Part 1: Problem 4

Consider the following models where ${e_t}$ is a $0$ mean white noise process with variance $σ^2$. \begin{enumerate} \item [i] $Y_t = 1.9 Y_{t−1} − 0.9 Y_{t−2} + e_t − 0.5e_{t−1}$ \item [ii] $Y_t = 0.5 Y_{t−1} + e_t − 0.2e_{t−1} − 0.15e_{t−2}$ \end{enumerate}

Part (a)

\fbox{Write each model above using backshift notation.}

\begin{enumerate} \item [(i)] $(1 - 1.9 \backshift + 0.9 \backshift^2) Y_t = (1 - 0.5 \backshift) e_t$.

Working on both sides of the equation simultaneously, we use the following sequence of transformations to arrive at a canonical form. \begin{align*} (1 - 1.9 \backshift + 0.9 \backshift^2) Y_t &= (1 - \sfrac{1}{2} \backshift) e_t\ %%%%%%%%%%%\sfrac{10}{10}(1 - 1.9 \backshift + 0.9 \backshift^2) Y_t &= (1 - \sfrac{1}{2} \backshift) e_t\ \sfrac{1}{10}(10 - 19 \backshift + 9 \backshift^2) Y_t &= (1 - \sfrac{1}{2} \backshift) e_t\ \sfrac{1}{10}(1-\backshift)(10-9\backshift) Y_t &= (1 - \sfrac{1}{2} \backshift) e_t\ \sfrac{1}{100}(1-\backshift)(1-\sfrac{9}{10}\backshift) Y_t &= (1 - \sfrac{1}{2} \backshift) e_t\ (1-\backshift)(1-\sfrac{9}{10}\backshift) Y_t &= (1 - \sfrac{1}{2} \backshift) 100 e_t \end{align*} We let $\eta_t = 10^2 e_t$, which is white noise with a mean $0$ and variance $10^4 \sigma^2$, thus

$$ (1-\backshift)(1-\sfrac{9}{10}\backshift) Y_t = (1 - \sfrac{1}{2} \backshift) \eta_t. $$

\item [(ii)] $(1 - 0.5 \backshift) Y_t = (1 - 0.2 \backshift - 0.15 \backshift^2) e_t$. Working on both sides of the equation simultaneously, we use the following sequence of transformations to arrive at a canonical form. \begin{align*} (1 - 0.5 \backshift) Y_t &= (1 - 0.2 \backshift - 0.15 \backshift^2) e_t\ (1 - \sfrac{1}{2} \backshift) Y_t &= (1 - \sfrac{1}{5} \backshift - \sfrac{3}{20} \backshift^2) e_t\ (1 - \sfrac{1}{2} \backshift) Y_t &= (1 - \sfrac{1}{2}\backshift)(1 + \sfrac{3}{10}) e_t\ Y_t &= (1 + \sfrac{3}{10}) e_t \end{align*} \end{enumerate}

Part (b)

\fbox{\begin{minipage}{.9\textwidth} Characterize these models as models in the $\arima(p,d,q)$ family, that is, identify $p$, $d$ and $q$. \end{minipage}}

\begin{enumerate} \item[(i)] By the form, ${Y_t}$ is an $\arima(1,1,1)$ process, or equivalently, ${\nabla Y_t}$ is an $\arma(1,1)$ process. \item[(ii)] By the form, ${Y_t}$ is an $\arima(0,0,1)$ process, or equivalently, $\ma(1)$ process. \end{enumerate}

Part (c)

\fbox{Determine if each model corresponds to a stationary process or not.}

\begin{enumerate} \item[(i)] ${Y_t}$ is an $\arima(1,1,1)$ process, which means that its characteristic function has a unit root. Recall that $\arima(1,1,1)$ denotes a ${Y_t}$ is therefore a non-stationary process.\footnote{$\nabla {Y_t}$ is a stationary $\arma(1,1)$ process since it does not have a unit root.} \item[(ii)] ${Y_t}$ is an $\ma(1)$ process, which are necessarily stationary. \end{enumerate}