IID bootstrap and jackknife for smooth statistics — Examples

By [F2], ${\bar{X}}_{n,(-i)}={\bar{X}}_n-(X_i-{\bar{X}}_n)\mathrm{/}(n-1)$. Averaging over $i$ gives ${\bar{\theta }}_{n,(-)}={\bar{X}}_n$, since ${\sum }_i(X_i-{\bar{X}}_n)=0$.

Substitution into [F1] gives $p_{n,i}=X_i$ and $V_{J,n}=\frac{n-1}{n}{\sum }_i{\left(\frac{{\bar{X}}_n-X_i}{n-1}\right)}^2=\frac{1}{n(n-1)}{\sum }_i(X_i-{\bar{X}}_n{)}^2\mathrm{.}$ Thus $V_{J,n}=s_n^2\mathrm{/}n$, where $s_n^2=(n-1{)}^{-1}{\sum }_i(X_i-{\bar{X}}_n{)}^2$ is the unbiased sample variance.

By [F1], the number $K$ of ones in a size-three bootstrap sample has the binomial law with parameters $3$ and $1\mathrm{/}3$. The lower median is $1$ exactly when $K\ge 2$, so ${\mathbb{P}}^{\ast }(\text{bootstrap median}=1\mid 0,0,1)=\left(\genfrac{}{}{0px}{}{3}{2}\right)(1\mathrm{/}3{)}^2(2\mathrm{/}3)+(1\mathrm{/}3{)}^3=7\mathrm{/}27.$ The other $20$ of the $27$ equally likely index tuples give median $0$.

List the $27$ conditional median values with multiplicity: twenty zeros and seven ones. With $\alpha =1\mathrm{/}3$, [F2] gives $Q_{27}(1\mathrm{/}6)=0$ and $Q_{27}(5\mathrm{/}6)=1$. Thus [F3] returns $[0,1]$, while [F4] with observed median $0$ returns $[-1,0]$. These are the outputs under the specified tie and quantile conventions.

By [F2], $Z_n=\sqrt{n}{\bar{X}}_n\Rightarrow Z$ with $Z\sim N(0,1)$. Since absolute value is continuous, [F3] gives $\sqrt{n}\mathrm{\mid }{\bar{X}}_n\mathrm{\mid }\Rightarrow \mathrm{\mid }Z\mathrm{\mid }$. The limiting CDF of $\mathrm{\mid }Z\mathrm{\mid }$ is continuous at $0$ and equals $\mathbb{P}(\mathrm{\mid }Z\mathrm{\mid }\le 0)=0$.

Let $V_n^{\ast }=\sqrt{n}({\bar{X}}_n^{\ast }-{\bar{X}}_n)$. Its conditional CDF $F_n^{\ast }$ satisfies $D_n:={\sup }_t\mathrm{\mid }{F}_n^{\ast }(t)-\mathrm{\Phi }(t)\mathrm{\mid }\to 0$ in probability by [F1]. The bootstrap root for $g(x)=\mathrm{\mid }x\mathrm{\mid }$ is $R_n^{\ast }=\sqrt{n}(\mathrm{\mid }{\bar{X}}_n^{\ast }\mathrm{\mid }-\mathrm{\mid }{\bar{X}}_n\mathrm{\mid })=\mathrm{\mid }{Z}_n+V_n^{\ast }\mathrm{\mid }-\mathrm{\mid }{Z}_n\mathrm{\mid }\mathrm{.}$ For fixed data value $Z_n=z\ge 0$, the event $R_n^{\ast }\le 0$ is $-2z\le V_n^{\ast }\le 0$; for $z<0$, it is $0\le V_n^{\ast }\le -2z$.

For any interval $[u,v]$, the difference between its conditional bootstrap probability and its standard normal probability is at most $2D_n$: use $F_n^{\ast }(v)-F_n^{\ast }(u-)$ and take left limits in the uniform CDF bound, using continuity of $\mathrm{\Phi }$. Therefore $Q_n:={\mathbb{P}}^{\ast }(R_n^{\ast }\le 0\mid X_1,\dots ,X_n)=\mathrm{\Phi }(2\mathrm{\mid }{Z}_n\mathrm{\mid })-\frac{1}{2}+o_{\mathbb{P}}(1)\mathrm{.}$ Indeed, the normal mass of either interval in step 1.2 is $\mathrm{\Phi }(2\mathrm{\mid }z\mathrm{\mid })-1\mathrm{/}2$. By [F3] and [F4], $Q_n\Rightarrow \mathrm{\Phi }(2\mathrm{\mid }Z\mathrm{\mid })-1\mathrm{/}2$. This limit is positive with positive probability because $\mathbb{P}(\mathrm{\mid }Z\mathrm{\mid }\ge 1)>0$ and [F5] gives $\mathrm{\Phi }(2)-1\mathrm{/}2>0$. Thus $Q_n$ does not converge in probability to zero.

If the conditional bootstrap law consistently estimated the sampling law of $\sqrt{n}\mathrm{\mid }{\bar{X}}_n\mathrm{\mid }$, its conditional CDF at the continuity point $0$ would converge in probability to the sampling limit CDF there, namely zero by step 1.1 and [F5]. Step 2.1 contradicts this. The failure occurs because $g(x)=\mathrm{\mid }x\mathrm{\mid }$ is not differentiable at the true mean $0$.

Here $g(x)=x$ is Borel and differentiable everywhere with derivative $a=1$. Thus $a^T\mathrm{\Sigma }a=\mathrm{Var}(X_1)={\sigma }^2\in (0,\mathrm{\infty })$, so all assumptions of [F2] hold.

Since ${\widehat{\theta }}_n={\bar{X}}_n$ and ${\widehat{\theta }}_n^{\ast }={\bar{X}}_n^{\ast }$ by [F1], [F2] gives $\sqrt{n}({\bar{X}}_n-\mu )\Rightarrow N(0,{\sigma }^2)$ and the conditional law of $\sqrt{n}({\bar{X}}_n^{\ast }-{\bar{X}}_n)$ converges weakly in probability to the same normal law. This is first-order distributional validity only.