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\begin{document}
\textbf{Solution for stat 207 HW\#1}\\

\textbf{1.3}
\begin{align*}
r(s,t) &= E[(x_s - \mu_s)(x_t - \mu_t)] \\
&=E[x_sx_t - x_s \mu_t - x_t \mu_s + \mu_t \mu_s]\\
&=E[x_s x_s] - E[x_s] \mu_t - \mu_s E[x_t] + \mu_s \mu_t\\
&=E[x_s x_t] - \mu_s \mu_t
\end{align*}

\textbf{1.7} moving average process.
$$x_t = w_{t-1} + 2 w_t + w_{t+1}, \ \ \ \ w_t \stackrel{i.i.d}{\sim}
N(0, \sigma_w^2)$$
\begin{align*}
E[x_t] &= E [w_{t-1} + 2 w_t + w_{t+1}]\\
&= E [w_{t-1}] + 2E[ w_t] + E[w_{t+1}]\\
&=0 \ \ \ \ \ \ \ \ \ \ \ \ \mbox{for all $t$}
\end{align*}
\begin{align*}
r(h) &= E[(x_{t+h} - \mu_x)(x_t - \mu_x)]\\
&= E [x_{t+h} x_t]\\
&=E[(w_{t+h-1} + 2 w_{t+h} + w_{t+h+1})(w_{t-1} + 2 w_t + w_{t+1})]\\
&=E[w_{t+h-1}w_{t-1} + 2w_{t+h-1}w_{t} + w_{t+h-1}w_{t+1} + 2w_{t+h}w_{t-1}+ 4w_{t+h}w_{t}\\
&\ \ \ \ +2 w_{t+h}w_{t+1} + w_{t+h-1}w_{t-1}+ 2w_{t+h-1}w_{t}+
w_{t+h+1}w_{t+1}]\\
&=E[w_{t+h-1}w_{t-1}] + 2E[w_{t+h-1}w_{t}] + E[w_{t+h-1}w_{t+1}] +
2E[w_{t+h}w_{t-1}] + 4E[w_{t+h}w_{t}]\\
&\ \ \ \ +2E [w_{t+h}w_{t+1}] + E[ w_{t+h-1}w_{t-1}]+
2E[w_{t+h-1}w_{t}]+
E[w_{t+h+1}w_{t+1}]\\
\end{align*}
\begin{align*}
h=0, r(h) &= \sigma_{w}^2 + 4\sigma_w^2 + \sigma_w^2 = 6\sigma_w^2\\
|h|=1,r(h) &= 2\sigma_{w}^2 + 2\sigma_w^2 = 4\sigma_w^2\\
|h|= 2, r(h) &= \sigma_{w}^2\\
|h|\geq 3,r(h) &= 0
\end{align*}
$$\rho(h) = \frac{r(h)}{r(0)},$$
so
\begin{align*}
h=0, \rho(h) &= 1 \\
|h|=1,\rho(h) &= \frac{2}{3} \\
|h|= 2, \rho(h) &= \frac{1}{6}\\
|h|\geq 3,\rho(h) &= 0
\end{align*}

\textbf{1.10}
$$x_t = \beta_0 + \beta_1 t + w_{t}, \ \ \ \ w_t \stackrel{i.i.d}{\sim}
N(0, \sigma_w^2), \ \ \ \ \beta_0, \beta_1 \ \ \mbox{unknown fixed
constants}$$
\begin{description}
\item[a.] nonstationary:\\
\begin{align*}
E[x_t] &= E[\beta_0 + \beta_1 t + w_{t}]\\
&=\beta_0 + \beta_1 t + E[w_{t}]\\
&=\beta_0 + \beta_1 t
\end{align*}
so $x_t$ is nonstationary since the mean chages with the time.
\item[b.]
\begin{align*}
\nabla x_t &= x_t - x_{t-1}\\
&= \beta_0 + \beta_1 t + w_t - \beta_0 - \beta_1 (t-1) - w_{t-1}\\
&= \beta_1 + w_t - w_{t-1}
\end{align*}
stationary:
$$E[\nabla x_t] = E[\beta_1 + w_t - w_{t-1}] = \beta_1 = \mu \ \ \
\forall t$$
\begin{align*}
r(h) &= E[(\nabla x_t - \mu)(\nabla x_t -\mu)]\\
&= E[\nabla x_t \nabla x_t] - \mu^2\\
&= E[(\beta_1 + w_{t+u} - w_{t+h-1})(\beta_1 + w_{t} - w_{t-1})] -
\beta_1^2\\
&= E[\beta_1^2 + \beta_1 w_t - \beta_1 w_{t-1} + \beta_1 w_{t+u} +
w_{t+u}w_t \\
&\ \ \ \ - w_{t+u}w_{t-1} - \beta_1 w_{t+h-1} - w_{t+h-1}w_t +
w_{t+h-1}w_{t-1}] - \beta_1^2\\
&= E[w_{t+h}w_t] - E[w_{t+h}w_{t-1}] - E[w_{t+h-1}w_t] +
E[w_{t+h-1}w_{t-1}]
\end{align*}
\begin{align*}
h = 0, r(h) &= \sigma_w^2 + \sigma_w^2 = 2\sigma_w^2 \\
|h| = 1, r(h) &= -\sigma_w^2\\
|h|\geq 2, r(h) &= 0
\end{align*}
so $x_t$ is weakly stationary since its mean is constant and its
autocovariance function doesn't depend on the time.
\end{description}

\rm
\end{document}
