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\begin{document}

\section{Exercises 4a p.103}
\begin{enumerate}
\setcounter{enumi}{2}
\item \textbf{Done} If $\hat{\lambda}_{H}$ is the least squares estimate of the Lagrange multiplier associated with the constraints $A\beta = c$ show that
\[RSS_{H} - RSS = \sigma^{2} \hat{\lambda}_{H}^{T} var(\hat{\lambda}_{H})^{-1} \hat{\lambda}_{H}\]

\noindent \textit{Proof}\\
From section 3.8.1 we know that
\[\hat{\lambda}_{H} = 2\left[A(X'X)^{-1}A'\right]^{-1}(A\hat{\beta} - c)\]
Then we can calculate $var(\hat{\lambda}_{H})^{-1}$.
\begin{eqnarray}
var(\hat{\lambda}_{H})^{-1} & = & var(2\left[A(X'X)^{-1}A'\right]^{-1}(A\hat{\beta} - c))^{-1} \\
 & = & var(2\left[A(X'X)^{-1}A'\right]^{-1}A\hat{\beta})^{-1} \\
 & = & \frac{1}{4}\left[A(X'X)^{-1}A'\right]A^{-1} var(\hat{\beta})^{-1} \left(\left[A(X'X)^{-1}A'\right]A^{-1}\right)' \\
% & = & \frac{1}{4\sigma^{2}}\left[A(X'X)^{-1}A'\right]A^{-1} X'X A^{-1}\left[A(X'X)^{-1}A'\right] \\
\varphi & = & \frac{1}{4\sigma^{2}}\left[A(X'X)^{-1}A'\right]\left(A (X'X)^{-1} A'\right)^{-1}\left[A(X'X)^{-1}A'\right]
\end{eqnarray}
Then our new expression is
\begin{eqnarray}
RSS_{H} - RSS & = & 4\sigma^{2}(A\hat{\beta} - c)\left[A(X'X)^{-1}A'\right]^{-1} \varphi \left[A(X'X)^{-1}A'\right]^{-1}(A\hat{\beta} - c) \\
& = & (A\hat{\beta} - c) \left(A(X'X)^{-1}\right)^{-1} (A\hat{\beta} - c)' \\
%RSS_{H} - RSS & = & (A\hat{\beta} - c) \left(A(X'X)^{-1}\right)^{-1} (A\hat{\beta} - c)'
\end{eqnarray}
So that we have
\begin{eqnarray}
RSS_{H} - RSS &=& \sigma^{2} \hat{\lambda}_{H}^{T} var(\hat{\lambda}_{H})^{-1} \hat{\lambda}_{H}
\end{eqnarray}\\\\


\setcounter{enumi}{4}
\item \textbf{Wrong} Consider the full rank model $X\beta = (X_{1},X_{2})(\beta_{1}',\beta_{2}')' $ where $X \sim n\times q$.
\begin{enumerate}
\item Obtain a test statistic for $\beta_{2} = 0$. \label{pt.a}

Let
\begin{eqnarray}
RSS_{H} - RSS& = & (A\hat{\beta} - c) \left(A(X'X)^{-1}A'\right)^{-1} (A\hat{\beta} - c)' \label{rss.h.rss}
\end{eqnarray}
Then we need to find the matrix $A$ for the hypothesis $H_{0} : \beta_{2} = 0$.  Then we want to let $c = 0$ from (\ref{rss.h.rss}) and the for the matrix $A$ we have
\begin{eqnarray}
A\beta & = & A_{1}\beta_{1} + A_{2}\beta_{2} \\
&=& A_{2}\beta_{2} = 0
\end{eqnarray}
Since $\beta_{2} \sim q\times 1$, and for our particular $H_{0}$, we have $A_{1} = 0$ and $A_{2} \sim (q-1) \times q = I_{(q-1)\times q}$.  Then we have 

\item Find $E[RSS_{H} - RSS]$. \label{pt.b}

\end{enumerate}
\end{enumerate}

\section{Exercises 4b p.109}
\begin{enumerate}
\item \textbf{Done} Let $Y_{i} = \beta_{0} + \beta_{1}x_{i,1} + \cdots + \beta_{p-1}x_{i,p-1} + \varepsilon_{i}$, $i=1,2,\ldots,n$, where the $\varepsilon_{i}$ are independent $N(0,\sigma^{2})$.  Prove that the $F$-statistic for testing the hypothesis $H : \beta_{r} = \beta_{r+1} = \cdots = \beta_{p-1} = 0$ $(0 < r \leq p-1)$ is unchanged if a constant, $c$, is subtracted from each $Y_{i}$.\\

\noindent \textit{Proof}\\
Since $r > 0$, we could rewrite the model
\begin{eqnarray}
Y_{i} - c = \beta_{0} + \beta_{1}x_{i,1} + \cdots + \beta_{p-1}x_{i,p-1}
\end{eqnarray}
as
\begin{eqnarray}
Y_{i} & =& (\beta_{0} + c) + \beta_{1}x_{i,1} + \cdots + \beta_{p-1}x_{i,p-1}\\
Y_{i} &=& \beta_{0}^{*} + \beta_{1}x_{i,1} + \cdots + \beta_{p-1}x_{i,p-1}
\end{eqnarray}
and we are done.
%We will use matrix notation for the model mentioned above, so that we have $Y = X\beta + \varepsilon$.  Then we have the following definitions
%\begin{eqnarray}
%F & = & \frac{RSS_{H} - RSS / q}{ RSS / (n-p)} \label{rssh.rss}\\
%RSS_{H} - RSS & = & (A\hat{\beta} - c)\left[A(X'X)^{-1}A'\right](A\hat{\beta} - c)'\\
%RSS & = & Y'Y - \hat{\beta}'X'X\hat{\beta} \label{rss}\\
%\hat{\beta} & = & (X'X)^{-1}X'Y
%\end{eqnarray}
%And we note that if $\Omega = \mathcal{C}(X)$ and $P = X(X'X)^{-1}X'$ is the projection matrix into $\Omega$, then $(I_{n} - P)$ is the projection matrix into $\Omega^{\bot}$.  Since $1_{n}$ is in $\mathcal{C}(X)$  we have $(I_{n} - P)c1_{n} = 0$ so that $(Y-c)'(I_{n} - P)(Y-c) = Y'(I_{n} - P)Y$.  
%We also know from (\ref{rss}) that $RSS = Y'(I_{n} - P)Y$, and therefore (\ref{rssh.rss}) is not affected by subtracting a constant from $Y$.
%If we choose $r=0$, we have $A = I_{n}$ and $c = 0$ and we can rewrite (\ref{rssh.rss}).  
%\begin{eqnarray}
%F & = & \frac{Y'PY}{Y'(I_{n} - P)Y}\left(\frac{n-p}{q}\right)
%\end{eqnarray}
%We now only need that shifting $Y$ does not affect the projection onto $\Omega$ using the matrix $P$.  But we have
%\begin{eqnarray}
%(Y-c)'P(Y-c) & = & (Y-c)'(PY - Pc) \\
%& = & (Y-c)'(PY - c) \\
%& = & Y'PY - c'PY - Y'c + c'c\\
%& = & Y'PY - 2Y'c + c'c
%\end{eqnarray}

\item \textbf{Done} Let $Y_{i} = \beta_{0} + \beta_{1}x_{i} + \varepsilon_{i}$, $(i = 0,1,\cdots,n)$. where $\varepsilon_{i} \sim N(0,\sigma^{2})$.
\begin{enumerate}
\item Show that the correlation coefficient of $\hat{\beta_{0}}$ and $\hat{\beta_{1}}$ is $-n\bar{x} / (\sqrt{n\sum x_{i}^{2}})$.\\

\noindent \textit{Proof}\\
We know that $var(\hat{\beta}) = \sigma^{2}(X'X)^{-1}$.  Where
\begin{eqnarray}
(X'X)^{-1} &= &\frac{1}{\sum (x_{i} - \bar{x})^{2}}
\left( \begin{array}{cc}
	\frac{1}{n}\sum x_{i}^{2} & -\bar{x} \\
	-\bar{x} & 1
	\end{array} 
\right) \label{xtxinv} \\
\rho_{\hat{\beta_{0}},\hat{\beta_{1}}} & = & \frac{cov(\hat{\beta_{0}},\hat{\beta_{1}})}{\sqrt{var(\hat{\beta_{0}})var(\hat{\beta_{1}})}}
\end{eqnarray}
Then if
\begin{eqnarray}
\gamma &=& \sum_{i=0}^{n} (x_{i} - \bar{x})^{2} \\
\end{eqnarray}
We have
\begin{eqnarray}
\rho & = & \frac{\sigma^{2}(-\bar{x})/\gamma}{\sqrt{\sigma^{4}(\frac{1}{n}\sum x_{i}^{2})/\gamma)(1/\gamma)}}\\
 & = & \frac{-\bar{x}}{\sqrt{(\frac{1}{n}\sum x_{i}^{2})}}\\
 & = & \frac{-n\bar{x}}{\sqrt{n\sum x_{i}^{2})}}
\end{eqnarray}
and we are done.

\item Derive an $F$-statistic for testing $H : \beta_{0} = 0$.\\

\noindent \textit{Proof}\\
Here we will use equation (\ref{rss.h.rss}) with the following definitions:
\begin{eqnarray}
A &=& \left( \begin{array}{ll}
	1 & 0
	\end{array} \right) \\
c &=& 0
\end{eqnarray}
Then the $F$-statistic is defined as
\begin{eqnarray}
F & = & \frac{\hat{\beta_{0}}\sum x_{i}^{2}}{S^{2} \sum(x_{i} - \bar{x})^{2}}
\end{eqnarray}

\end{enumerate}
\item \textbf{Partial} Given $\bar{x} = 0$, derive an $F$-statistic for $H : \beta_{0} = \beta_{1}$.  And show it is equivalent to a certain $t$-test.\\

Let $X \sim n\times 2$, $A \sim 1\times 2$ where
$A = \left( \begin{array}{ll} 1 & -1 \end{array} \right), c = 0$.
The $F$-statistic will be defined as
\begin{eqnarray}
F &=& \frac{RSS_{H} - RSS / q}{RSS / (n-p)} \label{fstat}
\end{eqnarray}
Where $RSS_{H} - RSS$ is defined as in (\ref{rss.h.rss}).  And we have
\begin{eqnarray}
(X'X)^{-1} & = &
\left( \begin{array}{ll} 
	\frac{1}{n} & 0 \\
	0 & \frac{1}{\sum x_{i}^{2} }
\end{array} \right) \\
A(X'X)^{-1}A' &=& \frac{1}{n} + \frac{1}{\sum x_{i}^{2}} = \frac{\sum x_{i}^{2} + n}{n\sum x_{i}^{2}} \\
A\hat{\beta} - c &=& \hat{\beta_{0}} - \hat{\beta_{1}}\\
\frac{RSS}{n-p} &=& S^{2}
\end{eqnarray}
Then the $F$-statistic is
\begin{eqnarray}
F &=& \frac{(\hat{\beta_{0}} - \hat{\beta_{1}})^{2}n\sum x_{i}^{2} }{S^{2}(\sum x_{i}^{2} + n)}
\end{eqnarray}

\item \textbf{Done} Let
\begin{eqnarray}
Y = \left( \begin{array}{rr}
	1 & 1 \\
	0 & 2 \\
	-1 & 1
	\end{array} \right) \theta + \varepsilon
\end{eqnarray}
And find an $F$-statistic for $H : \theta_{1} = 2\theta_{2}$

Then we take the $F$-statistic as defined in (\ref{fstat}) with $A = \left( \begin{array}{ll} 1 & -2 \end{array} \right)$ and $c = 0$.  Where
\begin{eqnarray}
(A\hat{\theta} - c) &=& \theta_{1} - 2\theta_{2} \\
(X'X)^{-1} &=& \left( \begin{array}{cc} 
	1/2 & 0 \\
	0 & 1/6
	\end{array} \right) \\
A(X'X)^{-1}A &=& 7/6 \\
F &=& \frac{(\hat{\theta_{1}} - 2\hat{\theta_{2}})^{2}}{(7/6)S^{2}}
\end{eqnarray}

\item \textbf{Done (Ryan)} Given $Y = \theta + \varepsilon$, where $\varepsilon \sim N(0,\sigma^{2}I_{4})$ and $\theta_{1} +\theta_{2} +\theta_{3} +\theta_{4} = 0$, show that the $F$-statistic for testing $H : \theta_{1} = \theta_{3}$ is
\begin{equation*}
\frac{2(Y_{1} - Y_{3})^{2}}{(Y_{1} + Y_{2} + Y_{3} + Y_{4})^{2}}
\end{equation*}

According to Dr. Horvath, what they mean by $\mathbf{Y} = \boldsymbol{\theta} + \boldsymbol{\varepsilon}$ is the following:
\[
	\begin{array}{l}
	Y_1 = \theta_1 + \varepsilon_1\\
	Y_2 = \theta_2 + \varepsilon_2\\
	Y_3 = \theta_3 + \varepsilon_3\\
	Y_4 = \theta_4 + \varepsilon_4
	\end{array}
\]
But with the restriction $\theta_1+\theta_2+\theta_3+\theta_4=0$ we can write it as:
\[
	\begin{array}{l}
	Y_1 = \theta_1 + \varepsilon_1\\
	Y_2 = \theta_2 + \varepsilon_2\\
	Y_3 = \theta_3 + \varepsilon_3\\
	Y_4 = -(\theta_1+\theta_2+\theta_3) + \varepsilon_4
	\end{array}
\]
So we can write the model as $\mathbf{Y} = \mathbf{X} \boldsymbol{\theta} + \boldsymbol{\varepsilon}$, where:
\[
\mathbf{X}= \left(
	\begin{array}{ccc}
	1&0&0 \\
	0&1&0 \\
	0&0&1 \\
	-1&-1&-1
	\end{array}
\right)
\qquad \mathrm{ and } \qquad
\boldsymbol{\theta}= \left(
	\begin{array}{c}
	\theta_1\\
	\theta_2\\
	\theta_3\\
	\end{array}
\right)
\]
Now with $\mathbf{A}=(1,0,-1)$ you can find the $F$ statistic for the test and you'll get the answer in the book.  I didn't use the Lagrange multiplier hint (or at least I didn't realize I was using it if I did), but doing it this way requires inverting a 3x3 matrix and I don't know if he'll ask one that requires that.
\end{enumerate}

\section{Exercises 4c p.113}
\begin{enumerate}
\item \textbf{Done} Suppose that $\beta_{1} = \cdots = \beta_{p-1} = 0$.
Find the distribution of $R^{2}$ and hence prove that $E[R^{2}] = (p-1)/(n-1)$.

From example $4.8$ in the book, we know that under $H : \beta_{1} = \cdots = \beta_{p-1} = 0$, we have the $F$-statistic
\[F = \frac{R^{2}}{1-R^{2}}\frac{n-p}{p-1}\]
Which we rewrite
\begin{equation}
R^{2} = \frac{\frac{p-1}{n-p}F}{1 + \frac{p-1}{n-p}F}
\end{equation}
Which is distributed like $Beta((p-1)/2,(n-p)/2))$, and has mean $(p-1)/(n-1)$.

\item \textbf{Partial,Most} For the general linear full-rank regression model, prove that $R^{2}$ and the $F$-statistic for testing $H : \beta_{j} = 0$ $(j \neq 0)$ are independent of the units in which the $Y_{i}$ and the $x_{ij}$ are measured.

This means that we have the new model
\begin{eqnarray}
Y/c &=& XK\beta + \varepsilon
\end{eqnarray}
Where the matrix $K = diag(1,k_{1},\ldots,k_{p-1})$
and we write
\begin{eqnarray}
Y^{*} &=& X^{*}\beta^{*} + \varepsilon^{*}
\end{eqnarray}
Then, to calculate the $F$-statistic we use (\ref{fstat}) with the matrix $A \sim 1 \times p$ where the $j^{th}$ element is $1$ and the rest are $0$ and $c = 0$.  We also note that the matrix $K$ is diagonal and, therefore, symmetric.  Then we use the following definitions to calculate (\ref{fstat})
\begin{eqnarray}
z_{j} &=& [(X'X)^{-1}]_{jj}
\end{eqnarray}
\begin{eqnarray}
(A(X'^{*}X^{*})^{-1}A')^{-1} &=& (A(KX'XK)^{-1}A')^{-1}\\
&=& (AK^{-1}(X'X)^{-1}K^{-1}A')^{-1} \\
&=& \frac{k_{j}^{2}}{z_{j}}
\end{eqnarray}
\begin{eqnarray}
(A\hat{\beta}^{*} - c) & = & \frac{\hat{\beta}_{j}}{ck_{j}}
\end{eqnarray}
\begin{eqnarray}
RSS_{H} - RSS / q &=& \frac{\hat{\beta}_{j}}{c^{2}z_{j}}
\end{eqnarray}
\begin{eqnarray}
\end{eqnarray}
\begin{eqnarray}
\hat{Y}^{*} & = & XK(KX'XK)^{-1}KX'Y/c \\
 & = & XKK^{-1}(X'X)^{-1}K^{-1}KX'Y/c \\
 & = & X(X'X)^{-1}X'Y/c \\
 & = & \hat{Y}/c
\end{eqnarray}
\begin{eqnarray}
RSS & = & (\hat{Y}^{*} - Y^{*})'(\hat{Y}^{*} - Y^{*}) \\
 & = & \frac{RSS}{c^{2}}
\end{eqnarray}
\begin{eqnarray}
\frac{RSS_{H} - RSS / q}{RSS / (n-p)} & = & \frac{\frac{\hat{\beta}_{j}}{c^{2}z_{j}}}{RSS / (n-p)c^{2}} \\
 & = & \frac{\hat{\beta}_{j}}{z_{j}S^{2}}
\end{eqnarray}
Which proves that the $F$-statistic is not affected by how the design points are measured.

\end{enumerate}

\section{Miscellaneous exercises 4 p.117}
\begin{enumerate}
\setcounter{enumi}{1}
\item \textbf{Done} Given the two regression lines
\[Y_{ki} = \beta_{k}x_{i} + \varepsilon_{ki} \hspace*{1.5cm} (k=1,2; i = 1,2,\ldots,n)\]
show that the $F$-statistic for testing $H : \beta_{1} = \beta_{2}$ can be put in the form
\[F = \frac{(\hat{\beta_{1}} - \hat{\beta_{2}})^{2}}{2S^{2}(\sum x_{i}^{2})^{-1}}\]
Obtain $RSS$ and $RSS_{H}$ and verify that
\[RSS_{H} - RSS = \frac{\sum x_{i}^{2} (\hat{\beta_{1}} - \hat{\beta_{2}})^{2}}{2}\]

First we define a combined model
\[Y = X\beta + \varepsilon\]
Where we define
\begin{eqnarray}
Y & = & \left( \begin{array}{c}
	Y_{1,1} \\
	Y_{1,2} \\
	\vdots \\
	Y_{1,n} \\
	Y_{2,1} \\
	Y_{2,2} \\
	\vdots \\
	Y_{2,n}
	\end{array}\right) \\
X & = & \left( \begin{array}{cc}
	x_{1} & 0 \\
	x_{2} & 0 \\
	\vdots & \vdots \\
	x_{n} & 0 \\
	0 & x_{1} \\
	0 & x_{2} \\
	\vdots & \vdots \\
	0 & x_{n} \\
	\end{array}\right) \\
\beta & = & \left( \begin{array}{c}
	\beta_{1} \\
	\beta_{2}
	\end{array} \right)
\end{eqnarray}
Then we follow the regular procedure of using (\ref{fstat}) to define the $F$-statistic.  In this case, use $A = \left( \begin{array}{ll} 1 & -1 \end{array}\right)$ and $c=0$.  Then we have
\begin{eqnarray}
(A(X'X)^{-1}A')^{-1} & = & \frac{\sum x_{i}}{2} \\
(A\hat{\beta} - c) & = & (\hat{\beta}_{1} - \hat{\beta}_{2})
\end{eqnarray}
Which gives us
\begin{eqnarray}
F & = & \frac{\sum x_{i}(\hat{\beta}_{1} - \hat{\beta}_{2})}{2S^{2}}
\end{eqnarray}
And we are done since the second part of this question follows from the first.

\setcounter{enumi}{3}
\item \textbf{Done, Ryan} A series of $n+1$ observations $Y_{i}$ $(i = 1,2,\ldots,n+1)$ are taken from a normal distribution with unknown variance $\sigma^{2}$.  After the first $n$ observations it is suspected that there is a sudden change in the mean of the distribution.  Derive a test statistic for testing the hypothesis that the $(n+1)^{st}$ observation has the same population mean as the previous observations.

For this one, to set it up like a linear model, you can look at the model:
\[
	\begin{array}{l}
	Y_1 = \mu + \varepsilon_1\\
	Y_2 = \mu + \varepsilon_2\\
	\qquad \vdots \\
	Y_n = \mu + \varepsilon_3\\
	Y_{n+1} = \lambda + \varepsilon_{n+1}
	\end{array}
\]
Which we can write as $\mathbf{Y} = \mathbf{X} \boldsymbol{\beta} + \boldsymbol{\varepsilon}$, where:
\[
\mathbf{X}= \left(
	\begin{array}{ccc}
	1&0 \\
	1&0 \\
	\vdots & \vdots \\
	1&0 \\
	0&1
	\end{array}
\right)
\qquad \mathrm{ and } \qquad
\boldsymbol{\beta}= \left(
	\begin{array}{c}
	\mu \\
	\lambda
	\end{array}
\right)
\]
So we can test $H: \mu = \lambda$ using $\mathbf{A} = (1,-1) $.  We get:
\[
(\mathbf{X}^T\mathbf{X})^{-1} = 
	\left(
	\begin{array}{cc}
	1/n&0 \\
	 0&1
	\end{array}
	\right)
\quad \mathrm{ so } \quad
\mathbf{\hat{\beta}} = 
	\left(
	\begin{array}{c}
	\bar{Y}_n \\
	Y_{n+1}
	\end{array}
	\right)
\quad \mathrm{ and } \quad
A(\mathbf{X}^T\mathbf{X})^{-1}A^T= \frac{1}{n}+1
\]
And
\[
RSS/(n-1) = \frac{1}{n-1} 
\left[  
	\left( 
		\begin{array}{c}
		 Y_1 \\
		\vdots \\
		Y_n \\
		 Y_{n+1}
		\end{array}	
	\right) - 
	\left(
		\begin{array}{c}
		\bar{Y}_1 \\
		\vdots \\
		\bar{Y}_n \\
		 Y_{n+1}
		\end{array}
	\right)  
\right]^T
\left[  
	\left( 
		\begin{array}{c}
		 Y_1 \\
		\vdots \\
		Y_n \\
		 Y_{n+1}
		\end{array}	
	\right) - 
	\left(
		\begin{array}{c}
		\bar{Y}_1 \\
		\vdots \\
		\bar{Y}_n \\
		 Y_{n+1}
		\end{array}
	\right)
\right] = S_n^2
\]
So I get:
\[
F=\frac{(\bar{Y}_n - Y_{n+1})^2}{S_n^2 \left(1 + \frac{1}{n} \right)}
\]

Now if you take the square root, it kind of looks like the answer in the back of the book.  If you can get it into a t-distribution, let me know.  It seems like it wouldn't be so bad, but I'm a little burned out.

\end{enumerate}
\section{Miscellaneous exercises 5 p.136}
\begin{enumerate}
\setcounter{enumi}{1}
\item \textbf{Not Done} Prove that $(1 - \alpha/k)^{k} > 1 - \alpha$  $(k > 1)$.
\setcounter{enumi}{5}
\item \textbf{Done} Let $Y_{I} = \beta_{0} + \beta_{1}x_{i} + \varepsilon_{i} (i = 1,2,\ldots,n)$, where the $\varepsilon_{i}$ are independently distributed as $N(0,\sigma^{2})$.  Obtain a set of multiple confidence intervals for all linear combinations $a_{0}\beta_{0} + a_{1}\beta_{1}$ ($a_{0},a_{1}$ not both zero) such that the overall confidence for the set is $100(1-\alpha)\%$.

Use equation (5.22) from the book which gives a confidence interval as
\begin{equation}
x'\hat{\beta} \pm (pF_{p,n-p}^{\alpha})^{1/2}S(x'(X'X)^{-1}x)^{1/2}
\end{equation}
with $x = (a_{0},a_{1})$ and recongizing that $(X'X)^{-1}$ can be defined as in (\ref{xtxinv}).  Then we get
\begin{eqnarray}
x'(X'X)^{-1}x & = & \frac{1}{\varphi}(a_{0}^{2}\sum(x_{i}^{2}/n) - 2a_{0}a_{1}\bar{x} + a_{1}^{2} \\
\varphi & = & \sum(x_{i} - \bar{x})^{2}
\end{eqnarray}
And we are done recognizing that $x'\beta = a_{0}\beta_{1} + a_{1}\beta_{2}$.

\item \textbf{Not Done} In constructing simultaneous confidence intervals for all $x'\beta$, explain why setting $x_{0} \equiv 1$ does not affect the theory.  What modifications to the theory are needed if $\beta_{0} = 0$?
\end{enumerate}

\end{document}
