Cameron-Martin theorem for a Gaussian measure Created 2026-10-05 Updated 2026-10-06
For with injective covariance operator of a Gaussian measure, its translated law is an equivalent probability measure to if and only if , the Cameron-Martin space of a Gaussian measure. For such ,The series has mean-square convergence and converges almost surely under , with normal distribution . Its exponential density has expected value one by the moment-generating function of a normal distribution. Finite-dimensional projections give the formula by ratios of multivariate normal densities; convergence of these likelihood ratios gives the infinite-dimensional result. The expression is only formal when the sample is outside . For , translation gives mutually singular measures.
Gaussian Bayes classifier Created 2026-09-24 Updated 2026-10-14
For two multivariate normal densities, the Bayes log posterior odds are a quadratic polynomial. Equal covariance matrices cancel the quadratic term and give linear discriminant analysis; unequal covariance matrices give quadratic discriminant analysis.
Evaluate Bayes' theorem at the posterior mode :The multivariate normal density in the Laplace approximation has height at its mean. ThereforeEquivalently, the approximate log marginal likelihood isThe determinant factor accounts for the volume of the locally plausible parameter region, which the peak likelihood alone does not measure.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 42 1 i Solution Created 2026-10-03 Updated 2026-10-14
Use the divisor- 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 , the product of the multivariate normal densities givesTo separate the sample mean from the sample covariance matrix, expand . The cross terms sum to zero because . Also by the trace identity. ThereforeThe printed expression suppresses the parameter-independent constant ; dropping this constant has no effect on maximum likelihood estimation or the likelihood-ratio test.
Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 43 1 ii Solution Created 2026-10-03 Updated 2026-10-14
Put , the precision matrix, with scalar and containing all remaining coordinates. Holding in the multivariate normal density, the terms involving in its exponent areCompleting the square gives a normal distribution with mean andEquivalently, the block matrix inverse identity gives . The Gaussian conditional variance from precision is a reciprocal diagonal entry of the precision matrix, rather than the marginal variance .
Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 43 2 i Solution Created 2026-10-03 Updated 2026-10-14
Use the usual sample mean and the divisor- sample covariance matrix,The residual sum is zero, so expanding givesMultiplying the independent multivariate normal densities and taking logarithms yields the log-likelihoodHere is the determinant and denotes the trace. The maximum-likelihood estimators areIf instead denotes the divisor- sample covariance matrix, replace by and then . 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.
Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 43 5 i Solution Created 2026-10-03 Updated 2026-10-14
Under equal costs for either wrong assignment, the Bayes classifier chooses the class of greatest posterior probability. With positive prior probabilities, it assigns class one when ; ties can be resolved arbitrarily. For the two multivariate normal densities with common positive-definite covariance matrix , their normalizing factors cancel. The log posterior odds areTherefore the Gaussian Bayes classifier has the required linear form, withA larger prior probability for class one decreases , expanding its decision region. The prior-dependent Gaussian discriminant boundary is the hyperplane .
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 46 1 a Solution Created 2026-10-03 Updated 2026-10-14
Let . Independence of the standard normal distributions gives , , and joint densityChoose a matrix with , for example the lower-triangular factor from the Cholesky decomposition. It exists and is nonsingular because is a symmetric positive-definite matrix. Then the required simulation isIt has mean and covariance matrix , and every linear combination of its coordinates is Gaussian, so it has the required multivariate normal distribution.
For the multivariate normal density, apply the change of variables formula to . Its absolute Jacobian determinant is , andHenceThe nonsingularity assumption is needed for a density with respect to -dimensional Lebesgue measure.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 46 2 a Solution Created 2026-10-03 Updated 2026-10-14
The inequality in the printed rule is reversed. With the stated definition of , a larger score favors group one. For example, in one dimension take , , , equal priors and . The score is , and , whereas the printed rule assigns group two. The correct Bayes classifier follows from minimizing the error cost as follows.
Let be the region assigned to group one, let be its multivariate normal density, and let each misclassification cost , with zero cost for correct classification. The expected cost isThe permitted integral-minimization result therefore selects . Points of equality may be assigned to either group without changing the risk. Equivalently, the Bayes classifier chooses the larger posterior probability.
For positive priors and a common positive-definite covariance matrix, the normalization constants cancel in the likelihood ratio. Expanding the two quadratic forms givesThus the correct Gaussian Bayes classifier, choosing group one on ties, isThis proves the intended classification result with the necessary correction. Equal error costs remove from the decision, while prior odds determine the threshold.
Past exam of the mathematics course of the University of Cambridge 2010 iii Paper 36 3 b Solution Created 2026-10-03 Updated 2026-10-14
Under the usual regular, identifiable, fixed-dimensional setting, the maximum-likelihood estimator is an interior point with vanishing log-likelihood gradient, and the observed information is positive definite. DefineThe second-order Taylor expansion of the Bayesian deviance isWith a prior density approximately constant over the likelihood's concentration region, the posterior density is therefore proportional toRecognizing the multivariate normal density givesTypically grows at rate and the posterior concentrates in a region of radius , making higher-order terms negligible under the regularity assumptions. The local flatness assumption concerns this region, not necessarily the entire parameter space. Boundary maxima, unidentified directions, or persistent multiple modes can invalidate this approximation.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 32 4 Solution Created 2026-10-03 Updated 2026-10-06
Put , the precision matrix. The conditional independence graph has an edge between and exactly when and are not conditionally independent given all the remaining components. For the multivariate normal distribution,Indeed, the conditional density of the pair given the others has its quadratic term determined by the corresponding two-by-two principal submatrix of ; its cross term vanishes exactly when , making the two conditional Gaussian variables independent. This is a Gaussian graphical model.
To derive the conditional law of one component, isolate in the multivariate normal density and complete the square. The terms involving it areThereforeEquivalently, its mean is and its variance is the Schur complement . These forms agree because the block of gives , and its block then identifies the variance. This proves the relevant conditional multivariate normal distribution formula.
In particular, the population regression of component on the other components has coefficientsintercept , and a Gaussian residual independent of the predictors with variance . These are the conditional regression coefficients from a Gaussian precision matrix. The coefficient is zero exactly when the corresponding edge is absent. Its counterpart has the same zero pattern and sign, but generally a different magnitude.
This motivates nodewise regression: for each , regress on , estimating an intercept as well. For sparse graph estimation, use the Nodewise Lasso criterionThe Lasso penalty encourages many fitted coefficients to be zero and remains usable when the number of predictors exceeds the sample size. Ordinary unpenalized sample regressions would usually give a dense estimated graph. Separate regressions can disagree about a pair, so an undirected graph can use the OR rule for nodewise graph selection, retaining an edge if either fitted direction is nonzero, or the more conservative AND rule for nodewise graph selection, retaining it only if both are nonzero. These are choices for combining estimated neighbourhoods, rather than identities forced by finite-sample fits.
The printed joint objective uses a Group Lasso penalty on the unordered pairsThis is paired group Lasso for Gaussian graph estimation. Each group represents a single undirected edge, coupling its two directional regressions. A sum of group norms encourages entire pairs to be zero; a squared group norm would give smooth ridge-type shrinkage instead of edge deletion. The constraint removes self-prediction, and the intercepts are not penalized.
The free symbols called in the objective are regression intercepts, not generally the marginal means of the original Gaussian vector. Profiling them out givesand replaces all columns by their centred versions in the squared-error terms. This is profiling intercepts in nodewise regression. Scaling predictors to comparable units can also help interpret a common penalty, although the displayed objective itself is well defined on the given data.
The grouped KKT conditions make the sparsity mechanism precise. With centred residuals , defineThe subdifferential of the Euclidean norm at zero is its closed unit ball, and away from zero is . Hence a minimizer satisfiesThese are the paired group Lasso optimality conditions. They jointly threshold the two directional scores, matching the population fact that both directions represent the same edge.
Estimate the graph by including exactly when . This graph is undirected by construction. The penalty does not impose , nor does it mathematically require both coordinates of every selected pair to be nonzero. That distinction is appropriate because the population regression coefficients themselves need not have equal magnitudes. The fitted coefficient matrix is not automatically a symmetric positive definite precision matrix; its pair supports, rather than its entries as a precision estimate, supply the requested conditional independence graph.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 202 4 ii Solution Created 2026-10-03 Updated 2026-10-05
Let , with three independent standard Brownian motions, and set and . The radial stochastic process is a three-dimensional Bessel process, which never hits zero by the Hitting-zero classification for a Bessel process. In three dimensions is a harmonic function away from zero: direct differentiation gives .
For the stopping times , , the Itô formula givesThe coefficients are bounded by , so each stopped Itô integral is a true martingale on every finite horizon. The stopping times form a localizing sequence; as permitted, its divergence need not be proved here. Thus is a nonnegative continuous local martingale.
To see that is not a true martingale, integrate the three-dimensional multivariate normal density of in spherical coordinates. The angular integral yieldswhere is the standard normal distribution function. Therefore the reciprocal of a three-dimensional Bessel process is a strict local martingale.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 216 5 Solution Created 2026-10-03 Updated 2026-10-05
Let be the standard multivariate normal density. Independence gives the proposal distribution densityFor an orthogonal matrix ,The isotropic Gaussian density is unchanged by this transformation. Since has the same distribution as ,Thus this is an orthogonal-mixture Metropolis proposal with a symmetric density; inversion invariance, rather than any assumption of a uniform distribution on orthogonal matrices, is what is needed.
With , the accepted off-diagonal transitions satisfyThe rejection part is supported on the diagonal and satisfies detailed balance automatically. Values of the acceptance rule at starting points where can be specified separately; they do not affect stationarity under . Therefore the Metropolis–Hastings algorithm kernel is reversible with invariant density , proving
Past exam of the mathematics course of the University of Cambridge 2018 ii Paper 2 28K a Solution Created 2026-09-24 Updated 2026-10-03
The risk function of an estimator under the stated quadratic loss isThe multivariate normal density is proportional, as a function of , toso maximum likelihood estimation gives . Writing with ,for every .
Past exam of the mathematics course of the University of Cambridge 2018 ii Paper 3 13C f Solution Created 2026-09-24 Updated 2026-10-03
Comparison with the multivariate normal density shows that is its precision matrix. Since the mean is zero,
Multiplying on the right by gives . Taking the transpose and using the symmetry of and gives . Hence the stationary covariance obeys the Continuous Lyapunov equationThe same equation follows directly by setting the time derivative of the second-moment equation from part (c) to zero.
Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 219 2 b Solution 2026-09-28
Let have entries and letIndependent Gaussian measurement errors give the multivariate normal density
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 3 28K a Solution Created 2026-09-24 Updated 2026-09-25
Let be the multivariate normal density of . Under zero-one loss, the Bayes classifier chooses the class with the larger posterior probability. ThusEquivalently, it chooses class one when the log posterior oddsis nonnegative. The decision boundary is .
If , the terms cancel, leavingan affine function. This is the linear boundary of linear discriminant analysis. If , the quadratic part iswhich is nonzero, so the boundary is a possibly degenerate quadratic hypersurface, as in quadratic discriminant analysis. This is the Gaussian Bayes classifier.
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 207 3 c iii Solution Created 2026-09-24 Updated 2026-10-14
Bayes' theorem turns the categorical distribution prior probabilities and the component multivariate normal densities intoThese probabilities define the label update in the Gibbs sampler.