Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 39 6 Solution Created 2026-10-03 Updated 2026-10-14
Let and let an approximate experimental design specify the proportions of observations allocated to points of the design region. Its normalized information matrix of an experimental design isFor independent equal-variance observations realizing those proportions, , and the least-squares estimator has covariance matrix . A D-optimal design maximizes , equivalently minimizing the determinant of this covariance at fixed . A G-optimal design minimizes the largest variance of a fitted regression mean over the design region, equivalentlywhere is the design sensitivity function. This is the variance of the estimated mean multiplied by , not the variance of a new noisy observation.
For continuous regressors spanning a -dimensional parameter space on a compact design region, the general equivalence theorem for optimal design states that, among approximate designs with nonsingular information matrices, the following are equivalent: being a D-optimal design, being a G-optimal design, andAt every support point of an optimal design, . To see the criterion, the sensitivity average is , so the supremum is at least . Adding an infinitesimal allocation at has directional derivative for . At a D-optimum each derivative is nonpositive. Conversely, if all sensitivities are at most , the tangent bound for the concave function shows that no competing design increases it. A nonsingular D-optimum exists by the spanning assumption and compactness of the information-matrix set, so the smallest possible G-value is also . Equality at support points follows from the average. The theorem refers to the full approximate-design class; constrained integer allocations or mandated augmentations require separate comparison.
For the linear mean , take . Equal replication at the four corners givesAll odd corner sums and the sum of vanish, while each squared regressor sum is . If is the mean of the observations at , , the least-squares normal equations giveIn particular the intercept is the average of the four cell means, and each slope is the average of its sign-weighted cell means. Independence and equal error variance suffice for this covariance formula; normal errors are not needed for it.
The normalized corner design has , so its sensitivity iswith equality at all four corners. Since , the general equivalence theorem for optimal design proves that the equally weighted corner design is D-optimal and G-optimal.
For the augmentation, all added observations share the regressor . The enlarged design matrix satisfiesBy the matrix determinant lemma,For the permitted nine points, the determinant is at the center, at an edge midpoint and at a corner. ThereforeAll four corners tie for the best mandated single-point augmentation. This is optimality among the specified augmented designs, not unrestricted D-optimality with observations: the resulting normalized determinant is , whereas equal corner proportions have determinant one. The restriction that the extra runs all occur at one point is what prevents recovering equal proportions.
Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 39 5 Solution Created 2026-10-03 Updated 2026-10-14
Code each factor level by . A nonconstant factorial contrast indexed by a nonempty factor set has sign column . Flipping any coordinate in reverses that sign and pairs the treatment combinations, so exactly have each sign. Allocating the two signs to two experimental blocks therefore divides the full factorial design into equal experimental blocks. The contrast is confounded with experimental blocks because its column is constant within each experimental block: changing its coefficient and compensating by the two experimental block intercepts leaves every fitted value unchanged. It has no information separate from experimental block differences. Randomize the allocation of the two groups to days and the run order within days.
Write for temperature, pressure and catalyst, respectively, with low levels and catalyst one coded minus. In the supplied allocation, on day one and on day two. Thus the three-factor interaction is confounded with days. Every other nonconstant factorial column is orthogonal to the day contrast, because its product with is a nonconstant sign column on the full cube and consequently sums to zero.
If all interactions are negligible, the fitted model contains an intercept, one day contrast and the three main effects. The corrected total statistical degrees of freedom partition asThe three residual statistical degrees of freedom are the unused two-factor interaction columns , pooled as error under the stated negligible-interaction assumption. There is no independent pure-error replication of treatment combinations.
For four days with two runs each, all three main effects can be estimated. One regular allocation pairs each treatment combination with its complete opposite:The block-generating columns can be and ; their product is also constant in experimental blocks. Thus span the three experimental block contrasts, while and vary within experimental blocks. The main effects remain mutually orthogonal and separate from the experimental block effects; with an additive main-effect model there is one residual degree of freedom, represented by .
For the further request about , the distinction between regular confounding and unrestricted pairing is important. A regular four-block allocation cannot retain all three main effects and . Its three confounded columns form a rank-two defining subgroup. To preserve main effects, these columns must lie in . If the subgroup includes and a two-factor column, their product is a main-effect column, which is forbidden. Hence the only permissible nonidentity subgroup is , and it necessarily confounds . This proves impossibility in the usual regular factorial-blocking framework, rather than relying on a count of available statistical degrees of freedom.
The literal request does not explicitly restrict arrangements to regular experimental blocks. If arbitrary pairings are permitted, a nonregular construction can estimate together, assuming the other interactions are omitted from the model. An explicit allocation isEvery treatment combination occurs exactly once. Fit one unrestricted intercept per day and the four coefficients . By estimability from within-block differences, subtracting the second run from the first in each experimental block removes all day effects and yields the coefficient matrixThus all four coefficients are uniquely determined from the four day differences. In this broader design class the answer is yes, under the reduced four-effect model. The resulting fit is saturated: four experimental block intercepts plus four treatment coefficients use all eight observations, leaving no residual estimate of error variance. Replication or external error information would be needed for inference. This does not contradict the impossibility for regular orthogonal confounding, and it does not claim simultaneous unbiased estimation when arbitrary additional interactions are retained.
Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 39 4 Solution Created 2026-10-03 Updated 2026-10-14
A randomized complete block design groups similar experimental units into experimental blocks and places every treatment once in each experimental block, independently randomizing treatment allocation within experimental blocks. Blocking removes background between-block variation from within-block treatment comparisons. A balanced incomplete block design has treatments and experimental blocks of size , with each treatment occurring in experimental blocks and every distinct treatment pair together in experimental blocks. Counting treatment occurrences and co-occurrences gives and .
A Latin square of order places each of treatment symbols once in every row and every column of a array. This controls two nuisance directions. A Graeco-Latin square superposes two mutually orthogonal Latin squares: each square is Latin, and each ordered pair of symbols from the two squares occurs exactly once. The two symbol factors can then be fitted together with row and column effects. Random permutations of row labels, column labels and treatment symbols provide a suitable randomized allocation without destroying these balances.
For the operator experiment, fit the additive normal linear modelHere account for the two fabric directions, and is the operator effect. Additivity, independent equal-variance errors and the normal assumption supply the exact F-test; there is only one observation per cell, so an unrestricted interaction model is not available.
In the analysis of variance for a Latin square, the three centered factor spaces each have dimension . Together with the intercept the model has rank ten, leaving residual statistical degrees of freedom. The corrected total has . The completed calculation isUnder the null hypothesis , the operator F-statistic isThe upper five-percent critical value for is , so the operator effect is not significant at five percent in this model.
The decision is independent of the order of fitting rows, columns and operators. To prove the required orthogonality, let be any centered row coefficients and any centered operator coefficients. Their observation-space inner product is , since every operator occurs once in each row. The same calculation applies to columns and operators, and all row-column pairs occur once as well. Thus the three centered spaces are mutually orthogonal; their projections commute and sequential sums of squares equal adjusted sums of squares in every fitting order.
With machines included, the layout is a Graeco-Latin square: the machine symbols are Latin, and all sixteen operator-machine pairs occur exactly once. Fitin addition to the earlier constraints. The machine space contributes three new statistical degrees of freedom and is orthogonal to the previous factor spaces, so the operator sum of squares remains . The residual sum of squares and statistical degrees of freedom becomeHenceThis exceeds but not . Therefore the operator effect is now significant at five percent, but not at one percent. Adjustment for machines has removed substantial residual variation without altering the operator numerator. This illustrates why a balanced nuisance factor can change precision and significance even when it does not change the estimated treatment contrasts.
Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 39 2 Solution Created 2026-10-03 Updated 2026-10-14
The calculation is principal component analysis on a correlation matrix, rather than on an unscaled covariance matrix. For each of the eleven rating variables, subtract its column mean and divide by its column standard deviation. If is the resulting matrix, use the divisor convention of the fitted object to writeHere the scaling standard deviations use divisor , consistently with this convention. Using sample standard deviations with divisor gives the same correlation matrix and directions, with the corresponding score scale. Every diagonal entry of is one, so .
A principal component is the linear combination of the standardized ratings along . Its principal component score for judge is , and its variance is . The displayed component standard deviations are ; the explained variance of a principal component is . In particular,The first component explains about of the standardized variation, and the first two jointly explain . Components three, four and five add about , and . The cumulative line adds these proportions successively; it is not the proportion explained by that component alone. Retaining two components discards about of the total standardized squared variation.
The displayed principal component loadings are the unit eigenvector coefficients . The first direction has positive, broadly similar weights for all ratings. It measures an overall favorable-rating tendency: a judge high on most ratings has a high first score. The second direction compares integrity and demeanour, with large positive coefficients, against such attributes as decision promptness, case-flow management and physical ability, with negative coefficients. Thus among similarly rated judges, a positive second score indicates relatively stronger integrity/demeanour compared with those operational ratings. This is a description of a data direction, not a causal interpretation. Multiplying a whole eigenvector and its scores by minus one reflects the display but changes nothing substantive.
These principal component loadings are not themselves variable-component correlations. With standardized variables,For example, the integrity correlation with component one is approximately . Blank loadings in the printed second column indicate suppressed small entries, not exact zeros. The complete unrounded loading vectors in the fitted object must be used to calculate scores.
A two-dimensional representation plots the 43 pairswith each point labeled by its judge. In the fitted software object these are the first two columns of the score matrix. Nearby points have similar projections of their standardized ratings. The rank-two reconstruction is the orthogonal projection onto the leading principal plane and minimizes the total squared reconstruction error over all two-dimensional linear subspaces. To see this, for any orthonormal pair of directions the captured variance is the sum of its two Rayleigh quotients, at most by diagonalizing ; the remaining variance is the total minus that captured sum. Distances in the plot omit the smaller components, so a close projected pair need not be equally close in every original rating. The actual 43 coordinates cannot be recovered from the summary alone; the supplied data matrix or fitted score matrix is needed.
Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 38 5 Solution Created 2026-10-03 Updated 2026-10-14
Discarding zero observations and resolving ties is unnecessary with probability one under the stated independent continuous model. Rank the absolute observations from to , let be the rank of , and form the positive Wilcoxon signed-rank statisticReject the null hypothesis for large . Under symmetry about zero, each sign is a fair independent binary variable, independent of all the absolute values. To see this independence, for every measurable positive set , symmetry gives . Independence of observations then gives independence of the whole sign vector from the absolute-value vector. Conditional on those absolute values, relabeling by rank therefore givesThe null distribution is consequently independent of the unknown symmetric shape. Its probability generating function is , so the coefficient of gives the exact probability of . Choose an upper critical value with null tail probability at most the desired level, or randomize at the boundary to obtain that level exactly. The observed upper-tail probability is an exact one-sided p-value. A positive translation raises each Walsh average, as established below, so this upper-tail rule has the correct direction against positive centers for every fixed symmetric shape.
The independent-sign representation provesThe asymptotic null distribution, stated without proof, isThis supplies the usual upper normal distribution critical value when exact calculation is inconvenient; for the discrete upper tail , use in the continuity correction.
For exact equivalence with the proposed statistic, reorder the observations so . Partition the unordered pairs, including diagonals, by the member with larger absolute value. The block belonging to consists of the pairs with an earlier member and its diagonal pair. For an earlier member , the strict inequality means the sign of is the sign of ; the diagonal sum has that same sign. Thus the entire block supplies exactly positive averages when , and none otherwise. Summing the blocks provesThe two tests therefore have exactly the same statistic, critical values and p-values, not merely the same limiting distribution. The Walsh averages are dependent, so their positivity indicators must not instead be treated as an independent binomial sample. For a translation of any null-symmetric sample, every Walsh average increases by ; their positive count is nondecreasing, which also verifies the direction of the one-sided test.
Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 38 4 Solution Created 2026-10-03 Updated 2026-10-14
A statistical functional is a map defined on a class of probability distributions. Its associated functional statistic is , where is the empirical distribution and each is a Dirac measure. For example, 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 is differentiable in a sense strong enough to linearize it at the empirical distribution, and that its derivative in a signed direction is . Averaging the directions over gives the zero direction, so linearity implies . The resulting asymptotic linear representation isIf , the central limit theorem and the negligible remainder giveThus the asymptotic variance is . For a vector functional the integral is . 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 . Under contamination , the first-order bias is , whose worst absolute coefficient is ; bounded influence therefore bounds infinitesimal gross-error bias. The local-shift sensitivity is , equal to 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 , 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 , a unique solution , and a density continuous and positive at . Define quantiles of the contaminated distribution by generalized inversion, since a distribution with a point mass need not attain as an equality. If , its indicator is constant near for sufficiently small contamination, and differentiatingat zero gives . HenceAt the exceptional point , the generalized-inverse quantile remains exactly : below the contaminated cumulative distribution function is below , and its value at exceeds . 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 and gross-error sensitivity . Its jump makes local-shift sensitivity infinite, and its nonzero constant tails make its rejection point infinite. If or the quantile is not unique, this regular derivation does not apply.
Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 38 3 Solution Created 2026-10-03 Updated 2026-10-14
A transformation model consists of a group acting on the sample space and on the set of statistical parameters, with the compatibility propertyIn words, transforming a sample from produces a sample from . A transitive transformation model is generated from one reference distribution by the group action. For densities, the compatibility property includes the Jacobian determinant of the sample transformation; it is not merely equality of density values at transformed points.
A maximal invariant is constant on orbits of a group action and separates them: , and implies for some . An equivariant estimator of the parameter satisfies . Equivariance means the estimate follows the transformed parameter, whereas invariance means the reported value stays unchanged.
First suppose the set of statistical parameters can be identified with the group, as happens for a simply transitive action. An equivariant estimator then provides a group-valued fit with . Normalize the sample by this fit:Indeed . Conversely, if , then , placing in the same orbit of a group action. This establishes maximality, rather than only invariance.
For a transitive action with a nontrivial stabilizer of a reference parameter , a parameter-valued equivariant estimator does not specify a unique normalizing transformation. Choose a section with and form . Under ,Thus the -orbit of is invariant. It is maximal: if with , then . Appropriate measurable sections and orbit of a group action coordinates are assumed when treating this as a statistic. This qualification explains why an arbitrary equivariant statistic alone is not automatically a complete set of invariant coordinates.
For the location-scale family, write with and a fixed joint distribution of . The group consists of maps with , acting on parameters by . For a nonconstant sample of size , take the equivariant estimatorIt gives the maximal invariantIts coordinates satisfy and . They stay unchanged under location and positive-scale transformations. If two samples have the same standardized vector, thenso a single group transformation maps the first sample onto the second. The vector retains the observation labels: sorting it would discard information not removed by this location-scale group. The construction is algebraic and does not require the underlying distribution to have finite moments. Constant samples can be assigned a separate invariant value; they form a single exceptional orbit of a group action and have probability zero under an independent continuous base distribution.
Finally, in a transitive transformation model every invariant statistic is ancillary. Choose taking to and use ; invariance gives . Thus the standardized residual configuration provides a natural conditioning variable, while the equivariant fit contains the location and scale information.
Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 38 2 Solution Created 2026-10-03 Updated 2026-10-14
For an interest parameter and nuisance parameter , the profile likelihood is , where maximizes the likelihood at fixed . Its maximum occurs at the joint maximum-likelihood estimate of . It preserves the optimized likelihood ratios for different interest values and is unchanged by one-to-one reparametrization of the nuisance at each fixed . It generally is neither a marginal sampling density nor a conditional likelihood, and eliminating nuisance parameters by maximization can introduce appreciable small-sample distortion. Differentiating the constrained score equation gives the curvature formulaat a joint stationary point. The corresponding expected-information expression is the efficient information for the interest parameter. Under the regular conditions of Wilks theorem, twice the profile log-likelihood drop has an asymptotic chi-squared distribution with the dimension of the interest parameter as its degrees of freedom.
For scalar nuisance, the Barndorff-Nielsen modified profile likelihood may be writtenwhere the sample-coordinate derivative holds and the ancillary statistic fixed. For vector nuisance the last two quantities are determinants. An equivalent form is , since differentiating the constrained nuisance score gives . The semicolon denotes differentiation in sample coordinates, not another ordinary parameter derivative. This Jacobian determinant adjustment is essential to the definition.
In the present model setThe log-likelihood is . ConsequentlyThe profile score is , and its derivative is . The expression in parentheses is the variance of the fixed design values under positive weights proportional to . Thus the profile log-likelihood is strictly concave for a nonconstant design. Its score changes from to as goes from to , giving a unique finite fit. If all coincide, only the mean is identifiable and inference on alone is not defined; that exceptional design must be excluded.
The specified ancillary coordinates give . At fixed , therefore,Meanwhile the constrained observed information isSubstitution provesEquivalently the sample derivative of the nuisance score is , which equals at the constrained fit and gives the same cancellation. Thus the modified profile likelihood for exponential regression is proportional to the ordinary profile likelihood. This conclusion uses the given ancillary; no proof of its ancillarity is required.
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