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\title{Prediction Intervals\footnote{See last slide for copyright information.}}
\subtitle{STA 302 Fall 2020}
\date{} % To suppress date

\begin{document}

\begin{frame}
  \titlepage
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\begin{frame}
\frametitle{Prediction} \pause
%\framesubtitle{} 
\begin{itemize}
    \item You have a data set, and you fit a regression model. \pause
    \item Now you obtain a \emph{new} observation, independently sampled from the same population. \pause
    \item You have $(1,x_{n+1,1}, \ldots, x_{n+1,k})^\prime$. \pause Or $(x_{n+1,1}, \ldots, x_{n+1,k})^\prime$. \pause
    \item Want to predict $y_{n+1}$. \pause
    \item For example, based on the \texttt{cars} data, you want to predict litres per kilometer for a Japanese car 4.52 metres long, weighing 1,295 kilograms. \pause
    \item I wish we could write $(1,x_{n+1,1}, \ldots, x_{n+1,k})^\prime = \mathbf{x}_{n+1}$. \pause
    \item But we will follow the book's notation and call it $\mathbf{x}_0$. 
\end{itemize}
\end{frame}

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\begin{frame}
\frametitle{New observation: $y_0 = \mathbf{x}_0^\prime\boldsymbol{\beta} + \epsilon_0$} \pause
%\framesubtitle{} 
\begin{itemize}
    \item $E(y_0) = \mathbf{x}_0^\prime\boldsymbol{\beta}$. \pause
    \item Estimate $E(y_0)$ with $\mathbf{x}_0^\prime\widehat{\boldsymbol{\beta}}$. \pause
    \item That's a reasonable \emph{prediction} of $y_0$, too. \pause
    \item But the intervals are different. \pause
    \item Prediction intervals are not the same as confidence intervals.  
\end{itemize}
\end{frame}


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\begin{frame}
\frametitle{Prediction intervals versus confidence intervals} \pause
\framesubtitle{Based on $y_0 = \mathbf{x}_0^\prime\boldsymbol{\beta} + \epsilon_0$} 
\begin{itemize}
    \item A confidence interval tries to trap the unknown constant $\mathbf{x}_0^\prime\boldsymbol{\beta}$ with high probability, say $1-\alpha = 0.95$ \pause
    \item Have $\mathbf{a}^\prime \widehat{\boldsymbol{\beta}}  \pm t_{\alpha/2} \, 
\sqrt{\mbox{\small\emph{MSE}} \, \mathbf{a}^\prime 
             (\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{a}}$. \pause Let $\mathbf{a} = \mathbf{x}_0$. \pause
    \item A prediction interval seeks to trap $y_0$, a random variable. \pause
    \item It makes sense that the prediction interval should be wider. \pause
    \item We will have $\mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}}  \pm t_{\alpha/2} \, 
\sqrt{\mbox{\small\emph{MSE}} \, (1 + \mathbf{x}_0^\prime 
             (\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{x}_0)}$.
\end{itemize}
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\begin{frame}
\frametitle{Theorem} 
%\framesubtitle{} 
\begin{itemize}
    \item Assume the usual linear regression model with normal error terms. \pause
    \item Let $y_0 = \mathbf{x}_0^\prime\boldsymbol{\beta} + \epsilon_0$, where 
          $\epsilon_0 \sim N(0,\sigma^2)$, independently of $\epsilon_1, \ldots, \epsilon_n$. \pause
\end{itemize} \vspace{3mm}

A $(1-\alpha)100\%$ prediction interval for $y_0$ is given by

{\LARGE
\begin{displaymath}
    \mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}}  \pm t_{\alpha/2} \, 
\sqrt{\mbox{\emph{MSE}} \, (1 + \mathbf{x}_0^\prime 
             (\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{x}_0)}
\end{displaymath} 
} % End size

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\begin{frame}
\frametitle{Proof will use $t =  \frac{z}{\sqrt{w/\nu}}$} \pause
%\framesubtitle{} 
\begin{itemize}
    \item Predict $y_0$ with $\mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}}$ \pause
          $\sim N(\mathbf{x}_0^\prime \boldsymbol{\beta},\pause \sigma^2
          \mathbf{x}_0^\prime (\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{x}_0)$. \pause 
    \item $y_0 \sim N(\mathbf{x}_0^\prime \boldsymbol{\beta},\sigma^2)$. \pause
    \item And $y_0$ is independent of $\mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}}$. Why? \pause
    \item Error of prediction: $y_0 - \mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}} \pause
    \sim N( \pause 0,  \pause \sigma^2 + 
    \sigma^2 \mathbf{x}_0^\prime (\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{x}_0)$. \pause
    % Okay, it's minus the error of prediction, but so what?
    \item Standardize this to get the $z$ in the numerator of $t$. \pause
\begin{displaymath}
    z = \frac{y_0 - \mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}}}
             {\sqrt{\sigma^2(1 + \mathbf{x}_0^\prime 
             (\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{x}_0)}} \pause
             \sim N(0,1)
\end{displaymath}
    \item And $w = \frac{\small{\mbox{\emph{SSE}}}}{\sigma^2}$ (what else?) \pause
    \item $z$ and $w$ are independent. Why?
\end{itemize}

\end{frame}

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\begin{frame}
\frametitle{$t = \frac{z}{\sqrt{w/(n-k-1)}} \sim t(n-k-1)$}
\framesubtitle{With $z = \frac{y_0 - \mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}}}
             {\sqrt{\sigma^2(1 + \mathbf{x}_0^\prime 
             (\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{x}_0)}}$ 
             and $w = \frac{{\mbox{\footnotesize\emph{SSE}}}}{\sigma^2}$} 
\begin{eqnarray*}
    t & = & \frac{z}{\sqrt{w/\nu}} \\ \pause
      & = & \frac{y_0 - \mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}}}
             {\sqrt{\sigma^2(1 + \mathbf{x}_0^\prime 
             (\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{x}_0)}}
             \left/
             \sqrt{\frac{{\color{blue} \small \mbox{\emph{SSE}} }}{\sigma^2}/({\color{blue}n-k-1})}
             \right.  \\ \pause
      & = &  \frac{y_0 - \mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}}}
             {\sqrt{ {\color{blue} \small \mbox{\emph{MSE}}}(1 + \mathbf{x}_0^\prime 
             (\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{x}_0)}} 
             \pause \sim t(n-k-1)
\end{eqnarray*} 
\end{frame}


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\begin{frame}
\frametitle{Deriving the prediction interval }
%\framesubtitle{} 
%\begin{center}
%\includegraphics[width=3in]{CI-curve}
% \end{center}
{\small
\begin{eqnarray*}
    1-\alpha & = & P\{ -t_{\alpha/2} < t < t_{\alpha/2} \} \\ \pause
    & = & P\left\{ -t_{\alpha/2} < 
\frac{y_0 - \mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}}}
             {\sqrt{ \footnotesize \mbox{\emph{MSE}} \, (1 + \mathbf{x}_0^\prime 
             (\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{x}_0)}}
           < t_{\alpha/2} \right\} \\ \pause
    && \vdots \\ \pause
    & = & P\left\{  \mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}} 
      - t_{\alpha/2} \sqrt{ \small \mbox{\emph{MSE}} \, (1 + \mathbf{x}_0^\prime 
             (\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{x}_0)}
      < y_0 \right. \\ 
    && \hspace{12mm} <  \left. \mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}} 
      + t_{\alpha/2} \sqrt{ \small \mbox{\emph{MSE}} \, (1 + \mathbf{x}_0^\prime 
             (\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{x}_0)}
      \right\} \pause
\end{eqnarray*} 
} % End size

Or, $\mathbf{x}_0^\prime \widehat{\boldsymbol{\beta}}  \pm t_{\alpha/2} \, 
\sqrt{\mbox{\small\emph{MSE}} \, (1 + \mathbf{x}_0^\prime 
             (\mathbf{X}^\prime \mathbf{X})^{-1}\mathbf{x}_0)}$. \pause 
\hspace{5mm} $\blacksquare$
\end{frame}


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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 Statistical Sciences, 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}}

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