Jackknife bias correction (${\tilde{T}}_J=n{T}_n-(n-1){\overline{T}}_J$)

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].

For observations $X_1,\dots ,X_n$, the unbiased sample variance isFor independent identically distributed observations with finite variance ${\sigma }^2$, it converges in probability to ${\sigma }^2$.

The sample standard deviation is the nonnegative square root of the usual sample variance for $n\ge 2$. The displayed convention uses divisor $n-1$. Although the sample variance is an unbiased estimator of the population variance under independent identically distributed sampling with finite variance, taking its square root does not in general give an unbiased estimator of the population standard deviation.

For a normal sample of variance ${\sigma }^2>0$, define $S^2=(n-1{)}^{-1}{\sum }_i(X_i-\overline{X}{)}^2$. The exact chi-squared distribution identity $(n-1)S^2/{\sigma }^2\sim {\chi }_{n-1}^2$ and the central limit theorem for independent squared standard normal random variables implyThe delta method for $g(v)=\sqrt{v}$, with derivative $g^{\prime }({\sigma }^2)=1/(2\sigma )$, therefore gives the displayed limit for $S$. Exact independence of the normal sample mean and sample variance also makes $S$ independent of $\overline{X}$. The resulting joint limit supports combining location and scale estimates when a model relates their parameters.

Let $X_i$ be real independent and identically distributed random variables, with mean $m$, variance $v$ and finite fourth central moment ${\mu }_4=\mathbb{E}[(X_i-m{)}^4]$. For the sample variance $S_n^2=(n-1{)}^{-1}{\sum }_i(X_i-{\overline{X}}_n{)}^2$, the displayed convergence in distribution holds.

Indeed $n^{-1}{\sum }_i(X_i-{\overline{X}}_n{)}^2=n^{-1}{\sum }_i(X_i-m{)}^2-({\overline{X}}_n-m{)}^2$. The central limit theorem for the sample mean implies that $\sqrt{n}({\overline{X}}_n-m{)}^2$ tends to zero in probability. Changing the divisor from $n$ to $n-1$ also has negligible effect at this scale, by the law of large numbers. Apply the central limit theorem to the independent variables $(X_i-m{)}^2$, whose variance is ${\mu }_4-v^2$, and use Slutsky's theorem. Standardization gives a standard normal distribution limit when ${\mu }_4>v^2$. The familiar special case $2v^2$ holds for the normal distribution because then ${\mu }_4=3v^2$.

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.

For two independent normal samples of sizes $m,n$ with a common unknown variance, pooling their within-sample sums of squares gives the unbiased estimator $S_p^2=(S_{XX}+S_{YY})/(m+n-2)$. The numerator divided by the common population variance has a chi-squared distribution with $m+n-2$ degrees of freedom and is independent of both sample means. Therefore $(\bar{X}-\bar{Y}-({\mu }_X-{\mu }_Y))/(S_p\sqrt{1/m+1/n})$ has a Student t-distribution. This exact construction requires a positive residual number of degrees of freedom and the common-variance assumption.