%&LaTeX
\documentclass[12pt]{report}

\usepackage{latexsym}
\usepackage[centertags]{amsmath}
\usepackage{amsfonts}
\usepackage{newlfont}



\setlength{\oddsidemargin}{-0.4cm}

\setlength{\evensidemargin}{0cm}

\setlength{\topmargin}{-0.75cm}

\setlength{\textheight}{8.9in}

\setlength{\textwidth}{6.5in}
\renewcommand{\baselinestretch}{1}
\setlength{\parskip}{3mm}
\setlength{\parindent}{0mm} % indented paragraphs just don't

\begin{document}
\textbf{2.3}
\begin{description}
\item[a]
\begin{align*}
& x_t = 0.8x_{t-1} - 0.15 x_{t-2} + w_t - 0.3 w_{t-1}\\ & x_t -
0.8x_{t-1} + 0.15 x_{t-2} = w_t - 0.3 w_{t-1}\\ &(1 - 0.8B + 0.15
B^2) x_t = (1-0.3B)w_t\\ &(1 - 0.3B)(1- 0.5 B) x_t = (1-0.3B)w_t\\
&(1 - 0.3B) x_t = w_t\\ &x_t - 0.5 x_{t-1} = w_t\\ &x_t = 0.5
x_{t-1} + w_t \ \ \ \ \ \ \ \ \ ARMA(1,0)\\ &\mbox{it's causal:}\
\ (1-0.5B) = 0 \ \ \mbox{or}\ \ B=2 > 1.\\ &\mbox{invertible:}\ \
\mbox{Yes.}\ \ w_t = x_t - 0.5x_{t-1}.
\end{align*}

\item[b]
\begin{align*}
& x_t = x_{t-1} - 0.5 x_{t-2} + w_t - w_{t-1}\\ & x_t - x_{t-1} +
0.5 x_{t-2} = w_t - w_{t-1}\\ &(1 - B + 0.5 B^2) x_t = (1- B)w_t\\
&(1 - 0.5B)(1- 0.5 B) x_t = (1-0.3B)w_t\ \ \ \ \ ARMA(2,1)\\
&\mbox{it's causal:}\ \ (1-B-0.5B^2) = 0\\ &\mbox{roots:}\ \ 1 \pm
i \ \ \mbox{with module greater than $1$}\\ &\mbox{not
invertible:} (1-B) = 0, \ \ \mbox{root lies on unit circle}
\end{align*}
\end{description}

\textbf{2.6}
\begin{description}
\item[a]
$$x_t + 1.6x_{t-1} + 0.64x_{t-2} = w_t,\ \ \phi_1 =
    -1.6, \ \ \phi_2 = -0.64$$
$$(1+1.6B + 0.64B^2) x_t = w_t$$ $$\phi(B) x_t = w_t$$ $$\psi(B)
\phi(B) x_t = \psi(B) w_t$$ $$(\psi_0 + \psi_1B + \psi_2B^2+
\ldots)(1 + 1.6B + 0.64B^2) =1$$

\begin{align*}
&B^0: \phi_0&&= 1\\ &B^1: \phi_1 + 1.6\phi_0&&= 0\\ &B^j: \phi_j +
1.6\phi_{j-1} + 0.64\phi_{j-2} &&= 0,\ \ j\geq 2
\end{align*}
ACF:

$$ \rho(h) - \phi_1\rho(h-1) - \phi_2\rho(h-2) = 0, h =
1,2,\ldots$$ $$ \rho(0) = 1 \ \ \ \ \mbox{see example 2, page
103}$$ $$\rho(-1) = \frac{\phi_1}{1 - \phi_2}, \ \ \rho(1) =
\rho(-1)$$ plot:
\begin{align*}
\rho(0)&=1\\ \rho(1)&=-\frac{1.6}{1+0.64} = -0.9756\\
\rho(2)&=-1.6\times (-0.9756) - 0.64 = -0.9210 \\
\rho(3)&=-1.6\times 0.9210 - 0.64\times (-0.9756) = -0.8492 \\
\rho(4)&=-1.6\times (-0.8492) - 0.64\times (0.9210) = -0.7095
\end{align*}

\item[b]
$$x_t - 0.40x_{t-1} - 0.45x_{t-2} = w_t,\ \ \phi_1 =
    0.40, \ \ \phi_2 = 0.45$$
$$(1 - 0.4B - 0.45 B^2) x_t = w_t$$ $$\phi(B) x_t = w_t$$
$$\psi(B) \phi(B) x_t = \psi(B) w_t$$

$$B^j: \phi_j - 0.4\phi_{j-1} -0.45\phi_{j-2} = 0,\ \
j=0,1,2,\ldots$$

ACF:\\ $$ \rho(h) - \phi_1\rho(h-1) - \phi_2\rho(h-2) = 0, h =
1,2,\ldots$$ $$ \rho(0) = 1 \ \ \ \ \mbox{see example 2, page
103}$$ $$\rho(-1) = \frac{\phi_1}{1 - \phi_2}, \ \ \rho(1) =
\rho(-1)$$ see plots.
\item[c] same idea as $a \& b.$


\end{description}

\textbf{2.7} ARMA(1,1) $$x_t = \phi x_{t-1} + \theta w_{t-1} +
w_t, \ \ \ \ |\phi| <1$$ from (2.43) $$r(h) - \phi r(h-1) = 0, \ \
\ \ \ h = 2,3,\ldots$$ so the general solutions is $r(h) = c
\phi^n, \ \ \ \ h = 1,2,\ldots$\\

solve for $c$ using (2.44)
\begin{align*}
r(0) &= \phi r(1) + \sigma_w^2 [\theta_0\psi_0 + \theta_1\psi_1] \
\ \ \ \theta_0 = \psi_0 =1\\ r(0) &= \phi r(1) + \sigma_w^2
[1+\theta(\phi+ \theta)] \ \ \ \ \mbox{see example $2.10, \psi_1 =
\phi+\theta$}\\ r(0) &= \phi r(1) + \sigma_w^2 [1+\theta\phi+
\theta^2] \\ r(0) &= \phi r(1) + \sigma_w^2 [1+\theta\phi+
\theta^2] \\ r(1) &= \phi r(0) + \sigma_w^2 \theta_1 \phi_0\\ r(1)
&= \phi r(0) + \sigma_w^2 \theta\\
\end{align*}
solving for $r(0) \& r(1),$
\begin{align*}
r(0) & = \sigma_w^2 \frac{1+2\theta\phi + \theta^2}{1 - \phi^2}\\
r(1) & = \sigma_w^2 \frac{(1+\theta\phi)(\phi + \theta)}{1 -
\phi^2}
\end{align*}

because $r(1) = c r(0) \Longrightarrow c = \frac{r(1)}{\phi}$ so:
$$ r(h) = r(1) \phi^{h-1}$$ or $$ \rho(h) =
\frac{(1+\theta\phi)(\phi + \theta)}{1
+2\theta+\theta^2}\phi^{h-1}, \ \ \ \ h\geq 1$$ ARMA(1,1), AR(1)\\
page $92,\rho(h) = \phi^h, \ \ \ h \geq 0,$\\

ARMA(0,1), MR(1)\\ page $96$,
\begin{align*}
\rho(h) &=\frac{\theta}{1+\theta^2}&h=1 \\ &=0&h>1
\end{align*}
plots $\phi = 0.6, \theta = 0.9$ see $2.8.$

\textbf{2.12} Cardiovascular mortality:
\begin{align*}
&M_t = 11.45 + 0.43M_{t-1} + 0.44M_{t-2} + w_t\\ &Var(w_t) =
32.5\\ &M^n_{n+m} = 11.45 + 0.45M^{n}_{n+m-1} + 0.44M^{n}_{n+m-2}
\end{align*}
using 2.78 $$P^n_{n+m} = \sigma_{w}^2 \sum_{j=0}^{m-1} \psi_j^2.$$


\textbf{2.13}
\begin{description}
\item[a]
$$E[(y-g(x) )^2] = E[E[(y-g(x) )^2|x]]$$
\begin{align*}
E[(y-g(x) )^2|x] &= E[(y-E[y|x]+ E[y|x] - g(x) )^2|x]\\ & =
E[(y-E[y|x])^2|x] E[(E[y|x]-g(x))^2|x] +
2E[(y-E[y|x])(E[y|x]-g(x))|x]
\end{align*}
$ E[y|x] - g(x ) \ \ \ \mbox{is a constant},$ so it's easy to
show: $$ E[(y-E[y|x])(E[y|x]-g(x))|x] = 0.$$ and as a result:
$$E[(y-g(x) )^2|x] \geq E[(y-E[y|x] )^2|x]$$ or $$ E[(y-g(x) )^2]
\geq E[(y-E[y|x] )^2]$$ and $E[(y-g(x))^2]$ is minimized when
$g(x) = E[y|x].$

\item[b]
$$ y = x^2 + z$$ To minimize MSE, set:
\begin{align*}
g(x) &= E[x^2 + z|x]\\ &=E[x^2] + E[z|x]\\ &=1
\end{align*}

$$ E(y) = E(x^2) + E(y) = var(x) + 0 =1$$

\begin{align*}
\mbox{MSE} &= E[(y-g(x))^2]\\ &= E[(y-1)^2]\\ &=var(y) \\
&=var(x^2 + y) \\ &=var(x^2) + var(y) \\ &=1
\end{align*}

\item[c]
\begin{align*}
\mbox{MSE} &= E[(y-a-bx)^2]\\ &=E[y^2] - 2aE[y] - 2bE[xy] + a^2 +
2abE[x] + b^2E[x^2]
\end{align*}

differentiate by $a$ and $b,$ and set both partial derivatives to
zeros, we get: $$ a =E[y]-bE[x]= 1,\ \ \ \ \
b=\frac{E[xy]}{E[x^2]}=0$$ so $MSE = E[(y-1)^2].$


\end{description}

\rm
\end{document}
