Sample-mean CLT for stable linear processes

The weighted summability implies ${\sum }_j\mathrm{\mid }{\psi }_j\mathrm{\mid }<\mathrm{\infty }$. By [F2], the series defining $X_t-\mu$ is well-defined almost surely and in $L^2$, and the filtered process is weakly stationary. Thus for a causal ARMA the weak-stationarity premise of [F4] is met, and [F4] gives $\mathrm{\mid }{\psi }_j\mathrm{\mid }\le C{r}^{-j}$ for some $C<\mathrm{\infty }$ and $r>1$. The ratio of consecutive terms $\sqrt{j+2}{r}^{-(j+1)}\mathrm{/}(\sqrt{j+1}{r}^{-j})$ tends to $1\mathrm{/}r<1$, so ${\sum }_j\sqrt{j+1}\mathrm{\mid }{\psi }_j\mathrm{\mid }<\mathrm{\infty }$ as required. Absolute summability also makes $\mathrm{\Psi }$ finite.

For $j\ge 0$ let $D_{n,j}:={\sum }_{t=1}^n{ϵ}_{t-j}-{\sum }_{t=1}^n{ϵ}_t$, so $D_{n,0}=0$. For finite $J$, summing the first $J$ filter coefficients over $t=1,\dots ,n$ gives ${\sum }_{t=1}^n{\sum }_{j=0}^J{\psi }_j{ϵ}_{t-j}=\left({\sum }_{j=0}^J{\psi }_j\right){\sum }_{t=1}^n{ϵ}_t+{\sum }_{j=0}^J{\psi }_j{D}_{n,j}\mathrm{.}$ As $J\to \mathrm{\infty }$, the left side converges in $L^2$ to ${\sum }_{t=1}^n(X_t-\mu )$ by [F2], and the first term on the right converges in $L^2$ because ${\sum }_j\mathrm{\mid }{\psi }_j\mathrm{\mid }<\mathrm{\infty }$. We pass the finite identity to its limit once the remainder convergence has been established.

For $1\le j\le n$, the two sums defining $D_{n,j}$ cancel on their overlap and leave $j$ distinct positive-sign and $j$ distinct negative-sign innovations. For $j>n$, they leave $n$ of each. These innovations are centered, mutually independent, and have variance ${\sigma }^2$, so $\mathrm{\parallel }{D}_{n,j}{\mathrm{\parallel }}_2=\sigma \sqrt{2\min (j,n)}\le \sigma \sqrt{2j}\mathrm{.}$ The variance calculation uses the zero covariance of distinct innovations from the white-noise condition in [F1] (also implied by iid in [F3]). For finite sums, expanding the square and using Cauchy–Schwarz [F10] gives the $L^2$ triangle bound. Therefore ${\sum }_j\mathbb{E}\mathrm{\mid }{\psi }_j{D}_{n,j}\mathrm{\mid }\le \sigma \sqrt{2}{\sum }_j\mathrm{\mid }{\psi }_j\mathrm{\mid }\sqrt{j}<\mathrm{\infty }$; Tonelli [F10] defines an almost-surely absolutely convergent remainder $R_n:={\sum }_{j\ge 0}{\psi }_j{D}_{n,j}$. Applying the finite-sum $L^2$ bound to each tail and Fatou [F10] gives $\mathrm{\parallel }{R}_n{\mathrm{\parallel }}_2\le \sigma {\sum }_{j\ge 0}\mathrm{\mid }{\psi }_j\mathrm{\mid }\sqrt{2j}<\mathrm{\infty },$ uniformly in $n$, and the same bound for tails tends to zero. Thus the partial remainders converge to $R_n$ in $L^2$. The uniform bound follows from the given weighted summability.

Passing to the $L^2$ limit in step 1.2 yields ${\sum }_{t=1}^n(X_t-\mu )=\mathrm{\Psi }{\sum }_{t=1}^n{ϵ}_t+R_n\mathrm{.}$ Step 1.3 gives $\mathbb{E}[R_n^2]\mathrm{/}n\to 0$. For every $\delta >0$, [F11] applied to $R_n^2$ gives $\mathbb{P}(\mathrm{\mid }{R}_n\mathrm{\mid }\mathrm{/}\sqrt{n}\ge \delta )\le \mathbb{E}[R_n^2]\mathrm{/}(n{\delta }^2)\to 0$. Thus $R_n\mathrm{/}\sqrt{n}{\to }_{\mathbb{P}}0$.

By [F5], $(\sigma \sqrt{n}{)}^{-1}{\sum }_{t=1}^n{ϵ}_t\Rightarrow N(0,1)$; multiplying by $\sigma$ gives $n^{-1\mathrm{/}2}{\sum }_{t=1}^n{ϵ}_t\Rightarrow N(0,{\sigma }^2)$. Multiplying by $\mathrm{\Psi }$ (also when $\mathrm{\Psi }=0$) and adjoining the remainder $R_n\mathrm{/}\sqrt{n}{\to }_{\mathbb{P}}0$ from step 2.1, [F6] gives $\sqrt{n}({\bar{X}}_n-\mu )\Rightarrow N(0,{\sigma }^2{\mathrm{\Psi }}^2)\mathrm{.}$ This also covers $\mathrm{\Psi }=0$, where the limit is the point mass $N(0,0)$.

Now assume $\tau =\sigma \mathrm{\mid }\mathrm{\Psi }\mathrm{\mid }>0$ and ${\widehat{\tau }}_n^2{\to }_{\mathbb{P}}{\tau }^2$. By [F7], ${\widehat{\tau }}_n=\sqrt{{\widehat{\tau }}_n^2}{\to }_{\mathbb{P}}\tau$. The map $g(x,s):=x\mathrm{/}\max (s,\tau \mathrm{/}2)$ is continuous everywhere; by [F6] and step 3.1, $g(\sqrt{n}({\bar{X}}_n-\mu ),{\widehat{\tau }}_n)\Rightarrow N(0,1)$. The event ${\widehat{\tau }}_n<\tau \mathrm{/}2$ has probability tending to zero by convergence in probability to $\tau$. On its complement this transformed variable equals $T_n$, so $T_n\Rightarrow N(0,1)$.

For fixed $c>0$, on $\{{\widehat{\tau }}_n>0\}$ membership of $\mu$ in the stated interval is equivalent to $\mathrm{\mid }{T}_n\mathrm{\mid }\le c$. The complementary event has probability tending to zero by step 4.1. The boundary $\{-c,c\}$ has $N(0,1)$ probability zero by [F8], so [F9] gives limiting coverage $\mathbb{P}(\mathrm{\mid }N(0,1)\mathrm{\mid }\le c)$. The conclusion is conditional on the separately proved consistency of ${\widehat{\tau }}_n^2$; it does not assert that naming an estimator makes it consistent.