Spectral density of a stationary ARMA solution

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].