Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 219 1 c Solution 2026-10-03
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 Cramer-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.
Precision matrix 2026-09-24
The precision matrix is the inverse of a nonsingular covariance matrix. For a multivariate normal distribution, a zero off-diagonal precision entry encodes conditional independence of the corresponding coordinates given all remaining coordinates.
Pseudo-covariance 2026-10-06
For a centered complex random variable , the pseudo-covariance is . Together with it determines the real covariance matrix: , , and .
Rank-one covariance spike Created 2026-10-06 Updated 2026-10-07
For a unit vector and , a rank-one covariance spike is the covariance matrix . Its eigenvalue is in the direction and one on the orthogonal complement. It arises by adding an independent scalar normal distribution signal to a standard Gaussian random vector.
Restricted maximum likelihood 2026-10-05
Restricted maximum likelihood estimates covariance matrix parameters from linear combinations of the data whose distribution does not involve the fixed mean coefficients. It differs from a restricted maximum-likelihood estimator obtained by maximizing an ordinary likelihood function under a null hypothesis.
Sudakov-Fernique inequality 2026-10-06
For centered multivariate normal distributions, if for every pair, then the displayed comparison holds. For a quick proof, make independent and set . Apply Gaussian integration by parts to . With softmax function weights and , the derivative of is . Since , let . Singular covariance matrices follow by adding identical small independent normal noise to both vectors and taking a limit.
Treatment contrast 2026-10-07
A treatment contrast is a linear contrast of cell means comparing treatment responses, with coefficients adding to zero. Pairwise differences and marginal factorial differences are examples. Its variance depends on the true allocation units and their covariance matrix.
Treatment interaction contrast in a Poisson regression Created 2026-10-05 Updated 2026-10-06
In a Poisson regression with , the treatment logarithm of the mean ratio at baseline is . The main treatment coefficient tests the contrast at . An overall no-treatment test requires both and , with their joint covariance matrix for a Wald test.
Two-stage residual mixture fitting 2026-10-06
Fitting a mixture model to regression residuals treats the fitted mean adjustment as fixed. Even with independent homoscedastic errors, ordinary least-squares residuals have covariance matrix and satisfy fitted linear constraints, so they are not exactly independent observations. A joint mixture regression with shared slopes or a bootstrap of the whole fitting process better accounts for the first-stage estimation uncertainty.