\documentclass[11pt]{article} \usepackage{fancyhdr} \usepackage{lastpage} \usepackage{amsfonts,amssymb,amsmath} \usepackage{undertilde} \usepackage{../MATH545} \usepackage{enumitem} \usepackage{booktabs} \usepackage{bm} \usepackage{hyperref} \parindent 0in \pagestyle{empty} \def\ExNumber{4} \lfoot{MATH 545 EXERCISES \ExNumber~Solutions} \cfoot{ } \rfoot{{\small{\em Page \thepage \ of \pageref{LastPage}}}} \renewcommand{\headrulewidth}{0.0pt} \renewcommand{\footrulewidth}{0.0pt} \def\ctitle{MATH 545: Introduction to Time Series Analysis} \def\ttitle{Winter 2025} \def\stitle{Exercises \ExNumber: Solutions} \begin{document} \pagestyle{fancy} \maketitle <>= library(knitr) library(kableExtra) suppressMessages(library(dplyr)) # global chunk options opts_chunk$set(cache=TRUE, autodep=TRUE) options(scipen=999) options(repos=c(CRAN="https://cloud.r-project.org/")) @ <>= set.seed(11) myrnd <- function(x,d) {formatC(x, format="f", digits=d)} @ \begin{enumerate}[label=\arabic*.,itemsep=0.2in] \item \textit{As quoted in lectures, an absolutely summable function $\gamma(h)$ is the ACVF for a stationary process if and only if it is even and \[ f(\lambda) = \frac{1}{2 \pi} \sum_{h=-\infty}^\infty e^{-\imagi h \lambda} \gamma(h) \geq 0 \] for all $\lambda$ in the interval $(-\pi,\pi]$. Use this result to assess whether the function \[ \gamma(h) = \left\{ \begin{array}{rl} 1 & h=0 \\[6pt] -0.5 & h= \pm 2\\[6pt] -0.25 & h= \pm 3\\[6pt] 0 & \text{otherwise} \end{array} \right. \] is the ACVF of a stationary process.} \textbf{Solution:} By direct calculation \begin{align*} f(\lambda) = \frac{1}{2 \pi} \sum_{h=-\infty}^\infty e^{-\imagi h \lambda} \gamma(h) & = \frac{1}{2 \pi} \left[ 1 - 0.5 (e^{-\imagi 2 \lambda} + e^{\imagi 2 \lambda}) - 0.25 (e^{-\imagi 3 \lambda} + e^{\imagi 3 \lambda}) \right] \\[6pt] & = \frac{1}{2 \pi} \left[ 1 - \cos (2 \lambda) - 0.5 \cos(3 \lambda) \right] \end{align*} This function is not non-negative as (for example) $f(0) = -0.5$. Therefore, $\gamma(.)$ is not the ACVF of a stationary process. <>= lam<-seq(-pi,pi,length=1000) flam<-(1-cos(2*lam)-0.5*cos(3*lam))/(2*pi) par(mar=c(4,4,2,2)) plot(lam,flam,type='l',ylim=range(-0.5,0.5), xlab=expression(lambda),ylab=expression(f(lambda))) abline(h=0,lty=3) @ If instead we had \[ \gamma(h) = \left\{ \begin{array}{rl} 1 & h=0 \\[6pt] -0.25 & h= \pm 2\\[6pt] -0.125 & h= \pm 3\\[6pt] 0 & \text{otherwise} \end{array} \right. \] then the equivalent calculation leads to \[ f(\lambda) = \frac{1}{2 \pi} \left[ 1 - 0.5 \cos (2 \lambda) - 0.25 \cos(3 \lambda) \right] \] which is non-negative and even, so this is a valid ACVF. Using a discrete approximation, we can simulate a process with this ACVF/SDF. Let \[ \{\omega_j = \pi (j/J) \qquad j=0,\ldots,J \} \] and \[ \sigma_j^2 = \frac{1}{2 \pi} \left[ 1 - 0.5 \cos (2 \omega_j) - 0.25 \cos(3 \omega_j) \right] \] and then define \[ X_t = \int_{-\pi}^\pi [\cos(t \lambda) + \imagi \sin (t \lambda)] \ dZ(\lambda) \] <>= lam<-seq(-pi,pi,length=1000) flam<-(1-0.5*cos(2*lam)-0.25*cos(3*lam))/(2*pi) par(mar=c(4,4,2,0)) plot(lam,flam,type='l',ylim=range(-0.5,0.5), xlab=expression(lambda),ylab=expression(f(lambda))) abline(h=0,lty=3) @ <>= k<-500 lam<-seq(-pi,pi,by=0.01) omega<-c(0:k)*pi/k sigsq<-(1/k)*(1-0.5*cos(2*omega)-0.25*cos(3*omega)) sum(sigsq[2:k])+0.5*(sigsq[1]+sigsq[k+1]) hvec<-c(0:25) gammah<-0.5*sigsq[1]*cos(omega[1]*hvec) for(j in 2:k){ gammah<-gammah+sigsq[j]*cos(omega[j]*hvec) } gammah<-gammah+0.5*sigsq[k+1]*cos(omega[k+1]*hvec) round(gammah,6) @ <>= Uvec<-rnorm(k+1)*sqrt(sigsq) Vvec<-rnorm(k+1)*sqrt(sigsq) tvec<-seq(0,500,by=1) Xt<-tvec*0 for(j in 1:(k+1)){ Xt<-Xt+Uvec[j]*cos(omega[j]*tvec)+Vvec[j]*sin(omega[j]*tvec) } par(mar=c(4,4,2,2)) plot(tvec,Xt,type='l',xlab='t',ylab=expression(X[t]),ylim=range(-3,3)) acf(Xt) points(hvec,gammah,pch=19,cex=0.7,col='red') @ \pagebreak \item \textit{Suppose $X_{1,t}$ and $X_{2,t}$ are independent stationary processes with ACVFs $\gamma_1(h)$ and $\gamma_2(h)$, and corresponding spectral density functions (SDFs) $f_1(\lambda)$ and $f_2(\lambda)$, respectively.} \medskip \textit{Derive the SDF of \[ X_1 = X_{1,t} + X_{2,t}. \]} \textbf{Solution:} We have that \[ \gamma_X(h) = \E[X_t X_{t+h}] = \E[(X_{1,t} + X_{2,t})(X_{1,t+h} + X_{2,t+h})] = \E[X_{1,t} X_{1,t+h}] + \E[X_{2,t} X_{2,t+h}] \] by independence. Hence \[ \gamma_X(h) = \gamma_1(h) + \gamma_2(h) \] and \begin{align*} f_X(\lambda) & = \frac{1}{2 \pi} \sum_{h=-\infty}^\infty e^{-\imagi h \lambda} \gamma(h) \\[6pt] & = \frac{1}{2 \pi} \sum_{h=-\infty}^\infty e^{-\imagi h \lambda} ( \gamma_1(h) + \gamma_2(h) ) \\[6pt] & = \frac{1}{2 \pi} \sum_{h=-\infty}^\infty e^{-\imagi h \lambda} \gamma_1(h) + \frac{1}{2 \pi} \sum_{h=-\infty}^\infty e^{-\imagi h \lambda} \gamma_2(h) \\[6pt] & = f_1(\lambda) + f_2(\lambda). \end{align*} \item\textit{ Derive the SDF of the stationary process $\{X_t\}$ defined by \[ X_t = A \cos (\omega t) + B \sin (\omega t) + Z_t + 2 Z_{t-1} \] where \begin{itemize} \item $A$ and $B$ are zero mean random variables with variance 1; \item $\omega \in (0,\pi/2)$ is a fixed constant; \item $\{Z_{t}\} \sim \WN(0,\sigma_Z^2)$; \item $A$, $B$ and $\{Z_t\}$ are mutually uncorrelated. \end{itemize}} \textbf{Solution:} Using the result from Q2, with \[ f_1(\lambda) = \left\{ \begin{array}{cl} \dfrac{1}{2} & \lambda = -\omega\\[12pt] \dfrac{1}{2} & \lambda = \omega\\[12pt] 0 & \text{otherwise} \end{array} \right. \] and \[ f_2(\lambda) = \frac{\sigma_Z^2}{2 \pi} |(1+2 e^{-\imagi \lambda})|^2 = \frac{\sigma_Z^2}{2 \pi}[1 + 4(1+ \cos(\lambda))] \] which is the SDF of the $\MA(1)$ process with $\theta=2$. Then \[ f_X(\lambda) = f_1(\lambda) + f_2(\lambda) \] is the SDF of $\{X_t\}$. It is an SDF of mixed type. \item \textit{Suppose that \[ X_t = 0.99 X_{t-3} + Z_t \] where $\{Z_t\} \sim \WN(0,1)$. Find and plot the SDF for $\{X_t\}$, and the SDF of \[ Y_t = \frac{1}{3} (X_{t-1} + X_t + X_{t+1}). \]} \textbf{Solution:} Here we have an $\AR(3)$ process with $\phi_1 = \phi_2 = 0$. The $\AR$ polynomial is \[ \Phi(z) = 1 - 0.99 z^3 \] which has reciprocal roots less than 1 in modulus so the process is stationary and has SDF \[ f_X(\lambda) = \frac{1}{2 \pi} \frac{1}{|1-0.99 e^{-\imagi 3 \lambda}|^2} \] <>= Mod(polyroot(c(1,0,0,-0.99))) #Roots 1/Mod(polyroot(c(1,0,0,-0.99))) #Reciprocal roots lam<-seq(-pi,pi,length=1000) flam<-(1/Mod(1-0.99*exp(-1i*3*lam))^2)/(2*pi) par(mar=c(4,4,1,0)) plot(lam,flam,type='l',xlab=expression(lambda),ylab=expression(f(lambda))) @ Now let $\{Y_t\}$ be the filtered version of $\{X_t\}$ given by \[ Y_t = \frac{1}{3} (X_{t-1} + X_t + X_{t+1}) = \sum_{j=-\infty}^\infty \psi_j X_{t-j} = \Psi(B) X_t \] where $\psi_j = 1/3$ for $j=-1,0,1$, and $\psi_j = 0$ otherwise. Then \[ f_Y(\lambda) = |\Psi(e^{-\imagi \lambda})|^2 f_X(\lambda) = \left(\frac{1}{3}\right)^2 |1 + e^{\imagi \lambda} + e^{-\imagi \lambda}|^2 f_X(\lambda) \] <>= lam<-seq(-pi,pi,length=1000) flam<-(1/Mod(1-0.99*exp(-1i*3*lam))^2)/(2*pi) flam<-flam*(1/9)*Mod(1+exp(1i*lam)+exp(-1i*lam))^2 par(mar=c(4,4,2,0)) plot(lam,flam,type='l',xlab=expression(lambda),ylab=expression(f(lambda))) @ <>= set.seed(3) n<-500 Xt<-arima.sim(n,model=list(ar=c(0,0,0.99))) par(mar=c(4,4,1,0)) plot(Xt,type='l') acf(Xt,lag.max=50) Yt<-Xt Yt[1]<-(Xt[1]+Xt[2])/2; Yt[n]<-(Xt[n-1]+Xt[n])/2 for(i in 2:(n-1)){Yt[i]<-mean(Xt[c((i-1):(i+1))])} par(mar=c(4,4,3,0)) plot(Yt,type='l') acf(Yt,lag.max=50) spectrum(Xt,demean=F,taper=0,main='Periodogram for Xt') abline(v=1/3,col='red') spectrum(Yt,demean=F,taper=0,main='Periodogram for Yt') @ \item \textit{The fractional white noise process, $\{W_t\}$, defined in terms of the white noise process $\{Z_t\}$ by \[ W_t = (1-B)^{-d} Z_t \] for $-0.5 < d < 0.5$ has SDF \[ f_W(\lambda) = \frac{\sigma_Z^2}{2 \pi} \frac{1}{|1-e^{-\imagi \lambda}|^{2d}}. \] Comment on the absolute summability of the corresponding ACVF, $\gamma_W(h)$.} \medskip \textbf{Solution:} From lectures, we have the formula that for large $h > 0$ \[ \gamma_W(h) \approx C h^{2d - 1} \] for some constant $C$. The value $2d - 1$ lies in the range $(-2,0)$, so it is evident that if $d < 0$, then $2d - 1 < -1$, and $\gamma_W(h)$ is absolutely summable, whereas if $d > 0$, $ -1 < 2 d - 1 < 0$ it is not; this follows as the sum \[ \sum_{h=1}^\infty h^{-\alpha} \] is convergent if and only if $\alpha > 1$. \medskip Note that for $h$ large enough \[ \log \gamma_W(h) = \log C + (2d-1) \log h \] so plotting $\log \gamma_W(h)$ versus $\log h$ will lead to a straight line with slope $2d - 1$. For $d=0.4$, <>= H<-1000 hvec<-c(1:H) d<-0.4 gamma0<-gamma(1-2*d)/gamma(1-d^2) gammah<-gamma0*exp(lgamma(hvec+d)+lgamma(1-d)-lgamma(hvec-d+1)-lgamma(d)) gammah[H] sum(abs(c(gamma0,gammah))) cumsum(gammah)[100*c(1:10)] par(mar=c(4,4,2,2)) plot(log(hvec), log(gammah),pch=19,cex=0.2, xlab='log(h)',ylab=expression(log(gamma[X](h)))) lm(log(gammah)~log(hvec)) @ Here the slope is $-0.2$. For $d=-0.1$, the slope is -1.2. <>= d<--0.1 gamma0<-gamma(1-2*d)/gamma(1-d^2) gammah<-gamma0*exp(lgamma(hvec+d)+lgamma(1-d)-lgamma(hvec-d+1)-lgamma(d)) gammah[H] sum(abs(c(gamma0,gammah))) cumsum(gammah)[100*c(1:10)] par(mar=c(4,4,2,2)) plot(log(hvec), log(gammah),pch=19,cex=0.2, xlab='log(h)',ylab=expression(log(gamma[X](h)))) lm(log(gammah)~log(hvec)) @ Therefore for $d > 0$, the ACVF converges slow enough so that the sum of its absolute values is divergent. This renders the use of the frequency-domain approach to inference based on the SDF is more palatable than time-domain approach. \end{enumerate} \end{document}