= Solution
A valid <covariance matrix> must be real symmetric and a <positive semidefinite matrix>, with $C_{ii}=\sigma_i^2\ge0$. 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 $\mathbf1$ 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 $x=(\widehat\mu_i)$, $\Lambda=C^{-1}$ and $S=\mathbf1^T\Lambda\mathbf1>0$. Up to a constant, the <log-likelihood> is $-\frac12(x-\mu\mathbf1)^T\Lambda(x-\mu\mathbf1)$. Its <score function> and second derivative are
$$
\ell'(\mu)=\mathbf1^T\Lambda(x-\mu\mathbf1),\qquad\ell''(\mu)=-S<0.
$$
The <correlated Gaussian common-mean estimator> is therefore
$$
\boxed{\widehat\mu_{\rm MLE}=\frac{\mathbf1^T\Lambda x}{\mathbf1^T\Lambda\mathbf1}=\frac{\sum_{i,j}\Lambda_{ij}\widehat\mu_j}{\sum_{i,j}\Lambda_{ij}}}.
$$
Its expectation is $\mu$, and direct <covariance> propagation gives
$$
\boxed{\operatorname{Bias}(\widehat\mu_{\rm MLE})=0,\qquad\operatorname{Var}(\widehat\mu_{\rm MLE})=\frac{\mathbf1^T\Lambda C\Lambda\mathbf1}{S^2}=S^{-1}}.
$$
The <Fisher information> is $-\mathbb E\ell''(\mu)=S$, 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 $\mathbf1$ 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 $u^T\mathbf1\ne0$, and $u^Tx=\mu u^T\mathbf1$ determines $\mu$ exactly with zero <variance>. The usual nonsingular Cramér–Rao calculation then does not apply.
Back to article page
