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The expected process variance in the Bühlmann model averages within-risk conditional variance over the population of risks. It measures variation that repeated observations can average out and contributes $v/n$ to the variance of the sample mean.

An expected value is the probability-weighted arithmetic mean of a real-valued random variable. On a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, its general definition is the Lebesgue integralwhere ${\mu }_X$ is the probability distribution of $X$. For an integrable random variable, this is a finite real number. With a probability mass function $p_X$, use ${\sum }_{x}x{p}_X(x)$; with a probability density function $f_X$, use ${\int }_{\mathbb{R}}x{f}_X(x)dx$. These are two ways of computing the same weighted average, rather than different definitions for different experiments.

For example, suppose $X$ takes the values $-1,2,8$ with probabilities $1/2,3/10,1/5$. ThenThe most likely outcome is $-1$, and $1.7$ is not a possible outcome at all. The expected value summarizes all outcomes and their weights; it need not predict the result of one trial. Geometrically, placing masses equal to the probability masses at the outcome positions gives a center of mass at $1.7$, since ${\sum }_x{p}_X(x)(x-1.7)=0$. This balance-point interpretation also explains the connection to a weighted arithmetic mean. ^[ref].

For the dimensionless random variable $Y$ with probability density function $f_Y(y)=2y$ on $[0,1]$, the expected value is $\mathbb{E}[Y]={\int }_0^{1}y(2y)dy=2/3$. Integrating the probability density function itself gives one; to compute the expected value, multiply by the outcome first. ^[ref].

Why is this useful? Linearity of expectation lets us compute an average total by adding average contributions, even for dependent random variables. And for independent and identically distributed integrable random variables, the strong law of large numbers says their sample mean approaches their common expected value almost surely. The theorem supplies the link to repeated observations; the weighted-average definition itself does not promise that any particular finite run will be close to the expected value.

There is a useful historical connection to valuing uncertain rewards. In his 1657 De ratiociniis in ludo aleae, Christiaan Huygens assigns the value $(a+b)/2$ to equally likely rewards $a,b$, and $(pa+qb)/(p+q)$ when they have $p,q$ equally likely chances. These are the finite weighted-average formulas for expected value. His opening discussion asks how to divide the stakes when a game stops before completion, connecting probability to a concrete question about an unfinished contest. ^[ref].

Extreme value theory studies limiting probability distributions of normalized sample maxima and related extremes. The standard limits for maxima include the Fréchet distribution, negative Weibull distribution, and Gumbel distribution.

The question is often practical: if a batch finishes only when its slowest task finishes, its completion time is a sample maximum. A typical individual delay or a sample mean does not answer that question. The central limit theorem describes suitably normalized sums and sample means under its assumptions; it does not give a limit law for sample maxima. Extreme value theory instead describes how the largest observation changes as the number of opportunities for a large observation grows.

For independent and identically distributed random variables $X_1,X_2,\dots$ with distribution function $F$, define the sample maximumThe event $M_n\le x$ means that every observation is at most $x$, so independence givesExtreme value theory asks whether there are constants $a_n>0$ and $c_n\in \mathbb{R}$ for which $(M_n-c_n)/a_n$ has convergence in distribution to a non-degenerate probability distribution. Equivalently, $F(c_n+a_{n}z{)}^n$ must tend to a limiting distribution function at its continuity points. The extremal types theorem says that any such limit belongs, up to a change of location and positive scale, to the Gumbel distribution, Fréchet distribution, or negative Weibull distribution. It does not guarantee that a limit exists for every $F$. ^[ref].

The maximum of independent exponential random variables gives a particularly transparent worked example: its center grows as $(\log n)/\lambda$, and its fluctuations have a Gumbel distribution limit on the scale $1/\lambda$. Here and below $\log$ denotes the natural logarithm. This normalization depends on the exponential distribution. Other survival functions can lead to the Fréchet distribution or negative Weibull distribution, with different constants $a_n,c_n$; the Gumbel distribution and the exponential example's normalization are not universal.

Independence is essential to the product calculation. If instead $X_1=X_2=\cdots =X$ for one rate-$\lambda$ exponential random variable, then $M_n=X$ for every $n$. Its distribution function stays $1-e^{-\lambda x}$ for $x\ge 0$ and zero below zero, and its median stays $(\log 2)/\lambda$. Perfect dependence creates no logarithmic growth: repeating the same random variable supplies no new chance of a larger value.

A historical starting point is R. A. Fisher and L. H. C. Tippett's 1928 paper on limiting laws of the largest or smallest sample member, published in volume 24, issue 2, pages 180–190 of the Proceedings of the Cambridge Philosophical Society. Its abstract studies the possible limiting forms and emphasizes that approach to a limit can be slow. ^[ref].

For uniform sampling without replacement from a population of $n>1$ objects, the finite population correction accounts for the reduced spread caused by drawing distinct objects. For the marked-object count $K$ in a hypergeometric distribution, put $p=c/n$. ThenRelative to the corresponding binomial distribution, the variance is multiplied by $f^2=(n-r)/(n-1)$ and the standard deviation by $f$. The square-root convention for the correction is used in Philip Stark's discussion of sample sums.^[ref]

The factor has a direct interpretation. For $r=1$, replacement makes no difference and $f=1$. For $r=n$, the entire population is observed and the marked count is exactly $c$, so $f=0$. In the ten-object, five-draw example, $f^2=5/9$ and $(6/5)(5/9)=2/3$. The exceptional population size $n=1$ is deterministic and should be handled directly rather than inserting it into a zero denominator.

A short explanation uses an indicator random variable $I_j$ for a marked object at draw $j$. Each has expected value $p$ and variance $p(1-p)$. For distinct draw positions $i,j$,Since $K={\sum }_{j=1}^r{I}_j$, the variance of a sum includes $r$ individual terms and $r(r-1)$ ordered covariance terms, giving the displayed correction. Linearity of expectation still gives $rp$ despite this dependence.

For arbitrary fixed real population values, use $N>1$ for population size and $1\le n\le N$ for sample size. Write the finite population mean $\bar{Y}=N^{-1}{\sum }_i{y}_i$, the finite population variance $S^2=(N-1{)}^{-1}{\sum }_i(y_i-\bar{Y}{)}^2$ and ${\sigma }_N^2=N^{-1}{\sum }_i(y_i-\bar{Y}{)}^2$. Under simple random sampling without replacement, the sample mean satisfiesThus $1-n/N$ is the variance factor relative to $S^2/n$, while $(N-n)/(N-1)$ is the exact variance ratio to independent uniform sampling with replacement, whose sample mean has variance ${\sigma }_N^2/n$. The standard error uses the square root of the corresponding factor. These formulas agree because ${\sigma }_N^2=(N-1)S^2/N$; the denominator convention must be specified. The sample mean under simple random sampling gives a derivation and an exact six-value example.

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.

Here $d_H$ counts the observations replaced. This curve measures the largest change after at most $m$ arbitrary replacements. Under the first-unbounded convention, the replacement breakdown point is $n^{-1}\min \{m:b_n(m;y)=\infty \}$. The largest fraction still guaranteed bounded is one grid step smaller; those conventions must not be confused. A sample mean breaks after one replacement, while a scalar sample median with a fixed ordinary tie convention needs about half the sample replaced.

Fitted values are the predictions at the predictor values used to fit a regression model. In linear regression with an explicit regression intercept,Here $X$ is the slope design matrix, and $r$ is the regression residual vector. The fitted values estimate the regression function at the observed predictors; they need not equal the observed responses, which may also contain random errors. Prediction at a new predictor vector $x_{\ast }$ uses $\hat{\alpha }+x_{\ast }^T\hat{\beta }$ with the same centering and scaling used in fitting. A prediction interval addresses uncertainty in a new response rather than just displaying this point prediction.

For the five constructed observations in linear regression, ordinary least squares gives fitted values $(4/5,9/5,14/5,19/5,24/5)$. The ridge regression fit at $\lambda =10$ instead gives $(9/5,23/10,14/5,33/10,19/5)$. Both fitted vectors have sample mean $14/5$ because the regression intercept is unpenalized; their regression residuals differ and their residual sums of squares are $4/5$ and $33/10$, respectively.

For ordinary least squares, the fitted values are the orthogonal projection of $Y$ onto the column space of the complete design matrix, including its regression intercept column. They remain unique even in rank-deficient ordinary least squares, although the regression coefficients can be nonunique. The regression residuals are then orthogonal to every column.^[ref] With centered slopes, ridge regression instead satisfies $X_c^{T}r=\lambda {\hat{\beta }}_{\lambda }$, so the regression residuals need not be orthogonal to the slope columns. Their sum is still zero because the regression intercept remains free.

For fixed predictors and scales and a fixed positive penalty, ridge regression gives a linear smoother,For $\lambda >0$ with at least one nonzero slope direction, this matrix is not an orthogonal projection matrix: applying it twice shrinks that direction again. Its effective degrees of freedom are $1+{\sum }_{j=1}^r{d}_j^2/(d_j^2+\lambda )$, where $d_j$ are the positive singular values of $X_c$; the extra $1$ counts the free regression intercept. If the penalty is selected using $Y$, the resulting fitting rule need not be a linear smoother, since its chosen $H_{\lambda }$ also depends on $Y$.

For inverse-gamma distribution observations of known shape $k$ and scale $\Theta$, their likelihood function in $\Theta$ is proportional to ${\Theta }^{nk}\exp (-\Theta {\sum }_i{x}_i^{-1})$. A shape-$\alpha$, rate-$\lambda$ gamma distribution conjugate prior therefore gives the displayed Bayesian posterior. The Bayes estimator under squared error loss of the conditional claim expected value $\Theta /(k-1)$ is $(\alpha +nk)/[(k-1)(\lambda +{\sum }_i{x}_i^{-1})]$. It depends on reciprocals, rather than the sample mean of the claims.

William Sealy Gosset worked at Guinness as a chemist and brewer and developed methods of statistical inference for small samples. He published under the pseudonym "Student". His employment as a chemist is independently recorded by MacTutor; the Dictionary of Irish Biography identifies him as an applied statistician and brewer and explains the pseudonym policy. ^{[ref] [ref]}

The 1908 paper The Probable Error of a Mean appeared in Biometrika 6(1), pages 1–25, with the byline "Student". Its introduction identifies the uncertainty introduced by estimating a standard deviation from few observations and explicitly assumes a normal distribution. Sections II and III develop the relation between sample spread and the standardized sample mean. Its notation uses a sample spread with divisor $n$ and calls the ratio $z$; with that convention, the modern $T$ equals $\sqrt{n-1}z$. The modern exact independence argument is given under studentized normal sample mean. ^[ref]

Section VI describes a physical sampling check: 3,000 cards carrying existing paired measurements were shuffled, then divided into 750 groups of four. He compared observed frequencies of sample spread and standardized sample mean with theoretical frequencies. This was a check using finite, grouped, approximately normal distribution data; it did not prove exact calibration for arbitrary populations. ^[ref]

The Hoeffding inequality turns finite interval bounds and independence into exponentially small tail probabilities for a sum. It requires neither a normal distribution nor identical probability distributions. Let $X_1,\dots ,X_n$ be independent essentially bounded real random variables, with deterministic finite bounds $a_i\le X_i\le b_i$ almost surely. Define the centered sum and its squared-width budget byFor $V>0$ and $t>0$,The union bound therefore gives $\mathbb{P}(|S_n|\ge t)\le \min \{1,2e^{-2t^2/V}\}$. If $V=0$, every $X_i$ is constant almost surely, so $S_n=0$ and both positive-threshold tail probabilities vanish. The equal-width version is stated in the authored Stanford notes.^[ref]

The quantity $V$ is a sum of squared interval widths, not the variance of $S_n$. Centering changes the endpoints to $a_i-\mathbb{E}{X}_i$ and $b_i-\mathbb{E}{X}_i$, leaving each width unchanged. For example, independent random variables in $[-2,1]$ and $[4,5]$ have $V=3^2+1^2=10$, whatever their expected values or probability distributions. A deviation at least three above their total expected value has probability at most $e^{-18/10}\approx 0.165299$. If all widths equal $b-a>0$, the sample mean ${\overline{X}}_n=n^{-1}{\sum }_i{X}_i$ satisfies

Independence is essential to this general guarantee. Take one Bernoulli random variable $B$ with success probability $1/2$ and set every $X_i=B$. Each lies in $[0,1]$, but for every $n$ the sample mean is either zero or one. Thus $\mathbb{P}(|{\overline{X}}_n-1/2|\ge 1/4)=1$, whereas incorrectly inserting these dependent observations into the formula would give $2e^{-n/8}$, less than one for $n\ge 6$. Some special dependent models have their own valid concentration inequalities; arbitrary statistical dependence cannot be ignored.

For the short proof, apply Markov's inequality to $e^{\lambda S_n}$ with $\lambda >0$, factor the expectation using independence, and apply the Hoeffding lemma to each factor:Minimizing this quadratic polynomial at $\lambda =4t/V$ gives the upper tail probability; applying it to $-S_n$ gives the lower one. This is the Chernoff bound method. The original 1963 article is Probability Inequalities for Sums of Bounded Random Variables by Wassily Hoeffding; its publisher record identifies the article, but its full text was unavailable during this source check. The argument above is a direct derivation rather than a claim to have inspected that article's proof.

Let $B_1,\dots ,B_n$ be independent and identically distributed Bernoulli random variables with success probability $p$, and let ${\hat{p}}_n=n^{-1}{\sum }_i{B}_i$ be their observed frequency, a sample mean. Since every interval width is one, $V=n$. Substituting $t=n\epsilon$ into the Hoeffding inequality givesFor tolerance $0<\epsilon <1$ and failure probability $0<\delta <1$, solving $2e^{-2n{\epsilon }^2}\le \delta$ yields the sufficient sample sizeHere $\log$ is the natural logarithm. For $\epsilon =0.05$ and $\delta =0.05$, $n=738$ guarantees an error smaller than $0.05$ with probability at least $0.95$, for every $p\in [0,1]$. The two-sided bound at this sample size is approximately $0.0499440$. A specified single direction instead needs $n\ge \lceil \log (1/\delta )/(2{\epsilon }^2)\rceil$, or $600$ for these choices. Halving the tolerance multiplies the unrounded sufficient sample size by four; reducing the failure probability tenfold adds $\log (10)/(2{\epsilon }^2)$ before rounding.

This is a conservative sufficient sample size, not a necessary minimum. For a specified $p$, the count $K={\sum }_i{B}_i$ has a binomial distribution, whose exact upper tail probability isFor $k>np$, the Hoeffding inequality bounds this by $\exp [-2(k-np{)}^2/n]$. An arbitrary frequency threshold uses $k=\lceil n(p+\epsilon )\rceil$; if $k>n$, the event is impossible. The discreteness of the count and its variance $np(1-p)$ can make the exact tail probability substantially smaller. At $n=100$, $p=1/2$, $k=60$, the exact upper tail probability is about $0.0284440$, while the bound is $e^{-2}\approx 0.135335$. At $p=1/10$, $k=20$, the same deviation $\epsilon =0.1$ gives the same bound but an exact upper tail probability about $0.00197856$.

For independent multivariate normal observations with positive-definite covariance, this statistic is the maximum squared studentized mean contrast. With unbiased sample covariance matrix $S$ and $n>p$, its null distribution obeys $(n-p)T^2/(p(n-1))\sim F_{p,n-p}$. This follows from independence of the Gaussian sample mean and the residual Wishart distribution.

The Huber location estimator solves ${\sum }_i{\psi }_k(x_i-\theta )=0$, where ${\psi }_k$ clips its argument to $[-k,k]$.

More generally, fix a known scale $s>0$ and a dimensionless threshold $k>0$ before fitting. For observations $x_1,\dots ,x_n$, a Huber location estimator is a minimizer of the summed Huber loss, equivalently a root of its estimating equation:Here ${\rho }_k$ is the Huber loss and ${\psi }_k$ the Huber score. The physical clipping distance is $ks$, in the observations' units; the initial formula uses $s=1$. A large observation keeps contributing a bounded signed score rather than being discarded. The original clipped location criterion appears in Peter J. Huber's 1964 paper.^[ref]

For the original data $(0,1,2,3,4,40)$, take $s=1$ and $k=2$. At $\theta =2.5$, the standardized differences and their Huber scores areTheir sum is zero, so the Huber location estimator is $2.5$. Four scores remain in the central linear region, making the root unique. By comparison, the sample mean is $50/6\simeq 8.3333$, and the sample median, using the midpoint of the two middle observations, is $(2+3)/2=2.5$. Agreement with the sample median is a feature of these data, rather than an identity between the two estimators.

The geometry explains both existence and possible nonuniqueness. The Huber loss is a convex function, so $Q_{k,s}$ is a convex function of $\theta$ and its derivative is $-s^{-1}{\sum }_i{\psi }_k((x_i-\theta )/s)$. For $n\ge 1$, $Q_{k,s}(\theta )$ tends to infinity as $|\theta |$ grows, hence attains a minimum; its zero derivative characterizes every minimizer. Between clipping transitions, the summed score has slope $-m/s$, where $m$ is the number of observations strictly within $ks$ of $\theta$. A root with $m>0$ is unique. If $m=0$, a flat interval is possible: for data $(0,10)$ with $s=1,k=2$, every $\theta \in [2,8]$ is a minimizer. A complete single-valued estimator then needs a selection rule, such as the interval's midpoint.

A bounded Huber score limits one observation's contribution to the estimating equation. The influence function of a Huber location estimator additionally divides by the population's central mass, so the size of its bound depends on the reference probability distribution. Clipping does not by itself make the bias of an estimator zero or prove a universal breakdown point; the separate local influence versus replacement breakdown argument specifies what fixed scale permits. Estimating $s$ from the same data changes the procedure and requires separate analysis.

Hold $(0,1,2,3,4)$ fixed and move only the sixth observation $M$. For $M\ge 5$, the sample mean is $(10+M)/6$, the sample median is $2.5$, and the Huber location estimator with fixed $s=1,k=2$ stays at $2.5$. Once $M-2.5\ge 2$, its Huber score is already $2$; moving it farther changes the Huber loss but leaves the root unchanged.

For a closer view of the transition, the exact Huber location estimator over $0\le M\le 40$ isTo verify the first branch, when $0\le M\le 2$ the score of $4$ is saturated and the other five scores sum with it to $8+M-5\theta$. When $2\le M\le 4.5$, the score of $0$ is saturated instead, giving the same expression. At $M=4.5$, the moving observation saturates too, and the equation becomes $10-4\theta =0$. The piecewise formula is continuous at the transition.

The rising sample mean comes from an unbounded linear score, the flat distant-outlier response from a bounded Huber score. This is a finite-data response curve with a specified scale and threshold, rather than a population influence function or a general claim about arbitrary numbers of outliers.

The Huber loss is quadratic near zero and linear beyond a fixed cutoff; its derivative is the clipped Huber score.

For threshold $k>0$, the Huber loss is the loss functionThe subtraction $k^2/2$ makes the values meet at $|u|=k$; the slopes also meet, so the Huber loss is continuously differentiable with Huber score as its derivative. It is a convex function, since that derivative is a monotone function. These quadratic and linear pieces define the original clipped location criterion.^[ref]

For $k=2$, ${\rho }_2(1)=1/2$, ${\rho }_2(2)=2$, and ${\rho }_2(5)=8$, whereas the squared criterion $u^2/2$ at $u=5$ is $12.5$. Increasing a standardized difference from $5$ to $6$ raises the Huber loss by $2$, but raises the squared criterion by $5.5$. The compromise preserves quadratic sensitivity to small differences while limiting the slope for large ones. This makes the Huber loss useful when ordinary observations should retain their detailed distances yet isolated outliers should exert limited pull.

For a fixed finite dataset and scale, taking $k$ large enough that every difference from the sample mean lies within $ks$ recovers that sample mean. Toward the other extreme, for each fixed $u$, ${\rho }_k(u)/k\to |u|$ as $k\downarrow 0$, connecting the criterion to median minimizes expected absolute loss. The normalization is essential: putting $k=0$ into the unscaled Huber loss gives zero everywhere, not a usable median criterion.

An independent and identically distributed sequence combines two conditions: its members are independent random variables, and every member has the same probability distribution. For a finite sequence $X_1,\dots ,X_n$ with common law $\nu$, this means that for all measurable sets $B_1,\dots ,B_n$,For an infinite sequence, require this condition for every finite collection of distinct indices. Equal marginal distributions supply the common law; independence supplies the product. Neither condition supplies the other. Grinstead and Snell call the combination an independent trials process.^[ref]

Choose a Bernoulli distribution with success probability $1/2$. Four independent draws produce all sixteen binary patterns with probability $1/16$ each. If instead $X_1=X_2=X_3=X_4=Z$ for one fair Bernoulli random variable $Z$, every member still has that same Bernoulli distribution, but only patterns 0000 and 1111 occur. Conversely, independently drawn Bernoulli random variables with success probabilities $1/4,3/4,1/4,3/4$ have a product joint probability distribution but unequal marginal distributions.

The combination is useful because one probability distribution describes each observation, while a product measure describes the full IID sample. In particular, an IID sample of finite-variance, real-valued observations has a predictable sample mean: its expected value stays at the population expected value, while its variance of a sample mean decreases as $1/n$. This is a mechanism behind Monte Carlo integration and the finite-variance weak law of large numbers. Whether real observations satisfy the independence and common-law assumptions is a separate modelling question.

For the statistical functional $T(F)=\mu ={\mathbb{E}}_{F}X$, assuming ${\mathbb{E}}_F|X|<\infty$, the contaminated mean is exactlyThus the influence function of the mean is $x-\mu$, and its gross-error sensitivity is infinite on an unbounded sample space. The linear approximation is exact here: any fixed positive contamination fraction allows an arbitrarily large change by moving $x$ far enough. No finite second moment is needed for this derivative, although a finite variance is needed for the usual central limit theorem for the sample mean.

For comparison, the influence function of the sample median iswhen the median $m$ is unique and the probability density function $f$ is positive and continuous at $m$. The sign function records which side of $m$ receives the contaminating mass. The influence function of a quantile supplies the short derivation. With the generalized-inverse quantile convention, contamination exactly at $x=m$ leaves the median at $m$, giving derivative zero there. If $f(m)$ is very small, this bounded influence function can still be large; if $f(m)=0$, the displayed formula is unavailable.

For independent identically distributed observations with finite fourth moment, the sample mean of their squares has sampling variance $\mathrm{Var}(X^2)/n$. Its jackknife variance estimator is exactly the displayed unbiased estimator, since its leave-one-out values differ from their average by $-(x_i^2-\overline{x^2})/(n-1)$. The conditional bootstrap variance of a sample mean is ${\hat{v}}_B=n^{-2}{\sum }_i(x_i^2-\overline{x^2}{)}^2=(n-1){\hat{v}}_J/n$. Thus the two methods have a known finite-sample scaling difference, despite sharing the same large-sample target.

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 delete-one form of jackknife resampling recalculates a statistic after deleting each observation in turn. It uses one observed statistical sample to investigate how sensitive an estimator is to its individual observations, and supplies both a jackknife variance estimator and jackknife bias correction. The calculations are deterministic once the data are fixed: no random draws are needed.

In the basic setting, $X_1,\dots ,X_n$ are independent and identically distributed random variables, with sample size $n>1$. Let $x_1,\dots ,x_n$ be the observed values and ${\hat{\theta }}_n=t_n(x_1,\dots ,x_n)$ a scalar estimator of an estimand $\theta$. Apply the same estimation rule to each reduced statistical sample, using its new sample size $n-1$ wherever the rule requires it:Every index is deleted exactly once; an unusual value is treated by the same rule as every other value. The mean ${\overline{\theta }}_J$ summarizes the deletion estimates. They overlap in $n-2$ observations, so they are generally not independent random variables representing fresh repetitions of the original experiment. Cosma Shalizi's ^[ref] explain this construction and its scaling.

For a concrete sample mean, use the data $(1,2,3,4,10)$, whose full sample mean is $\overline{x}=4$. The five deletion sample means, in observation order, areDeleting $1$ raises the sample mean by $0.75$, while deleting $10$ lowers the sample mean by $1.5$. Thus $10$ has the largest deletion effect in this statistical sample. This is a descriptive sensitivity calculation, not a rule for removing that observation: the full sample mean remains $4$.

Early literature includes M. H. Quenouille's ^[ref] and John W. Tukey's one-paragraph abstract ^[ref], within the issue's collected abstracts. Shalizi's ^[ref] discuss the later development of jackknife bias correction and the jackknife variance estimator.