Pointwise resampling misses the long-run variance of an AR(1) mean

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.