\documentstyle[11pt]{article}
\topmargin 0pt \headheight 0pt \headsep 0pt \textheight 225mm
\textwidth 160mm \footheight 0pt \oddsidemargin 0pt
\evensidemargin 0pt \marginparwidth 0pt \marginparsep 0pt
\begin{document}
\begin{center}
\bf {\underline{ Stat 207, Summer 01. Solutions to Midterm}}
\end{center}

\noindent \bf 1. (a)\rm
\[ E[x_t]=E[\alpha_0+\alpha_1 t+\alpha_2 t^2+w_t]=\alpha_0+\alpha_1
t+\alpha_2 t^2,\]
so $\{x_t\}$ is nonstationary.\\

\noindent \bf (b) \rm Consider $\nabla^2$.
\[ \nabla x_t=x_t-x_{t-1}=\alpha_1+2\alpha_2 t-\alpha_2+w_t-w_{t-1},\]
\begin{eqnarray*}
 \nabla^2 x_t &=& \nabla(x_t-x_{t-1}) \\
              &=& (x_t-x_{t-1})-(x_{t-1}-x_{t-2}) \\
              &=& w_t-2w_{t-1}+w_{t-2}+2\alpha_2 \\
              &=& \nabla^2 w_t+2\alpha_2.
\end{eqnarray*}
\[ E[\nabla^2 x_t]=E[w_t-2w_{t-1}+w_{t-2}+2\alpha_2]=2\alpha_2=\mu_x.\]
\begin{eqnarray*}
\gamma_{\nabla^2 x_t}(h) &=& E[(\nabla^2 x_{t+h}-\mu_x)(\nabla^2
x_{t}-\mu_x)]\\
                         &=& E[\nabla^2 w_{t+h}\nabla^2 w_t] \\
                         &=& E[(w_{t+h}-2w_{t+h-1}+w_{t+h-2})(w_t-2w_{t-1}+w_{t-2})]\\
                         &=& \left\{\begin{array}{rl}
                             6\sigma_w^2, & h=0,\\
                             -4\sigma_w^2, & h=\pm 1,\\
                             \sigma_w^2, & h=\pm 2,\\
                             0, & |h|\geq 3 .
                   \end{array}
             \right.
\end{eqnarray*}
Hence $\nabla^2 x_t$ is stationary since the mean and
autocovariance
do not depend on time $t$.\\

\noindent \bf 2. \rm In the following, to check if it is causal,
we want to see
if the roots of the AR operator are outside the unit circle.\\
\noindent \bf (a) \rm ARMA(2,0).
\[ 1-1.05B+0.4B^2=0\ \ \Longrightarrow \ \ B=1.3125\pm 0.62188i.\]
So
\[|B|=\sqrt{1.3125^2+0.62188^2}=1.45>1.\]
Since the roots are outside the unit circle, the model is causal.\\

\noindent \bf (b) \rm ARMA(1,0).
\[ 1-1.05B=0\ \ \Longrightarrow \ \ B=0.9524<1,\]
so it is not causal.\\

\noindent \bf (c) \rm ARMA(1,1).
\[ 1+0.8B=0\ \ \Longrightarrow \ \ B=-1.25,\ |B|>1,\]
so it is causal. In order to check whether the model is
invertible, we want to see if the roots of the MA operator are
outside the unit circle.
\[ 1-0.25B=0 \ \ \Longrightarrow \ \ B=4,\]
so it is invertible.\\

\noindent \bf 3. (a) \rm
\[ x_t=x_{t-1}+w_t-\theta w_{t-1},\]
\[(1-B)x_t=(1-\theta B)w_t.\]
\[ 1-\theta B=0\ \Longrightarrow\ |B|=|\frac{1}{\theta}|>1,\]
so it is invertible.
\[w_t= \frac{1-B}{1-\theta B}x_t=(1-B)(1+\theta
B+\theta^2B^2+\cdots)x_t,\] so
\begin{eqnarray*}
w_t=(1 &+& \theta B+\theta^2 B^2+\cdots\\
       &-& B-\theta B^2-\cdots)x_t.
\end{eqnarray*}
Therefore
\[ \pi_j=-(1-\theta)\theta^{j-1}.\]

\noindent \bf (b) \rm
\[ x_t=\sum_{j=1}^{\infty}(1-\theta)\theta^{j-1}x_{t-j}+w_t.\]

\noindent \bf (c) \rm
\[E[x_{n+1}|x_n,x_{n-1},\cdots]=E[\sum_{j=1}^{\infty}(1-\theta)\theta^{j-1}x_{n+1-j}+w_{n+1}|x_n,x_{n-1},\cdots].\]
\begin{eqnarray*}
{\tilde x}_{n+1} &=&\sum_{j=1}^{\infty}(1-\theta)\theta^{j-1}x_{n+1-j} \\
                 &=&(1-\theta)x_n+\sum_{j=2}^{\infty}(1-\theta)\theta^{j-1}x_{n+1-j}\\
                 &=&(1-\theta)x_n+\theta\sum_{k=1}^{\infty}(1-\theta)\theta^{k-1}x_{n-k}\\
                 &=& (1-\theta)x_n+\theta {\tilde x}_{n}.
\end{eqnarray*}

\noindent \bf (d) \rm
\[ {\tilde x}_{n+1}^n =(1-\theta)x_n+\theta {\tilde x}_{n}^{n-1},\]
the truncated forecasts are computed by letting ${\tilde
x}_{1}^0=1$
and the use of the above formula. (see pages 143,144 for EWMA)\\

\noindent
\bf 4.\\
\noindent \bf (a) \rm MA(2).
$x_t=w_t+\theta_1w_{t-1}+\theta_2w_{t-2},\ \ \ \theta_1<0,\
\theta_2>0$.

\noindent \bf (b) \rm AR(2).
$x_t=\phi_1x_{t-1}+\phi_2x_{t-2}+w_t,\ \ \ \phi_1<0,\ \phi_2<0$.

\noindent \bf (c) \rm MA(1). $x_t=w_t+\theta_1w_{t-1},\ \ \
\theta_1>0$.

\noindent \bf (d) \rm AR(1). $x_t=\phi_1x_{t-1}+w_t,\ \ \
\phi_1>0$.

\end{document}
