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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 Cramer-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.