Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 218 4 c Solution Created 2026-10-03 Updated 2026-10-05
Under a correctly specified Poisson regression with adequate expected counts, the residual Poisson deviance is approximately chi-squared distribution with residual degrees of freedom. The observed givesThus the ordinary Poisson model gives a very poor fit under this goodness-of-fit approximation. This is consistent with overdispersion, although a wrong mean model or dependence can also cause lack of fit; quasi-Poisson variance inflation alone does not establish that the mean model is adequate.
For Quasi-Poisson regression, . The Pearson dispersion estimator isIt uses squared Pearson residuals, not the residual deviance divided by 102. The fitted coefficients are unchanged, and their estimated covariance matrix is multiplied by . With , standard errors increase by , so the main treatment coefficient hasThe main Progabide coefficient is not significant at 5% after this dispersion adjustment. R uses a Student's t-distribution when dispersion is estimated, as described in its GLM summary documentation; the large-sample normal approximation also gives the same decision.
Because an interaction term is present, this tests the treatment contrast at baseline zero. An overall absence of treatment influence requires the joint null . The printed marginal standard errors do not give their covariance, so they are insufficient to calculate that joint Wald test, or the treatment contrast test at an arbitrary nonzero baseline.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 219 1 v Solution Created 2026-10-03 Updated 2026-10-14
A valid covariance matrix must be real symmetric and a positive semidefinite matrix, with . Conversely, every such matrix is the covariance of a possibly degenerate multivariate normal distribution. The diagonal correlations are ; the printed strict inequality can only concern distinct indices. Pairwise bounds do not suffice: the by correlation matrix with diagonal and every off-diagonal entry has eigenvalue along and is invalid. Even strict pairwise inequalities allow singularity: off-diagonal entries give an eigenvalue zero.
For the inverse requested in the question, assume in addition that is a positive-definite matrix. Write , and . Up to a constant, the log-likelihood is . Its score function and second derivative areThe correlated Gaussian common-mean estimator is thereforeIts expectation is , and direct covariance propagation givesThe Fisher information is , so this variance attains the Cramér-Rao bound and the estimator is an efficient estimator.
If is singular, an ordinary Lebesgue density and are unavailable. When lies in the range of , the same formulas hold on that range with its Moore-Penrose inverse. Otherwise there is with , and determines exactly with zero variance. The usual nonsingular Cramér–Rao calculation then does not apply.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 335 1 i Solution Created 2026-10-03 Updated 2026-10-05
Use the time-harmonic wave convention and write , , with a deterministic incident wave envelope. The printed speed ratio is inconsistent with : the refractive index used below is . Substituting in the Helmholtz equation givesThe paraxial approximation discards . The usual weak-fluctuation model also linearizes , givingDropping the quadratic contrast is an additional weak-fluctuation assumption, not a consequence of small propagation angles. Although the linearized random potential generates attenuation of order , it does not retain every effect of that order in the literal finite-correlation index: the discarded quadratic contrast can also produce a mean wave phase shift. This linearization must precede a Gaussian white noise limit: the square of ideal white noise has no ordinary pointwise meaning.
The split-step Fourier method alternates free-space diffraction, , with a random phase screen,For jointly Gaussian random fields, Gaussian phase averaging gives . More generally, a product of fields and conjugate fields picks up , with signs or . Its screen average is determined entirely by the wave phase covariance matrix. The deterministic diffraction step acts on each coordinate, with opposite signs on factors formed by complex conjugation. This is the basis of the field-moment equations.
There is an important closure qualification. A stationary Gaussian random field need not have independent longitudinal increments. For finite-correlation fluctuations, the unlinearized parabolic wave equation instead givesIn the weak-fluctuation model, the exact first-moment equation isThe last term cannot in general be replaced by a constant times . An exact generic solution of the linearized model iswhere denotes time ordering of the propagation operators. For the unlinearized model, add inside the propagation generator. The covariance function of the medium is needed to evaluate this expression; One-point Gaussian distributions alone would not even determine the joint wave phase statistics.
For example, omit diffraction and take a longitudinal autocovariance function . Direct Gaussian phase averaging givesEven here, unit-variance stationary Gaussian random fields with different give different answers. In the general problem the diffraction and multiplication operators do not commute, so this scalar attenuation cannot simply be multiplied by without an additional approximation.
The standard closed answer uses the Markov approximation for a random medium, made explicit in part ii. Let the longitudinally integrated covariance kernel beReplace the medium by longitudinal Gaussian white noise with this strength. A screen of thickness then has and is independent of the incoming field. For the moment , expanding both steps to order givesIn particular, the coherent attenuation in a white-noise random medium and its Fresnel propagator solution areEquivalently, write , where is Brownian motion in with transverse covariance kernel . The Stratonovich integral formulation is . Its Itô integral form isThe mean of the Itô integral vanishes, independently confirming the attenuation drift. For in two transverse dimensions,Thus a unit plane wave has . For a general incident wave envelope, the Fresnel propagator supplies its spreading. The attenuation is redistribution between the coherent and diffuse wave fields, rather than wave absorption: for a field and its conjugate at the same point, the screen contribution in the second-moment equation cancels.
Past exam of the mathematics course of the University of Cambridge 2019 ia Paper 2 11F a iii Solution Created 2026-09-24 Updated 2026-09-29
The covariance matrix isIf , the moment-generating function in part (ii) factors into the product of the two marginal moment-generating functions. Thus and are independent random variables; this is the independence of uncorrelated jointly normal variables.
By bilinearity of covariance,ChooseThen is bivariate normal by part (i) and has zero covariance, so is a normal random variable independent of . Hence .
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 219 1 c Solution Created 2026-10-03 Updated 2026-10-14
Under homoskedasticity, writeThen and . Both are unbiased estimators, and their covariance matrix isIndeed the Fisher information isand its inverse is exactly the displayed covariance matrix. The estimators therefore attain the multivariate Cramér-Rao bound and are efficient estimators.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 349 2 Solution Created 2026-10-03 Updated 2026-10-05
Use a stationary, non-streaming tracer population with spherical symmetry in its full velocity distribution. In particular, the mean velocities vanish, the mixed covariances vanish, and the two tangential components have equal velocity dispersions. WriteHere is the variance of one tangential component. In the alternative convention using the sum of both tangential variances, the same velocity-anisotropy parameter is . These definitions agree; they must not be mixed by inserting an extra factor of two.
The time derivative and mean-velocity terms in the supplied Jeans equation vanish, as do its angular mixed-moment terms. Its remaining geometric term is . Thus the Spherical Jeans equation becomesThe no-streaming assumption excludes a steady radial flow; stationarity of the density alone would not remove its advective terms.
By the shell theorem, the gravitational potential of a spherical mass distribution satisfies . Expanding the derivative above, dividing by , and using logarithmic derivatives givesThe tracer density need not be the density of the gravitating matter. The circular speed is , whether or not the tracers themselves follow circular orbits.
Now let be constant, , and . With , the Spherical Jeans equation readsFor a power-law ansatz , , so . Therefore the scale-free spherical Jeans solution isThe positivity condition is necessary for this particular solution to be a physical velocity dispersion.
There is a boundary condition implicit in selecting this solution. The integrating factor gives, when ,The homogeneous term represents an additional radial-pressure boundary contribution. It is removed, for example, by requiring at infinity. Under that condition,and convergence requires , giving the boxed result. If , the solution instead has the logarithmic formso the quoted constant proportionality does not apply. A finite outer boundary can retain a homogeneous term even when .
To derive the heliocentric projection of a spherical velocity dispersion, put the Sun at vector of length , and a tracer at . Its heliocentric line of sight isIf is the angle between and the Galactocentric radial direction, thenAfter correction for the Sun's motion, the intrinsic covariance matrix has eigenvalues in the local spherical basis. Projection onto the line of sight therefore givesAt fixed , and are constant across the spherical shell. Averaging its squared line of sight velocity, with zero mean, givesFor a uniformly sampled Galactocentric shell, the solid angle measure makes uniform on , giving the geometric correction for heliocentric velocity dispersionTo evaluate it, set and , and divide the integrand:The odd term integrates to zero, while . For , , and , this givesThe removable limit at is . As , , while as , . If , and the observed velocity dispersion is exactly radial. For isotropic velocities, , the correction vanishes at every radius.
The averaging measure matters. If directions are instead uniform in heliocentric solid angle at a fixed Galactocentric radius , let be the angle between the line of sight and . Geometry gives , so that particular weighting gives exactly. The logarithmic expression above uses uniform tracer weighting on a Galactocentric shell. For an observational sample, use its actual angular selection weights in ; the boxed dispersion relation remains valid.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 350 1 b Solution Created 2026-10-03 Updated 2026-10-05
Write , so the data map is . The Gaussian likelihood and prior distribution giveCompleting the square, or using Gaussian conjugacy for a normal linear model, identifies a multivariate normal distribution with precision matrix . Since , its posterior mean and covariance matrix areThe parameters areFor example, , which checks the square completion.
The orthogonal vectorsare eigenvectors of the covariance matrix. Their respective variances areThey are independent under the posterior distribution, by independence of uncorrelated jointly normal variables. Thus uncertainty is different in the two directions even for positive noise: the data constrain , whereas spans the null space of the data map.
As , the posterior distribution converges to the law ofThis is supported on , and its posterior mean is the minimum-norm least-squares solution. The prior distribution still determines the distribution along the unobserved null space; it has not disappeared. This example of a Gaussian posterior in the zero-noise limit distinguishes exact recovery of an observed component from recovery of the whole unknown.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 350 2 a Solution Created 2026-10-03 Updated 2026-10-05
The conditional mean estimator is the posterior meanprovided the first moment is finite. A maximum a posteriori estimate is any maximizer of the posterior density relative to Lebesgue measure:Existence of the posterior density alone does not guarantee a finite posterior mean or existence of a maximizing point.
For the stated multivariate normal distribution and independent standard Gaussian noise, the negative log of the posterior density, up to a constant, isIts precision matrix is a positive-definite matrix, so Gaussian conjugacy for a normal linear model givesThe two forms agree becauseThe covariance matrix is , and the maximum a posteriori estimate equals the posterior mean for this multivariate normal distribution.
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 1 30K a Solution Created 2026-09-24 Updated 2026-09-29
The time-zero budget constraint isso . The terminal wealth is thereforePutThenand the target mean is the linear constraint .
Since the invertible covariance matrix is positive definite, the Lagrange multiplier equation for minimizing isThusThe constraint determinesThis is the one-period Gaussian minimum-variance portfolio; the riskless holding is then .
Past exam of the mathematics course of the University of Cambridge 2020 ii Paper 3 28J ii Solution Created 2026-09-24 Updated 2026-09-29
Let be the likelihood function. Since the prior distribution is , the posterior density has the formwhere is the normalizing constant.
Choose and use the Gaussian autoregressive proposal reversible with respect to a standard normal distributionThuswhose covariance matrix is the required . Reversibility with respect to the standard normal distribution sayssoThe Metropolis–Hastings acceptance probability therefore reduces toThis is the Preconditioned Crank–Nicolson algorithm with proposal scale .
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 218 3 d Solution 2026-09-28
Principal component analysis centers the word-count vectors, diagonalizes their sample covariance matrix, and projects onto the eigenvectors with the largest eigenvalues. Fitting logistic regression to principal-component scores removes exact collinearity and yields a lower-dimensional design.
Each principal component is an unsupervised linear combination of many words chosen to explain predictor variance; it is not a word selected for association with spam. The dimension can be chosen from a scree plot or cumulative explained variance, but cross-validating the downstream classification loss better targets prediction.
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 219 1 b Solution 2026-09-28
Write the Cepheid measurement as , where , and define the calibrated absolute magnitude dataThus has a multivariate normal distribution with mean and covariance matrix
The Hubble law and the definition of distance modulus giveConsequently, withthe Hubble-flow observations are independent random variables satisfying . Apart from a factor independent of the parameters, the joint likelihood function isThe expression extends by continuity to a singular limiting covariance such as .
Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 218 6 c i Solution 2026-09-28
For and centered data, the objective is the Rayleigh quotientThis is the first-direction optimization in principal component analysis. Therefore is any unit eigenvector of the sample covariance matrix corresponding to its largest eigenvalue.
Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 339 1 d Solution 2026-09-28
Define the softmax weightsThey obey and . The gradient is their weighted mean,Differentiating once more gives the covariance-form Hessian matrixFor every unit vector ,where . The Hessian is a covariance matrix, so it is positive semidefinite; the displayed upper bound also gives in the Loewner order. Consequently has a Lipschitz gradient with
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 218 2 e Solution Created 2026-09-24 Updated 2026-09-25
Put , so and with independent noise vectors having covariance matrix . For any deterministic linear smoother ,whereas independence givesExpanding the first trace shows that the second expression exceeds the first by , proving the identity.
For ridge regression,Its effective degrees of freedom are . Thus training error is optimistically biased for independent-copy prediction error by . The graph exhibits exactly this effect: the dashed training curve keeps falling as decreases, while the solid test curve eventually rises through overfitting.
Let be the column-centered data matrix. Its sample covariance matrix is . The loadings matrix has orthonormal eigenvectors of as columns, ordered by nonincreasing eigenvalues:Each principal component loading specifies the coefficients of a principal direction in the original variables. Signs are arbitrary, and equal eigenvalues permit orthogonal changes of basis within their eigenspaces.
The least-squares estimators depend only on the observed responses and design, so the formulas in i still hold. Their variances and covariance need only the error covariance matrix, now . ThereforeFor example the first formula follows from , and the second from . The formulas remain the same as before with ; this calculation does not establish independence or a normal distribution for the estimators.
For the random vector , its covariance matrix isIts diagonal entries are variances, and its off-diagonal entries measure pairwise covariance. Finite second moments make every entry well defined.
The future mean response is . By invariance of maximum likelihood estimation, its estimator is the fitted meanIts sampling variance is the quadratic form of the coefficient covariance matrix with :This estimator is unbiased because each fitted coefficient is unbiased. The future error does not enter the estimator of the mean, so its independent variance is not part of this expression.
Write . Since ,The errors have mean zero and covariance matrix . ThereforeThus is an unbiased estimator. In addition, the multivariate normal distribution is preserved by linear maps, so is multivariate normal with the displayed mean and covariance matrix.