%  Weighted Least Squares and Generalized Least Squares for STA302

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% Comment this out for handout
\title{Centered Explanatory Variables\footnote{See last slide for copyright information.}}
\subtitle{STA302 Fall 2020}
\date{} % To suppress date

\begin{document}

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\begin{frame}
  \titlepage
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Overview}
\tableofcontents
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The Centered Model}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Center the explanatory variables }
\framesubtitle{By subtracting off the sample mean}  \pause
\begin{itemize}
    \item Replace $x_{ij}$ with $x_{ij}-\overline{x}_j$\pause, expressing each explanatory variable as a deviation from its mean. \pause
    \item Can be useful at times.
\end{itemize}
\end{frame}

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\begin{frame}
\frametitle{Simple Regression}
%\framesubtitle{} 
\begin{center}
\includegraphics[width=3in]{centershift1}
\end{center}
\end{frame}

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\begin{frame}
\frametitle{Simple Regression}
%\framesubtitle{} 
\begin{center}
\includegraphics[width=3in]{centershift2}
\end{center}
\end{frame}

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\begin{frame}
\frametitle{It looks like}
%\framesubtitle{} 
\begin{columns}  
\column{0.5\textwidth}
\begin{center}
\includegraphics[width=2.5in]{centershift3}
\end{center} \pause
\column{0.5\textwidth}
  \begin{itemize}
    \item Estimated slopes will be unaffected.
    \item Estimated intercepts \emph{will} be affected. \pause
    \item $\widehat{y}_i$ should be unaffected. \pause
    \item $\widehat{\epsilon}_i$ should be unaffected. \pause
    \item If so, prediction intervals and $R^2$ should be unaffected. \pause
    \item And tests for slopes should be unaffected.
  \end{itemize}
\end{columns}
\end{frame}

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\begin{frame}
\frametitle{Interpretation}
%\framesubtitle{} 
\begin{columns}  
\column{0.5\textwidth}
\begin{center}
\includegraphics[width=2.5in]{CenteredScatter}
\end{center}
\column{0.5\textwidth}
  \begin{itemize}
    \item Having the $y$ axis go through the data can make the intercept more meaningful. \pause
    \item Suppose $x$ is age, and $y$ is weight loss in an exercise program. \pause
    \item Question: Is any weight loss to be expected for a person of average age? \pause
    \item $H_0: \beta_0 = 0$ \pause is tested automatically. \pause
    \item Testing $H_0: \beta_0 + \beta_1\overline{x}= 0$ requires more effort.
  \end{itemize}
\end{columns}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{The Model for Simple Regression}
%\framesubtitle{} 
\begin{eqnarray*}
    y_i & = & \beta_0 + \beta_1 x_i + \epsilon_i \\ \pause
& = & \beta_0 + \beta_1 x_i {\color{red} - \beta_1\overline{x} + \beta_1\overline{x}} 
    + \epsilon_i \\ \pause
& = & (\beta_0 + {\color{red}\beta_1\overline{x}})
    + \beta_1 (x_i - {\color{red}\overline{x}}) + \epsilon_i \\ \pause
& = &  ~~~~ \, \alpha_0 ~~~~~~+ \alpha_1 (x_i-\overline{x}) + \epsilon_i
\end{eqnarray*} 

\vspace{3mm}
The intercept is affected by centering, but the slope is not.
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Center all the predictor variables} \pause
%\framesubtitle{}
%\framesubtitle{} 
\begin{eqnarray*}
    y_i & = & \beta_0 + \beta_1 x_{i,1} + \cdots + \beta_k x_{i,k} + \epsilon_i \\ \pause
        & = &  \beta_0 + \beta_1 \overline{x}_1 + \cdots + \beta_{k} \overline{x}_k \\
        &   &   + \beta_1 (x_{i,1}-\overline{x}_1) + \cdots 
                + \beta_{k} (x_{i,k}-\overline{x}_{k}) + \epsilon_i \\  \pause
        & = & \alpha_0 + \alpha_1 (x_{i,1}-\overline{x}_1) + \cdots 
                + \alpha_{k} (x_{i,k}-\overline{x}_{k}) + \epsilon_i  
\end{eqnarray*}  \pause
with
\begin{itemize}
    \item[] $\alpha_0 = \beta_0 + \beta_1 \overline{x}_1 + \cdots + \beta_{k} \overline{x}_k$.  \pause
    \item[] $\alpha_j=\beta_j$ for $j = 1, \ldots, k$ \pause
\end{itemize}

\vspace{3mm}
\begin{itemize}
    \item The intercept is affected by centering, but the slopes are not. \pause
    \item You don't have to center all the $x$ variables.
\end{itemize}




\end{frame}

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\begin{frame}
\frametitle{Dummy Variable Regression}
\framesubtitle{Just center the covariate(s)} \pause
\begin{eqnarray*}
    y_i & = & \beta_0 + \beta_1 x_i  
              + \beta_2 d_{i,1} +  \beta_3 d_{i,2} + \epsilon_i \\ \pause
& = & \beta_0 + \beta_1 x_i {\color{red} - \beta_1\overline{x} + \beta_1\overline{x}} 
              + \beta_2 d_{i,1} +  \beta_3 d_{i,2} + \epsilon_i \\ \pause
& = & (\beta_0 + {\color{red}\beta_1\overline{x}})
    + \beta_1 (x_i - {\color{red}\overline{x}}) 
    + \beta_2 d_{i,1} +  \beta_3 d_{i,2} + \epsilon_i \\ \pause
& = &  ~~~~ \, \alpha_0 ~~~~~~+ \alpha_1 (x_i-\overline{x}) 
              + \alpha_2 d_{i,1} +  \alpha_3 d_{i,2} + \epsilon_i \\ \pause
\end{eqnarray*} 

\vspace{3mm}
Slopes are not affected by centering.
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Parallel Regression Lines}
%\framesubtitle{} 
\begin{center}
%{\footnotesize
\begin{tabular}{|c|c|c|c|} \hline
Drug    & $d_1$ & $d_2$ &  $E(y|\mathbf{x}) = \alpha_0+\alpha_1(x-\overline{x}) +     
                            \alpha_2d_1+\alpha_3d_2$\\ \hline
A       &   1   &   0   &  $(\alpha_0+\alpha_2)
                                     +\alpha_1(x-\overline{x})$    \\ \hline
B       &   0   &   1   &  $(\alpha_0+\alpha_3)
                                     +\alpha_1(x-\overline{x})$    \\ \hline
Placebo &   0   &   0   &   ~~~~~$\alpha_0$~~~~
                                     $+\alpha_1(x-\overline{x})$    \\ \hline
\end{tabular} \pause
%} % End size
\end{center} 
\vspace{3mm}
Could describe the estimated intercepts as  ``adjusted means," or ``corrected means."

\end{frame} 

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\begin{frame}
\frametitle{Interactions}
%\framesubtitle{} 
\begin{center}
\includegraphics[width=3in]{Cross3}
\end{center}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Interactions}
%\framesubtitle{} 
\begin{center}
\includegraphics[width=3in]{Cross3b}
\end{center}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Interactions}
{\Large
\begin{center}
\begin{tabular}{|c|c|c|c|} \hline
Group & $d_1$ & $d_2$ &  $E(y|\mathbf{x})$ \\ \hline
1     &   1 &     0  & $(\beta_0+\beta_2) + (\beta_1+\beta_4) (x-\overline{x})$ \\ \hline
2     &   0 &     1  & $(\beta_0+\beta_3) + (\beta_1+\beta_5) (x-\overline{x})$ \\ \hline
3     &   0 &     0  & $~~~~~\beta_0 ~~~~+ ~~~~~\beta_1 ~~~~(x-\overline{x})$ \\ \hline
\end{tabular}
\end{center} \pause
} % End size
\begin{itemize}
    \item What happens at $x = \overline{x}$? \pause
    \item If you are interested in estimating or testing for differences at some other point, it might be easiest to subtract that value from $x$ instead.
\end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Estimation and Testing}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Estimation and Testing} \pause
%\framesubtitle{} 
\begin{itemize}
    \item Have 
        \begin{itemize}
            \item[] $\alpha_0 = \beta_0 + \beta_1 \overline{x}_1 + \cdots + \beta_{k} \overline{x}_k$.  
            \item[] $\alpha_j=\beta_j$ for $j = 1, \ldots, k$
        \end{itemize} \pause
    \item Will have 
        \begin{itemize}
            \item[] $\widehat{\alpha}_0 = \widehat{\beta}_0 + \widehat{\beta}_1 \overline{x}_1 + \cdots + \widehat{\beta}_{k} \overline{x}_k$.  
            \item[] $\widehat{\alpha}_j=\widehat{\beta}_j$ for $j = 1, \ldots, k$ \pause
        \end{itemize}
    \item $\widehat{\mathbf{y}}$ will be unaffected. \pause
    \item $\widehat{\boldsymbol{\epsilon}}$ will be unaffected.  \pause
    \item Prediction intervals and $R^2$ will be unaffected. \pause
    \item Tests for slopes will be unaffected. 
\end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Re-parameterization} \pause
%\framesubtitle{} 
\begin{itemize}
    \item The mapping 
        \begin{itemize}
            \item[] $\alpha_0 = \beta_0 + \beta_1 \overline{x}_1 + \cdots + \beta_{k} \overline{x}_k$.  
            \item[] $\alpha_j=\beta_j$ for $j = 1, \ldots, k$
        \end{itemize}
is a one-to-one re-parameterization. \pause
    \item Furthermore, it's linear. \pause
    \item Write as matrix multiplication.
\end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Matrix Multiplication}
\framesubtitle{To get $\alpha_0 = \beta_0 + \beta_1 \overline{x}_1 + \cdots + \beta_{k} \overline{x}_k$ and $\alpha_j=\beta_j$ for $j = 1, \ldots, k$} 
\begin{displaymath}
\left(\begin{array}{ccccc}
1      & \overline{x}_1 & \overline{x}_2 & \cdots & \overline{x}_k \\
0      & 1              & 0              & \cdots & 0 \\
0      & 0              & 1              & \cdots & 0 \\
\vdots & \vdots         & \vdots         & \ddots & \vdots \\
0      & 0              & 0              & \cdots & 1
\end{array} \right)
\left(\begin{array}{c}
\beta_0 \\ \beta_1 \\ \beta_2 \\ \vdots \\ \beta_k
\end{array} \right)  
=
\left(\begin{array}{c}
\beta_0 + \beta_1 \overline{x}_1 + \cdots + \beta_{k} \overline{x}_k \\ 
\beta_1 \\ \beta_2 \\ \vdots \\ \beta_k
\end{array} \right)
\end{displaymath}  \pause
This matrix $\uparrow$ definitely has an inverse.
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Inverse}
%\framesubtitle{} 
\begin{columns}  
\column{1.2\textwidth}
{\scriptsize
\begin{displaymath}
\left(\begin{array}{ccccc}
1      & -\overline{x}_1 & -\overline{x}_2 & \cdots & -\overline{x}_k \\
0      & 1              & 0              & \cdots & 0 \\
0      & 0              & 1              & \cdots & 0 \\
\vdots & \vdots         & \vdots         & \ddots & \vdots \\
0      & 0              & 0              & \cdots & 1
\end{array} \right) 
\left(\begin{array}{ccccc}
1      & \overline{x}_1 & \overline{x}_2 & \cdots & \overline{x}_k \\
0      & 1              & 0              & \cdots & 0 \\
0      & 0              & 1              & \cdots & 0 \\
\vdots & \vdots         & \vdots         & \ddots & \vdots \\
0      & 0              & 0              & \cdots & 1
\end{array} \right) 
= 
\left(\begin{array}{ccccc}
1      & 0 & 0 & \cdots & 0 \\
0      & 1              & 0              & \cdots & 0 \\
0      & 0              & 1              & \cdots & 0 \\
\vdots & \vdots         & \vdots         & \ddots & \vdots \\
0      & 0              & 0              & \cdots & 1
\end{array} \right)  
\end{displaymath}  \pause
} % End size
\end{columns}

% Unfortunate visual formatting. Array did not fit on the slide.
\hspace{15mm}
$\mathbf{A}$ \hspace{35mm} $\mathbf{A}^{-1}$  \hspace{11mm} = \hspace{12mm} $\mathbf{I}$   \pause

\vspace{-3mm}
\begin{eqnarray*}
\mathbf{y} & = & \mathbf{X} \boldsymbol{\beta} + \boldsymbol{\epsilon} \\ \pause
& = & \mathbf{X} \mathbf{AA}^{-1}\boldsymbol{\beta} + \boldsymbol{\epsilon} \\ \pause
& = & (\mathbf{X} \mathbf{A})(\mathbf{A}^{-1}\boldsymbol{\beta}) + \boldsymbol{\epsilon} \\ \pause
& = & ~~~\mathbf{W} ~~~~~ \boldsymbol{\alpha} ~~~ \, + \boldsymbol{\epsilon}\pause,
\end{eqnarray*} 
Where $\mathbf{W}$ is the centered $\mathbf{X}$ matrix.
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Does the matrix $\mathbf{A}$ really center the $\mathbf{X}$ matrix?}
\framesubtitle{Just look at row $i$ of $\mathbf{XA}$} \pause 
\begin{eqnarray*}
&&\left( \begin{array}{ccccc}
1 & x_{i1} & x_{i2} & \cdots &  x_{ik} 
\end{array}\right)
\left(\begin{array}{ccccc}
1      & -\overline{x}_1 & -\overline{x}_2 & \cdots & -\overline{x}_k \\
0      & 1              & 0              & \cdots & 0 \\
0      & 0              & 1              & \cdots & 0 \\
\vdots & \vdots         & \vdots         & \ddots & \vdots \\
0      & 0              & 0              & \cdots & 1
\end{array} \right) \\ \pause 
&& 
\rule{0mm}{10mm} =     % Write in by hand and talk through.
\left(\begin{array}{ccccc}
1 & x_{i1}-\overline{x}_1 & x_{i2}-\overline{x}_2 & \cdots &  x_{ik}-\overline{x}_k 
\end{array}\right)
\end{eqnarray*}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{One-to-one linear transformation}
%\framesubtitle{}
The point is that centering the explanatory variables is a one-to-one linear transformation of $\mathbf{X}$ matrix: $\mathbf{W} = \mathbf{AX}$. \pause

\begin{displaymath}
\mathbf{A} =  
\left(\begin{array}{ccccc}
1      & -\overline{x}_1 & -\overline{x}_2 & \cdots & -\overline{x}_k \\
0      & 1              & 0              & \cdots & 0 \\
0      & 0              & 1              & \cdots & 0 \\
\vdots & \vdots         & \vdots         & \ddots & \vdots \\
0      & 0              & 0              & \cdots & 1
\end{array} \right)
\end{displaymath} 
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Centering Just Some of the Variables}
%\framesubtitle{} 

\begin{displaymath}
 \left( \begin{array}{ccccc}
1 & x_{i1} & x_{i2} & \cdots &  x_{ik} 
\end{array}\right)  
\left(\begin{array}{ccccc}
1      & -\overline{x}_1 & -\overline{x}_2 & \cdots & -\overline{x}_k \\
0      & 1              & 0              & \cdots & 0 \\
0      & 0              & 1              & \cdots & 0 \\
\vdots & \vdots         & \vdots         & \ddots & \vdots \\
0      & 0              & 0              & \cdots & 1
\end{array} \right)
\end{displaymath} \pause
\begin{itemize}
    \item To leave variable $j$ uncentered, replace $\overline{x}_j$ with zero. \pause
    \item Rows are still linearly independent. 
\end{itemize} 

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{We have been here before}
\framesubtitle{See Assignment 9, Problem 4}  \pause
\begin{eqnarray*}
     &  & \mathbf{y} = \mathbf{X} \boldsymbol{\beta} + \boldsymbol{\epsilon} \\
    & \iff &  \mathbf{y} = \mathbf{XA A}^{-1} \boldsymbol{\beta} + \boldsymbol{\epsilon} \\
    & \iff &  \mathbf{y} = \mathbf{W} \boldsymbol{\alpha} + \boldsymbol{\epsilon} \pause
\end{eqnarray*}

\begin{eqnarray*}
    \widehat{\boldsymbol{\alpha}} & = & (\mathbf{W}^\prime \mathbf{W})^{-1} \mathbf{W}^\prime \mathbf{y} \\ \pause
    & = & \left((\mathbf{XA})^\prime \mathbf{XA}\right)^{-1} (\mathbf{XA})^\prime \mathbf{y} \\  \pause
    & = & \left( \mathbf{A}^\prime \mathbf{X}^\prime \mathbf{XA}\right)^{-1} 
                 \mathbf{A}^\prime \mathbf{X}^\prime \mathbf{y} \\  \pause
    & = & \mathbf{A}^{-1} (\mathbf{X}^\prime \mathbf{X})^{-1} \mathbf{A}^{\prime-1}
                 \mathbf{A}^\prime \mathbf{X}^\prime \mathbf{y} \\  \pause
    & = & \mathbf{A}^{-1} (\mathbf{X}^\prime \mathbf{X})^{-1} \mathbf{X}^\prime \mathbf{y} \\  \pause
    & = & \mathbf{A}^{-1} \widehat{\boldsymbol{\beta}}
\end{eqnarray*} 
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{$\widehat{\boldsymbol{\alpha}} = \mathbf{A}^{-1} \widehat{\boldsymbol{\beta}}$}
\framesubtitle{Same form as $\boldsymbol{\alpha} = \mathbf{A}^{-1} \boldsymbol{\beta}$: Invariance} \pause
{\small
\begin{displaymath} \hspace{-8mm}
\left(\begin{array}{c}
\widehat{\alpha}_0 \\ \widehat{\alpha}_1 \\ \widehat{\alpha}_2 \\ \vdots \\ \widehat{\alpha}_k
\end{array} \right)  
= 
\left(\begin{array}{ccccc}
1      & \overline{x}_1 & \overline{x}_2 & \cdots & \overline{x}_k \\
0      & 1              & 0              & \cdots & 0 \\
0      & 0              & 1              & \cdots & 0 \\
\vdots & \vdots         & \vdots         & \ddots & \vdots \\
0      & 0              & 0              & \cdots & 1
\end{array} \right)
\left(\begin{array}{c}
\widehat{\beta}_0 \\ \widehat{\beta}_1 \\ \widehat{\beta}_2 \\ \vdots \\ \widehat{\beta}_k
\end{array} \right)  \pause
=
\left(\begin{array}{c}
\widehat{\beta}_0 + \widehat{\beta}_1 \overline{x}_1 + \cdots + \widehat{\beta}_{k} \overline{x}_k \\ 
\widehat{\beta}_1 \\ \widehat{\beta}_2 \\ \vdots \\ \widehat{\beta}_k
\end{array} \right)
\end{displaymath}
} % End size
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Predicted $\mathbf{y}$ for $\mathbf{y} = \mathbf{W} \boldsymbol{\alpha} + \boldsymbol{\epsilon}$}
%\framesubtitle{} 
{\LARGE
\begin{eqnarray*}
    \mathbf{W} \widehat{\boldsymbol{\alpha}} \pause
    &=&  (\mathbf{XA})(\mathbf{A}^{-1} \widehat{\boldsymbol{\beta}}) \\ \pause
    &=& \mathbf{X} \widehat{\boldsymbol{\beta}} \\ \pause 
    &=& \widehat{\mathbf{y}}
\end{eqnarray*} 
} % End size
\begin{itemize}
    \item So $\widehat{\mathbf{y}}$ is unchanged by centering. \pause
    \item This means $\widehat{\boldsymbol{\epsilon}}$, \emph{SSE}, \emph{MSE} and $R^2$ are also unchanged. 
\end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Prediction Intervals are unchanged}
\framesubtitle{$\mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}}  \pm t_{\alpha/2} \, 
\sqrt{\mbox{\small\emph{MSE}} \, (1 + \mathbf{x}_0^\prime 
             (\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{x}_0)}$} \pause
The key is that you need to give it a vector of centered $x$ variables\pause:
$\mathbf{x}_0^{*\prime} = \mathbf{x}_0^\prime\mathbf{A} \pause 
\iff \mathbf{x}_0^* = \mathbf{A}^\prime \mathbf{x}_0$. \pause
\begin{eqnarray*}
&&
\mathbf{x}_0^{*\prime} \widehat{\boldsymbol{\alpha}}  \pm t_{\alpha/2} \, 
\sqrt{\mbox{\small\emph{MSE}} \, (1 + \mathbf{x}_0^{*\prime} 
(\mathbf{W}^\prime \mathbf{W})^{-1}\mathbf{x}_0^*)} \\ \pause
& = &
(\mathbf{x}_0^\prime\mathbf{A})(\mathbf{A}^{-1} \widehat{\boldsymbol{\beta}}) \pause
\pm t_{\alpha/2} \,
\sqrt{\mbox{\small\emph{MSE}} \, (1 + \mathbf{x}_0^\prime\mathbf{A} 
(\mathbf{W}^\prime \mathbf{W})^{-1}\mathbf{A}^\prime \mathbf{x}_0)} \\ \pause
& = &
\mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}} \pause
\pm t_{\alpha/2} \,
\sqrt{\mbox{\small\emph{MSE}} \, (1 + \mathbf{x}_0^\prime\mathbf{A} 
\left((\mathbf{XA})^\prime \mathbf{XA}\right)^{-1}
\mathbf{A}^\prime \mathbf{x}_0)} \\ \pause
& = &
\mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}}
\pm t_{\alpha/2} \,
\sqrt{\mbox{\small\emph{MSE}} \, (1 + \mathbf{x}_0^\prime\mathbf{A} 
\left( \mathbf{A}^\prime \mathbf{X}^\prime \mathbf{XA}\right)^{-1}
\mathbf{A}^\prime \mathbf{x}_0)} \\ \pause
& = &
\mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}}
\pm t_{\alpha/2} \,
\sqrt{\mbox{\small\emph{MSE}} \, (1 + \mathbf{x}_0^\prime\mathbf{A} 
\mathbf{A}^{-1} (\mathbf{X}^\prime \mathbf{X})^{-1} \mathbf{A}^{\prime-1}
\mathbf{A}^\prime \mathbf{x}_0)} \\ \pause
& = &
\mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}}  \pm t_{\alpha/2} \, 
\sqrt{\mbox{\small\emph{MSE}} \, (1 + \mathbf{x}_0^\prime 
             (\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{x}_0)}
\end{eqnarray*} 
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Hypothesis tests: $H_0: \mathbf{C}\boldsymbol{\beta} = \mathbf{t}$}
\framesubtitle{Using $F^* = \frac{(\mathbf{C}\widehat{\boldsymbol{\beta}} -\mathbf{t})^\prime
            (\mathbf{C}(\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{C}^\prime)^{-1}
            (\mathbf{C}\widehat{\boldsymbol{\beta}} - \mathbf{t})}
           {q \, \mbox{\footnotesize\emph{MSE}}}$} \pause 
{\small
\begin{eqnarray*}
    \mathbf{C}\boldsymbol{\beta} = \mathbf{t} 
    & \iff & (\mathbf{CA})(\mathbf{A}^{-1}\boldsymbol{\beta}) = \mathbf{t} \\ \pause
    & \iff & (\mathbf{CA}) \boldsymbol{\alpha} = \mathbf{t} \\ 
\end{eqnarray*} \pause

\vspace{-6mm}

Look at the numerator of $F^*$ for the centered data. \pause
\begin{eqnarray*}
&& 
(\mathbf{CA}\widehat{\boldsymbol{\alpha}} -\mathbf{t})^\prime
            (\mathbf{CA}
            (\mathbf{W}^\prime\mathbf{W})^{-1} 
            (\mathbf{CA})^\prime)^{-1}
            (\mathbf{CA}\widehat{\boldsymbol{\alpha}} - \mathbf{t}) \\ \pause
&=& 
(\mathbf{CA}\mathbf{A}^{-1} \widehat{\boldsymbol{\beta}} -\mathbf{t})^\prime
            (\mathbf{CA}
            { \color{red}(\mathbf{W}^\prime\mathbf{W})^{-1} }
            \mathbf{A}^\prime \mathbf{C}^\prime)^{-1}
            (\mathbf{CA}\mathbf{A}^{-1} \widehat{\boldsymbol{\beta}} - \mathbf{t}) \\ \pause
&=& 
(\mathbf{C}\widehat{\boldsymbol{\beta}} -\mathbf{t})^\prime
            (\mathbf{CA}
            { \color{red}\mathbf{A}^{-1} (\mathbf{X}^\prime \mathbf{X})^{-1} \mathbf{A}^{\prime-1} }
            \mathbf{A}^\prime \mathbf{C}^\prime)^{-1}
            (\mathbf{C} \widehat{\boldsymbol{\beta}} - \mathbf{t}) \\ \pause
&=&
(\mathbf{C}\widehat{\boldsymbol{\beta}} -\mathbf{t})^\prime
            (\mathbf{C}(\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{C}^\prime)^{-1}
            (\mathbf{C}\widehat{\boldsymbol{\beta}} - \mathbf{t})
\end{eqnarray*} \pause

\vspace{-5mm}

\begin{itemize}
    \item This is the numerator for the uncentered data, so the test statistics are equal. \pause
    \item If the hypothesis does not involve $\alpha_0$, you don't need to transform $\mathbf{C}$.
\end{itemize}
} % End size
\end{frame}

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\begin{frame}
\frametitle{Summary}
\framesubtitle{A simple story, in spite of all the technical details}  \pause
\begin{itemize}
    \item Centering some or all of the explanatory variables can be helpful. \pause
    \item Only the intercept is affected. \pause
    \item There is no effect on predicted $y$, residuals, $R^2$, or prediction intervals. \pause
    \item There is no effect on tests and confidence intervals, unless the intercept is involved. 
\end{itemize}
\end{frame}

% I guess I could ask about CI in homework or quiz.

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\begin{frame}
\frametitle{Copyright Information}

This slide show was prepared by  \href{http://www.utstat.toronto.edu/~brunner}{Jerry Brunner},
Department of Statistics, University of Toronto. It is licensed under a 
\href{http://creativecommons.org/licenses/by-sa/3.0/deed.en_US}
     {Creative Commons Attribution - ShareAlike 3.0 Unported License}. Use any part of it as you like and share the result freely. The \LaTeX~source code is available from the course website:
\href{http://www.utstat.toronto.edu/~brunner/oldclass/302f20} {\small\texttt{http://www.utstat.toronto.edu/$^\sim$brunner/oldclass/302f20}}

\end{frame}


\end{document}

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# Generating Centershift1.pdf and Centershift2.pdf -- apparent motion?

rm(list=ls())
set.seed(9999)
n = 12; mu = 75; sig=6
x1 = round(rnorm(n,mu,sig)); x2 = round(rnorm(n,mu,sig)); x3 = round(rnorm(n,mu,sig))
eps1 = round(rnorm(n,0,sig)); eps2 = round(rnorm(n,0,sig)); eps3 = round(rnorm(n,0,sig))
y1 = round(-mu + 2*x1 + eps1)
y2 = round(x2 + eps2); y2[x2<mu] = mu + eps2[x2<mu]
y3 = round(x3 + eps3)
x = c(x1,x2,x3); y = c(y1,y2,y3) + 2

plot(x,y,xlim=c(-25,100),ylim=c(-10,125),pch=20,frame.plot=F, 
axes=F, xlab='', ylab='')
title(expression(paste('Centering x by subtracting off ',bar(x))))
# Draw new axes; length is the length of the arrowhead.
arrows(x0=-25,y0=0,x1=100,y1=0,length=0.1,code=3,col='red') # Draw x axis
arrows(x0=0,y0=-5,x1=0,y1=125,length=0.1,code=2,col='red') # Draw y axis
# Label axes
text(102,0,'x',col='red'); text(0,128,'y',col='red')
# Plot regression line for uncentered model
ucmod = lm(y~x)
xpts = data.frame(x=c(50,100))
lines(c(50,100),predict(ucmod,xpts))
# Indicate location of x-bar
mx = c(mean(x),mean(x)); my = c(-5,115)
lines(mx,my,col='red',lty=2)
text(mean(x),-10,expression(bar(x)),col='red')
text(0,-10,0,col='red')
# Now save centershift1.pdf

# Repeating but producing empty plot
plot(x,y,xlim=c(-25,100),ylim=c(-10,125),pch=..,frame.plot=F, 
axes=F, xlab='', ylab='')
title(expression(paste('Centering x by subtracting off ',bar(x))))
# Draw new axes; length is the length of the arrowhead.
arrows(x0=-25,y0=0,x1=100,y1=0,length=0.1,code=3,col='red') # Draw x axis
arrows(x0=0,y0=-5,x1=0,y1=125,length=0.1,code=2,col='red') # Draw y axis
# Label axes
text(102,0,'x',col='red'); text(0,128,'y',col='red')
# Indicate location of x-bar
mx = c(mean(x),mean(x)); my = c(-5,115)
lines(mx,my,col='red',lty=2)
text(mean(x),-10,expression(bar(x)),col='red')
text(0,-10,0,col='red')
# Plot shifted points and regression line
cx = x-mean(x)
points(cx,y,pch=20 )
cmod = lm(y~cx)
cxpts = data.frame(cx=c(-25,25))
lines(c(-25,25),predict(cmod,cxpts))
# Now save centershift2.pdf


# This generates CenterShift3.pdf
rm(list=ls())
set.seed(9999)
n = 12; mu = 75; sig=6
x1 = round(rnorm(n,mu,sig)); x2 = round(rnorm(n,mu,sig)); x3 = round(rnorm(n,mu,sig))
eps1 = round(rnorm(n,0,sig)); eps2 = round(rnorm(n,0,sig)); eps3 = round(rnorm(n,0,sig))
y1 = round(-mu + 2*x1 + eps1)
y2 = round(x2 + eps2); y2[x2<mu] = mu + eps2[x2<mu]
y3 = round(x3 + eps3)
x = c(x1,x2,x3); y = c(y1,y2,y3) + 2

plot(x,y,xlim=c(-25,100),ylim=c(-10,125),pch=20,frame.plot=F, 
axes=F, xlab='', ylab='')
title(expression(paste('Centering x by subtracting off ',bar(x))))
# Draw new axes; length is the length of the arrowhead.
arrows(x0=-25,y0=0,x1=100,y1=0,length=0.1,code=3,col='red') # Draw x axis
arrows(x0=0,y0=-5,x1=0,y1=125,length=0.1,code=2,col='red') # Draw y axis
# Label axes
text(102,0,'x',col='red'); text(0,128,'y',col='red')
# Plot regression line for uncentered model
ucmod = lm(y~x)
xpts = data.frame(x=c(50,100))
lines(c(50,100),predict(ucmod,xpts))
# Indicate location of x-bar
mx = c(mean(x),mean(x)); my = c(-5,115)
lines(mx,my,col='red',lty=2)
text(mean(x),-10,expression(bar(x)),col='red')
text(0,-10,0,col='red')
# Plot shifted points and regression line
cx = x-mean(x)
points(cx,y,pch=20 )
cmod = lm(y~cx)
cxpts = data.frame(cx=c(-25,25))
lines(c(-25,25),predict(cmod,cxpts))

# This generates CenteredScatter.pdf
# Based on same data
rm(list=ls())
set.seed(9999)
n = 12; mu = 75; sig=6
x1 = round(rnorm(n,mu,sig)); x2 = round(rnorm(n,mu,sig)); x3 = round(rnorm(n,mu,sig))
eps1 = round(rnorm(n,0,sig)); eps2 = round(rnorm(n,0,sig)); eps3 = round(rnorm(n,0,sig))
y1 = round(-mu + 2*x1 + eps1)
y2 = round(x2 + eps2); y2[x2<mu] = mu + eps2[x2<mu]
y3 = round(x3 + eps3)
x = c(x1,x2,x3)
# Transform y to make it more like weight loss
y = -c(y1,y2,y3)/8 + 10.5
cx = x-mean(x) # Center x
plot(cx,y,pch=20, frame.plot=F, xlim=c(-15,15), # ylim=c(-6,3),
xlab=expression(paste(x-bar(x))),ylab='Weight Loss in Pounds'); abline(v=0)
creg = lm(y~cx); abline(reg=creg,lty=2)

###### Plots: Crossing lines

# Cross in data range. (Also there's truly a curve for blue.)
# This generates Cross3.pdf and Cross3b.pdf
rm(list=ls())
set.seed(9999)
n = 12; mu = 75; sig=6
x1 = round(rnorm(n,mu,sig)); x2 = round(rnorm(n,mu,sig)); x3 = round(rnorm(n,mu,sig))
eps1 = round(rnorm(n,0,sig)); eps2 = round(rnorm(n,0,sig)); eps3 = round(rnorm(n,0,sig))
y1 = round(-mu + 2*x1 + eps1)
y2 = round(x2 + eps2); y2[x2<mu] = mu + eps2[x2<mu]
y3 = round(x3 + eps3)

x = c(x1,x2,x3); y = c(y1,y2,y3) + 2
train = c(numeric(n)+1,numeric(n)+2,numeric(n)+3)

plot(x,y, pch=' ')
points(x1,y1,pch=16,col='red')
points(x2,y2,pch=16,col='blue')
points(x3,y3,pch=16,col='green')

reg1 = lm(y1~x1); abline(reg=reg1,col='red') # abline is so much better
reg2 = lm(y2~x2); abline(reg=reg2,col='blue')
reg3 = lm(y3~x3); abline(reg=reg3,col='green')
title('Effect of Group Depends on x')

# Add a legend. Give it a vector to have multiple lines of legend. Beautiful.
legendtext = c('Group 1','Group 2','Group 3')
textcol = c('red','blue','green'); linetype = c(1,1,1) 
pointchar = c(16,16,16); ptlinecol = c('red','blue','green')
legend('topleft',legendtext,text.col=textcol, ,lty=linetype, 
pch=pointchar, col=ptlinecol)
# Now save Cross3.pdf

# Use lines rather than abline. The line needs to be shorter.
xx = c(mean(x),mean(x)); yy = c(0,110)
lines(xx,yy,lty=2)
text(mean(x),111,expression(bar(x)))
# Save Cross3b.pdf