ARMA roots, spectra and inference

Weak stationarity makes $\mathbb{E}[Z_t^2]=\gamma (0)$ for every $t$. Applying [F3] to $Z_t$ and the constant one gives $\mathbb{E}\mathrm{\mid }{Z}_t\mathrm{\mid }\le \sqrt{\gamma (0)}$, while applying it to the centered variables $Z_{t+h},Z_t$ gives $\mathrm{\mid }\gamma (h)\mathrm{\mid }\le \gamma (0)$. Consequently ${\sum }_{j\in \mathbb{Z}}\mathbb{E}\left[\mathrm{\mid }{c}_j{Z}_{t-j}\mathrm{\mid }\right]\le \sqrt{\gamma (0)}{\sum }_{j\in \mathbb{Z}}\mathrm{\mid }{c}_j\mathrm{\mid }<\mathrm{\infty }\mathrm{.}$ Tonelli [F5] on counting measure over $\mathbb{Z}$ times the probability measure shows that the random series of absolute values has finite expectation, hence is finite almost surely. Thus $Y_t$ is defined by an absolutely convergent sum.

For any finite $J\subset \mathbb{Z}$, the finite filter $Y_t^J={\sum }_{j\in J}{c}_j{Z}_{t-j}$ satisfies, by [F4], $\mathbb{E}[(Y_t^J{)}^2]={\sum }_{j,k\in J}{c}_j{c}_k\gamma (k-j)\le \gamma (0){\left({\sum }_{j\in J}\mathrm{\mid }{c}_j\mathrm{\mid }\right)}^2\mathrm{.}$ The equality is a finite covariance expansion; the inequality follows from $\mathrm{\mid }\gamma (k-j)\mathrm{\mid }\le \gamma (0)$ from step 1.1. Applying the same bound to a finite tail $J$ gives $\mathrm{\parallel }{\sum }_{j\in J}{c}_j{Z}_{t-j}{\mathrm{\parallel }}_2\le \sqrt{\gamma (0)}{\sum }_{j\in J}\mathrm{\mid }{c}_j\mathrm{\mid }$.

Enumerate $\mathbb{Z}$ as $0,1,-1,2,-2,\dots$ and let $Y_t^N$ be the partial sum through the first $N$ terms. The absolute almost-sure convergence from step 1.1 gives $Y_t-Y_t^N={\lim }_{M\to \mathrm{\infty }}(Y_t^M-Y_t^N)$ almost surely. Fatou [F7] and the finite-tail bound of step 2.1 therefore give $\mathbb{E}\mathrm{\mid }{Y}_t-Y_t^N{\mathrm{\mid }}^2\le \gamma (0){\left({\sum }_{j\text{ omitted after }N}\mathrm{\mid }{c}_j\mathrm{\mid }\right)}^2⟶0.$ Thus $Y_t^N\to Y_t$ in $L^2$ and $Y_t$ is square-integrable.

Each finite filter is centered. The $L^2$ convergence in step 3.1 implies $L^1$ convergence by [F3] applied to the difference and the constant one; hence $\mathbb{E}[Y_t]=0$. For finite $J$, covariance bilinearity [F4] gives $\mathrm{Cov}(Y_{t+h}^J,Y_t^J)={\sum }_{j,k\in J}{c}_j{c}_k\gamma (h-j+k)\mathrm{.}$ As $J$ increases through the enumeration, the left side converges to $\mathrm{Cov}(Y_{t+h},Y_t)$ by $L^2$ convergence and [F3]. The right side converges absolutely because ${\sum }_{j,k\in \mathbb{Z}}\mathrm{\mid }{c}_j{c}_k\gamma (h-j+k)\mathrm{\mid }\le \gamma (0){\left({\sum }_j\mathrm{\mid }{c}_j\mathrm{\mid }\right)}^2<\mathrm{\infty },$ with order of summation justified by [F6]. Its limit is independent of $t$, so $Y$ is weakly stationary with the stated autocovariance.

Let $c,d\in {\mathrm{\ell }}^1(\mathbb{Z})$. The triangle inequality and [F6] give ${\sum }_{m\in \mathbb{Z}}\mathrm{\mid }(c\ast d{)}_m\mathrm{\mid }\le {\sum }_{j,k\in \mathbb{Z}}\mathrm{\mid }{c}_j\mathrm{\mid }\mathrm{\mid }{d}_k\mathrm{\mid }=\left({\sum }_j\mathrm{\mid }{c}_j\mathrm{\mid }\right)\left({\sum }_k\mathrm{\mid }{d}_k\mathrm{\mid }\right)<\mathrm{\infty },$ so $c\ast d\in {\mathrm{\ell }}^1(\mathbb{Z})$. Tonelli [F5] and the uniform bound from step 1.1 give $\mathbb{E}\text{\hspace{0.17em}⁣}\left[{\sum }_{j,k\in \mathbb{Z}}\mathrm{\mid }{c}_j{d}_k{Z}_{t-j-k}\mathrm{\mid }\right]={\sum }_{j,k\in \mathbb{Z}}\mathrm{\mid }{c}_j{d}_k\mathrm{\mid }\mathbb{E}\mathrm{\mid }{Z}_{t-j-k}\mathrm{\mid }\le \sqrt{\gamma (0)}\left({\sum }_j\mathrm{\mid }{c}_j\mathrm{\mid }\right)\left({\sum }_k\mathrm{\mid }{d}_k\mathrm{\mid }\right)<\mathrm{\infty }\mathrm{.}$ Hence this double random series is absolutely convergent almost surely. Reindexing the absolutely convergent terms by $m=j+k$ gives ${\sum }_j{c}_j\left({\sum }_k{d}_k{Z}_{t-j-k}\right)={\sum }_m\left({\sum }_j{c}_j{d}_{m-j}\right)Z_{t-m}={\sum }_m(c\ast d{)}_m{Z}_{t-m}\mathrm{.}$ Both sides are the filter limits defined in the Statement, and the equality holds almost surely and in $L^2$ by the bounds of steps 2.1 and 3.1. This proves the composition formula.

Suppose a causal representation exists, with $\mathrm{\Psi }(z)={\sum }_{j\ge 0}{\psi }_j{z}^j$ and ${\sum }_j\mathrm{\mid }{\psi }_j\mathrm{\mid }<\mathrm{\infty }$. Applying the finite polynomial $\varphi (B)$ and using the ARMA equation gives $(\varphi (B)\mathrm{\Psi }(B)-\theta (B)){ϵ}_t=0.$ The coefficient sequence $a$ of $\varphi \mathrm{\Psi }-\theta$ is in ${\mathrm{\ell }}^1$ because $\varphi ,\theta$ are finite polynomials and $\psi \in {\mathrm{\ell }}^1$. By [F3], the variance of this zero random variable is ${\sigma }^2{\sum }_{j\ge 0}{a}_j^2$. Since ${\sigma }^2>0$, every $a_j=0$, so $\varphi (z)\mathrm{\Psi }(z)=\theta (z)$ on the closed unit disk; the series for $\mathrm{\Psi }$ converges uniformly there because ${\sum }_j\mathrm{\mid }{\psi }_j\mathrm{\mid }<\mathrm{\infty }$.

Suppose $\varphi$ has degree $m\ge 1$ and roots $r_1,\dots ,r_m$ counted with multiplicity. None is zero because $\varphi (0)=1$. By [F4] and normalization, $\varphi (z)={\prod }_{\mathrm{\ell }=1}^m(1-z\mathrm{/}{r}_{\mathrm{\ell }})\mathrm{.}$ If every $\mathrm{\mid }{r}_{\mathrm{\ell }}\mathrm{\mid }>1$, then for $\mathrm{\mid }z\mathrm{\mid }\le 1$ each factor has the geometric expansion $(1-z\mathrm{/}{r}_{\mathrm{\ell }}{)}^{-1}={\sum }_{k\ge 0}{r}_{\mathrm{\ell }}^{-k}{z}^k\mathrm{.}$ The coefficient of $z^j$ in $1\mathrm{/}\varphi (z)$ has absolute value at most $\left(\genfrac{}{}{0px}{}{j+m-1}{m-1}\right)r_{\ast }^{-j}$, where $r_{\ast }={\min }_{\mathrm{\ell }}\mathrm{\mid }{r}_{\mathrm{\ell }}\mathrm{\mid }>1$, since there are $\left(\genfrac{}{}{0px}{}{j+m-1}{m-1}\right)$ nonnegative $m$-tuples summing to $j$. Choose $1<r<r_{\ast }$. The polynomial factor $\left(\genfrac{}{}{0px}{}{j+m-1}{m-1}\right)(r\mathrm{/}{r}_{\ast }{)}^j$ is bounded in $j$, so these coefficients are bounded by $C{r}^{-j}$. Multiplication by the finite polynomial $\theta$ preserves an exponential bound. If $m=0$, $\varphi =1$ and $\theta \mathrm{/}\varphi$ is finite, which also has an exponential bound after enlarging $C$. Thus $\eta (z):=1\mathrm{/}\varphi (z)$ and $\mathrm{\Psi }(z):=\eta (z)\theta (z)=\theta (z)\mathrm{/}\varphi (z)$ have absolutely summable one-sided coefficients, and those of $\mathrm{\Psi }$ have the stated exponential decay.

Suppose an invertible representation exists, ${ϵ}_t=\pi (B)(X_t-\mu )$ with $\pi \in {\mathrm{\ell }}^1({\mathbb{N}}_0)$. Apply the finite filter $\varphi (B)$ and commute the absolutely summable and polynomial filters by [F3] on the weakly stationary process [F2]: $\varphi (B){ϵ}_t=\varphi (B)\pi (B)(X_t-\mu )=\pi (B)\varphi (B)(X_t-\mu )=\pi (B)\theta (B){ϵ}_t\mathrm{.}$ The coefficient sequence of $\varphi -\pi \theta$ is in ${\mathrm{\ell }}^1$. Its filter of white noise is zero, so [F3] and ${\sigma }^2>0$ imply every coefficient is zero. The absolutely summable series $\mathrm{\Pi }$ converges uniformly on the closed unit disk, so the coefficient identity yields $\varphi (z)=\mathrm{\Pi }(z)\theta (z)$ for $\mathrm{\mid }z\mathrm{\mid }\le 1$.

If $\varphi (z_0)=0$ for some $\mathrm{\mid }{z}_0\mathrm{\mid }\le 1$, step 1.1 gives $\theta (z_0)=\varphi (z_0)\mathrm{\Psi }(z_0)=0$. This is a common root, contrary to the coprime condition in [F1]. Therefore causality implies that $\varphi$ has no zero in the closed unit disk.

Let $\eta (z)=1\mathrm{/}\varphi (z)$ be the absolutely summable causal series from step 1.2. Apply the filter $\eta (B)$ to the given equation. Since $X-\mu$ is weakly stationary, [F2] and [F3] justify composition and $\eta \ast \varphi$ is the constant coefficient sequence $1$. Therefore $X_t-\mu =\eta (B)\varphi (B)(X_t-\mu )=\eta (B)\theta (B){ϵ}_t=\mathrm{\Psi }(B){ϵ}_t\mathrm{.}$ This is the required causal representation of the given solution.

If $\theta (z_0)=0$ for some $\mathrm{\mid }{z}_0\mathrm{\mid }\le 1$, step 1.3 gives $\varphi (z_0)=\mathrm{\Pi }(z_0)\theta (z_0)=0$, contradicting coprimality. Hence invertibility implies that $\theta$ has no zero in the closed unit disk.

If $\theta$ has no zero in $\mathrm{\mid }z\mathrm{\mid }\le 1$, the same geometric-series argument as in step 1.2 applied to its roots gives an absolutely summable causal series $\eta (z)=1\mathrm{/}\theta (z)$. Set $\mathrm{\Pi }(z)=\eta (z)\varphi (z)$; its coefficients are absolutely summable. Applying $\eta (B)$ to the ARMA equation and using filter composition [F3] on the weakly stationary process [F2] gives $\mathrm{\Pi }(B)(X_t-\mu )=\eta (B)\varphi (B)(X_t-\mu )=\eta (B)\theta (B){ϵ}_t={ϵ}_t\mathrm{.}$ Thus the model is invertible.

Steps 1.1, 1.2, 2.1, and 2.2 prove both directions of the causal criterion and identify its coefficients; steps 1.3, 2.3, and 2.4 prove both directions of the invertibility criterion and identify its coefficients. The proof uses no choice axiom.

By [F6], factor $\varphi$ over $\mathbb{C}$ as $\varphi (z)={\prod }_{\mathrm{\ell }=1}^m(1-z\mathrm{/}{r}_{\mathrm{\ell }})$, with multiplicities and $\varphi (0)=1$; if $m=0$, take the empty product $1$. None of the roots satisfies $\mathrm{\mid }{r}_{\mathrm{\ell }}\mathrm{\mid }=1$. For a root with $\mathrm{\mid }{r}_{\mathrm{\ell }}\mathrm{\mid }>1$ use $(1-z\mathrm{/}{r}_{\mathrm{\ell }}{)}^{-1}={\sum }_{j\ge 0}{r}_{\mathrm{\ell }}^{-j}{z}^j$ on the unit circle. For a root with $\mathrm{\mid }{r}_{\mathrm{\ell }}\mathrm{\mid }<1$ use $(1-z\mathrm{/}{r}_{\mathrm{\ell }}{)}^{-1}=-{\sum }_{j\ge 1}{r}_{\mathrm{\ell }}^j{z}^{-j}$ there. Each coefficient sequence is in ${\mathrm{\ell }}^1(\mathbb{Z})$, so their finite convolution and multiplication by $\theta$ give a two-sided sequence $h\in {\mathrm{\ell }}^1(\mathbb{Z})$ with ${\sum }_{k\in \mathbb{Z}}{h}_k{z}^k=\frac{\theta (z)}{\varphi (z)},\mathrm{\mid }z\mathrm{\mid }=1.$ Conjugate roots occur in pairs for real $\varphi ,\theta$. For each conjugate pair the finite convolution coefficient is unchanged by complex conjugation after reversing its summation index; real roots have real coefficients. Multiplication by real $\theta$ preserves reality.

The angular measure $d\lambda$ is a finite Borel measure of total mass $2\pi$ by [F2]. Normalize it to $P=d\lambda \mathrm{/}(2\pi )$, a Borel probability by [F11]. By [F7], $\mathbb{T}$ is Polish, so [F8] gives compact inner approximations for every Borel set under $P$; scaling by $2\pi$ gives the same property for $d\lambda$. For any Borel $E$, apply this inner approximation to $\mathbb{T}\setminus E$: for every $\eta >0$ there is compact $K\subseteq \mathbb{T}\setminus E$ with $d\lambda ((\mathbb{T}\setminus E)\setminus K)<\eta$. Then $U=\mathbb{T}\setminus K$ is open, contains $E$, and $d\lambda (U\setminus E)<\eta$. Thus $d\lambda$ is outer regular on Borel sets and inner regular on open sets; it is finite on compact sets because its total mass is finite. Hence $d\lambda$ is Radon by [F4].

Write $a(z)=1\mathrm{/}\varphi (z)$ for the two-sided absolutely convergent series constructed in step 1.1. Its product with $\varphi$ is the constant series $1$. Applying filter composition [F3] to the ARMA equation gives $X_t-\mu =a(B)\varphi (B)(X_t-\mu )=a(B)\theta (B){ϵ}_t={\sum }_{k\in \mathbb{Z}}{h}_k{ϵ}_{t-k}\mathrm{.}$ Thus every weakly stationary solution under the stated root condition is the same two-sided filtered white-noise process.

Define $f(\lambda ):=\frac{{\sigma }^2}{2\pi }{\mid {\sum }_{k\in \mathbb{Z}}{h}_k{e}^{-ik\lambda }\mid }^2=\frac{{\sigma }^2}{2\pi }\frac{\mathrm{\mid }\theta (e^{-i\lambda }){\mathrm{\mid }}^2}{\mathrm{\mid }\varphi (e^{-i\lambda }){\mathrm{\mid }}^2}\mathrm{.}$ The series is uniformly convergent because $h\in {\mathrm{\ell }}^1$, so $f$ is continuous, periodic, and nonnegative. It is bounded on the compact circle; therefore $F(E):={\int }_{E}fd\lambda$ is a finite positive Borel measure by [F9] and [F10]. For any Borel $E$ and $\epsilon >0$, outer regularity of $d\lambda$ gives an open $U\supset E$ with $d\lambda (U\setminus E)<\epsilon \mathrm{/}(\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}+1)$. Since $F(U)=F(E)+F(U\setminus E)$ and $F(U\setminus E)\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}d\lambda (U\setminus E)$, this proves outer regularity of $F$. For any open $O$, inner regularity of $d\lambda$ gives a compact $K\subset O$ with $d\lambda (O\setminus K)<\epsilon \mathrm{/}(\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}+1)$; the same bound on $F(O\setminus K)$ proves inner regularity of $F$ on open sets. Thus $F$ is Radon by [F4].

Applying the covariance formula in [F3] to the representation in step 2.1 and using $\mathrm{Cov}({ϵ}_s,{ϵ}_t)={\sigma }^2$ when $s=t$ and $0$ otherwise gives $\gamma (h)={\sigma }^2{\sum }_{k\in \mathbb{Z}}{h}_{k+h}{h}_k\mathrm{.}$ The sum is absolutely convergent because $h\in {\mathrm{\ell }}^1(\mathbb{Z})$.

Put $H_N(\lambda ):={\sum }_{\mathrm{\mid }k\mathrm{\mid }\le N}{h}_k{e}^{-ik\lambda }$. Since $h\in {\mathrm{\ell }}^1$, $H_N$ converges uniformly to $H(\lambda ):={\sum }_k{h}_k{e}^{-ik\lambda }$. The sequence $H_N$ is uniformly bounded, so $\mathrm{\mid }{H}_N{\mathrm{\mid }}^2$ converges uniformly to $\mathrm{\mid }H{\mathrm{\mid }}^2$. Expanding each finite square gives $\mathrm{\mid }{H}_N(\lambda ){\mathrm{\mid }}^2={\sum }_{\mathrm{\mid }j\mathrm{\mid },\mathrm{\mid }k\mathrm{\mid }\le N}{h}_j{h}_k{e}^{-i(j-k)\lambda }$, and [F5] gives ${\sum }_{j,k}\mathrm{\mid }{h}_j{h}_k\mathrm{\mid }=({\sum }_k\mathrm{\mid }{h}_k\mathrm{\mid }{)}^2<\mathrm{\infty }$. Hence the finite squares also converge uniformly to the full double series. Integration against $e^{ih\lambda }$ preserves this uniform limit because the circle has finite measure. For each integer $h$, ${\int }_{\mathbb{T}}{e}^{ih\lambda }f(\lambda )d\lambda ={\sigma }^2{\sum }_{j,k}{h}_j{h}_k\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{i(h-j+k)\lambda }d\lambda ={\sigma }^2{\sum }_k{h}_{k+h}{h}_k=\gamma (h),$ where the middle integral is $1$ when $h-j+k=0$ and $0$ otherwise, and the last equality is step 3.1. Hence $F$ represents the autocovariance and $f$ is a spectral density under [F2].

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.

Fix $a\in {\mathbb{R}}^d$ and $t\ge 1$. For any $A\in {\mathcal{F}}_{t-1}$, the variables $a^T{Z}_t{1}_A$ and $(a^T{Z}_t{)}^2{1}_A$ are ${\mathcal{F}}_{t-1}$-measurable. Past-independence, [F5], and $\mathbb{E}[{ϵ}_t]=0$ give $\mathbb{E}[a^T{Z}_t{ϵ}_t{1}_A]=0$; past-independence, [F5], and $\mathbb{E}[{ϵ}_t^2]={\sigma }^2$ give $\mathbb{E}[(a^T{Z}_t{)}^2{ϵ}_t^2{1}_A]={\sigma }^2\mathbb{E}[(a^T{Z}_t{)}^2{1}_A]$. By [F4] these event-integral identities say $\mathbb{E}[a^T{Z}_t{ϵ}_t\mid {\mathcal{F}}_{t-1}]=0,\mathbb{E}[(a^T{Z}_t{ϵ}_t{)}^2\mid {\mathcal{F}}_{t-1}]={\sigma }^2(a^T{Z}_t{)}^2\mathrm{.}$ Factorization also gives $\mathbb{E}[(a^T{Z}_t{ϵ}_t{)}^2]={\sigma }^2{a}^T\mathrm{\Gamma }a<\mathrm{\infty }$, so each score increment is square-integrable.