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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.

The estimator $U=X_1{X}_2$ is unbiased for $p^2$ because the independent random variables $X_1,X_2$ satisfy $\mathbb{E}[X_1{X}_2]=p^2$. Given $T=t$, all placements of the $t$ successes are equally likely, soThus the Rao-Blackwellized estimator isIt is unbiased by the tower property. Since $n\ge 3$ and $p\in (0,1)$, the event $T=2$ has positive probability, and conditionally on it $X_1{X}_2$ takes both zero and one with positive probability. Hence $\mathbb{E}[\mathrm{Var}(X_1{X}_2\mid T)]>0$, so the new estimator has strictly smaller variance.

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.

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.

Write the linear mean model asThe errors are independent but heteroscedastic. Minimizing the unweighted residual sum of squaresgives the normal equationThus, provided ${\sum }_i{x}_i^2>0$,Since $\mathbb{E}{Y}_i=\theta x_i$,Hence it is an unbiased estimator. This is the least-squares part of the estimators for a Poisson exposure model.

Because $I-H$ is an orthogonal projection of rank $n-p$,It follows that Mallows Cp satisfieswhere the last equality is part (i). Thus $C_p$ is an unbiased estimator of the independent-copy prediction error, as summarized by the unbiased prediction-error identity for ordinary least squares.

For a correctly specified full-rank fixed-design Gaussian linear regression with $p<n$, dividing the residual sum of squares by the residual degrees of freedom $n-p$ gives an unbiased estimator of the noise variance. The projection removes the whole deterministic mean, so the numerator's expectation is $(n-p){\sigma }^2$. A misspecified subset model need not have this property.

In a full-rank normal linear model with $n$ observations and $p$ mean coefficients, the residual standard error estimates the common error standard deviation. Its square is the unbiased estimator $\mathrm{RSS}/(n-p)$; the square root itself is not generally unbiased. It measures unexplained response variation, whereas a regression coefficient's standard error measures uncertainty in that coefficient.

For $Y\sim N_n(\mu ,{\sigma }^{2}I)$ and a fixed orthogonal projection matrix $P$ of rank $k$, the displayed expression is an unbiased estimator of the mean-vector prediction risk whenever ${\hat{\sigma }}^2$ is an unbiased estimator of ${\sigma }^2$. Independence of its two terms is unnecessary. Comparing fixed models yields the Mallows Cp penalty, but minimizing unbiased estimates does not preserve unbiasedness after selection.

The uncentered empirical second-moment matrix averages the outer products of the observations. Its expected value is $\mathrm{Cov}(X)+\mathbb{E}[X]\mathbb{E}[X{]}^{\top }$. It is an unbiased estimator of the covariance matrix when the population expected value is known to be zero. It differs from a sample covariance matrix centered at the sample mean.

For a normal distribution $N_n(\mu ,{\sigma }^{2}I)$ with unrestricted $\mu \in {\mathbb{R}}^n$ and unknown ${\sigma }^2$, an integrable data-only unbiased estimator of the mean-vector prediction risk of a fixed rank-$k$ orthogonal projection matrix exists exactly when $2k=n$. To prove necessity, randomize the mean by independent Gaussian variance $v$: conditioning adds $(n-k)v$ to the squared bias term, whereas the marginal variance identity would add $kv$. When $2k=n$, the residual sum of squares itself is unbiased.

A U-statistic averages a symmetric function of $m$ distinct sample observations over all unordered $m$-tuples. For independent and identically distributed observations it is an unbiased estimator of $\mathbb{E}H(X_1,\dots ,X_m)$. Omitting repeated indices can remove diagonal contributions in quadratic plug-in estimators.