Moving and tapered block bootstrap for dependent data — Examples

At the first position of a selected block, the conditional average over starts is $(0+1+2)\mathrm{/}3=1$. At the second position it is $(1+2+3)\mathrm{/}3=2$.

The output mean is the average of two first-position values and two second-position values, so its conditional expectation is $(1+2)\mathrm{/}2=3\mathrm{/}2$, which equals the observed sample mean. The resampled blocks preserve each adjacent pair inside its chosen block; only the join between blocks can alter adjacency.

Substitution of $r=1,2,3$ gives $w_3=(1\mathrm{/}3,1,1\mathrm{/}3)$, so $W_3^2=1\mathrm{/}9+1+1\mathrm{/}9=11\mathrm{/}9$ and $a_3=3\sqrt{3\mathrm{/}11}$.

The scaled within-block weights are therefore $(\sqrt{3\mathrm{/}11},3\sqrt{3\mathrm{/}11},\sqrt{3\mathrm{/}11})$. Their squared sum is $\frac{3+27+3}{11}=3$, equal to $b$, as required by the normalization $a_b^2{W}_b^2=b$.

Since $n^{1\mathrm{/}3}\le b_n<n^{1\mathrm{/}3}+1$, $b_n\to \mathrm{\infty }$, and $b_n$ is a positive integer. For every integer $n\ge 1$, $n^{1\mathrm{/}3}\le n$; since $n$ itself is an integer, taking the least integer above $n^{1\mathrm{/}3}$ gives $b_n\le n$. Thus the block length is admissible at every positive sample size.

Dividing the same bounds by $\sqrt{n}$ gives $0\le b_n\mathrm{/}\sqrt{n}<n^{-1\mathrm{/}6}+n^{-1\mathrm{/}2}\to 0$. Thus $b_n^{-1}+b_n{n}^{-1\mathrm{/}2}\to 0$.

For $J\ge 1$ let $X_{t,J}={\sum }_{j=0}^{J-1}{2}^{-j}{\epsilon }_{t-j}$. For $K>J$, independence, centering and [F12] give $E\mathrm{\mid }{X}_{t,K}-X_{t,J}{\mathrm{\mid }}^2={\sigma }^2{\sum }_{j=J}^{K-1}{4}^{-j}$, so the partial sums are $L^2$-Cauchy and have a limit by [F13]. Also [F9] and [F2], [F6] give $E{\sum }_{j\ge 0}{2}^{-j}\mathrm{\mid }{\epsilon }_{t-j}\mathrm{\mid }=2E\mathrm{\mid }{\epsilon }_0\mathrm{\mid }<\mathrm{\infty }$, so the series converges absolutely almost surely. The $L^2$ limit and the almost-sure series are both probability limits of these partial sums: [F14] applied to squared $L^2$ errors proves the first assertion, and [F10] applied to indicators of tail events proves the second. For each $ϵ>0$, the triangle inequality bounds the probability that the two limits differ by more than $ϵ$ by the sum of the two partial-sum error probabilities at threshold $ϵ\mathrm{/}2$; letting $J\to \mathrm{\infty }$ gives zero, so the limits agree almost surely. By [F6], the expectations of the partial sums converge to $E{X}_0=0$, and continuity of the $L^2$ norm gives $v:=E{X}_0^2={\lim }_{J}E{X}_{0,J}^2={\sigma }^2{\sum }_{j\ge 0}{4}^{-j}=4{\sigma }^2\mathrm{/}3$. For any finite set of times, the finite truncation vector has the same law after a common time shift by [F1]; its coordinates converge almost surely to the series vector, so bounded uniformly continuous test expectations pass to the limit by [F10]. [F15] gives equality of the limiting vector laws, proving strict stationarity in the sense of [F16].

The characteristic function of $X_{0,J}$ is $\exp \{-t^2{\sigma }^2{\sum }_{j=0}^{J-1}{4}^{-j}\mathrm{/}2\}$ by [F3] and [F2]. By the a.s. convergence in step 1.1 and [F10], these characteristic functions converge to $\exp (-t^{2}v\mathrm{/}2)$; [F4] identifies $X_0\sim N(0,v)$. Thus [F5] gives $E{X}_0^4=3v^2$.

For $h\ge 1$, the recursion gives $X_h=2^{-h}{X}_0+{\eta }_h$, where ${\eta }_h={\sum }_{j=0}^{h-1}{2}^{-j}{\epsilon }_{h-j}$. The variable $X_0$ is measurable with respect to the past innovation sigma-algebra and ${\eta }_h$ with respect to the disjoint sigma-algebra generated by ${\epsilon }_1,\dots ,{\epsilon }_h$, so [F11] makes them independent. [F12] gives $E{\eta }_h=0$, $E{\eta }_h^2=v(1-4^{-h})$, and $E[X_0^2{\eta }_h^2]=E{X}_0^{2}E{\eta }_h^2$. Thus $\gamma (h):=E(X_0{X}_h)=2^{-h}v$. Expanding the product of squares, the cross term vanishes because $E{\eta }_h=0$ and independence, and $E{X}_0^4=3v^2$ by step 2.1; by the covariance definition [F7], $\mathrm{Cov}(X_0^2,X_h^2)=2v^2{4}^{-h}=2\gamma (h{)}^2$.

By [F8], conditional on the observations each of the $n$ draws is uniform on the observed values, so its mean is ${\bar{X}}_n$ and its variance is $n^{-1}{\sum }_{t=1}^n(X_t-{\bar{X}}_n{)}^2$. [F12] factors the cross terms in the variance of their mean, giving ${\mathrm{Var}}^{\ast }\text{\hspace{0.17em}⁣}(\sqrt{n}({\bar{X}}_n^{\ast }-{\bar{X}}_n))=n^{-1}{\sum }_{t=1}^n(X_t-{\bar{X}}_n{)}^2$. Stationarity and step 3.1 give $\mathrm{Var}(n^{-1}{\sum }_t{X}_t^2)\le \frac{2}{n}(v^2+2{\sum }_{h\ge 1}\gamma (h{)}^2)\to 0$: the lag-zero covariance is $E{X}_0^4-v^2=2v^2$, and the positive-lag covariances are $2\gamma (h{)}^2$. Also $E{\bar{X}}_n^2=\frac{1}{n}{\sum }_{\mathrm{\mid }h\mathrm{\mid }<n}(1-\mathrm{\mid }h\mathrm{\mid }\mathrm{/}n)\gamma (\mathrm{\mid }h\mathrm{\mid })\to 0$ because ${\sum }_{h\in \mathbb{Z}}\mathrm{\mid }\gamma (h)\mathrm{\mid }<\mathrm{\infty }$. Markov [F14] applied to squared deviations gives $n^{-1}{\sum }_t{X}_t^2\to v$ in probability and ${\bar{X}}_n^2\to 0$ in probability. Therefore $n^{-1}{\sum }_t(X_t-{\bar{X}}_n{)}^2=n^{-1}{\sum }_t{X}_t^2-{\bar{X}}_n^2\to v$ in probability.

Counting pairs at each lag gives $\mathrm{Var}(\sqrt{n}{\bar{X}}_n)={\sum }_{\mathrm{\mid }h\mathrm{\mid }<n}(1-\mathrm{\mid }h\mathrm{\mid }\mathrm{/}n)\gamma (\mathrm{\mid }h\mathrm{\mid })$. Absolute summability of $\gamma$ gives the limit $\gamma (0)+2{\sum }_{h\ge 1}\gamma (h)=v(1+2{\sum }_{h\ge 1}{2}^{-h})=3v=4{\sigma }^2$, whereas the conditional bootstrap variance tends to $v=4{\sigma }^2\mathrm{/}3$ by step 4.1. The finite, unequal limits refute the universal variance-consistency claim. This calculation makes no broader assertion about every possible distributional metric.

Here $M=7$, ${\bar{Y}}_P=(6+52)\mathrm{/}7=58\mathrm{/}7$, $N_1=2$, $N_2=3$, and $k_1=k_2=2$. The first residual series is $(-2,0,2)$; its candidate blocks are $(-2,0)$ and $(0,2)$, with position averages $(-1,1)$. After subtracting those averages the two candidates are $(-1,-1)$ and $(1,1)$. The second residual series is $(-3,-1,1,3)$; its candidates are $(-3,-1),(-1,1),(1,3)$, also with position averages $(-1,1)$, and become $(-2,-2),(0,0),(2,2)$.

Let $V_1,V_2$ be independent uniforms on $\{-1,1\}$ and $W_1,W_2$ independent uniforms on $\{-2,0,2\}$, all four independent. The outputs are $(58\mathrm{/}7+V_1,58\mathrm{/}7+V_1,58\mathrm{/}7+V_2)$ and $(58\mathrm{/}7+W_1,58\mathrm{/}7+W_1,58\mathrm{/}7+W_2,58\mathrm{/}7+W_2)$. The first output retains only the first entry of its final block. Since all four offsets have mean zero, $E^{\ast }{\bar{Y}}_1^{\ast }=E^{\ast }{\bar{Y}}_2^{\ast }=58\mathrm{/}7$, explicitly realizing [F2] despite the unequal observed means and unequal sample sizes.

The filter $f(s_0,s_1,\dots )=s_{0}e$ is measurable: it is the coordinate projection followed by a map from the finite innovation space to $H$. We have $E{X}_t=0$ and $E\mathrm{\parallel }{X}_t{\mathrm{\parallel }}_H^2=1$. For every $m\ge 1$, the retained current innovation gives $X_{t,m}=X_t$ exactly, so ${\delta }_m=0$. Thus both ${\sum }_{m\ge 1}{\delta }_m<\mathrm{\infty }$ and $m{\delta }_m\to 0$ hold. Independence and identical laws follow directly because each $X_t$ depends only on its own IID innovation.

The integer lengths satisfy $1\le \lceil n^{1\mathrm{/}3}\rceil \le n$, $b_n\to \mathrm{\infty }$ and $b_n\mathrm{/}\sqrt{n}\le n^{-1\mathrm{/}6}+n^{-1\mathrm{/}2}\to 0$. The window vanishes off $[0,1]$, takes values in $[0,1]$, is positive near $1\mathrm{/}2$, symmetric, and nondecreasing on $[0,1\mathrm{/}2]$. Every midpoint weight equals one, so $W_{b_n}=\sqrt{b_n}>0$ and $a_{b_n}=1$. All assumptions of both theorems are therefore met, giving their stated conditional law and covariance limits for this IID example.