Spectral measures and periodogram bias

For $z\in \mathbb{T}$, define ${\chi }_z(n):=z^n$. By [F4], ${\chi }_z(m+n)=z^{m+n}=z^m{z}^n={\chi }_z(m){\chi }_z(n)$, so ${\chi }_z$ is a group homomorphism by [F3]. It is continuous because its domain $\mathbb{Z}$ is discrete by [F2]. Hence ${\chi }_z\in \widehat{\mathbb{Z}}$ by [F1], and $\mathrm{\Psi }(z)={\chi }_z$ is well defined.

Multiplication on $\mathbb{T}$ is continuous. For $z,w,z_0,w_0\in \mathbb{T}$, $\mathrm{\mid }zw-z_0{w}_0\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }w-w_0\mathrm{\mid }+\mathrm{\mid }{w}_0\mathrm{\mid }\mathrm{\mid }z-z_0\mathrm{\mid }=\mathrm{\mid }w-w_0\mathrm{\mid }+\mathrm{\mid }z-z_0\mathrm{\mid }\mathrm{.}$ Thus a product-neighborhood restricting both coordinates within $\epsilon \mathrm{/}2$ of $(z_0,w_0)$ maps inside the $\epsilon$-ball about $z_0{w}_0$. Inversion is continuous because $z^{-1}=\bar{z}$ by [F6] and $\mathrm{\mid }\bar{z}-\bar{w}\mathrm{\mid }=\mathrm{\mid }z-w\mathrm{\mid }$ by conjugation preserving modulus [F6].

The map $\mathrm{\Psi }$ is a group homomorphism: for $z,w\in \mathbb{T}$ and $n\in \mathbb{Z}$, $(zw{)}^n=z^n{w}^n$ because $\mathbb{T}$ is abelian, by [F4] and [F6]. It is injective since $\mathrm{\Psi }(z)(1)=z$. If $\chi \in \widehat{\mathbb{Z}}$, put $z:=\chi (1)\in \mathbb{T}$. First $\chi (0)=\chi (0{)}^2$; since $\chi (0)\in \mathbb{T}$ is invertible, $\chi (0)=1$. Repeatedly applying the homomorphism law then gives $\chi (n)=z^n$ for $n\ge 0$. Thus for $n>0$, $\chi (n)\chi (-n)=\chi (0)=1$, so $\chi (-n)=z^{-n}$; hence $\chi =\mathrm{\Psi }(z)$ and $\mathrm{\Psi }$ is surjective.

Each coordinate map $z\mapsto \mathrm{\Psi }(z)(n)=z^n$ is continuous: for $n=0$ it is constant; for $n>0$ it is a finite composite of the continuous multiplication map from step 1.2 and the identity; for $n<0$ it is a finite composite followed by the continuous inverse map from step 1.2. Composite continuity follows from [F10]. A basic open set in the product ${\mathbb{T}}^{\mathbb{Z}}$ restricts only finitely many coordinates by [F8], so its preimage is a finite intersection of open coordinate preimages. Thus $\mathrm{\Psi }$ is continuous into the product, and [F7] gives continuity into the subspace $\widehat{\mathbb{Z}}$.

The inverse map sends $\chi$ to $\chi (1)$. Evaluation at coordinate $1$ is continuous as the restriction of a product projection by [F7]–[F8], and $\mathrm{\Psi }(\chi (1))=\chi$ by step 2.1. Thus the inverse is continuous; $\mathrm{\Psi }$ is a homeomorphism by [F9], and with step 2.1 it is the asserted group isomorphism.

By [F12], every $z\in \mathbb{T}$ has a unique angle $\lambda$ modulo $2\pi$ with $z=e^{i\lambda }$, using [F11]. The exponential addition law [F11] and integer-power laws [F4] give $\mathrm{\Psi }(e^{i\lambda })(n)=(e^{i\lambda }{)}^n=e^{in\lambda }$ for every $n\in \mathbb{Z}$.

Let $g:[-\pi ,\pi ]\to \mathbb{R}$ be continuous and bounded by $B\ge 0$. For each $N\ge 1$, partition the interval into the fixed equal pieces $J_{j,N}=[-\pi +2\pi j\mathrm{/}N,-\pi +2\pi (j+1)\mathrm{/}N)$, with the endpoint $\pi$ added to the last piece. On each piece take the infimum and supremum of $g+B$ on its closed hull and use them as the constant values of simple step functions ${\mathrm{\ell }}_N$ and $u_N$. They satisfy $0\le {\mathrm{\ell }}_N\le g+B\le u_N$. By [F3]–[F4], under [A1], their Lebesgue integrals with respect to ${\lambda }_1$ are exactly the lower and upper Darboux sums of $g+B$ for this partition. Given $\eta >0$, uniform continuity [F2] gives $\delta >0$ such that the oscillation on any interval of length below $\delta$ is less than $\eta$. Choose $N$ with $2\pi \mathrm{/}N<\delta$ by [F11]; then every oscillation is below $\eta$, and the gap between the sums is at most $2\pi \eta$. Hence that gap tends to zero. Monotonicity of the Lebesgue integral places ${\int }_{[-\pi ,\pi )}(g+B)d{\lambda }_1$ between the same sums; the Darboux integral of $g+B$ lies between them as well. Therefore the Lebesgue and Darboux integrals of $g+B$ agree. By linearity [F5] and ${\lambda }_1([-\pi ,\pi ))=2\pi$ from [F1], the two integrals of $g$ agree.

If $k=0$, the integrand is the constant $1$, so its integral is $2\pi$ by [F1] under [A1].

If $k\mathrm{\ne }0$, apply step 1.1 to the continuous real functions $g(\lambda )=\cos (k\lambda )$ and $g(\lambda )=\sin (k\lambda )$. Their antiderivatives are respectively $\sin (k\lambda )\mathrm{/}k$ and $-\cos (k\lambda )\mathrm{/}k$ by [F6] and the chain rule. Their endpoint differences vanish by [F7], so both Lebesgue integrals on $[-\pi ,\pi )$ are zero. These functions are $2\pi$-periodic; by the angular integral definition [F1], their integrals on $\mathbb{T}$ are those same integrals over $[-\pi ,\pi )$. Euler's identity and the complex-integral convention [F8] give ${\int }_{\mathbb{T}}{e}^{ik\lambda }d\lambda =0$.

The cases $k=0$ and $k\mathrm{\ne }0$ exhaust the integers, and steps 1.2 and 2.1 give the stated formula.

For any finite integers $h_1,\dots ,h_m$ and complex numbers $c_1,\dots ,c_m$, finite second moments make ${\sum }_j{c}_j(X_{h_j}-\mu )$ square-integrable. Expanding its squared modulus and using stationarity gives ${\sum }_{j,k=1}^m{c}_j\overline{c_k}\gamma (h_j-h_k)=\mathbb{E}{\mid {\sum }_{j=1}^m{c}_j(X_{h_j}-\mu )\mid }^2\ge 0.$ Thus $\gamma$ is positive definite on the additive group $\mathbb{Z}$ by [F1]–[F3].

Now assume absolute summability and define $f(\lambda )=(2\pi {)}^{-1}{\sum }_{h\in \mathbb{Z}}\gamma (h)e^{-ih\lambda }$. For every $N$, the tail is bounded uniformly in $\lambda$ by $(2\pi {)}^{-1}{\sum }_{\mathrm{\mid }h\mathrm{\mid }>N}\mathrm{\mid }\gamma (h)\mathrm{\mid }$, which tends to zero; finite partial sums are continuous, so the series converges uniformly to a continuous function $f$ by [F14].

For each $N\ge 0$, set $f_N(\lambda ):=\frac{1}{2\pi }{\sum }_{\mathrm{\mid }h\mathrm{\mid }\le N}(1-\frac{\mathrm{\mid }h\mathrm{\mid }}{N+1})\gamma (h)e^{-ih\lambda }\mathrm{.}$ Using $c_j=e^{-ij\lambda }$ in [F3] for $j=0,\dots ,N$ and grouping pairs by $h=j-k$ gives $f_N(\lambda )=\frac{1}{2\pi (N+1)}{\sum }_{j,k=0}^N{e}^{-i(j-k)\lambda }\gamma (j-k)\ge 0.$ Moreover, $2\pi \mathrm{\mid }{f}_N(\lambda )-f(\lambda )\mathrm{\mid }\le {\sum }_{\mathrm{\mid }h\mathrm{\mid }>N}\mathrm{\mid }\gamma (h)\mathrm{\mid }+{\sum }_{\mathrm{\mid }h\mathrm{\mid }\le N}\frac{\mathrm{\mid }h\mathrm{\mid }}{N+1}\mathrm{\mid }\gamma (h)\mathrm{\mid },$ and the right side tends to zero: first make the absolutely summable tail small, then let $N\to \mathrm{\infty }$ on the remaining finite sum. Thus $f_N\to f$ uniformly, so $f$ is real and nonnegative.

Give $\mathbb{Z}$ the discrete topology. Every singleton is compact and open, distinct integers have disjoint open singleton neighborhoods, and maps from a discrete space are continuous; hence $\mathbb{Z}$ is a locally compact Hausdorff abelian topological group and $\gamma$ is continuous. By [F4] and step 1.1, there is a unique finite positive Radon measure $M$ on $\widehat{\mathbb{Z}}$ with $\gamma (h)={\int }_{\widehat{\mathbb{Z}}}\chi (h)M(d\chi ),M(\widehat{\mathbb{Z}})=\gamma (0)\mathrm{.}$ Use the homeomorphism $\mathrm{\Psi }:\mathbb{T}\to \widehat{\mathbb{Z}}$ from [F5] and define $F(B):=M(\mathrm{\Psi }(B))$ for Borel $B\subseteq \mathbb{T}$. Images of Borel sets are Borel by [F7], so this is a finite positive Borel measure. It is finite on compact sets because $\mathrm{\Psi }$ maps compact sets to compact sets. For Borel $E\subseteq \mathbb{T}$, outer regularity of $M$ gives an open $V\supseteq \mathrm{\Psi }(E)$ with $M(V)<M(\mathrm{\Psi }(E))+\epsilon$; then $U={\mathrm{\Psi }}^{-1}(V)$ is open, contains $E$, and $F(U)=M(V)$. For open $O\subseteq \mathbb{T}$, inner regularity gives compact $C\subseteq \mathrm{\Psi }(O)$ with $M(C)>M(\mathrm{\Psi }(O))-\epsilon$; then $K={\mathrm{\Psi }}^{-1}(C)$ is compact in $O$ and $F(K)=M(C)$. Thus $F$ is Radon by [F6]. For any nonnegative Borel $q$ on $\widehat{\mathbb{Z}}$, ${\int }_{\mathbb{T}}q(\mathrm{\Psi }(z))F(dz)={\int }_{\widehat{\mathbb{Z}}}q(\chi )M(d\chi )\mathrm{.}$ Indeed the identity holds for indicators by the definition of $F$, hence for nonnegative simple functions by finite addition, and for nonnegative Borel functions by increasing simple approximation and monotone convergence [F8]. It also holds for bounded complex Borel $q$ by applying that argument to its positive and negative real and imaginary parts; these are integrable by [F9] because both measures are finite. Taking $q(\chi )=\chi (h)$ and using [F5] proves $\gamma (h)={\int }_{\mathbb{T}}{e}^{ih\lambda }F(d\lambda )$. Also $F(\mathbb{T})=M(\widehat{\mathbb{Z}})=\gamma (0)$. If another finite positive Radon measure $G$ on $\mathbb{T}$ represents $\gamma$, let $\tilde{G}(E):=G({\mathrm{\Psi }}^{-1}(E))$ for Borel $E\subseteq \widehat{\mathbb{Z}}$. It is finite on compact sets because ${\mathrm{\Psi }}^{-1}$ maps compact sets to compact sets. For Borel $E$, outer regularity of $G$ gives open $U\supseteq {\mathrm{\Psi }}^{-1}(E)$ with $G(U)<G({\mathrm{\Psi }}^{-1}(E))+\epsilon$; then $\mathrm{\Psi }(U)$ is open, contains $E$, and $\tilde{G}(\mathrm{\Psi }(U))=G(U)$. For open $V\subseteq \widehat{\mathbb{Z}}$, inner regularity gives compact $K\subseteq {\mathrm{\Psi }}^{-1}(V)$ with $G(K)>G({\mathrm{\Psi }}^{-1}(V))-\epsilon$; then $\mathrm{\Psi }(K)$ is compact in $V$ and $\tilde{G}(\mathrm{\Psi }(K))=G(K)$. Thus $\tilde{G}$ is Radon by [F6] and [F7]. The same indicator/simple/monotone argument gives ${\int }_{\widehat{\mathbb{Z}}}q(\chi )\tilde{G}(d\chi )={\int }_{\mathbb{T}}q(\mathrm{\Psi }(z))G(dz)$ for bounded Borel $q$. Taking $q(\chi )=\chi (h)$ shows $\tilde{G}$ represents $\gamma$; uniqueness in [F4] gives $\tilde{G}=M$. Hence $G(B)=\tilde{G}(\mathrm{\Psi }(B))=M(\mathrm{\Psi }(B))=F(B)$ for every Borel $B\subseteq \mathbb{T}$. The integrands $\chi \mapsto \chi (h)$ are Borel by [F15], so the representing integrals are defined.

Put $s_N(\lambda ):=(2\pi {)}^{-1}{\sum }_{\mathrm{\mid }k\mathrm{\mid }\le N}\gamma (k)e^{-ik\lambda }$. Absolute summability gives $\mathrm{\parallel }{s}_N-f{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. For $N\ge \mathrm{\mid }h\mathrm{\mid }$, the finite sum and [F11] give ${\int }_{\mathbb{T}}{e}^{ih\lambda }{s}_N(\lambda )d\lambda =\frac{1}{2\pi }{\sum }_{\mathrm{\mid }k\mathrm{\mid }\le N}\gamma (k){\int }_{\mathbb{T}}{e}^{i(h-k)\lambda }d\lambda =\gamma (h)\mathrm{.}$ If $q_N=e^{ih\lambda }(f-s_N)$ and ${\tau }_N=\mathrm{\parallel }f-s_N{\mathrm{\parallel }}_{\mathrm{\infty }}$, then $\mathrm{\mid }\mathrm{Re}{q}_N\mathrm{\mid },\mathrm{\mid }\mathrm{Im}{q}_N\mathrm{\mid }\le {\tau }_N$. By [F9], [F13], and $d\lambda (\mathbb{T})=2\pi$, the integrals of the real and imaginary parts each lie between $-2\pi {\tau }_N$ and $2\pi {\tau }_N$: subtracting either part from the constant ${\tau }_N$ or adding it to that constant gives a nonnegative function, so monotonicity and linearity apply. The complex integral is the real-part integral plus $i$ times the imaginary-part integral [F9], hence its modulus is at most the sum of the two absolute values, $4\pi {\tau }_N$. This tends to zero. Hence ${\int }_{\mathbb{T}}{e}^{ih\lambda }f(\lambda )d\lambda =\gamma (h)\mathrm{.}$ By [F12], $\nu (B):={\int }_{B}fd\lambda$ is a measure; it is finite and positive since $0\le f\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $d\lambda (\mathbb{T})=2\pi$. It is Radon: for a Borel $E$, outer regularity of $d\lambda$ [F10] gives open $U\supseteq E$ with $$\begin{gathered} d\lambda (U \\ setminusE)<\epsilon \mathrm{/}(\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}+1) \end{gathered}$$, hence $$\begin{gathered} \nu (U \\ setminusE)<\epsilon \end{gathered}$$. Countable additivity [F12] then yields $\nu (U)<\nu (E)+\epsilon$. For open $O$, inner regularity gives compact $K\subseteq O$ with $$\begin{gathered} d\lambda (O \\ setminusK)<\epsilon \mathrm{/}(\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}+1) \end{gathered}$$, hence $$\begin{gathered} \nu (O \\ setminusK)<\epsilon \end{gathered}$$ and $\nu (O)<\nu (K)+\epsilon$ by [F12]. These are the regularity clauses in [F6]. Thus $\nu$ is a finite positive Radon measure representing $\gamma$, and uniqueness in [F4] gives $F=fd\lambda$, with the stated density.

Expand the finite squared modulus in $I_n(\lambda )$. By [F2] and [F3], $\mathbb{E}[(X_t-\mu )(X_s-\mu )]=\gamma (t-s)$, so $\mathbb{E}[I_n(\lambda )]=\frac{1}{2\pi n}{\sum }_{t=1}^n{\sum }_{s=1}^n\gamma (t-s)e^{-i(t-s)\lambda }\mathrm{.}$ All sums are finite and all terms integrable.

Since $K$ is continuous and $K(0)>0$, it is positive on some interval $[-r,r]$ with $r>0$. The nearest point of the Fourier grid is within $\pi \mathrm{/}n$ of ${\lambda }_0$; because $n{b}_n\to \mathrm{\infty }$, eventually $\pi \mathrm{/}(n{b}_n)<r$, so the denominator defining the weights is positive. Every positive weight is supported at a grid point within circular distance $R{b}_n$ of ${\lambda }_0$, where $[-R,R]$ contains the support of $K$; this distance tends to zero.

For each $h$ with $\mathrm{\mid }h\mathrm{\mid }<n$, exactly $n-\mathrm{\mid }h\mathrm{\mid }$ ordered pairs $(t,s)\in \{1,\dots ,n{\}}^2$ have $t-s=h$. Grouping the sum in step 1.1 by $h$ gives the asserted finite Fourier sum.

If a spectral measure $F$ is supplied, substitute $\gamma (t-s)={\int }_{\mathbb{T}}{e}^{i(t-s)\theta }F(d\theta )$ into the finite double sum and interchange the finite sum and integral. The resulting integrand is $\frac{1}{2\pi n}{\mid {\sum }_{t=1}^n{e}^{-it(\lambda -\theta )}\mid }^2={\mathrm{\Phi }}_n(\lambda -\theta ),$ which proves the Fejér convolution formula. When $F=fd\theta$, substitute that density in the same integral.