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Antithetic variates deliberately couple unbiased estimators of the same expectation with negative covariance. Their average can have smaller variance than an average using independent draws. For example, evaluating a monotone integrand at $U$ and $1-U$, where $U$ has a uniform distribution, gives opposite fluctuations while preserving both marginal distributions.

Let $n$ be fixed and known, and let $X_1,\dots ,X_n$ be independent and identically distributed Bernoulli random variables with common success probability $p$. Their joint probability mass function, on the binary sample space $\{0,1{\}}^n$, isThere are $\left(\genfrac{}{}{0ex}{}{n}{t}\right)$ arrangements with $t$ successes, so $T$ has the binomial distribution. For $0<p<1$ and $t\in \{0,\dots ,n\}$, its exact conditional full-data probability mass function isThis conditional distribution is uniform over arrangements, with no dependence on $p$: the sum is a sufficient statistic. ^[ref]

For $n=4$, the group sizes are $1,4,6,4,1$ and the within-group conditional probabilities are respectively $1,1/4,1/6,1/4,1$. For example, $T=2$ leaves exactly $0011,0101,0110,1001,1010,1100$, each with conditional probability $1/6$. At $p=1/4$, each such sequence has unconditional probability $9/256$ and the group has probability $54/256$. At $p=1/2$, these values are $1/16$ and $6/16$; at $p=3/4$ they return to $9/256$ and $54/256$. The conditional weights stay $1/6$ in all three cases.

At the boundary $p=0$, only $T=0$ has positive probability; at $p=1$, only $T=n$ does. Dividing by $P_p(T=t)$ on the other groups would divide by zero. Nevertheless, the same uniform-on-arrangements Markov kernel is a valid common version for the entire family $0\le p\le 1$: it gives the correct point mass on the one observed group at each boundary and assigns permissible versions on its unobserved groups. Define the joint probability mass function there by the original product of the one-trial Bernoulli probability mass functions, or use the limiting convention $0^0=1$ in the compact formula.

The sample mean $T/n$ is an unbiased estimator of $p$ and retains the sufficient statistic because $n$ is known. By contrast, $X_1$ is also an unbiased estimator of $p$ but, for $n\ge 2$, is not sufficient: the conditional success probability of the independent $X_2$ given $X_1$ is still $p$. Thus unbiasedness does not imply sufficiency.

Sufficiency also does not imply completeness of a statistic. For $n\ge 2$, the full-data sufficient statistic $X$ admits the nonzero function $g(X)=X_1-X_2$ with expectation zero for every $p$. Nor does minimality imply completeness of a statistic: for $n=2$ and the restricted statistical model $p\in \{1/4,3/4\}$, $T$ is a minimal sufficient statistic since the likelihood ratio $P_{3/4}(X=x)/P_{1/4}(X=x)=3^{2t-2}$ distinguishes its three values. Yet $g(0)=3,g(1)=-5,g(2)=3$ has expectation $(27-30+3)/16=0$ at both parameter values and is never zero. Changing the parameter family can change completeness of a statistic even while preserving sufficiency.

The Chapman estimator is a correction intended to reduce the small-sample bias of the Lincoln-Petersen estimator. Under the fixed-size hypergeometric sampling assumptions it is exactly an unbiased estimator when $M+C\ge N$; it can have nonzero bias otherwise. This qualification and the formula are stated in Barker, §1.1.1, equation (1.3), citing Chapman's work through Seber. For the toy observations,It is an alternative estimator, not a maximum-likelihood estimate for this capture-recapture model. The Chapman estimator remains finite when $R=0$, but a finite output does not supply the missing overlap information or repair unequal catchability, lost marks, or population change.

Split independent observational blocks into two folds. Use one fold to estimate the combination weight for the two sample means in the other fold, and conversely. Each held-out combination is an unbiased estimator conditionally on its training fold because its two held-out sample means have the same expectation. A deterministic average of the two combinations remains unbiased. Under consistency of the moment estimates and balanced fold sizes, the weights approach those of the minimum-variance combination of unbiased sample means. Paired observations must stay together when splitting or bootstrapping so their covariance is preserved.

The empirical distribution function counts the fraction of observations at or below a threshold. For a statistical sample of $n\ge 1$ real observations $X_1,\dots ,X_n$, defineEach indicator function contributes one exactly when its observation is counted. Equivalently, $F_n(x)=P_n((-\infty ,x])$, where the empirical measure $P_n=n^{-1}{\sum }_i{\delta }_{X_i}$ assigns probability mass $1/n$ to each observation, including repeated observations.^[ref]

Choose the illustrative statistical sample $[1,1,2,4,6]$. Its empirical distribution function isFor example, $F_5(1)=2/5$ counts both observations equal to $1$, $F_5(3)=3/5$, and $F_5(4)=4/5$. There is no observation at $3$, so there is no jump there. The five observations produce only four distinct jump locations.

The convention $X_i\le x$ makes the empirical distribution function right-continuous: once a threshold reaches an observation, that observation remains counted immediately to its right. For a finite statistical sample, the count is constant from any $x$ up to the next larger observed value, proving right continuity. Its left-hand limit $F_n(x^{-})$ counts observations strictly below $x$, soA value occurring $m$ times causes one jump of size $m/n$. At $1$ the jump is $2/5$; at $2$, $4$, and $6$ each jump is $1/5$. All jumps sum to one. Thus the empirical distribution function is a nondecreasing step function with limits zero and one, and is itself a valid cumulative distribution function.

This turns many questions about data into simple cumulative distribution function operations. The observed fraction in $(a,b]$ is $F_n(b)-F_n(a)$: for $(1,4]$ it is $4/5-2/5=2/5$, counting $2$ and $4$ but excluding both $1$s. Including the left endpoint gives $F_n(4)-F_n(1^{-})=4/5$. A sample quantile reads the same staircase in reverse: the smallest threshold reaching $1/2$ is $2$. Unlike a histogram, this representation requires no choice of bins; its jumps retain every observation and its multiplicity.

The exact description of a finite statistical sample must be distinguished from a statistical model for how it was obtained. The empirical distribution function is defined for any finite list, with no independence assumption. Once the list is observed, its counts are fixed. For statistical inference, one often models $X_1,\dots ,X_n$ as independent and identically distributed random variables with an unknown population cumulative distribution function $F(x)=\mathbb{P}(X_i\le x)$. Then $F$ describes the sampling law, while $F_n$ varies across repeated statistical samples. The value $F_5(1)=0.4$ is an observed fraction and an estimate of $F(1)$; it does not establish that the population probability is $0.4$. Even when $F$ is continuous, every finite $F_n$ is discrete.

For any threshold $x$ fixed before sampling, the indicator functions are independent Bernoulli random variables with success probability $F(x)$. Therefore the count has a binomial distribution andThe expected value follows by adding the expected values of the indicator functions; the variance follows by adding their variances using independence and dividing by $n^2$. Thus $F_n(x)$ is an unbiased estimator of $F(x)$ at a fixed threshold. Controlling the error at every threshold at once requires more than a fixed-threshold calculation: the DKW inequality supplies such control, and the Glivenko-Cantelli theorem describes its uniform convergence almost surely as $n$ grows.

Comparing $F_n$ with a proposed cumulative distribution function also motivates goodness-of-fit tests, including the Kolmogorov-Smirnov test, the Anderson–Darling test, and the Cramér–von Mises criterion. A standard Kolmogorov-Smirnov test calibration assumes a fully specified continuous population law; fitting its statistical parameters from the same statistical sample changes that calibration.^[ref]

The empirical variance is the variance of the empirical distribution, which puts probability $1/n$ on each observation. It uses divisor $n$, whereas the usual sample variance uses divisor $n-1$. For independent and identically distributed random variables with finite second moment, jackknife bias correction converts the empirical variance into the unbiased estimator $s_n^2$ exactly.

Prepare a fresh copy of the same quantum state for each trial and measure the same observable. If preparations and readouts are independent, the outcomes $X_i$ are independent and identically distributed random variables with the Born rule probability distribution. Their sample mean is an unbiased estimator of the expectation value:The law of large numbers explains convergence of the sample mean, while the standard error describes its finite-sample spread. It takes four times as many independent trials to halve that standard error. This reduces estimation error; it does not reduce the quantum uncertainty of a single prepared copy.

For the spin expectation value, let $n_{+}$ and $n_{-}$ be the two outcome counts, with $N=n_{+}+n_{-}$. ThenThe binomial distribution gives the exact sampling model. For $\theta =\pi /3$, the standard error is $\hslash \sqrt{3}/(4\sqrt{N})$: approximately $0.1083\hslash$ for $N=16$ and $0.03827\hslash$ for $N=128$. A run can fluctuate away from the predicted mean even as its typical spread decreases.

Repeatedly measuring one surviving system is a different protocol. Under the ideal Lüders rule, outcome $a_r$ leaves $P_r|\psi ⟩/\sqrt{p_r}$ when $p_r>0$. With no intervening change of observable or dynamics moving the state out of that eigenspace, the next projective measurement returns $a_r$ with certainty. Such a record is correlated and does not sample the original Born rule distribution independently. The quantum nondemolition measurement discussion describes this repeatability; fresh quantum state preparations supply the ensemble needed for the estimator above.

The finite population mean for $N\ge 1$ is the arithmetic mean of all fixed values in the target population. Under simple random sampling, the sample mean is an unbiased estimator of it.

In the Gaussian white noise model, estimate a Haar scaling function coefficient by $\int {\varphi }_{J,k}dY$. The resulting Haar projection estimator is an unbiased estimator with independent coefficient errors of variance $1/n$. On each cell it is the observed path increment divided by the cell length. This gives a finite-dimensional estimator without imposing smoothness on the drift.

For independent unbiased estimators $X_i$ of a common mean with known positive variances $v_i$, the linear unbiased estimator with smallest variance has weights $w_i=v_i^{-1}/{\sum }_j{v}_j^{-1}$ and variance $({\sum }_j{v}_j^{-1}{)}^{-1}$. Unbiasedness requires ${\sum }_i{w}_i=1$. Minimizing ${\sum }_i{v}_i{w}_i^2$ by a Lagrange multiplier gives the weights, and the strictly positive diagonal Hessian matrix proves a unique global minimum. This conclusion requires no normality assumption.

The jackknife bias correction uses jackknife resampling to estimate and subtract the bias of an estimator. Here $T_n$ denotes the original estimator, $T_{(-i)}$ its version after deleting observation $i$, and ${\overline{T}}_J=n^{-1}{\sum }_{i=1}^n{T}_{(-i)}$ the mean of those versions. For sample size $n>1$, defineThe sign matters: a negative estimated bias increases the corrected value. The original estimator and all deletion versions must target the same estimand and use the same estimation rule at their respective sample sizes. The definition alone does not make ${\tilde{T}}_J$ an unbiased estimator.

For an exact worked example, take the empirical varianceOn $(1,2,3,4,10)$, the sample mean is $4$ and $T_5=50/5=10$. Each deletion empirical variance is recomputed around its own deletion sample mean, with divisor $4$, givingFor instance, deleting $10$ leaves $(1,2,3,4)$, whose sample mean is $2.5$ and whose empirical variance is $1.25$. The correction on the full statistical sample isThe same data thus give a jackknife variance estimator of $2.5$ for the sample mean, and a corrected empirical variance of $12.5$; these are different estimands.

The exact equality with the usual sample variance is not an accident of these numbers. For any data with $n>1$, write $Q={\sum }_i(x_i-\overline{x}{)}^2=n{T}_n$. Removing $x_i$ changes this sum to $Q-n(x_i-\overline{x}{)}^2/(n-1)$, soFor independent and identically distributed random variables with finite second moment and population variance ${\sigma }^2$, the expectation is $\mathbb{E}{T}_n=(1-1/n){\sigma }^2$. Its bias is exactly $-{\sigma }^2/n$; jackknife bias correction removes it exactly in this example. Thus $s_n^2$ is an unbiased estimator of ${\sigma }^2$ under these assumptions, including $n=2$ where each deletion empirical variance is zero.

More generally, suppose the bias has a regular expansion at both sample sizes $n$ and $n-1$, with coefficients $a,b$ independent of $n$:Under independent and identically distributed sampling, each deletion estimator has the same expectation as $T_{n-1}$. Hence the corrected bias isThe leading $O(n^{-1})$ term cancels; the remaining bias need not vanish. The $O$ notation is big O notation. This argument presumes the expansion, rather than proving it for every statistic; the regularity conditions discussed for the jackknife variance estimator and statistical dependence still require care. Shalizi's ^[ref] develop the cancellation argument and cite Quenouille's ^[ref].

The jackknife variance estimator estimates the sampling variance of the full-data estimator, using the deletion estimates and their mean from jackknife resampling:The square root is the estimated standard error. The factor $n-1$ multiplies the divisor-$n$ variance of the deletion estimates: those estimates vary less than estimates from new independent experiments because almost all their observations are shared. Ryan Tibshirani's ^[ref] give the same scaling in terms of the divisor-$(n-1)$ sample variance of the deletion estimates.

For the sample mean, this formula has an exact algebraic simplification. With $s_n^2$ denoting the usual sample variance,and thereforeUnder the independent and identically distributed assumption and a finite second moment, $s_n^2$ is an unbiased estimator of the population variance ${\sigma }^2$. Consequently ${\hat{V}}_J$ is an unbiased estimator of the sample mean's sampling variance ${\sigma }^2/n$. The equality ${\hat{V}}_J=s_n^2/n$ itself holds for any observed data with $n>1$; its sampling interpretation uses the stated assumptions.

For $(1,2,3,4,10)$, the sum of squared deviations from the sample mean is $50$, giving sample variance $s_5^2=50/4=12.5$. The deletion sample means have squared deviations totaling $3.125$, soHere $12.5$ estimates the variance of individual observations, while $2.5$ estimates the sampling variance of their sample mean. These answer different questions.

For more general statistics, validity requires regularity beyond the deletion formula, such as a sufficiently smooth function of sample means with suitable moments. Ordinary delete-one jackknife resampling can fail for nonsmooth quantiles, including the median; see Larry Wasserman's ^[ref]. With statistical dependence, deleting individual observations need not reproduce the relevant sampling variance; the deletion scheme must reflect that statistical dependence. Shalizi's ^[ref] distinguish the basic independent and identically distributed case from extensions.

Suppose $X_1,\dots ,X_n$, $n\ge 1$, are independent and identically distributed with Laplace distribution $\mathrm{Laplace}(\mu ,b)$, where $b>0$. For realized values $x_1,\dots ,x_n$, the likelihood function and log-likelihood function areFor fixed $b$, maximum likelihood estimation of the location therefore minimizes the sum of absolute losses. This connects the Laplace distribution to robust statistics: each observation contributes a cost growing linearly with its distance, whereas squared-error loss grows quadratically.

Why does a sample median minimize $S$? Between successive order statistics, moving $a$ a little to the right increases each distance to a value on the left and decreases each distance to a value on the right. Hence the slope is the number on the left minus the number on the right. It changes from negative to positive at the middle. More precisely, if $x_{(1)}\le \cdots \le x_{(n)}$, all minimizing locations areFor an even sample size, the objective is flat throughout the interval between the two central order statistics. The midpoint convention for the sample median selects one solution; it does not prove uniqueness. ^[ref] This is the finite-sample counterpart of median minimizes expected absolute loss.

If $S(\hat{\mu })>0$, maximizing over $b>0$ givesThe derivative is positive below this value and negative above it. Moreover the maximized likelihood function is proportional to $S(\mu {)}^{-n}$ wherever $S(\mu )>0$, so the same minimizing locations give the joint maximum-likelihood estimates. ^[ref] The scale estimate divides by $n$; this formula is not an assertion that it is an unbiased estimator.

The boundary case matters: if all observations equal $c$, choose $\mu =c$. Then $S(c)=0$ and $L(c,b)=(2b{)}^{-n}\to \infty$ as $b\downarrow 0$. There is no finite joint maximum-likelihood estimate in the parameter space $b>0$. In particular, this always occurs for $n=1$. Writing $\hat{b}=0$ does not produce an admissible Laplace distribution.

For two unbiased estimators $A,B$ of the same parameter with variances $V_X,V_Y$ and covariance $C$, the fixed-weight estimator $wA+(1-w)B$ is unbiased and has variance $w^2{V}_X+(1-w{)}^2{V}_Y+2w(1-w)C$. Completing the square gives $w_{\ast }=(V_Y-C)/(V_X+V_Y-2C)$ and minimum variance $(V_X{V}_Y-C^2)/(V_X+V_Y-2C)$ when the denominator is positive. These are optimal among fixed linear unbiased combinations. Substituting weights estimated from the same observations need not preserve unbiasedness.

Nested Monte Carlo approximates an outer expected value whose integrand contains another expected value. A nonlinear transformation, such as a logarithm, of an inner sample average usually introduces finite-sample bias. Increasing both sample sizes requires suitable integrability controls to establish consistency; an unbiased estimator of the inner quantity does not make its logarithm unbiased.

If a design matrix $X$ has a nonzero null vector $z$, the normal linear model has identical observation distributions at $\beta$ and $\beta +z$. Any estimator has the same expectation at those two parameters, and therefore cannot be an unbiased estimator of both coefficient vectors. Full column rank is required for unbiased estimation of the entire unconstrained coefficient vector; identifiable linear combinations can still be estimated without it.

For independent and identically distributed random variables $X_1,\dots ,X_n\sim N(\mu ,{\sigma }^2)$, the maximum-likelihood estimators areThe first is unbiased, while $\mathbb{E}{\hat{\sigma }}^2=(n-1){\sigma }^2/n$.

In the normal mean proportional to standard deviation model, both $\overline{X}$ and $S$ estimate the positive statistical parameter $\mu$. Their independent limiting normal distributions have rescaled variances ${\mu }^2$ and ${\mu }^2/2$, respectively, by asymptotic normality of a normal sample standard deviation. Thus for fixed $w\in \mathbb{R}$,The bracket is $3(w-1/3{)}^2/2+1/3$, minimized uniquely at $w=1/3$. Hence $(\overline{X}+2S)/3$ has leading sampling variance ${\mu }^2/(3n)$, the inverse expected Fisher information. The result is first-order optimality within these fixed-weight combinations and does not assert that $S$ is an exactly unbiased estimator.

For a linear regression design matrix $X$ with full column rank, the ordinary least squares coefficient vector is $\hat{\beta }=(X^{T}X{)}^{-1}{X}^{T}Y$. In a normal linear model with error covariance matrix ${\sigma }^{2}I$, it is an unbiased estimator with covariance matrix ${\sigma }^2(X^{T}X{)}^{-1}$. Replacing ${\sigma }^2$ by the residual variance estimate gives the reported coefficient standard errors.

In simple linear regression with a nonconstant predictor,

For the zero-truncated Poisson distribution, $21_{\{Y\text{ even}\}}$ is the unique integrable nonrandomized unbiased estimator of $p=1-e^{-\theta }$. Indeed unbiasedness gives ${\sum }_{y\ge 1}t(y){\theta }^y/y!=e^{\theta }+e^{-\theta }-2$, and power series coefficients force the parity values. Its variance is $2p-p^2$; clipping it to $1_{\{Y\text{ even}\}}$ gives mean squared error $p/2$, strictly smaller for $0<p<1$.