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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.

Set $Y_i=X_i^2$, $\overline{Y}=n^{-1}{\sum }_i{Y}_i$ and ${\tau }^2={\mathrm{Var}}_F(Y_1)={\mathbb{E}}_F{X}_1^4-{\theta }^2$. The fourth-moment assumption makes ${\tau }^2$ finite, so the variance of a sample mean is $v={\tau }^2/n$. Assume $n\ge 2$, as required for leave-one-out estimation.

The Jackknife variance estimator recomputes the original statistic on each sample with one observation removed. HereBy the definition of the Jackknife variance estimator,For completeness, expanding the centered sum givesConsequently $\mathbb{E}{\hat{v}}_{\mathrm{JACK}}={\tau }^2/n=v$. The jackknife variance estimator is unbiased for this statistic. This calculation uses only the finite fourth moment, not normality of the original observations.

Given the observed data, put $y_i=x_i^2$ and $\overline{y}=n^{-1}{\sum }_i{y}_i$. The empirical distribution assigns mass $1/n$ to each observed $x_i$, including repeated values with their multiplicities. A bootstrap sample consists of $n$ independent draws from this distribution, with replacement. The nonparametric bootstrap variance estimate is the conditional varianceIndependence of the bootstrap draws gives the exact conditional bootstrap variance of a sample mean:Thus its finite-sample expectation is $(n-1)v/n$, in contrast to the unbiased jackknife estimate. This is the jackknife and bootstrap variance of a sample second moment scaling difference.

To approximate it by Monte Carlo simulation, independently generate $B$ bootstrap samples. For replicate $b$, choose $n$ independent uniform indices $I_{b1},\dots ,I_{bn}\in \{1,\dots ,n\}$ and compute $T_b^{\ast }=n^{-1}{\sum }_j{x}_{I_{bj}}^2$. With ${\overline{T}}^{\ast }=B^{-1}{\sum }_b{T}_b^{\ast }$, useFor $B>1$ this is conditionally unbiased for the bootstrap variance and converges to it as $B\to \infty$. Uniform indices can be obtained by $I=\lceil nU\rceil$ for uniforms on $(0,1]$.

Let $G_{\ast }(t)=P_{\ast }(T^{\ast }\le t)$ be the conditional bootstrap distribution of the statistic. A nominal level $1-\alpha$ percentile bootstrap confidence interval isReuse the bootstrap replicates, sort them as $T_{(1)}^{\ast }\le \cdots \le T_{(B)}^{\ast }$, and take the order statistics with ranks $\lceil B\alpha /2\rceil$ and $\lceil B(1-\alpha /2)\rceil$ as empirical quantiles. These are quantiles of the statistic itself, rather than reflected quantiles of its centered error. Increasing $B$ reduces simulation error; it does not turn an approximate confidence interval into an exact finite-sample one.

An analytic alternative follows from the central limit theorem. If ${\tau }^2>0$, then $\sqrt{n}({\hat{\theta }}^n-\theta )\Rightarrow N(0,{\tau }^2)$, and the sample variance $s_Y^2=(n-1{)}^{-1}{\sum }_i(x_i^2-\overline{y}{)}^2$ converges to ${\tau }^2$ by the strong law of large numbers. The Slutsky theorem therefore gives the large-sample normal approximation intervalwhere $z_{1-\alpha /2}$ is the standard normal quantile. Its lower endpoint may be truncated to zero because the parameter is nonnegative. If ${\tau }^2=0$, $X_i^2=\theta$ almost surely and the statistic identifies the parameter exactly; the interval degenerates to that value. The nondegenerate normal interval is asymptotic, not an exact Student interval under arbitrary $F$.

An unordered bootstrap sample is determined by the multiplicities $N_i\ge 0$ of the $n$ distinct observations, with ${\sum }_i{N}_i=n$. Conversely each such vector specifies exactly one sample up to rearrangement. Place $n$ identical marks into $n$ boxes using $n-1$ separators. The stars and bars argument therefore givesThese bootstrap count vectors are not equally likely: the probability of a vector is $n!/(n^n{\prod }_i{N}_i!)$, counting its possible ordered sequences. The ordered index sequences themselves number $n^n$.

The first simulation uses independent standard Cauchy distributions and returnsThe tail probability and estimator moments areThe indicator count has a binomial distribution with parameters $n,\theta$; the infinite variance of a Cauchy observation is irrelevant because this estimator averages bounded indicators.

For the next code block, each row is an independent sample of size $n$ from the empirical distribution of the original $x_i$. Its entry in the vector $v$ is the corresponding bootstrap indicator mean ${\hat{\theta }}_b^{\ast }$. The last expression is the variance of these $B=199$ replicate estimates, with denominator $B-1$:Condition on the original data and write a star on bootstrap expectations. Every resampled indicator has a Bernoulli distribution with success probability $\hat{\theta }$, and the $n$ draws within a row are independent. Thus the conditional bootstrap variance of a sample mean givesIt estimates the sampling variance of the tail-probability estimator, not the variance of the raw Cauchy data. Its square root is an estimated standard error. Finite $B$ leaves bootstrap simulation noise. Averaging over the original data, its expectation is $(n-1)\theta (1-\theta )/n^2$, so it is not exactly unbiased for the true variance $\theta (1-\theta )/n$ at finite $n$.

In the importance-sampling block, the transformation $Y=2/(1-U)$ gives, for $y\ge 2$,The variable $Y$ is a proposal draw supported entirely in the integration region. The deterministic value computed from it is the importance sampling weightConsequentlyThe code uses ordinary importance sampling of a Cauchy tail, not self-normalization; $Y$ itself is neither an observation from the target Cauchy law nor the estimator's summand. The weight compensates for sampling from its proposal law.

The final block bootstraps these same proposal observations $Y_i$ and recomputes the weight average in each row. Since $W$ is a deterministic function of $Y$, this is equivalent to resampling the observed weights $W_i$. If ${\tilde{\theta }}_b^{\ast }$ denotes a replicate, the final output is its sample variance over the $B$ replicates. Its conditional expectation isIts expectation over the original proposal sample is $(n-1)\mathrm{Var}(W)/n^2$.

The variance reduction can be calculated, rather than inferred just from the appearance of the weights. Write $A=\arctan (1/2)$. Thenbecause an antiderivative of the integrand without its prefactor is $\frac{1}{2}(\arctan y-y/(1+y^2))$. HenceBy comparison, $\theta (1-\theta )\approx 0.125803$. For $n=100$ the true estimator variances are approximately $9.55253\times 10^{-7}$ and $1.25803\times 10^{-3}$ respectively. The importance estimator has about 1/1,317 of the indicator estimator's variance, because all proposals land in the tail and the weights remain in the narrow interval $[2/(5\pi ),1/(2\pi ))$.

Thus the last bootstrap output should usually be much smaller than the first, and its expected value is smaller by the same variance ratio. This is not a deterministic ordering of two finite-run outputs: if all the original Cauchy draws happen to be at most two, the first bootstrap variance is zero, while unequal importance weights can still give a positive second variance.