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Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 301 3 i Solution by
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Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 301 2 iv Solution by
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Every scalar-fermion vertex contracts the scalar momentum with the Dirac current. Between on-shell external spinors,and the analogous particle-antiparticle identity also vanishes. Equivalently, the field redefinition in the previous part turns the theory into a free theory. Consequently every putative tree channel for has zero amplitude and
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 301 2 iii Solution by
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The tree amplitude is, up to an overall convention-dependent sign,The Clifford algebra gives the momentum-space Dirac equation and its adjoint:ThereforeAlthough permits the relativistic two-body decay kinematically, the matrix element vanishes. Thus
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 301 2 ii Solution by
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In four spacetime dimensions , , and . HenceThe coupling is an irrelevant coupling by power counting in quantum field theory, so it is a nonrenormalizable interaction that would ordinarily define only an effective theory with a cutoff. Here it is also a redundant operator: integration by parts gives up to a boundary term, and the Dirac current is conserved. This derivative coupling to a conserved current is removed exactly by the local phase redefinition .
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 301 2 i Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-05
With all momenta incoming and Fourier convention , differentiating the scalar contributes . The sole interaction vertex is thereforewhere enters on the scalar line; reversing the convention reverses the irrelevant overall sign. The free internal lines use the Dirac propagatorand the scalar Feynman propagator . Momentum is conserved at the vertex.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 301 1 i Solution by
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A spacetime translation gives, by Noether theorem, the canonical stress-energy tensorThe Klein-Gordon equation implies . With canonical momentum , the conserved physical three-momentum isThe minus sign follows from for metric signature .
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 219 4 d Solution by
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Every move can be drawn from a full conditional distribution, producing a Gibbs sampler with acceptance probability one. Write and . First update independentlyand thenLet have rows and . Update the linear regression coefficients jointly byand updateFinally, the flat positive variance priors give the following full conditionals, each an inverse-gamma distribution:A systematic sweep in the displayed order, using the newly sampled values immediately, defines the chain. The shapes are positive for ; full column rank of is also required.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 219 4 c Solution by
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For each , place and inside a plate replicated times. The directed edges are represented byThe shaded observed nodes are ; the unshaded nodes are latent; and lie outside the plate. This is the probabilistic graphical model encoded by the joint factorization.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 219 4 b Solution by
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With the stated flat priors, the full joint density, up to a constant, isThe priors on and contribute constants on , while those on the two variances contribute constants on . These are improper priors, so posterior propriety must be checked; the full-rank, sufficiently large-data case used below is proper.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 219 4 a Solution by
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For one object, the probabilistic graphical model factorization isEach factor is the normal distribution density specified by the model. This factorization displays the conditional independences of the latent variables and the noisy observations .
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 219 3 d Solution by
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Write the target posterior density as and the proposal distribution density as . The Metropolis–Hastings algorithm accepts a proposed move withFor distinct states,which is symmetric in and . The rejection probability supplies the diagonal part, so the entire transition kernel satisfies detailed balance. Integrating the detailed-balance identity over the starting state proves . Hence the posterior is a stationary distribution; an irreducible Markov chain that is also an aperiodic Markov chain converges uniquely to it.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 219 3 c Solution by
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Use broad proper uniform priors for , , and over physically plausible ranges, and broad log-uniform priors for the positive scales and . ThenA Random-walk Metropolis algorithm can update with a multivariate Gaussian proposal distribution. Initialize several dispersed chains near plausible cross-correlation delays and near the marginal-likelihood optimum; reject proposals outside the prior bounds; discard warm-up while adapting only the proposal scale and covariance; then freeze the kernel and retain a long run. Evaluate trace plots, autocorrelations, acceptance rates, between-chain agreement, and the effective sample size of a Markov chain. Posterior predictive quasar light curves provide a model check.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 219 3 b Solution by
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At the fitted parameters, let and . For prediction times defineThe microlensing processes and measurement errors contribute no cross-covariance with the latent quasar light curve. The Gaussian process regression posterior is thereforeThe requested pointwise posterior variances are the diagonal entries of the latter matrix.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 219 3 a Solution by
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Stack the observations as and setLet and denote matrices obtained by evaluating the two Gaussian process covariance kernels. Independence of the quasar light curve, gravitational microlensing, and Gaussian noise processes giveswhere . Thus is a multivariate normal distribution and its Gaussian-process marginal likelihood isThe off-diagonal blocks are essential: both images contain the same delayed Ornstein-Uhlenbeck process.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 219 2 e Solution by
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Let and draw independently from an importance density . The unbiased estimatorof has one-sample second momentBy the Cauchy-Schwarz inequality,Equality holds precisely when , giving the optimal importance density for a single integralThis is circular in practice: constructing and normalizing requires detailed knowledge of the posterior and the expectation of . Here log masses are positive, so the unknown normalizer is the posterior mean being estimated. It is also optimal only for this one integral, not for general posterior summaries.
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