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\begin{document}
\begin{center}
\bf {\underline{ Stat 207, Summer 01. Solutions to Homework 2}}
\end{center}

\noindent
\bf 1. (1.8) (a)\rm
\[ E[x_t]=E[\sum_{k=1}^{t}w_k]=\sum_{k=1}^{t}E[w_k]=0.\]
\begin{eqnarray*}
\gamma(s,t) &=& E[(x_s-\mu)(x_t-\mu)] \\
            &=& E[x_sx_t] \\
            &=& E[\sum_{k=1}^{s}w_k\sum_{l=1}^{t}w_l] \\
            &=& E[w_1(\sum_{l=1}^{t}w_l)+\cdots+w_s(\sum_{l=1}^{t}w_l)] \\
            &=& E[w_1^2]+\cdots+E[w_{\min(s,t)}^2] \\
            &=& \sigma_{w}^2\min(s,t),
\end{eqnarray*}
since $E[w_mw_n]=0$ if $m\not=n$.\\

\noindent
\bf (b) \rm
The series is nonstationary since the autocovariance function
\[ \gamma(s,t)=\sigma_{w}^2\min(s,t)\]
depends on $t$.  So $x_t$ is not weakly stationary ($\Longrightarrow$
not strictly stationary).\\

\noindent
\bf (c) \rm
Consider the difference
\[ \nabla x_t=x_t-x_{t-1}=\sum_{k=1}^{t}w_k-\sum_{k=1}^{t-1}w_k=w_t.\]
We have
\[ E[\nabla x_t]=E[w_t]=0,\]
and
\begin{eqnarray*}
\gamma(h) &=& E[(\nabla x_{t+h}-\mu)(\nabla x_{t}-\mu)] \\
          &=& E[\nabla x_{t+h}\nabla x_{t}] \\
          &=& E[w_{t+h}w_t]\\
          &=& \left\{\begin{array}{cc}
                             \sigma_w^2, & h=0,\\
                             0, &  |h|>0 .
                   \end{array}
             \right.
\end{eqnarray*}
Hence $\nabla x_t$ is stationary since its mean is constant and its
autocovariance function does not depend on $t$.\\

\noindent
\bf 2. (1.9) \rm
\begin{eqnarray*}
E[x_t] &=& E[u_1 \sin(2\pi\nu_0 t)+u_2 \cos(2\pi\nu_0 t)] \\
       &=& E[u_1] \sin(2\pi\nu_0 t)+ E[u_2] \cos(2\pi\nu_0 t) \\ 
       &=& 0.
\end{eqnarray*}

In the following, we will use 
\begin{eqnarray*}
\sin (A+B) &=& \sin A\cos B+\cos A\sin B, \\
\cos (A-B) &=& \cos A\cos B+\sin A\sin B.
\end{eqnarray*}
\begin{eqnarray*}
\gamma (h) &=& E[(x_{t+h}-\mu)(x_t-\mu)] \\
           &=& E[x_{t+h}x_t] \\
           &=& E[\{u_1 \sin(2\pi\nu_0 (t+h))+u_2 \cos(2\pi\nu_0
           (t+h))\}\cdot \{u_1 \sin(2\pi\nu_0 t)+u_2 \cos(2\pi\nu_0
           t)\}] \\
           &=& E[u_1^2 \sin(2\pi\nu_0 (t+h))\sin(2\pi\nu_0 t)+u_1 u_2
           \sin(2\pi\nu_0 (t+h))\cos(2\pi\nu_0 t) \\
           & & +u_2 u_1
           \cos(2\pi\nu_0 (t+h))\sin(2\pi\nu_0 t)+u_2^2 \cos(2\pi\nu_0
           (t+h))\cos(2\pi\nu_0 t)] \\
           &=& E[u_1^2 \sin(2\pi\nu_0 (t+h))\sin(2\pi\nu_0 t)+u_1 u_2
           \sin(2\pi\nu_0 (t+h)+2\pi\nu_0 t) \\
           & & +u_2^2 \cos(2\pi\nu_0
           (t+h))\cos(2\pi\nu_0 t)] \\
           &=& E[u_1^2] \sin(2\pi\nu_0 (t+h))\sin(2\pi\nu_0 t)+E[u_1]E[ u_2]
           \sin(2\pi\nu_0 (t+h)+2\pi\nu_0 t)\\
           & & +E[u_2^2] \cos(2\pi\nu_0
           (t+h))\cos(2\pi\nu_0 t) \\
           &=& \sigma^2[\sin(2\pi\nu_0 (t+h))\sin(2\pi\nu_0 t)+\cos(2\pi\nu_0
           (t+h))\cos(2\pi\nu_0 t)] \\
           &=& \sigma^2\cos(2\pi\nu_0 (t+h)-2\pi\nu_0 t) \\
           &=& \sigma^2\cos(2\pi\nu_0 h).
\end{eqnarray*}
So $x_t$ is weakly stationary since its mean is constant and its
autocovariance function does not depend on $t$.\\

\noindent
\bf 3. (1.14) (a)\rm
\[ E[y_t]=E[w_t-\theta w_{t-1}+u_t]=E[w_t]-\theta E[w_{t-1}]+E[u_t]=0\ (=\mu_{y}).\]
\begin{eqnarray*}
\gamma_{y}(h) &=& E[(y_{t+h}-\mu_y)(y_{t}-\mu_y)] \\
              &=& E[y_{t+h}y_{t}] \\
              &=& E[(w_{t+h}-\theta w_{t+h-1}+u_{t+h})(w_t-\theta
              w_{t-1}+u_t)] \\
              &=& E[ w_{t+h}w_{t}-\theta
              w_{t+h}w_{t-1}+w_{t+h}u_{t}-\theta
              w_{t+h-1}w_{t}+\theta^2 w_{t+h-1}w_{t-1}\\
              & &-\theta
              w_{t+h-1}u_{t}+u_{t+h}w_{t}-\theta
              u_{t+h}w_{t-1}+u_{t+h}u_{t}] \\
              &=& E[w_{t+h}w_{t}]-\theta E[w_{t+h}w_{t-1}]-\theta
              E[w_{t+h-1}w_{t}]+\theta^2
              E[w_{t+h-1}w_{t-1}]+E[u_{t+h}u_{t}],
\end{eqnarray*}
since $w_t$ is independent of $u_t$.\\
When $h=0$, 
\[ \gamma_y(0)=E[w_t]+\theta^2E[w_{t-1}]+E[u_{t}^2]=(1+\theta^2)\sigma_{w}^2+\sigma_u^2;\]
When $h=\pm 1$,
\[ \gamma_y(h)=-\theta E[w_{t}^2]=-\theta\sigma_{w}^2;\]
When $|h|\geq 2$, 
\[\gamma_y(h)=0.\]
So the ACF 
\[ \rho_{y}(h)=\left\{\begin{array}{cc}
                                  1, & h=0,\\
                             -\frac{\theta\sigma_w^2}{(1+\theta^2)\sigma_{w}^2+\sigma_u^2}, & |h|=1,\\
                             0, &  |h|\geq 2 .
                   \end{array}
             \right.
\]

\noindent
\bf (b)\rm
\[ E[x_t]=E[w_t]=0\ (=\mu_x).\]
\begin{eqnarray*}
\gamma_x(h) &=& E[(x_{t+h}-\mu_x)(x_{t}-\mu_x)] \\
            &=& E[x_{t+h}x_{t}] \\
            &=& E[w_{t+h}w_{t}] \\
            &=& \left\{\begin{array}{cc}
                             \sigma_w^2, & h=0,\\
                             0, &  |h|\geq 1 .
                   \end{array}
             \right.
\end{eqnarray*}
\begin{eqnarray*}
\gamma_{xy}(h) &=& E[(x_{t+h}-\mu_x)(y_{t}-\mu_y)] \\
               &=& E[x_{t+h}y_{t}] \\
               &=& E[w_{t+h}(w_t-\theta w_{t-1}+u_t)] \\
               &=& E[w_{t+h}w_t]-\theta E[w_{t+h}w_{t-1}] \ \ \
               (\mbox{since}\ w_t\ \mbox{and}\ \mbox{are
               independent})\\
               &=& \left\{\begin{array}{cc}
                             \sigma_w^2, & h=0,\\
                             -\theta \sigma_w^2, & h=-1,\\
                             0, &  \mbox{otherwise} .
                   \end{array}
             \right.
\end{eqnarray*}
So the CCF 
\[ \rho_{xy}(h)=\frac{\gamma_{xy}(h)}{\sqrt{\gamma_{x}(0)}\sqrt{\gamma_{y}(0)}}=\left\{\begin{array}{cc}
                             \frac{\sigma_w}{\sqrt{(1+\theta^2)\sigma_{w}^2+\sigma_{u}^2}}, & h=0,\\
                           -\frac{\theta\sigma_w}{\sqrt{(1+\theta^2)\sigma_{w}^2+\sigma_{u}^2}}  , & h=-1,\\
                             0, &  \mbox{otherwise} .
                   \end{array}
             \right.
\]

\noindent
\bf (c)\rm
\[ E[x_t]=0\ \ \mbox{and}\ \ \gamma_{x}(h)=\left\{\begin{array}{cc}
                   \sigma_w^2, & h=0, \\
                    0, & |h|\geq 1, 
                   \end{array}
             \right.
\]
so $x_t$ is stationary.\\
\[ E[y_t]=0\ \ \mbox{and}\ \ \gamma_{y}(h)=\left\{\begin{array}{cc}      
       (1+\theta^2)\sigma_{w}^2+\sigma_{u}^2, & h=0, \\
       -\theta \sigma_{w}^2, & |h|=1, \\
          0, & |h|\geq 2,
                   \end{array}
             \right.
\]
so $y_t$ is stationary.
\[ \gamma_{xy}(h)=\left\{\begin{array}{cc}
                             \sigma_w^2, & h=0,\\
                             -\theta \sigma_w^2, & h=-1,\\
                             0, &  \mbox{otherwise} .
                   \end{array}
             \right.
\]
The series have constant means and the ACF's and CCF only depend on
$h$, so $x_t$ and $y_t$ are jointly stationary.

\end{document}

