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An unbiased estimator $T$ of a parameter $\theta$ satisfies ${\mathbb{E}}_{\theta }[T]=\theta$ for every permissible value of $\theta$. Since the sample mean is linear,so $\hat{\mu }$ is unbiased.

Use the identityThe two expectations on the right are $n{\sigma }^2$ and $n\mathrm{Var}(\overline{X})={\sigma }^2$, respectively. HenceThusThe unbiased sample variance instead divides the same sum of squares by $n-1$.

The sample mean isWrite the Gaussian sample vector as its orthogonal projection onto the span of $(1,\dots ,1)$ plus its projection onto the orthogonal complement. These two Gaussian projections are independent. Their squared standardized lengths give Cochran's theorem:

Because every column of $X$ has sample mean zero, every column $u_i$ of $U=XV$ also has sample mean zero. The spectral theorem for real symmetric matrices givesHence for $i\ne j$ the sample covariance is $n^{-1}{u}_i^T{u}_j=0$, while the sample variance of $u_i$ isThus the $u_i$ are pairwise uncorrelated sample principal components, ordered by decreasing sample variance.

A statistic $T$ is sufficient for $\theta$ if the conditional distribution of the full sample given $T$ does not depend on $\theta$. It is minimal sufficient if it is a function of every sufficient statistic.

The likelihood factors asso the Fisher-Neyman factorization theorem shows that $M$, and hence its one-to-one transform $\hat{\theta }=M/2$, is sufficient. Moreover, for two samples $\mathbf{x}$ and $\mathbf{y}$, the ratio $L(\theta ;\mathbf{x})/L(\theta ;\mathbf{y})$ is independent of $\theta$ exactly when their maxima agree. The likelihood-ratio criterion for minimal sufficiency therefore shows that $M$ and $\hat{\theta }$ are minimal sufficient.

For $n\ge 2$, the sample mean $\tilde{\theta }$ is not sufficient: two samples can have the same mean but different maxima, and their likelihood ratio then depends on $\theta$ through the support indicators. It is consequently not minimal sufficient. For the degenerate special case $n=1$, the sample mean and maximum coincide and both conclusions reverse.

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 Cramér-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

Assuming $g$ is measurable, applying it separately to independent and identically distributed random variables preserves those properties. Thus $W_i=g(X_i)$ have common mean ${\mu }_g$ and variance ${\sigma }_g^2$. For ${\sigma }_g^2>0$, the central limit theorem givesHence, for large $n$, the approximate probability distribution isThe second argument is the variance. If ${\sigma }_g^2=0$, this sample mean equals ${\mu }_g$ almost surely, so the appropriate description is a point mass at ${\mu }_g$.

The central limit theorem states that if $W_1,W_2,\dots$ are independent and identically distributed random variables with finite mean $\mu$ and finite positive variance ${\sigma }^2$, thenHere the arrow means convergence in distribution: for every real $z$, the probability that the expression on the left is at most $z$ tends to $\Phi (z)$, the standard normal distribution function. If ${\sigma }^2=0$, the variables equal $\mu$ almost surely and their sample mean is exactly $\mu$.

The recurrence relation at $k=3$ gives $I_5=2\theta I_3=2{\theta }^2$. ConsequentlyBy variance additivity for independent random variables, the variance of their sample mean is

The algebraic identity ${\hat{\beta }}_0-{\beta }_0=n^{-1}{\sum }_i{ϵ}_i$ does not require a normal distribution for the errors. Applying the central limit theorem to their sample mean givesThus, for large $n$,Its mean ${\beta }_0$ and variance ${\sigma }^2/n$ remain exact, even though its normal distribution is now only an approximation. With the error probability distribution unspecified, the formula is retained as an estimator; it is not being asserted to remain a maximum likelihood estimation formula.

The sample mean is exactly $N(\mu ,{\sigma }^2/n)$, so $(\overline{X}-\mu )/(\sigma /\sqrt{n})$ has the standard normal distribution. The given quantile gives probability $0.95$ to the interval $[-1.96,1.96]$. Inverting this pivot yields the exact $95\%$ confidence interval

For observations ${\mathbf{x}}_i\in {\mathbb{R}}^p$, K-means clustering seeks nonempty clusters in cluster analysis $C_1,\dots ,C_K$ and centroids ${\mathbf{m}}_k$ minimizingAt fixed centroids, assigning each point to a nearest centroid minimizes its squared Euclidean distance. At fixed assignments, the centroid sum-of-squares identity shows that the sample mean uniquely minimizes each group's contribution. These alternating steps cannot increase $W$.

For finite termination of K-means clustering, more than a lower bound on $W$ is needed. Keep an assignment unchanged when nearest centroids tie, and stop when assignments no longer change. Every change then strictly decreases $W$. After updating the centroids, $W$ is determined by the partition of a set of $n$ indices; there are only finitely many choices for that partition of a set, so no strictly improving partition of a set can recur. Empty clusters in cluster analysis require a convention, such as moving a point from a nonsingleton cluster in cluster analysis to become its own centroid, which cannot increase $W$. With such a consistent convention the algorithm terminates, although it may reach only a local minimum.

At sample size 100, keep essentially the same numerical confidence interval, but treat its calibration as approximate. For independent heights with finite, nonzero variance, the central limit theorem makes the sample mean approximately normally distributed, and the sample variance consistently estimates the population variance. Hence the studentized sample mean is approximately standard normal even without a normal distribution for individual heights.

Using the permitted cutoff 2 givesA Student t-distribution cutoff gives a similar approximation for large sample size, but its justification at fixed sample size is no longer exact. Extreme skewness or influential outliers could make 100 observations insufficient for accurate coverage probability; abandoning normality alone supplies no specific numerical correction. The sampling with replacement assumption still precludes a finite population correction.

For this rate parametrization of the Laplace distribution, $\mu$ is its mean and median. By symmetry, $\mathbb{E}(X-\mu )=0$, while $|X-\mu |$ has exponential distribution with rate $\lambda$, soThe integral follows by two applications of integration by parts. The central limit theorem therefore givesWith known $\lambda$ and the supplied cutoff 2, an approximate 95% confidence interval isThis known-rate Laplace location confidence interval uses the sample mean; the later maximum-likelihood estimator of location need not be used to construct this valid asymptotic confidence interval.

For equal one-second observation windows, the sample mean estimates the Poisson distribution mean:This matches the expected mean count; it is also the maximum-likelihood estimator for independent Poisson distribution observations. With $N=12169$ windows, fitted expected frequencies are $E_j=N{e}^{-\hat{\lambda }}{\hat{\lambda }}^j/j!$ for $j=0,\dots ,4$, and $E_{\ge 5}=N[1-{\sum }_{j=0}^4{e}^{-\hat{\lambda }}{\hat{\lambda }}^j/j!]$.

Each last-column entry is $(O_j-E_j{)}^2/E_j$, and the Pearson statistic is $\boxed{X^2=\sum _{j\in \{0,1,2,3,4,\ge 5\}}(O_j-E_j{)}^2/E_j}$. The displayed expected frequencies and contributions are rounded, so their printed values need not reproduce every final digit of a calculation using unrounded probabilities.

A $100(1-\alpha )\%$ confidence interval procedure produces random endpoints $L,U$ such that ${\mathbb{P}}_{\mu }(L\le \mu \le U)=1-\alpha$, under the model. In everyday terms, repeating the same sampling and interval construction many times would give intervals containing the true mean in about $100(1-\alpha )\%$ of repetitions.

For $n\ge 2$, define the sample mean $\bar{X}=n^{-1}{\sum }_i{X}_i$ and the sample variance $S^2=(n-1{)}^{-1}{\sum }_i(X_i-\bar{X}{)}^2$. For independent observations from a normal distribution with ${\sigma }^2>0$, the standard sampling results areThe first follows from the independent normal sample mean and sample variance. Let $q=t_{n-1,1-\alpha /2}$ be the positive Student quantile. Symmetry gives the central confidence intervalIts width is $W=2qS/\sqrt{n}$, so

For independent daily counts $X_1,\dots ,X_N$ modeled by a Poisson distribution, the log-likelihood isDifferentiating gives the maximum-likelihood estimator $\hat{\mu }=\overline{X}$. Here the observed frequencies sum to $N=1096$, and the expected frequencies are consistent with $\hat{\mu }\approx 2.157$ (their ratio $E_1/E_0$ estimates the same parameter). The final class is open-ended: its single observation is only reported as at least 9, so its exact contribution to the sample total cannot be recovered from this grouped table alone. The underlying ungrouped daily observations determine the exact sample mean.

Minimize $\Vert Y-X\theta {\Vert }^2$. Differentiation gives the normal equations $X^{T}X\hat{\theta }=X^{T}Y$. Full column rank makes $X^{T}X$ positive definite, soThe expectation uses the zero error mean; the covariance uses $\mathrm{Cov}(ϵ)={\sigma }^{2}I$. With independent Gaussian errors of common variance, maximizing likelihood minimizes exactly the same residual sum of squares, so this is also the maximum likelihood estimator, whether the common variance is known or estimated.

For the quadratic design, decompose each group's residual sum of squares into its within-group constant plus the group size times the squared residual of its sample mean. The three parameter combinations $\alpha$, $\alpha +\beta +\gamma$ and $\alpha -\beta +\gamma$ can fit all three group means exactly. ThereforeThe group means are independent normal variables, with variances ${\sigma }^2/m,{\sigma }^2/n,{\sigma }^2/n$. Hence their two stated linear combinations are jointly normal, andJoint normality therefore proves the two estimators are independent. Their variances are ${\sigma }^2/(2n)$ and ${\sigma }^2/(2n)+{\sigma }^2/m$, respectively. Equal replication at the two off-center temperatures is what cancels the covariance.

For each normal sample, its sample mean is independent of its sample variance, and the scaled residual sum of squares is chi-squared. Independence of the samples also makes their two residual sums independent. Adding their chi-squared distributions givesConsequentlyThe pooled variance is also independent of the difference of sample means.

For each normal sample, its sample mean is independent of its sample variance, and the scaled residual sum of squares is chi-squared. Independence of the samples also makes their two residual sums independent. Adding their chi-squared distributions givesConsequentlyThe pooled variance is also independent of the difference of sample means.

The sample mean has expectation $\theta$ and variance ${\theta }^2/n$. For $g(z)=\log z$, $g^{\prime }(\theta )=1/\theta$. The first-order delta-method expansion isTaking its mean and variance givesThese are large-sample approximations. Retaining the quadratic Taylor term would add the leading mean correction $-1/(2n)$, but it does not change the requested first-order expression.