Spectral measures and the known-mean periodogram

Assume Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)) for the angular Lebesgue measure clause below; under it ${\lambda }_1$ is a measure and intervals have their usual lengths (Assuming countable choice, $\mathcal{L}({\mathbb{R}}^n)$ is a sigma-algebra containing every elementary set and ${\lambda }_n$ is a complete measure extending elementary volume, A box in ${\mathbb{R}}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure ${\prod }_{i<n}(b_i-a_i)$, whichever of its faces are included).

Let $\mathbb{T}:=\{z\in \mathbb{C}:\mathrm{\mid }z\mathrm{\mid }=1\}$ be the multiplicative unit circle with its subspace topology from $\mathbb{C}$ (The multiplicative unit circle is a compact metrizable topological abelian group), identified with angular frequencies by $z=e^{i\lambda }$. Write $d\lambda$ for the finite angular Lebesgue Borel measure obtained from ordinary Lebesgue measure on $[-\pi ,\pi )$; thus $d\lambda (\mathbb{T})=2\pi$, and the endpoints $-\pi$ and $\pi$ denote the same point. The characters of discrete additive $\mathbb{Z}$ are $h\mapsto e^{ih\lambda }$ by The Pontryagin dual of the discrete integers is the circle.

For angular representatives $\lambda ,\theta \in \mathbb{R}$, define their circular distance by $d_{\mathbb{T}}(\lambda ,\theta ):={\min }_{k\in \mathbb{Z}}\mathrm{\mid }\lambda -\theta +2\pi k\mathrm{\mid }\mathrm{.}$ The minimum exists because the displayed absolute values tend to infinity with $\mathrm{\mid }k\mathrm{\mid }$, and changing either representative by $2\pi$ leaves it unchanged.

Precisely, for Borel $B\subseteq \mathbb{T}$ put $d\lambda (B):={\lambda }_1\{\theta \in [-\pi ,\pi ):e^{i\theta }\in B\}\mathrm{.}$ The map $\theta \mapsto e^{i\theta }$ is continuous by The complex exponential is entire and its complex derivative is itself and Complex differentiability at a point implies continuity there, so the preimage is Borel by A continuous map has Borel preimages of Borel sets. Preimages preserve disjoint unions, so this is a finite Borel measure under the stated Countable Choice premise. Integrating a Borel function of the circle angle means integrating its periodic lift over this half-open interval.

Let $X=(X_t{)}_{t\in \mathbb{Z}}$ be a real weakly stationary process with mean $\mu$ and autocovariance $\gamma (h)=\mathrm{Cov}(X_{t+h},X_t)$ (Strict and weak stationarity for integer-indexed time series, Autocovariance and sample autocovariance). A spectral measure for $X$ is a finite positive Radon Borel measure $F$ on $\mathbb{T}$ (Radon measure on an LCH space) satisfying $\gamma (h)={\int }_{\mathbb{T}}{e}^{ih\lambda }F(d\lambda )(h\in \mathbb{Z})\mathrm{.}$ Putting $h=0$ gives $F(\mathbb{T})=\gamma (0)$. A spectral density is a nonnegative measurable function $f$ with $F(B)={\int }_{B}f(\lambda )d\lambda$ for every Borel set $B\subseteq \mathbb{T}$; it exists only when $F$ is absolutely continuous with respect to $d\lambda$ and is unique up to $d\lambda$-null sets.

For a known mean $\mu$ and integer $n\ge 1$, the known-mean periodogram is the random function $I_n(\lambda ):=\frac{1}{2\pi n}{\mid {\sum }_{t=1}^n(X_t-\mu )e^{-it\lambda }\mid }^2,\lambda \in \mathbb{T}\mathrm{.}$ The centering uses the population mean, not the sample mean. The finite sum is square-integrable, so $I_n(\lambda )$ has finite expectation: for $Y_t=X_t-\mu$, the pointwise inequality $\mathrm{\mid }{\sum }_{t=1}^n{Y}_t{e}^{-it\lambda }{\mathrm{\mid }}^2\le n{\sum }_{t=1}^n\mathrm{\mid }{Y}_t{\mathrm{\mid }}^2$ bounds its expectation by $n{\sum }_{t=1}^n\mathbb{E}[Y_t^2]<\mathrm{\infty }$ (Moments, variance, and covariance on a probability space).

Define the normalized finite-sample Fejér kernel ${\mathrm{\Phi }}_n(x):=\frac{1}{2\pi n}{\mid {\sum }_{t=1}^n{e}^{-itx}\mid }^2,x\in \mathbb{T}\mathrm{.}$

For smoothing, let $K:\mathbb{R}\to [0,\mathrm{\infty })$ be continuous with compact support, ${\int }_{\mathbb{R}}K(u)du=1$, and $K(0)>0$. For bandwidth $b>0$, periodize it on the circle by $K_b(x):={\sum }_{k\in \mathbb{Z}}\frac{1}{b}K\text{\hspace{0.17em}⁣}\left(\frac{x+2\pi k}{b}\right)\mathrm{.}$ Compact support makes this a finite sum at each $x$, and changing variables over the intervals that tile $\mathbb{R}$ gives ${\int }_{\mathbb{T}}{K}_b(x)d\lambda (x)=1$. Fix a sequence of bandwidths $(b_n{)}_{n\ge 1}$ with $b_n>0$. For the Fourier grid ${\lambda }_{j,n}=2\pi j\mathrm{/}n$ ($0\le j<n$) and target ${\lambda }_0\in \mathbb{T}$, when the denominator is positive define $w_{n,j}({\lambda }_0):=\frac{K_{b_n}({\lambda }_0-{\lambda }_{j,n})}{{\sum }_{\mathrm{\ell }=0}^{n-1}{K}_{b_n}({\lambda }_0-{\lambda }_{\mathrm{\ell },n})},{\tilde{f}}_n({\lambda }_0):={\sum }_{j=0}^{n-1}{w}_{n,j}({\lambda }_0)I_n({\lambda }_{j,n})\mathrm{.}$ The weights are nonnegative and sum to one.

The finite-sum bound used in the finite-expectation paragraph follows directly. Put $Z_t:=Y_t{e}^{-it\lambda }$. Since $\mathrm{\mid }{e}^{-it\lambda }\mathrm{\mid }=1$ by $\exp (x+iy)=e^x(\cos y+i\sin y)$, $\mathrm{\mid }\exp (x+iy)\mathrm{\mid }=e^x$, and $e^{i\pi }+1=0$, ${\sum }_{t=1}^n\mathrm{\mid }{Z}_t{\mathrm{\mid }}^2={\sum }_{t=1}^n\mathrm{\mid }{Y}_t{\mathrm{\mid }}^2$. Expanding the squares gives $n{\sum }_{t=1}^n\mathrm{\mid }{Z}_t{\mathrm{\mid }}^2-{\mid {\sum }_{t=1}^n{Z}_t\mid }^2={\sum }_{1\le s<t\le n}\mathrm{\mid }{Z}_s-Z_t{\mathrm{\mid }}^2\ge 0,$ so $\mathrm{\mid }{\sum }_{t=1}^n{Y}_t{e}^{-it\lambda }{\mathrm{\mid }}^2\le n{\sum }_{t=1}^n\mathrm{\mid }{Y}_t{\mathrm{\mid }}^2$.