Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 219 (Astrostatistics)

For distinct sibling supernovae $s$ and $s^{\prime }$ in the same galaxy, the shared fluctuation is the only common random term. Independence and linearity of covariance therefore giveTheir correlation coefficient isThis is the intraclass correlation coefficient of the galaxy-level random intercept model.

Write the Cepheid measurement as $\hat{\mu }=\mu +{ϵ}_{\mu }$, where ${ϵ}_{\mu }\sim N(0,{\sigma }_{\mu }^2)$, and define the calibrated absolute magnitude dataThus $q=(q_1,\dots ,q_K{)}^T$ has a multivariate normal distribution with mean $M_{0}1$ and covariance matrix

The Hubble law and the definition of distance modulus giveConsequently, withthe Hubble-flow observations are independent random variables satisfying $x_i\sim N(M_0-\theta ,{\sigma }_{\mathrm{tot}}^2)$. Apart from a factor independent of the parameters, the joint likelihood function isThe expression extends by continuity to a singular limiting covariance such as ${\sigma }_I=0$.

Letand setThe within-galaxy contrasts $q_k-\bar{q}$ contain no $M_0$, so the parameter-dependent log-likelihood reduces toThe score equations give the maximum-likelihood estimatorsThe calibrator and Hubble-flow samples are independent, so both estimators are unbiased andThe Fisher information matrix for $(M_0,\theta )$ isHence $\mathrm{Var}(\hat{\theta })=({\mathcal{I}}^{-1}{)}_{22}$: $\hat{\theta }$ attains the multiparameter Cramer-Rao bound.

The invariance property of maximum likelihood estimation givesSince $\hat{\theta }-\theta \sim N(0,{\sigma }_{\theta }^2)$, the ratio $\hat{h}/h$ has a log-normal distribution. Its exact variance isand the lowest-order delta method approximation isFor fixed ${\sigma }_{\mathrm{tot}}$, case (i) has $A={\sigma }_{\mu }^2+{\sigma }_{\mathrm{tot}}^2$, whereas case (ii) has $A={\sigma }_{\mu }^2+{\sigma }_{\mathrm{tot}}^2/K$. Because $K\ge 2$ and $B$ is the same in both cases, independent supernova-level variation in case (ii) gives the smaller fractional variance: averaging reduces it, while a shared galaxy fluctuation does not average away.

The vector $(y_1,y_2,y_3)$ is Jointly Gaussian. Assuming $R>0$, Gaussian conditional independence givesThus the required condition is $R{R}_{13}=R_{12}{R}_{23}$. Under it, conditioning on $y_1$ supplies no further information after $y_2$, and the Gaussian process regression posterior isBoth the conditional expectation and conditional variance depend only on $y_2,t_2,t_3$; neither contains $y_1$ or $t_1$.

For $t_1<t_2<t_3$, the condition from part a becomesIt holds for arbitrary positive time gaps exactly when $\eta =1$. The resulting exponential covariance function is the covariance of a stationary Ornstein-Uhlenbeck process, hence has the Markov property.

Writing $a_{32}=e^{-(t_3-t_2)/\tau }$, the predictive law isAs $t_3\to \infty$, $a_{32}\to 0$, so the predictive mean tends to the stationary mean $\mu$ and the predictive variance tends to the stationary variance $1$.

SetThe Markov factorization and the conditional normal laws from part b give the fully univariate productwhere $\varphi (\cdot ;m,v)$ denotes the $N(m,v)$ density. This is a weighted least squares problem in $\mu$. Differentiating its log-likelihood givesEvery innovation in the numerator has expectation equal to its coefficient in the denominator times $\mu$. Therefore $\mathbb{E}\hat{\mu }=\mu$, so this maximum likelihood estimator is unbiased.

For a fixed observed colour ${\tilde{c}}_s$, Bayes theorem givesCompleting the square shows that this is a truncated normal distribution with variance ${\sigma }_c^2$, lower endpoint zero, and untruncated locationHenceConditioning first on $E_s$ and using the law of total expectation yields the dusty colour-magnitude relationAs ${\tilde{c}}_s\to -\infty$, the Inverse Mills ratio implies ${\overline{E}}_s\to 0$ with vanishing derivative, so the asymptotic slope is $\beta$. As ${\tilde{c}}_s\to +\infty$, ${\overline{E}}_s\sim {\tilde{c}}_s-c_0-{\sigma }_c^2/\tau$, so the asymptotic slope is $R$.

Let $c_s={\tilde{c}}_s-E_s$ and $M_s={\tilde{M}}_s-R{E}_s$. With the stated improper hyperpriors, the unnormalized posterior density isIts Directed acyclic graph has, for each star, the arrowsfollowed by the deterministic observation arrowsThe latent variables $(c_s,M_s,E_s)$ are repeated inside a plate indexed by $s=1,\dots ,N$; all hyperparameters lie outside that plate.

A Gibbs sampler draws each variable from its full conditional distribution, so every proposed update is accepted. At the start of a sweep define $c_s={\tilde{c}}_s-E_s$, $y_s={\tilde{M}}_s-R{E}_s$, andFirst update the reddenings independently asRecompute $c_s$ and $y_s$, then make the Gaussian updatesandWiththe remaining full conditionals, in the parameterization given in the question, are the scaled inverse chi-squared lawsRepeating these updates in the displayed order gives a complete Gibbs sweep. The formulas assume $N>2$ and nondegenerate sampled predictors; posterior propriety must be checked because the hyperpriors are improper.

Using Bayes theorem and the fact that the prior is proper,Thus $\hat{I}$ is an unbiased estimator of $Z^{-1}$, and the Harmonic mean estimator of Bayesian model evidence is $\hat{Z}=1/\hat{I}$. The reciprocal is not itself generally unbiased, though it is consistent when the strong law of large numbers applies.

Multiplying the Gaussian likelihood and prior and completing the square givesTherefore Normal-normal conjugacy gives

The evidence is the convolution of $N(0,{\tau }^2)$ and $N(0,{\sigma }^2)$, soPart i now gives the fully simplified expectation

Let $S={\sigma }^2+{\tau }^2$. For one posterior draw, the Gaussian quadratic-exponential moment is finite precisely when $\tau <\sigma$, and thenBecause the $m$ posterior draws are independent,For $\tau \ge \sigma$ the second moment, and hence the variance, is infinite. The Harmonic mean estimator of Bayesian model evidence is therefore unstable in the usual diffuse-prior regime: posterior sampling does not adequately control the reciprocal likelihood in the posterior tails.

Under $M_1$, Bayes theorem at the nested value $\psi =0$ givesSeparability of the prior and equality of the $\varphi$ priors implyRearranging proves the Savage-Dickey density ratiowhich is the Bayes factor in favor of the nested model.