Bootstrap failure, m-out-of-n methods, and subsampling

For $u\in (0,1)$ the quantile set $\{x:F(x)\ge u\}$ is nonempty and bounded below because the events $\{X\le k\}$ increase to the whole space and $\{X\le -k\}$ decrease to the empty set; [F2]–[F3] give the CDF limits $1$ and $0$ at the two tails. If $q=q_F(u)$, then for every $r>0$ there is $x<q+r$ with $F(x)\ge u$, so $F(q+r)\ge u$. The events $\{X\le q+1\mathrm{/}k\}$ decrease to $\{X\le q\}$; by [F2], $F(q)={\lim }_{k}F(q+1\mathrm{/}k)\ge u$. Also, for every $x<q$, $F(x)<u$ by the definition of infimum.

If ${\alpha }_2=0$, the event is vacuous because $q_G(1)=+\mathrm{\infty }$. Otherwise, if $1-{\alpha }_2-\epsilon \le 0$, the lower bound is nonpositive and is trivial. In the remaining case, every $x$ with $G(x)\ge 1-{\alpha }_2$ satisfies $F(x)\ge 1-{\alpha }_2-\epsilon$, so $q_F(1-{\alpha }_2-\epsilon )\le q_G(1-{\alpha }_2)$. Step 1.1 gives $F(q_F(1-{\alpha }_2-\epsilon ))\ge 1-{\alpha }_2-\epsilon$, and monotonicity of $F$ proves the first probability bound.

If ${\alpha }_1=0$, the event is vacuous because $q_G(0)=-\mathrm{\infty }$. If ${\alpha }_1+\epsilon \ge 1$, the lower bound is nonpositive and is trivial. Otherwise, each $x$ with $F(x)\ge {\alpha }_1+\epsilon$ has $G(x)\ge {\alpha }_1$, whence $q_G({\alpha }_1)\le q_F({\alpha }_1+\epsilon )$. For every $y<q_F({\alpha }_1+\epsilon )$, step 1.1 gives $F(y)<{\alpha }_1+\epsilon$. The events $\{X\le q_F({\alpha }_1+\epsilon )-1\mathrm{/}k\}$ increase to $\{X<q_F({\alpha }_1+\epsilon )\}$; [F3] therefore gives $\mathbb{P}\{X<q_F({\alpha }_1+\epsilon )\}\le {\alpha }_1+\epsilon$. This proves the second bound.

The stated replacement on $N$ preserves measurable evaluations and gives CDF paths everywhere. For $u\in (0,1)$, $q_{\widehat{G}}(u)$ is measurable because $\{q_{\widehat{G}}(u)<a\}={\bigcup }_{r\in \mathbb{Q},r<a}\{\widehat{G}(r)\ge u\}$; the endpoint quantiles are constant, so the coverage event is measurable. Also $E:=\{{\sup }_x\mathrm{\mid }\widehat{G}(x)-F(x)\mathrm{\mid }\le \epsilon \mathrm{/}2\}$ is measurable because $\mathrm{\mid }\widehat{G}-F\mathrm{\mid }$ is right-continuous for every outcome and its supremum over $\mathbb{R}$ equals its supremum over $\mathbb{Q}$. If ${\alpha }_1+{\alpha }_2+\epsilon +\delta \ge 1$, part 3 is the trivial nonnegative lower bound. Otherwise, on $E$, when ${\alpha }_1>0$ the pointwise order implies $q_{\widehat{G}}({\alpha }_1)\le q_F({\alpha }_1+\epsilon \mathrm{/}2)$; hence the lower-tail failure on $E$ is contained in $\{X<q_F({\alpha }_1+\epsilon \mathrm{/}2)\}$ and has probability at most ${\alpha }_1+\epsilon \mathrm{/}2$ by step 2.2. If ${\alpha }_1=0$, this tail is empty. Similarly, when ${\alpha }_2>0$, $q_F(1-{\alpha }_2-\epsilon \mathrm{/}2)\le q_{\widehat{G}}(1-{\alpha }_2)$; hence the upper-tail failure on $E$ is contained in $\{X>q_F(1-{\alpha }_2-\epsilon \mathrm{/}2)\}$, whose probability is at most ${\alpha }_2+\epsilon \mathrm{/}2$ by step 1.1. If ${\alpha }_2=0$, that tail is empty. A union bound and $\mathbb{P}(E^c)\le \delta$ give the claimed coverage bound; no independence between $E$ and $X$ is needed because each intersection with $E$ is bounded by the corresponding unconditional tail event.

The stated replacement on each measurable null set preserves measurable evaluations and their pointwise convergence in probability, and gives CDF paths everywhere. Consequently ${\sup }_x\mathrm{\mid }{G}_n(x)-F(x)\mathrm{\mid }$ equals the supremum over rational $x$ for every outcome, by right-continuity, and is measurable. Fix $0<\eta <1\mathrm{/}2$. The events defining a CDF increase to the whole space as $x\to +\mathrm{\infty }$ and decrease to the empty set as $x\to -\mathrm{\infty }$; [F3]–[F4] therefore give its tail limits. Choose $L<R$ with $F(L)<\eta$ and $1-F(R)<\eta$. By [F2], $F$ is uniformly continuous on $[L,R]$, so choose a finite partition $L=x_0<\cdots <x_k=R$ for which every increment $F(x_{j+1})-F(x_j)<\eta$.

Pointwise convergence at the finitely many grid points implies that, with probability tending to one, $\mathrm{\mid }{G}_n(x_j)-F(x_j)\mathrm{\mid }\le \eta$ for every $j$. On this event, monotonicity of both CDFs bounds their difference on each $[x_j,x_{j+1}]$ by $3\eta$. For $x<L$, both CDF values lie between $0$ and their values at $L$, so their difference is at most $2\eta$; for $x>R$, apply the same argument to the complementary tails using their values at $R$. Hence ${\sup }_x\mathrm{\mid }{G}_n(x)-F(x)\mathrm{\mid }\le 3\eta$ with probability tending to one. Since $\eta >0$ was arbitrary, the supremum converges to zero in probability.

If $p=q$, the laws coincide. Suppose $p\le q$. For each $i$, give $(B_i,C_i)\in \{0,1{\}}^2$ probabilities $p$ at $(1,1)$, $q-p$ at $(0,1)$, $1-q$ at $(0,0)$, and zero at $(1,0)$, and take $n$ independent copies of this finite pair. Then $B_i\sim \mathrm{Bernoulli}(p)$, $C_i\sim \mathrm{Bernoulli}(q)$, and $D_i:=C_i-B_i\sim \mathrm{Bernoulli}(q-p)$; the pairs are independent across $i$. Thus $S_{n,q}-S_{n,p}=n^{-1\mathrm{/}2}{\sum }_i(D_i-(q-p))$, whose mean is zero and whose variance, by [F2], is $(q-p)(1-q+p)\le q-p$. If $q<p$, interchange the two parameters and change the sign of the difference; its variance is $\mathrm{\mid }p-q\mathrm{\mid }(1-\mathrm{\mid }p-q\mathrm{\mid })\le \mathrm{\mid }p-q\mathrm{\mid }$.

Put $r=\mathrm{\mid }p-q\mathrm{\mid }$ and $\epsilon =r^{1\mathrm{/}3}$ for $r>0$. By [F3] and step 1.1, the coupling just constructed satisfies $\mathbb{P}(\mathrm{\mid }{S}_{n,p}-S_{n,q}\mathrm{\mid }>\epsilon )\le r\mathrm{/}{\epsilon }^2=\epsilon$. For any closed set $A$, the event $\{S_{n,p}\in A,\mathrm{\mid }{S}_{n,p}-S_{n,q}\mathrm{\mid }\le \epsilon \}$ is contained in $\{S_{n,q}\in A^{[\epsilon ]}\}$, so ${\mu }_{n,p}(A)\le {\mu }_{n,q}(A^{[\epsilon ]})+\epsilon$. Interchanging $p$ and $q$ gives the reverse inequality. Hence $\epsilon$ is admissible in [F4], so the infimum defining $\pi$ is at most $\epsilon$. For $r=0$ the laws coincide and their distance is zero by [F4]. This proves the bound for every endpoint case as well.

Almost surely $M_n\le 1$, so $R_n\le 0$ and $0\le h\le 1$ with support in $[-1\mathrm{/}2,1\mathrm{/}2]$. The uniform CDF gives $\mathbb{E}[h(R_n)]\le \mathbb{P}(\mathrm{\mid }{R}_n\mathrm{\mid }\le 1\mathrm{/}2)=\mathbb{P}(M_n\ge 1-1\mathrm{/}(2n))=1-(1-1\mathrm{/}(2n){)}^n$. By [F5], $\log (1-u)\mathrm{/}u\to -1$ as $u\downarrow 0$: the integral formula gives $-\log (1-u)\mathrm{/}u=u^{-1}{\int }_{1-u}^{1}dt\mathrm{/}t$, an average of $1\mathrm{/}t$ over an interval shrinking to $1$. Thus $n\log (1-c\mathrm{/}n)\to -c$ for each fixed $c>0$, and ${\mathrm{lim\hspace{0.17em}sup}}_n\mathbb{E}[h(R_n)]\le 1-e^{-1\mathrm{/}2}$.

The maximum observation is unique almost surely because the uniform law has no atoms. Conditional on the data, each bootstrap draw selects that maximal observation with probability $1\mathrm{/}n$, independently across draws, so ${\mathbb{P}}^{\ast }(R_n^{\ast }=0)=1-(1-1\mathrm{/}n{)}^n$. Since $h(0)=1$ and $h\ge 0$, ${\mathbb{E}}^{\ast }[h(R_n^{\ast })]\ge 1-(1-1\mathrm{/}n{)}^n\to 1-e^{-1}$.

Since $e^{-1\mathrm{/}2}>e^{-1}$, the lower limit in step 1.2 is strictly greater than the upper limit in step 1.1. Therefore the conditional test integral cannot converge in probability to the sampling test integral. By [F4], weak consistency would require that convergence for this particular bounded continuous $h$, so weak bootstrap consistency fails.

For a size-$k$ IID uniform sample, $\mathbb{P}\{k(M_k-1)\le x\}=0$ for $x<-k$, equals $(1+x\mathrm{/}k{)}^k$ for $-k\le x<0$, and is $1$ for $x\ge 0$. For fixed $x<0$, [F10] gives $k\log (1+x\mathrm{/}k)\to x$, so these CDFs converge to $F(x)=e^x$; for $x\ge 0$ they equal $1$. The limiting CDF is continuous, including at zero.

Fix $x<0$ and put $c=-x>0$, $D_n=1-M_n$, and $K_n(c)=\mathrm{\#}\{i:X_i>M_n-c\mathrm{/}m\}$. For fixed $t>0$ and all sufficiently large $m$, each indicator $1\{X_i>1-t\mathrm{/}m\}$ is Bernoulli$(t\mathrm{/}m)$, and the indicators are independent by [F2]; hence $N_n(t):=\mathrm{\#}\{i:X_i>1-t\mathrm{/}m\}$ is binomial by [F4]. By [F5]–[F6], $m{N}_n(t)\mathrm{/}n\to t$ in probability, since its mean is $t$ and its variance is $t(m\mathrm{/}n)(1-t\mathrm{/}m)\to 0$. Also $m{D}_n\to 0$ in probability, since for each $\eta >0$ and large $m$, $\mathbb{P}(m{D}_n>\eta )=(1-\eta \mathrm{/}m{)}^n$. By [F10], $\log (1-u)=-{\int }_{1-u}^{1}dt\mathrm{/}t\le -u$ for $0<u<1$, so this probability is at most $e^{-\eta n\mathrm{/}m}\to 0$.

For any $0<\eta <c$, on $m{D}_n\le \eta$ the interval from $1$ down to $M_n-c\mathrm{/}m$ has width between $c\mathrm{/}m$ and $(c+\eta )\mathrm{/}m$, so $N_n(c)\le K_n(c)\le N_n(c+\eta )$. Fix $\rho >0$ and choose $0<\eta <\min (c,\rho \mathrm{/}3)$. Step 1.2 gives $m{N}_n(c)\mathrm{/}n\to c$ and $m{N}_n(c+\eta )\mathrm{/}n\to c+\eta$, while $P(m{D}_n>\eta )\to 0$; the sandwich therefore gives $m{K}_n(c)\mathrm{/}n\to c$ and $K_n(c)\mathrm{/}n\to 0$ in probability. For $u_n=K_n(c)\mathrm{/}n$, the integral identity in [F10] gives $\mathrm{\mid }\log (1-u_n)+u_n\mathrm{\mid }\le u_n^2\mathrm{/}(1-u_n)$ when $u_n<1$. Since $m{u}_n\to c$ and $u_n\to 0$, $m{u}_n^2\mathrm{/}(1-u_n)\to 0$, hence $(1-u_n{)}^m\to e^{-c}$. By [F1], the conditional with-replacement CDF is $P^{\ast }(R_{m,n}^{\ast }\le x)=(1-K_n(c)\mathrm{/}n{)}^m\to e^{-c}=e^x$.

For subsampling, the conditional probability that every selected observation is at most $M_n-c\mathrm{/}m$ is $P^{\ast }(R_{m,n}^{\mathrm{s}\mathrm{u}\mathrm{b}}\le x)=\left(\genfrac{}{}{0px}{}{n-K_n(c)}{m}\right)\mathrm{/}\left(\genfrac{}{}{0px}{}{n}{m}\right)={\prod }_{j=0}^{m-1}(1-u_{n,j})$, where $u_{n,j}=K_n(c)\mathrm{/}(n-j)$; this formula holds on $n-K_n(c)\ge m$, an event whose probability tends to one by step 2.1 and $m\mathrm{/}n\to 0$.

We have ${\max }_{j<m}{u}_{n,j}\le (K_n(c)\mathrm{/}n)\mathrm{/}(1-m\mathrm{/}n)\to 0$ in probability, and ${\sum }_{j=0}^{m-1}{u}_{n,j}=(m{K}_n(c)\mathrm{/}n)\cdot m^{-1}{\sum }_{j=0}^{m-1}n\mathrm{/}(n-j)\to c$, because the average factor lies between $1$ and $(1-m\mathrm{/}n{)}^{-1}\to 1$. Therefore ${\sum }_j{u}_{n,j}^2\le ({\max }_j{u}_{n,j}){\sum }_j{u}_{n,j}\to 0$.

With probability tending to one, all $u_{n,j}<1$; [F10] and $\mathrm{\mid }\log (1-u)+u\mathrm{\mid }\le u^2\mathrm{/}(1-u)$ then show ${\sum }_j\log (1-u_{n,j})\to -c$, so the product in step 3.1 tends to $e^{-c}=e^x$.

For $x\ge 0$, both resampled maxima are at most $M_n$, so both conditional CDFs equal $1$. Step 1.1 gives the continuous sampling-root limit $F$, while steps 2.1 and 4.1 prove pointwise convergence for negative $x$; [F7] upgrades each method's conditional CDF convergence to uniform convergence in probability. Step 1.1 also identifies $F$ as the limiting CDF of the actual size-$m$ sampling root, which is the first-order target specified in the statement. The use of the actual size-$m$ scale is essential in steps 2.1–4.1.

By [F2], under each fixed $P$ the IID observations have joint sampling law $P^{\otimes n}$. By [F5], $J_{n,P}$ is the CDF of the actual root $R_{n,P}$ under that law. Fix $\eta >0$. The two root-law limsup conditions imply that, for all sufficiently large $n$ and every $P\in \mathcal{P}$, both one-sided differences between $H_{m,P}$ and $J_{n,P}$ are at most $\eta \mathrm{/}2$. The feasible-root condition gives numbers ${\delta }_n\to 0$ such that, uniformly in $P$, the event $E_{n,P}:=\{{\sup }_x\mathrm{\mid }{\widehat{H}}_{m,n}(x)-H_{m,P}(x)\mathrm{\mid }\le \eta \mathrm{/}2\}$ has probability at least $1-{\delta }_n$. On this event, the triangle inequality gives ${\sup }_x\mathrm{\mid }{\widehat{H}}_{m,n}(x)-J_{n,P}(x)\mathrm{\mid }\le \eta$.

To see that neither root-law comparison follows from the m-regime, take $\mathcal{P}=\{{\delta }_0\}$, $\theta (P)=0$, $a_k=k$, and the measurable constant statistic $T_k={\mathbf{1}}_{\{k\text{ is odd}\}}\mathrm{/}k$. Let $m_n=\lfloor \sqrt{n}\rfloor$. Then $a_{m_n}\mathrm{/}{a}_n=m_n\mathrm{/}n\to 0$, $J_{n,P}={\delta }_{{\mathbf{1}}_{\{n\text{ is odd}\}}}$, and $H_{m_n,P}={\delta }_{{\mathbf{1}}_{\{m_n\text{ is odd}\}}}$. For every sufficiently large even $m$, the block $m^2\le n<(m+1{)}^2$ contains odd $n$, giving $H_{m_n,P}={\delta }_0$, $J_{n,P}={\delta }_1$, and ${\sup }_x(H_{m_n,P}(x)-J_{n,P}(x))=1$. For every sufficiently large odd $m$, the same block contains even $n$, giving $H_{m_n,P}={\delta }_1$, $J_{n,P}={\delta }_0$, and ${\sup }_x(J_{n,P}(x)-H_{m_n,P}(x))=1$. Each root-law comparison can therefore fail under the stated m-regime, even with $a_m\mathrm{/}{a}_n\to 0$.

Apply [F4], part 3, with $F=J_{n,P}$, $\widehat{G}={\widehat{H}}_{m,n}$, CDF error $2\eta$, and exceptional-event probability ${\delta }_n$. Step 1.1 supplies the event on which the CDF error is at most $\eta$, so the lemma's pointwise tail inclusions bound the joint miss event without assuming independence of $R_{n,P}$ and that event. The root interval $[q_{{\widehat{H}}_{m,n}}({\alpha }_1),q_{{\widehat{H}}_{m,n}}(1-{\alpha }_2)]$ therefore misses $R_{n,P}$ with probability at most ${\alpha }_1+{\alpha }_2+2\eta +{\delta }_n$, uniformly in $P$.

Suppose only the first root-law limsup condition and the feasible-root condition hold. For any $\eta >0$, all sufficiently large $n$ satisfy $H_{m,P}(x)-J_{n,P}(x)\le \eta \mathrm{/}2$ uniformly in $P,x$, and on an event $E_{n,P}$ of probability at least $1-{\delta }_n$ we have ${\widehat{H}}_{m,n}(x)-H_{m,P}(x)\le \eta \mathrm{/}2$ for all $x$. Hence on $E_{n,P}$, ${\widehat{H}}_{m,n}-J_{n,P}\le \eta$. If ${\alpha }_2+\eta <1$, quantile ordering in [F4] gives $q_{J_{n,P}}(1-{\alpha }_2-\eta )\le q_{{\widehat{H}}_{m,n}}(1-{\alpha }_2)$. Thus the joint failure event $\{R_{n,P}>q_{{\widehat{H}}_{m,n}}(1-{\alpha }_2)\}\cap E_{n,P}$ is contained in $\{R_{n,P}>q_{J_{n,P}}(1-{\alpha }_2-\eta )\}$, whose probability is at most ${\alpha }_2+\eta$ by [F4]; adding $P(E_{n,P}^c)\le {\delta }_n$ bounds total failure. Letting $n\to \mathrm{\infty }$ and then $\eta \downarrow 0$ proves uniform asymptotic coverage at least $1-{\alpha }_2$ for the lower confidence bound; if ${\alpha }_2+\eta \ge 1$, the intermediate lower bound is trivial. For the second assertion, if only the second root-law condition holds, then on the same event $J_{n,P}-{\widehat{H}}_{m,n}\le \eta$. When ${\alpha }_1+\eta <1$, [F4] gives $q_{{\widehat{H}}_{m,n}}({\alpha }_1)\le q_{J_{n,P}}({\alpha }_1+\eta )$. The joint failure event $\{R_{n,P}<q_{{\widehat{H}}_{m,n}}({\alpha }_1)\}\cap E_{n,P}$ is contained in $\{R_{n,P}<q_{J_{n,P}}({\alpha }_1+\eta )\}$, whose probability is at most ${\alpha }_1+\eta$ by [F4]; adding the complement probability and letting $\eta \downarrow 0$ proves uniform asymptotic coverage at least $1-{\alpha }_1$ for the upper confidence bound. If ${\alpha }_1+\eta \ge 1$, the intermediate lower bound is trivial.