\documentclass[notitlepage]{article} \usepackage{../../Math545} \usepackage{listings} \usepackage{numprint} \usepackage{enumerate} \usepackage{bbm} \usepackage{bm} \usepackage{amsfonts,amsmath,amssymb} \newtheorem{alg}{Algorithm} \newtheorem{ex}{Example} \newtheorem{note}{Note} \newenvironment{proof}{\par\noindent{\normalfont{\bf{Proof }}}} {\hfill \small$\blacksquare$\\[2mm]} \setlength{\parindent}{0in} \def\E{\mathbb{E}} \def\Var{\text{Var}} \def\Cov{\text{Cov}} \def\Corr{\text{Corr}} \def\bW{\mathbf{W}} \newcommand{\bs}[1]{\bm{#1}} \newcommand{\Rho}{\mathrm{P}} \newcommand{\cif}{\text{ if }} \newcommand{\Ra}{\Longrightarrow} \newcommand{\ra}{\longrightarrow} \def\bphi{\boldsymbol \phi} \lstloadlanguages{R} \definecolor{keywordcolor}{rgb}{0,0.6,0.6} \definecolor{delimcolor}{rgb}{0.461,0.039,0.102} \definecolor{Rcommentcolor}{rgb}{0.101,0.043,0.432} \lstdefinestyle{Rsettings}{ basicstyle=\ttfamily, breaklines=true, showstringspaces=false, keywords={if, else, function, theFunction, tmp}, % Write as many keywords otherkeywords={}, commentstyle=\itshape\color{Rcommentcolor}, keywordstyle=\color{keywordcolor}, moredelim=[s][\color{delimcolor}]{"}{"}, } \lstset{basicstyle=\ttfamily, numbers=none, literate={~} {$\sim$}{2}} \def\ctitle{MATH 545: Introduction to Time Series Analysis} \def\ttitle{Winter 2025} \def\stitle{Vector Autoregression Models} \begin{document} <>= library(knitr) # global chunk options opts_chunk$set(cache=TRUE, autodep=TRUE) options(scipen=999) options(repos=c(CRAN="https://cloud.r-project.org/")) @ \maketitle Consider the model \begin{eqnarray*} X_{1,t} &=& \phi_{10} + \phi_{11} X_{1,t-1} + \phi_{12} X_{2,t-1} + Z_{1,t}\\ X_{2,t} &=& \phi_{20} + \phi_{21} X_{1,t-1} + \phi_{22} X_{2,t-1} + Z_{2,t} \end{eqnarray*} The elements of \[ \bPhi = \left[\begin{array}{cc} \phi_{11} & \phi_{12} \\ \phi_{21} & \phi_{22} \end{array} \right] \] define the dependence of current elements of $\bX_t$ on past elements. Let \[ \Sigma = \bL \bG \bL^\top \] be a Cholesky decomposition where $\bG$ is a diagonal matrix, and $\bL$ is a lower-triangular matrix, where the diagonal elements of $\bL$ are all equal to 1. Define $\bU_t = \bL^{-1} \bZ_t$. Then \[ \E[\bU_t] = \bm{0} \qquad \qquad \Var[\bU_t] = \bG \] so that the elements of $\bU_t$ are uncorrelated. Then \[ \bL^{-1} \bX_t = \bL^{-1} \bphi_0 + \bL^{-1} \bPhi \bX_{t-1} + \bL^{-1} \bZ_t \] so that \[ \bL^{-1} \bX_t = \bphi^\star + \bPhi^\star \bX_{t-1} + \bU_t \] say. Thus, as $\bL$ is lower-triangular, the $K$th row of this vector equation is \[ X_{K,t} + \sum_{k=1}^{K-1} L_{Kk} X_{k,t} = \phi_{K0}^\star + \sum_{k=1}^{K} \phi_{Kk}^\star X_{k,t-1} + U_{K,t} \] As the elements of $\bU_t$ are uncorrelated, this equation shows explicitly the concurrent dependence of $X_{K,t}$ on $X_{k,t}$, for $k=1,\ldots,K-1$. A similar expression can be found for each of the elements of $\bX_t$. \medskip \textbf{Example:} Let \[ \left[ \begin{array}{c} X_{1,t} \\ X_{2,t} \end{array} \right] = \left[ \begin{array}{c} 0.2 \\ 0.4 \end{array} \right] + \left[ \begin{array}{rr} 0.2 & 0.3 \\ -0.6 & 1.1 \end{array} \right] \left[ \begin{array}{c} X_{1,t-1} \\ X_{2,t-1} \end{array} \right] + \left[ \begin{array}{c} Z_{1,t} \\ Z_{2,t} \end{array} \right] \] with \[ \Sigma = \left[ \begin{array}{rr} 2 & 1 \\ 1 & 1 \end{array} \right] \] Then \[ \bL^{-1} = \left[ \begin{array}{rr} 1.0 & 0.0 \\ -0.5 & 1.0 \end{array} \right] \qquad \bG = \bL^{-1} \Sigma ( \bL^\top )^{-1} = \left[ \begin{array}{rr} 2.0 & 0.0 \\ 0.0 & 0.5 \end{array} \right] \] Using the above structural equation formulation: \begin{eqnarray*} X_{2,t} & = & 0.3 + 0.5 X_{1,t} -0.7 X_{1,t-1} + 0.95 X_{2,t-1} + U_{2,t} \end{eqnarray*} and \begin{eqnarray*} X_{1,t} & = & -0.2 + 1.0 X_{2,t} -0.8 X_{2,t-1} + 0.8 X_{1,t-1} + U_{1,t} \end{eqnarray*} where $U_{2,t} \sim \calN(0,0.5)$ and $U_{1,t} \sim \calN(0,1)$ are independent random variables. <>= library(mvnfast) n<-1000 Sigma<-matrix(c(2,1,1,1),nrow=2,byrow=T) Phi<-matrix(c(0.2,0.3,-0.6,1.1),nrow=2,byrow=T) Phi0<-c(0.2,0.4) Zt<-rmvn(n,c(0,0),Sigma) Xt<-matrix(0,nrow=n,ncol=2) for(i in 2:n){ Xt[i,]<-Phi0+Phi %*% Xt[i-1,] + Zt[i,] } par(mar=c(2,2,4,4)) acf(Xt) @ \medskip Now for the usual Cholesky decomposition, we have \[ \bC \bC^\top = \Sigma \] where $\bC$ is lower triangular, but here we want \[ \bL \bG \bL^\top = \Sigma. \] The diagonal elements of $\bC$ are square roots of the diagonal elements of $\Sigma$. Hence $\bC = \bL \bG^{1/2}$ so that $\bL = \bC \bG^{-1/2}$. We then have \[ \Var[\bU_t] = \Var[ \bL^{-1} \bZ_t] = \bL^{-1} \Var[ \bZ_t] \bL^{-\top} = \bL^{-1} \Sigma \bL^{-\top} = \bG. \] where $\bL^{-\top}$ is the transpose of $\bL^{-1}$. <>= Sigma Cmat<-t(chol(Sigma)) Cmat Gmat.root<-diag(diag(Cmat)) Gmat.root Lmat<-Cmat %*% solve(Gmat.root) Gmat<-Gmat.root^2 Lmat Gmat Linv<-solve(Lmat) Linv @ We can then compute \[ \bphi^\ast = \bL^{-1} \bphi_0 \] <>= Phi0star<-Linv %*% Phi0 Phi0star @ and \[ \bPhi^\ast = \bL^{-1} \bPhi \] <>= Phistar<-Linv %*% Phi Phistar @ To obtain the structural equation for $X_{2,t}$, we note that \[ \bL^{-1} \bX_t = \bphi^\star + \bPhi^\star \bX_{t-1} + \bU_t \] becomes \[ \begin{bmatrix} \Sexpr{Linv[1,1]} & \Sexpr{Linv[1,2]} \\[6pt] \Sexpr{Linv[2,1]} & \Sexpr{Linv[2,2]} \end{bmatrix} \begin{bmatrix} X_{1,t} \\[6pt] X_{2,t} \end{bmatrix} = \begin{bmatrix} \Sexpr{Phi0star[1]} \\[6pt] \Sexpr{Phi0star[2]} \end{bmatrix} + \begin{bmatrix} \Sexpr{Phistar[1,1]} & \Sexpr{Phistar[1,2]} \\[6pt] \Sexpr{Phistar[2,1]} & \Sexpr{Phistar[2,2]} \end{bmatrix} \begin{bmatrix} X_{1,t-1} \\[6pt] X_{2,t-1} \end{bmatrix} + \begin{bmatrix} U_{1,t} \\[6pt] U_{2,t} \end{bmatrix} \] from which we can deduce via the second equation that \[ -0.5 X_{1,t} + X_{2,t} = 0.3 - 0.7 X_{1,t-1} + 0.95 X_{2,t-1} + U_{2,t} \] that is \[ X_{2,t} = 0.5 X_{1,t} + 0.3 - 0.7 X_{1,t-1} + 0.95 X_{2,t-1} + U_{2,t}. \] Note that from the first equation, we have \[ X_{1,t} = 0.2 + 0.2 X_{1,t-1} + 0.3 X_{2,t-1} + U_{1,t} \] which shows the dependence of $\{X_{1,t}\}$ on lagged values only. \medskip To obtain the structural equation for $\{X_{1,t}\}$, we swap the positions of the two variables: we have \[ \left[ \begin{array}{c} X_{2,t} \\ X_{1,t} \end{array} \right] = \left[ \begin{array}{c} 0.4 \\ 0.2 \end{array} \right] + \left[ \begin{array}{rr} 1.1 & -0.6 \\ 0.3 & 0.2 \end{array} \right] \left[ \begin{array}{c} X_{1,t-1} \\ X_{2,t-1} \end{array} \right] + \left[ \begin{array}{c} Z_{2,t} \\ Z_{1,t} \end{array} \right] \] with now \[ \Sigma = \left[ \begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array} \right] \] <>= Sigma<-matrix(c(1,1,1,2),nrow=2,byrow=T) Phi<-matrix(c(1.1,-0.6,0.3,0.2),nrow=2,byrow=T) Phi0<-c(0.4,0.2) Sigma Cmat<-t(chol(Sigma)) Cmat Gmat.root<-diag(diag(Cmat)) Gmat.root Lmat<-Cmat %*% solve(Gmat.root) Gmat<-Gmat.root^2 Lmat Gmat Linv<-solve(Lmat) Linv Phi0star<-Linv %*% Phi0 Phi0star Phistar<-Linv %*% Phi Phistar @ so that \[ \begin{bmatrix} \Sexpr{Linv[1,1]} & \Sexpr{Linv[1,2]} \\[6pt] \Sexpr{Linv[2,1]} & \Sexpr{Linv[2,2]} \end{bmatrix} \begin{bmatrix} X_{2,t} \\[6pt] X_{1,t} \end{bmatrix} = \begin{bmatrix} \Sexpr{Phi0star[1]} \\[6pt] \Sexpr{Phi0star[2]} \end{bmatrix} + \begin{bmatrix} \Sexpr{Phistar[1,1]} & \Sexpr{Phistar[1,2]} \\[6pt] \Sexpr{Phistar[2,1]} & \Sexpr{Phistar[2,2]} \end{bmatrix} \begin{bmatrix} X_{2,t-1} \\[6pt] X_{1,t-1} \end{bmatrix} + \begin{bmatrix} U_{2,t} \\[6pt] U_{1,t} \end{bmatrix} \] yielding the structural equation \[ X_{1,t} = -0.2 + X_{2,t} + 0.8 X_{1,t-1} - 0.8 X_{2,t-1} + U_{1,t} \] where $\Var[U_{1,t}] = 1$, as note that here \[ \bG = \begin{bmatrix} 1 & 0 \\[6pt] 0 & 1 \end{bmatrix} \] \end{document}