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For $X\sim N(\mu 1,C)$ with a known positive-definite matrix $C$, the maximum-likelihood estimator of $\mu$ is $\hat{\mu }=(1^T{C}^{-1}X)/(1^T{C}^{-1}1)$. It is unbiased with variance $(1^T{C}^{-1}1{)}^{-1}$, attaining the Cramer-Rao bound. It is also the best linear unbiased estimator. Pairwise admissible correlation coefficients alone do not ensure that $C$ is a valid covariance matrix; the entire matrix must be symmetric and a positive semidefinite matrix. Inverse-based formulas require positive definiteness.

An efficient estimator attains the relevant lower bound on estimator variance, such as the Cramer-Rao bound in a regular parametric model.

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.

Under homoskedasticity, writeThen ${\hat{M}}_0=\bar{q}$ and $\hat{\theta }=\bar{q}-\bar{r}$. Both are unbiased estimators, and their covariance matrix isIndeed the Fisher information isand its inverse is exactly the displayed covariance matrix. The estimators therefore attain the multivariate Cramer-Rao bound and are efficient estimators.

PutThe central limit theorem gives $\sqrt{n}({\hat{\mu }}_n-\mu ({\theta }_0))\Rightarrow N(0,{\sigma }_h^2)$. Provided ${\mu }^{\prime }({\theta }_0)\ne 0$, the delta method applied to ${\mu }^{-1}$ yieldsStrict monotonicity alone does not exclude ${\mu }^{\prime }({\theta }_0)=0$; in that exceptional case this root-$n$ limit need not hold and higher-order analysis is required.

One does not necessarily have $\mathrm{Var}({\hat{\theta }}_n)\ge [nI({\theta }_0){]}^{-1}$ for every $n$, because ${\hat{\theta }}_n$ need not be unbiased. The Cramer-Rao bound for an estimator with bias $B_n(\theta )$ is insteadunder the corresponding regularity assumptions.

The assumed maximum-likelihood estimator is the sample mean of $h(X_i)$. Its central limit theorem isFor a regular model, the regular maximum-likelihood limit isThe two asymptotic variances must agree, soConsequentlyto first asymptotic order. The estimator therefore attains the Cramer-Rao bound asymptotically: its precision, the reciprocal variance, exhausts the information in the sample.

Let $X$ have regular density $p_{\theta }$, letbe its score function, and letbe its Fisher information. Regularity gives the mean-zero score identity ${\mathbb{E}}_{\theta }{S}_{\theta }=0$.

Suppose an estimator $T$ has mean ${\mathbb{E}}_{\theta }T=\psi (\theta )$. Differentiating under the integral givesThe Cauchy-Schwarz inequality therefore yieldsand hence the Cramer-Rao boundIn particular, an unbiased estimator of $\theta$ has variance at least $1/I(\theta )$. For an independent sample, the information is the sum of the individual informations.

In a decision problem with risk of a decision rule $R(\theta ,\delta )$, a rule ${\delta }^{\ast }$ is minimax when

Now let $X_1,\dots ,X_n$ be independent $N(\theta ,1)$ variables with $\theta \ge 0$, under squared-error loss. The sample mean is unbiased with variance $1/n$, sofor every $\theta \ge 0$. Thus the minimax value is at most $1/n$.

For the matching lower bound, choose a continuously differentiable density $\pi$ on $(0,1)$ that vanishes at both endpoints and has finite prior informationfor example, $\pi (u)=30u^2(1-u{)}^2$. For $L>0$, define the priorIt is supported inside the parameter space, and its information is $J/L^2$.

Here the likelihood score iswhose Fisher information is $n$. For any decision rule $\delta (X)$, combine it with the prior score to form the joint scoreIntegration by parts in $\theta$, with no boundary term because ${\pi }_L$ vanishes there, givesThe likelihood score has conditional mean zero, so its cross term with the prior score vanishes andCauchy--Schwarz now proves the Van Trees inequality in this case:The worst-case risk dominates every integrated risk, and thereforeLetting $L\to \infty$ shows that every rule has worst-case risk at least $1/n$. Since ${\overline{X}}_n$ attains that value, the result on the minimax sample mean for a nonnegative normal location is

The model factors into the normal distribution densitiesand, conditionally on $X_{i-1}$,Ignoring constants independent of $\theta$, the log likelihood isIts score isIn terms of the independent innovations,Variance additivity for independent random variables therefore yields

Since $I_1(\theta )=1$, the equality $I_n=n{I}_1$ holds for all $n$ exactly whennamely whenThus the information tensorizes only in the independent case.

Finally, the Cramer-Rao bound gives every unbiased estimator $\hat{\theta }$ the lower boundThis is the Fisher information in a stationary Gaussian autoregressive location model.

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 risk of ${\overline{X}}_n$ is constantly $1/n$. Suppose an estimator $\delta$ dominates it, and put
$m(\theta )={\mathbb{E}}_{\theta }\delta$ and $b(\theta )=m(\theta )-\theta$. The biased Cramer-Rao bound, using total information $n$, givesDomination would implyfor every real $\theta$. The standard differential-inequality argument shows that the only globally defined differentiable solution is $b\equiv 0$: a nonzero value forces $b^{\prime }$ to retain a sign and magnitude that makes $b$ leave the permitted bounded interval in one time direction.

Thus $m(\theta )=\theta$. The Cramer--Rao inequality now gives
${\mathrm{Var}}_{\theta }\delta \ge 1/n$, so domination forces equality everywhere. Equality in Cramer--Rao makes the centred estimator proportional to the Gaussian score:and hence $\delta ={\overline{X}}_n$ almost surely. No strict improvement exists, so ${\overline{X}}_n$ is admissible.

$I(\theta )=E_{\theta }[({\partial }_{\theta }\log f(X;\theta ){)}^2]$. The biased Cramer-Rao bound bound is ${\mathrm{Var}}_{\theta }\delta \ge (1+b^{\prime }(\theta ){)}^2/I(\theta )$.

Combining the risk assumption with Cramer-Rao bound gives $(1+b^{\prime }{)}^2\le \gamma -I{b}^2$. Taking the branch compatible with super-efficiency yields $b^{\prime }\le -[1-\sqrt{\gamma -I{b}^2}]\le -(1-\sqrt{\gamma })$, equivalent to the displayed sharper pointwise inequality.