Fourier Analysis of Periodic Weakly Stationary Processes: A Note on Slutsky’s Observation

Abstract

The periodic behavior of a specific weakly stationary stochastic process (w.s.p.) is examined from a viewpoint of classical Fourier analysis.(1) A w.s.p. has a spectral measure which is absolutely continuous with respect to the Lebesgue measure if and only if it is a moving average of a white noise. (2) A periodic or almost periodic w.s.p. must have a “discrete” spectral measure. Combining these two, we can conclude that any moving average of a white noise can neither be periodic nor almost periodic.However any w.s.p. can be approximated by a sequence of almost periodic w.s.p.’s in some specific sense.

The earlier draft of this paper was read at the 20th Conference of the International Federation of Operational Research Societies, which was held in Barcelona, (July 13–18, 2014). It is a pleasure for me to express my cordial gratitude to the late Professor Tatsuo Kawata for incessant encouragement, which led me to the field of Fourier analysis. I am much indebted to Professor Shigeo Kusuoka for his kind suggestions concerning probability theory, which I am not familiar with very well. Helpful comments by Dr. Yuhki Hosoya and Mr.Chaowen Yu are also gratefully acknowledged.

JEL Classification: CO2, E32

Mathematics Subject Classification (2010): 42A38, 42A82, 60G10

Appendix

We reviewed, in the first few sections, some basic concepts and results to be required in our main concerns (Theorems 1, 2, and 3). I expect this exposition to be a small help for the readers who are not familiar with them. There can be no communications without common language. Of course, these materials are, more or less, known to every mathematician working in this discipline. That is why we just stated necessary facts and dropped proofs in the text.

However it does not seem to be a waste of time to give here complete proofs of selected three propositions (Propositions 5’, 6’ and 7’) as exceptions, taking account of their indispensable roles in our problem.

Proof of Proposition 5’.

The proof is basically due to Kawata [19] pp. 56–57.¹⁹ If we define $\theta: \mathcal{B}(\mathbb{R}) \rightarrow \mathbb{R}$ by

$$\displaystyle{ \theta (E) = \frac{\nu (E)} {\nu (\mathbb{R})},\quad E \in \mathcal{B}(\mathbb{R}), }$$

then $\theta$ is a Radon probability measure. There exist independent real random variables Z(ω) and Y (ω) defined on some probability space $(\varOmega,\mathcal{E},P)$ such that the distribution of Z(ω) is equal to $\theta$, $\mathbb{E}Y (\omega ) = 0$, and finally $\mathbb{E}Y (\omega )^{2} = (1/\sqrt{2\pi })\nu (\mathbb{R})$. ²⁰ Define a stochastic process X(t, ω) by

$$\displaystyle{ X(t,\omega ) = Y (\omega )e^{iZ(\omega )t}. }$$

Then $\mathbb{E}X(t,\omega )$ and ρ(t + u, t) are calculated as follows.²¹

$$\displaystyle\begin{array}{rcl} \mathbb{E}X(t,\omega )& =& \mathbb{E}Y (\omega )\mathbb{E}e^{iZ(\omega )t} = 0\quad \text{(by independence)}, {}\\ \rho (t + u,t)& =& \mathbb{E}X(t + u,\omega )\overline{X(t,\omega )} = \mathbb{E}[\,y(\omega )^{2}e^{iZ(\omega )u}] {}\\ & =& \frac{1} {\sqrt{2\pi }}\nu (\mathbb{R})\int _{\mathbb{R}}e^{i\lambda u}d\theta (\lambda ) = \frac{1} {\sqrt{2\pi }}\int _{\mathbb{R}}e^{i\lambda u}d\nu (\lambda ). {}\\ \end{array}$$

Thus the covariance ρ(t + u, t) of X(t, ω) depends only upon u, and it is expressed as the Fourier transform of ν. Since the Fourier transform of ν is uniformly continuous, X(t, ω) is strongly continuous on $\mathbb{R}$ by Proposition 1. The measurability of X(t, ω) is obvious. Q.E.D.

Proof of Proposition 6’.

(i)$\Rightarrow$ (ii): Assume that ρ(u) is τ-periodic. We then prove that

$$\displaystyle{ \mathbb{E}\vert X(t+\tau,\omega ) - X(t,\omega )\vert ^{2} = 0 }$$

which is equivalent to (ii). By direct computation, we have

$$\displaystyle\begin{array}{rcl} & & \mathbb{E}\vert X(t+\tau,\omega ) - X(t,\omega )\vert ^{2} {}\\ & & \ = \mathbb{E}\vert X(t+\tau,\omega )\vert ^{2} + \mathbb{E}\vert X(t,\omega )\vert - 2\mathcal{R}e\,\mathbb{E}X(t+\tau,\omega )\overline{X(t,\omega )} {}\\ & & \ = 2\rho (0) - 2\mathcal{R}e\rho (\tau ) {}\\ & & \ = 2(\rho (0) -\mathcal{R}e\rho (\tau )) {}\\ & & \ = 0\quad (\text{by (i)}). {}\\ \end{array}$$

(ii)$\Rightarrow$ (i): Assume (ii). then we have

$$\displaystyle\begin{array}{rcl} \vert \rho (u+\tau ) -\rho (u)\vert ^{2}& =& \vert \mathbb{E}[X(u+\tau,\omega )\overline{X(0,\omega )} - X(u,\omega )\overline{X(0,\omega )}]\vert ^{2} {}\\ & =& 0.\quad (\text{by (ii)}) {}\\ \end{array}$$

(i)$\Rightarrow$ (iii): If ρ(u) is τ-periodic, then we have

$$\displaystyle{ 0 = 2\rho (0) -\rho (\tau ) -\rho (-\tau ) = \frac{2} {\sqrt{2\pi }}\int _{\mathbb{R}}(1 -\cos t\tau )d\nu (t). }$$

Taking account of $1 -\cos t\tau \geqq 0$, we must have ν(E) = 0 for any $E \in \mathcal{B}(\mathbb{R})$ such that

$$\displaystyle{ E \cap \{ t \in \mathbb{R}\vert 1 -\cos t\tau = 0\} =\emptyset, }$$

which is equivalent to

$$\displaystyle{ E \cap \{ 2k\pi /\tau \vert k \in \mathbb{Z}\} =\emptyset. }$$

(iii)$\Rightarrow$ (i): By definition of the spectral measure,

$$\displaystyle{ \rho (u) = \frac{1} {\sqrt{2\pi }}\int _{\mathbb{R}}e^{-iut}d\nu (t). }$$

So putting $a_{k} =\nu (\{2\pi k/\tau \})\;(k \in \mathbb{Z})$, we obtain, by (iii), that

$$\displaystyle{ \rho (u) = \frac{1} {\sqrt{2\pi }}\sum a_{k}e^{-iu\cdot 2\pi k/\tau }. }$$

This is clearly τ-periodic. Q.E.D.

The following well-known lemma²² will be used in the course of the proof of Proposition 7’.

Lemma.

Let μ be a Radon measure on $\mathbb{R}$ . If the Fourier transform $\hat{\mu }$ is almost periodic, then μ is discrete.

Proof of Proposition 7’.

(i)$\Rightarrow$(ii): Assume that ρ(μ) is almost periodic. Then we have

$$\displaystyle\begin{array}{rcl} \sup _{t\in \mathbb{R}}\mathbb{E}\vert X(t+\tau,\omega ) - X(t,\omega )\vert ^{2}& =& 2[\rho (0) -\mathcal{R}e\rho (\tau )] \\ & =& 2\mathcal{R}e[\rho (0) -\rho (\tau )] \leqq 2\vert \rho (0) -\rho (\tau )\vert \\ &\leqq & 2\sup _{u\in \mathbb{R}}\vert \rho (u+\tau ) -\rho (u)\vert. {}\end{array}$$

(1)

Since ρ(⋅ ) is almost periodic, there exists a number $\varLambda (\varepsilon /2,\rho ) > 0$, for each $\varepsilon > 0$, such that any interval the length of which is $\varLambda (\varepsilon /2,\rho )$ contains an $\varepsilon /2$- almost period, τ. Since

$$\displaystyle\begin{array}{rcl} \{\tau \in \mathbb{R}\vert & \sup _{u\in \mathbb{R}}& \vert \rho (u+\tau ) -\rho (u)\vert < \frac{\varepsilon } {2}\} {}\\ & \subset &\{\tau \in \mathbb{R}\vert \sup _{t\in \mathbb{R}}\mathbb{E}\vert X(t+\tau,\omega ) - X(t,\omega )\vert ^{2} <\varepsilon \} {}\\ \end{array}$$

by (1), we see that (ii) is satisfied by setting $\varGamma (\varepsilon,X) =\varLambda (\varepsilon /2,\rho )$.

(ii)$\Rightarrow$(i): Assume (ii). By a simple calculation, we obtain the inequality

$$\displaystyle\begin{array}{rcl} \vert \rho (u+\tau ) -\rho (u)\vert ^{2}& =& \vert \mathbb{E}[X(u+\tau,\omega )\overline{X(0,\omega )} - X(u,\omega )\overline{X(0,\omega )}]\vert ^{2} {}\\ & \leqq & \mathbb{E}\vert X(u+\tau,\omega ) - X(u,\omega )\vert ^{2}\mathbb{E}\vert X(0,\omega )\vert ^{2} {}\\ & & \qquad \qquad \text{(by Schwarz inequality)}. {}\\ \end{array}$$

It follows that

$$\displaystyle{ \vert \rho (u+\tau ) -\rho (u)\vert \leqq [\mathbb{E}\vert X(t+\tau,\omega ) - X(u,\omega )\vert ^{2}]^{1/2}\rho (0)^{1/2}. }$$

Hence we have

$$\displaystyle\begin{array}{rcl} & & \Big\{\tau \in \mathbb{R}\Big\vert \sup _{u\in \mathbb{R}}\mathbb{E}\vert X(u+\tau,\omega ) - X(u,\omega )\vert ^{2} < \frac{\varepsilon ^{2}} {\rho (0)}\Big\} {}\\ & & \phantom{\Big\{\tau \in \mathbb{R}\Big\vert \sup _{u\in \mathbb{R}}\mathbb{E}\vert X(u+\tau,\omega )-}\subset \{\tau \in \mathbb{R}\vert \sup _{u\in \mathbb{R}}\vert \rho (u+\tau ) -\rho (u)\vert <\varepsilon \}. {}\\ \end{array}$$

Thus (i) holds true by setting $\varLambda (\varepsilon,\rho ) =\varGamma (\varepsilon ^{2}/\rho (0),X)$.

(i)$\Rightarrow$ (iii): This is a direct consequence of the lemma. The covariance function ρ is represented by the Fourier transform of some positive Radon measure ν on $\mathbb{R}$ (Proposition 4’). Since $\rho (u) =\hat{\nu } (u)$ is almost periodic, ν must be discrete by the lemma.

(iii)$\Rightarrow$ (i): Assume that the spectral measure ν of X is discrete, say

$$\displaystyle{ \nu =\sum _{ n=1}^{\infty }a_{ n}\delta _{\xi _{n}}\quad (\delta _{\xi _{n}}: \text{Dirac measure}). }$$

Then the covariance can be expressed as

$$\displaystyle{ \rho (u) = \frac{1} {\sqrt{2\pi }}\int _{\mathbb{R}}e^{-i\lambda u}d\nu (\lambda ) = \frac{1} {\sqrt{2\pi }}\sum _{n=1}^{\infty }a_{ n}e^{-i\xi _{n}u}. }$$

(2)

Since $a_{n} \geqq 0$, the series (2) is absolutely and so uniformly convergent. Thus ρ(u) is the uniform limit of trigonometric polynomials,²³ and hence almost periodic.

Q.E.D.