Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 219 / 1

Linearity of expectation gives ${\mathbb{E}}_{\mu }[\hat{\mu }]=({\alpha }_1+{\alpha }_2)\mu$. Thus a linear unbiased estimator for every value of the distance modulus satisfies $\boxed{{\alpha }_1+{\alpha }_2=1}$. Requiring unbiasedness at the single value $\mu =0$ would not impose this constraint; an unbiased estimator must work throughout the parameter space.

Independence makes the variance of an estimator equal toWith ${\alpha }_2=1-{\alpha }_1$, differentiate to obtain $2{\alpha }_1{\sigma }_1^2-2(1-{\alpha }_1){\sigma }_2^2=0$. The second derivative is $2({\sigma }_1^2+{\sigma }_2^2)>0$, so the unique minimum isEquivalently, the inverse-variance weighted mean uses ${\alpha }_i=K{\sigma }_i^{\gamma }$ withWe assume positive measurement variances. A zero-variance observation already supplies the true value exactly.

Unbiasedness requires ${\sum }_i{\alpha }_i=1$. Let $S={\sum }_i{\sigma }_i^{-2}$ and minimize $V(\alpha )={\sum }_i{\sigma }_i^2{\alpha }_i^2$. The Lagrange multiplier equations for $V-\lambda ({\sum }_i{\alpha }_i-1)$ are $2{\sigma }_i^2{\alpha }_i=\lambda$, so normalization gives $\lambda =2/S$. HenceThe gradient at this point is $(2/S)1$, whose inner product with every feasible displacement $h$ satisfying $1^{T}h=0$ is zero. The Hessian matrix is $2\mathrm{diag}({\sigma }_1^2,\dots ,{\sigma }_N^2)$, and $h^T{\nabla }^{2}Vh=2{\sum }_i{\sigma }_i^2{h}_i^2>0$ for every nonzero feasible $h$. These are the first- and second-order constrained minimum conditions. Strict convexity also proves global uniqueness of this inverse-variance weighted mean; normal errors are not needed.

Equal variances make every weight $1/N$, so the inverse-variance weighted mean reduces to the arithmetic mean. Independence givesThus the variance scales as $N^{-1}$, whereas the uncertainty expressed as a standard deviation scales as $N^{-1/2}$.

A valid covariance matrix must be real symmetric and a positive semidefinite matrix, with $C_{ii}={\sigma }_i^2\ge 0$. Conversely, every such matrix is the covariance of a possibly degenerate multivariate normal distribution. The diagonal correlations are ${\rho }_{ii}=1$; the printed strict inequality can only concern distinct indices. Pairwise bounds do not suffice: the $3$ by $3$ correlation matrix with diagonal $1$ and every off-diagonal entry $-0.9$ has eigenvalue $-0.8$ along $1$ and is invalid. Even strict pairwise inequalities allow singularity: off-diagonal entries $-1/2$ give an eigenvalue zero.

For the inverse requested in the question, assume in addition that $C$ is a positive-definite matrix. Write $x=({\hat{\mu }}_i)$, $\Lambda =C^{-1}$ and $S=1^T\Lambda 1>0$. Up to a constant, the log-likelihood is $-\frac{1}{2}(x-\mu 1{)}^T\Lambda (x-\mu 1)$. Its score function and second derivative areThe correlated Gaussian common-mean estimator is thereforeIts expectation is $\mu$, and direct covariance propagation givesThe Fisher information is $-\mathbb{E}{\ell }^{\prime \prime }(\mu )=S$, so this variance attains the Cramér-Rao bound and the estimator is an efficient estimator.

If $C$ is singular, an ordinary Lebesgue density and $C^{-1}$ are unavailable. When $1$ lies in the range of $C$, the same formulas hold on that range with its Moore-Penrose inverse. Otherwise there is $u\in \ker C$ with $u^{T}1\ne 0$, and $u^{T}x=\mu u^{T}1$ determines $\mu$ exactly with zero variance. The usual nonsingular Cramér–Rao calculation then does not apply.