agree with the corresponding principal component scores up to these harmless choices of axes.

scan reads the numerical prices, including the three missing-value markers, into a vector. The data must be read row by row for the subsequent factor construction to match them. The sample size available for analysis is therefore $37$, rather than $40$. The numerical summary describes the observed prices: the sample median is £43.54 and the sample mean is £46.94. The upper tail, particularly the boots, raises the sample mean above the sample median. These summaries mix very different products and are not adjusted country comparisons.

The two character scans create labels. gl(8,5,length=40,labels=item) makes a regression factor with each of eight item labels repeated five times. gl(5,1,length=40,labels=country) cycles through the five country labels once for each item. For a regression factor and numerical response, the two plot commands produce grouped box plots. The country box plots suggest lower US prices, while the item box plots show particularly expensive boots and inexpensive cardigans. They are descriptive displays of this small basket, with different item compositions where prices are missing; they cannot separate country effects from item effects.

The first lm fits an additive two-way analysis of variance:The constraints are R's usual treatment coding. The normal linear model used for the printed tests assumes independent errors with common variance and approximately a normal distribution, and no item-by-country interaction in the mean. Missing prices are omitted, leaving $37$ responses and $1+4+7=12$ independent mean coefficients. Thus the residual degrees of freedom are $37-12=25$. This deletion is defensible for the conditional price model when missingness does not depend on the unobserved error after conditioning on the two factors; the output itself does not establish such a missing-data mechanism.

The anova tables report sequential sums of squares, rather than order-independent adjusted effects. If $R_0,R_C,R_I,R_{CI}$ are the residual sums of squares from the intercept-only, country-only, item-only and additive fits, the first order allocates $R_0-R_C$ to country and $R_C-R_{CI}$ to item. The second order allocates $R_0-R_I$ to item and $R_I-R_{CI}$ to country. Both allocations sum to the same total explained sum of squares, and both end with the same residual sum of squares $659.4$. The three missing combinations destroy balanced factorial orthogonality, so the allocation changes with order. To test country after allowing for items, use the country row in the second table; to test item after allowing for countries, use the item row in the first table. The first-entered rows do not supply those adjusted tests.

Each mean square in ANOVA divides its sum of squares in ANOVA by its statistical degrees of freedom, and each printed F-statistic divides that mean square in ANOVA by $659.4/25=26.38$. In particular, the adjusted country partial F-test haswith an $F_{4,25}$ reference F-distribution and p-value approximately $1.86\times 10^{-6}$. The adjusted item partial F-test gives $F=91.58$ on $7,25$ statistical degrees of freedom. Both factors contribute strongly under the additive normal linear model. The larger item sum of squares in ANOVA measures the much greater differences between these products, but is not evidence that every pair of items differs significantly.

The intercept $50.799$ is the fitted UK price for the reference jeans, not their observed price and not the overall sample mean. Each country regression coefficient is its adjusted price difference from the UK, assumed common across items: Sweden is £5.224 lower, France £10.708 lower, Germany £7.676 lower and the US £21.328 lower. Each item regression coefficient is its adjusted difference from the reference jeans, assumed common across countries. Thus the fitted UK boots price is $50.799+52.854=103.653$, and the fitted US boots price is $103.653-21.328=82.325$.

The coefficient standard errors come from $s^2(X^{T}X{)}^{-1}$, where $s^2=659.4/25$ and $X$ is the design matrix. Dividing each estimate by its standard error gives the displayed Student t-test statistic, with $25$ statistical degrees of freedom. The stars summarize the corresponding two-sided p-values. The individual tests compare each indicated country or item with its reference level, conditional on all other additive terms; they are neither tests of every pairwise comparison nor a simultaneous confidence interval procedure. In particular, Sweden's p-value $0.0626$ provides weaker evidence than the other country contrasts, and does not establish equality with the UK. The intercept test has little substantive relevance, and interpreting many individual stars also raises multiple hypothesis testing issues.

The residual standard error $s=5.136$ measures unexplained prices in pounds. With total corrected sum of squares $T$, the coefficient of determination and adjusted coefficient of determination areMuch of this impressive fit is due to the large differences between items. The overall F-test compares the additive fit with a constant mean: $F=((T-659.4)/11)/(659.4/25)=62.12$ on $11,25$ statistical degrees of freedom. It tests that all eleven non-intercept coefficients vanish, not that the additive model is adequate. Regression diagnostics should examine regression residuals, changing price variability across items and possible interaction. Percentage differences may be more plausible than common pound differences, motivating a natural logarithm of price. No random sample of countries or products, and no independent replication of a price within a cell, is supplied; broad claims about all designer goods or the causes of country differences require more than these conditional tests.

Finally, Item*Country expands to both main effects and their full interaction. A cell-means parameterization has one separate fitted mean for every observed cell. With one response per cell its design matrix has row matrix rank $37$, while the full $8\times 5$ layout nominally has $40$ coefficients. Three coefficient combinations are aliased because their cells are missing. This is saturation of an unreplicated two-factor regression: every observed price is fitted exactly and the residual degrees of freedom are zero. The final fit has zero residuals, $R^2=1$ and no residual-based estimate of the error variance or ordinary interaction significance test. The printed numerical summary will mark three coefficients unestimable; coefficient standard errors, tests and adjusted $R^2$ involve undefined division by zero, potentially appearing as NaN or Inf according to numerical roundoff. Perfect interpolation here is a consequence of a saturated statistical model, not evidence for a superior predictive model.

The profile likelihood for the parameter of interest replaces the nuisance parameter by its constrained best fit: $L_p(\psi )={\sup }_{\lambda }L(\psi ,\lambda )$, with ${\ell }_p(\psi )=\ell (\psi ,{\hat{\lambda }}_{\psi })$. It is not in general a normalized sampling density. A conditional likelihood instead uses the sampling law conditional on a statistic carrying the nuisance dependence. In the canonical exponential family here, conditioning on $S$ removes $\lambda$ because the factor involving $\lambda$ is constant over the conditioning fibre.

Let ${\bar{T}}_1=n^{-1}{\sum }_i{\tau }_1(y_i)$ and let the observed $S$ equal $s$. For the cumulant-generating function of ${\tau }_2(Y)$ at fixed $\psi$,The saddlepoint solves $K^{\prime }(\hat{t})=s$, equivalently $d_{\lambda }(\psi ,{\hat{\lambda }}_{\psi })=s$ with ${\hat{\lambda }}_{\psi }=\lambda +\hat{t}$. This is also the constrained likelihood equation, since ${\ell }_{\lambda }=n(s-d_{\lambda })$. Assume an interior saddle, $d_{\lambda \lambda }>0$, and the usual smooth nonlattice conditions under which $S$ has a density. Exponential tilting to ${\hat{\lambda }}_{\psi }$ makes $s$ the tilted mean. A local normal approximation gives tilted density $\{n/(2\pi d_{\lambda \lambda }(\psi ,{\hat{\lambda }}_{\psi })){\}}^{1/2}$ at that mean; undoing the tilt gives the saddlepoint density approximationThe factor $n^{1/2}$ is for the sample mean, rather than the sum. A transform-based derivation gives the same expression by expanding $K$ to second order at its stationary point.

The joint log-likelihood is $n\psi {\bar{T}}_1+n\lambda s-nd(\psi ,\lambda )-{\sum }_{i}Q(y_i)$. Subtracting $\log {\hat{f}}_S(s;\psi ,\lambda )$ cancels every appearance of the original $\lambda$, leavingAny change from the ambient density to a conditional density on $S=s$ contributes only a data-dependent Jacobian determinant. Thus, up to a constant independent of $\psi$, the saddlepoint conditional likelihood adjustment isEquivalently $B=\frac{1}{2}\log j_{\lambda \lambda }(\psi ,{\hat{\lambda }}_{\psi })$ up to the constant $-\frac{1}{2}\log n$, where $j_{\lambda \lambda }=n{d}_{\lambda \lambda }$ is the nuisance observed information. The adjustment has a positive sign. The printed exponential-family form alone does not guarantee a density for $S$: for example a continuous $Y$ can have ${\tau }_2(Y)=1_{\{Y>0\}}$. In a lattice case one must approximate the probability mass of the sum instead; the displayed density calculation uses the regular continuous interpretation.

A transformation model consists of a group $G$ acting measurably and invertibly on the sample space, an induced group action on the parameter space, and a family $\{P_{\theta }\}$ satisfying $Y\sim P_{\theta }\Rightarrow gY\sim P_{g\theta }$. In a generated, transitive model, every $P_{\theta }$ is the image of a reference distribution under some transformation. In the density formulation, the transformed density includes the appropriate sample-space Jacobian determinant; the model is a statement about whole probability laws, not merely about changing parameter names.

A maximal invariant $A$ has two properties: $A(gy)=A(y)$ for every $g$, and $A(y)=A(z)$ implies $z=gy$ for some $g$. It is therefore a coordinate for the group orbits, retaining exactly the information unaffected by the transformations. An equivariant estimator obeys $T(gy)=gT(y)$, where the right action is the induced parameter action.

For the normalization construction, suppose the estimator determines a group element $t(y)\in G$ with $t(gy)=gt(y)$. Define $A(y)=t(y{)}^{-1}y$. ThenIf $A(y)=A(z)$, rearranging gives $z=t(z)t(y{)}^{-1}y$. This is an actual transformation in $G$, so the normalized sample is a maximal invariant. This group-valued assumption matters: an estimator of a parameter with a nontrivial stabilizer does not necessarily determine a unique normalization. For a transitive parameter space $G/H$, choose a transformation carrying the estimated parameter to the reference value and take the resulting normalized sample modulo the remaining $H$-action. Different choices then have the same residual orbit. Under a free action $H$ is trivial and the simpler formula applies.

For a location-scale model, $g_{a,b}(y_1,\dots ,y_n)=(a+b{y}_1,\dots ,a+b{y}_n)$, $b>0$, with composition $(a,b)(c,d)=(a+bc,bd)$. Its parameter action is $(\mu ,\sigma )\mapsto (a+b\mu ,b\sigma )$. Choose location and scale estimates satisfyingFor instance, on nonconstant samples take the sample mean and the positive square root of $n^{-1}{\sum }_i(y_i-\bar{y}{)}^2$. The standardized vectoris invariant. Conversely, equality of these vectors gives $z_i=c+d{y}_i$ for all $i$, with $d=\hat{s}(z)/\hat{s}(y)>0$ and $c=\hat{m}(z)-d\hat{m}(y)$. Hence it is maximal, not just invariant. Degenerate constant samples have zero fitted scale and form a separate orbit; define a separate invariant value there. The construction is on labelled sample vectors: one must not discard their order unless permutations are also included in the transformation group.

At the standard normal distribution, symmetry and the odd Huber score give population root zero. For $0<b<\infty$, the derivative of the clipped score is $1_{\{|x|<b\}}$ except at its two corners, which have zero normal probability. Put $m_b=2\Phi (b)-1>0$ and write $\varphi$ for the standard normal density. The sandwich variance of an M-estimator has numeratorUsing ${\varphi }^{\prime }(z)=-z\varphi (z)$ and integration by parts,ThereforeAs checks, it tends to $1$ as $b\to \infty$, the sample mean limit, and to $\pi /2$ as $b\downarrow 0$, the sample median limiting variance. The estimator variance for a sample of size $n$ is approximately this quantity divided by $n$.

Population influence functions describe infinitesimal contamination; finite-sample analogues examine actual changes to an empirical sample. For its empirical distribution $F_n$, add an observation $z$ and define the sensitivity curveUnder suitable differentiability it approximates the influence function. A replacement version is ${\mathrm{RC}}_{n,i}(z)=n[T_n(y_1,\dots ,z,\dots ,y_n)-T_n(y)]$, approximately $\mathrm{IF}(z;T,F_n)-\mathrm{IF}(y_i;T,F_n)$. Their supremum magnitudes quantify finite one-observation sensitivity.

The empirical influence function instead substitutes $F_n$ directly in $\mathrm{IF}(z;T,F)$. For a differentiable M-estimator this gives $-\psi (z,T(F_n))/[n^{-1}{\sum }_i{\partial }_{\theta }\psi (y_i,T(F_n))]$. For quantile functionals, a discrete unsmoothed empirical law may not admit the required density derivative; a sensitivity curve or a smoothed empirical law is then preferable.

Global contamination is measured by finite-sample maximum bias $b_n(m;y)={\sup }_{z:d_H(z,y)\le m}|T_n(z)-T_n(y)|$, where $d_H$ counts replacements. Under the convention of the first fraction that causes unbounded bias,This is the finite-sample replacement breakdown point. The largest replacement fraction still guaranteed bounded is one grid step smaller, so the conventions must be specified. One replacement sends a sample mean arbitrarily far, giving $1/n$, while the ordinary scalar sample median has first-breaking fraction $\lceil n/2\rceil /n$ with the usual midpoint convention for even samples. Its limiting value is $1/2$. Local influence and global breakdown answer different questions; a bounded influence function is not by itself a proof of a high breakdown point.

For the sample, put $\bar{x}=n^{-1}{\sum }_i{x}_i$ and use the sample covariance matrixOrder its eigenvalues ${\hat{\lambda }}_j$ decreasingly and choose unit orthogonal eigenvectors ${\hat{\ell }}_j$. The sample principal components are the score columnsTheir sample means are zero and their sample covariance is diagonal, since ${\hat{\ell }}_j^{T}S{\hat{\ell }}_k={\hat{\lambda }}_j{1}_{\{j=k\}}$. The $j$th sample variance is ${\hat{\lambda }}_j$, and its explained variance of a principal component is ${\hat{\lambda }}_j/\mathrm{tr}(S)$. Using divisor $n$ instead changes the numerical variances by a common factor but leaves the component directions unchanged.

Standardizing means replacing a variable with positive sample variance by $(x_{ij}-{\bar{x}}_j)/s_j$. This gives principal component analysis on a correlation matrix, using $C=D^{-1/2}S{D}^{-1/2}$ with $D=\mathrm{diag}(S)$. It removes arbitrary differences of measurement units and prevents a large-variance coordinate from dominating solely through its scale. It also changes the question being optimized: small-variance and possibly noisy coordinates receive equal marginal weight. Principal component analysis based on the covariance matrix can be more appropriate when the variables have common meaningful units and absolute variability matters. Standardization is therefore a substantive choice, not an automatic improvement. A zero-variance variable must first be removed. In the word-rating example the variables share the same numerical range, but their across-word dispersions can differ; standardization gives each attribute equal initial variance.

For the standardized eight-variable analysis, total variance is $\mathrm{tr}(C)=8$. The first three fractions of explained variance of a principal component are approximately 59.6%, 19.1% and 10.1%, respectively. The first two account for about 78.8%, and the first three for 88.9%. The first direction is dominant but one direction alone loses substantial variation. Retaining two principal components is a plausible descriptive summary: their eigenvalues exceed $1$, whereas the third is below $1$. That cutoff is a heuristic, and the third principal component may still have interpretable content.

The reported coefficient rows are not unit-length vectors. The printed rounded rows have squared lengths $4.7514$, $1.5222$ and $0.8063$, consistent with their eigenvalues to the precision of the coefficient table. They are consistent with the scaled principal component loading convention $r_j=\sqrt{{\lambda }_j}{\ell }_j$. Such rows remain eigenvectors, but the score direction with unit length is ${\ell }_j=r_j/\sqrt{{\lambda }_j}$, or more accurately the row divided by its actual Euclidean norm when using rounded entries. Under this convention $r_{ij}$ is the correlation of standardized attribute $i$ with the variance-one score for component $j$:This explains both their sizes and why they must not be treated as the unit vectors from parts (i) and (ii).