Empirical variance Created 2026-10-14 Updated 2026-10-25
The empirical variance is the variance of the empirical distribution, which puts probability on each observation. It uses divisor , whereas the usual sample variance uses divisor . For independent and identically distributed random variables with finite second moment, jackknife bias correction converts the empirical variance into the unbiased estimator exactly.
Jackknife bias correction Created 2026-09-24 Updated 2026-10-14
The jackknife bias correction uses jackknife resampling to estimate and subtract the bias of an estimator. Here denotes the original estimator, its version after deleting observation , and the mean of those versions. For sample size , 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 an unbiased estimator.
For an exact worked example, take the empirical varianceOn , the sample mean is and . Each deletion empirical variance is recomputed around its own deletion sample mean, with divisor , givingFor instance, deleting leaves , whose sample mean is and whose empirical variance is . The correction on the full statistical sample isThe same data thus give a jackknife variance estimator of for the sample mean, and a corrected empirical variance of ; these are different estimands.
The exact equality with the usual sample variance is not an accident of these numbers. For any data with , write . Removing changes this sum to , soFor independent and identically distributed random variables with finite second moment and population variance , the expectation is . Its bias is exactly ; jackknife bias correction removes it exactly in this example. Thus is an unbiased estimator of under these assumptions, including where each deletion empirical variance is zero.
More generally, suppose the bias has a regular expansion at both sample sizes and , with coefficients independent of :Under independent and identically distributed sampling, each deletion estimator has the same expectation as . Hence the corrected bias isThe leading term cancels; the remaining bias need not vanish. The 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]
https://academic.oup.com/biomet/article-abstract/43/3-4/353/345190. Notes on Bias in Estimation, Biometrika 43(3–4), 353–360 (1956)
. Jackknife resampling 2026-10-14
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, are independent and identically distributed random variables, with sample size . Let be the observed values and a scalar estimator of an estimand . Apply the same estimation rule to each reduced statistical sample, using its new sample size 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 summarizes the deletion estimates. They overlap in observations, so they are generally not independent random variables representing fresh repetitions of the original experiment. Cosma Shalizi's [ref]
https://stat.cmu.edu/~cshalizi/490/23/2023-11-29-jackknife.pdf. Carnegie Mellon jackknife notes, Sections 1–3 and 5
explain this construction and its scaling.For a concrete sample mean, use the data , whose full sample mean is . The five deletion sample means, in observation order, areDeleting raises the sample mean by , while deleting lowers the sample mean by . Thus 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 .
Delete each observation once and recompute the sample mean
. The left panel shows all five reduced statistical samples from ; each crossed entry is omitted only for that row. The right panel shows their sample means against the full sample mean . The orange row makes the deletion effect of visible. Details.Early literature includes M. H. Quenouille's [ref]
https://academic.oup.com/jrsssb/article/11/1/68/7026551. Approximate Tests of Correlation in Time-Series, JRSS B 11(1), 68–84 (1949)
and John W. Tukey's one-paragraph abstract [ref]https://www.jstor.org/stable/2237363. Bias and Confidence in Not-Quite Large Samples, Annals of Mathematical Statistics 29(2), 614 (1958)
, within the issue's collected abstracts. Shalizi's [ref]https://stat.cmu.edu/~cshalizi/490/23/2023-11-29-jackknife.pdf. historical notes and bibliography
discuss the later development of jackknife bias correction and the jackknife variance estimator. Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 47 4 a Solution Created 2026-10-03 Updated 2026-10-14
For jackknife resampling, calculate the leave-one-out estimates by deleting observation , and their average . The jackknife variance estimator isThe factor compensates for the small differences between the overlapping samples. For sufficiently smooth statistics it estimates their sampling variance. The related estimated bias is , leading to the jackknife bias correction .
For the nonparametric bootstrap, replace the unknown distribution by its empirical distribution functionIndependently sample observations with replacement from the original observations, compute the estimator on that bootstrap sample, and repeat this for independently generated bootstrap samples. If the resulting values are , estimate the conditional bootstrap variance byIts interpretation as an approximation to the original sampling variance relies on the bootstrap approximating the sampling law of the statistic; increasing only reduces simulation error and cannot repair a failure of that approximation.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 47 4 c Solution Created 2026-10-03 Updated 2026-10-14
The first matrix contains the leave-one-out samples, and each element of
vect is their sample mean. For ,Thus R4a computes the jackknife variance estimator of the full sample mean:where is the unbiased sample variance. R5a computes the estimated bias used in jackknife bias correction. The identity for the average leave-one-out sample mean gives R5a = 0, apart from floating-point rounding.The second printed loop contains an indexing error: its row index is
i, although its loop variable is b. Literally, after the first block, i equals . When , every iteration overwrites row of the new matrix; the other rows remain missing. Hence vect contains one finite mean and 198 missing entries. R's default sorting removes missing entries, leaving a vector of length one, so selecting its fifth and 195th entries gives R7b = c(NA, NA). If , the matrix assignment is out of bounds; if the second block runs independently with i undefined, it fails even earlier. This is a source error, not a property of the bootstrap.Replacing just the row index by
b gives the intended algorithm: draw 199 independent bootstrap samples and sort their sample means as . The two indices are and , soThese are the endpoints of an approximate 95% percentile bootstrap confidence interval for the population mean. Their actual numerical values depend on the supplied data and random draws, which are not specified. They are not the reflected endpoints of a basic bootstrap confidence interval, and the choice of ranks does not guarantee exact finite-sample coverage. Past exam of the mathematics course of the University of Cambridge 2019 ii Paper 4 28J b iii Solution Created 2026-09-24 Updated 2026-10-03
For each , let be the same estimator computed after deleting , and put . The jackknife bias correction isIts bias isso the leading term has been removed and the bias is smaller than that of .
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 3 28K a Solution Created 2026-09-24 Updated 2026-09-25
Each leave-one-out statistic has the same distribution as , soConsequently the bias of the jackknife bias correction isThe assumed expansion givesandTheir difference is therefore , as required.

