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\mode<presentation>

\title{Background for the Personality and Job Satisfaction Example\footnote{See last slide for copyright information.}}
\subtitle{STA431 Winter/Spring 2017}
\date{} % To suppress date

\begin{document}

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

\section{Matrix Input}

\begin{frame}
\frametitle{Multivariate Normal Likelihood $L(\boldsymbol{\mu,\Sigma})$}
%\framesubtitle{} 

\begin{eqnarray*}
     &  & |\boldsymbol{\Sigma}|^{-n/2} (2\pi)^{-np/2} 
    \exp -\frac{n}{2}\left\{ tr(\boldsymbol{\widehat{\Sigma}\Sigma}^{-1}) +
    (\overline{\mathbf{d}}-\boldsymbol{\mu})^\top \boldsymbol{\Sigma}^{-1} 
    (\overline{\mathbf{d}}-\boldsymbol{\mu}) \right\} \\ \pause
    & = & |\boldsymbol{\Sigma}|^{-n/2} (2\pi)^{-np/2} 
    \exp -\frac{n}{2}\left\{ tr(\boldsymbol{\widehat{\Sigma}\Sigma}^{-1}) 
                     \right\} 
\end{eqnarray*} \pause
  \begin{itemize}
    \item All you need is $\widehat{\boldsymbol{\Sigma}}$ \pause and more rarely, $\overline{\mathbf{d}}$. \pause
    \item You don't need the raw data.  \pause
    \item $(\overline{\mathbf{d}},\widehat{\boldsymbol{\Sigma}})$ are \emph{sufficient statistics} for $(\boldsymbol{\mu,\Sigma})$. \pause
    \item Software should be able to fit a model based only on $\widehat{\boldsymbol{\Sigma}}$ and $n$.
  \end{itemize}
\end{frame}


\section{Auxiliary Information}

\begin{frame}
\frametitle{Auxiliary Information} \pause
%\framesubtitle{} 

  \begin{itemize}
    \item Auxiliary means extra, supporting.  \pause 
    \item Mostly used to identify parameters of the measurement model.  \pause
    \item For example, there might be double or gold standard measurement on just a subset of the sample.  \pause
    \item In a multi-group analysis, the groups can have overlapping sets of variables and overlapping sets of parameters.  \pause
    \item Or you can sometimes use published information from other studies. \pause
    \item Like (estimated) reliabilities.
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Typical Example}
% \framesubtitle{} \pause
\begin{eqnarray*}
    D_1 &=& F_1 + e_1 \\
    D_2 &=& F_2 + e_2 \\
    D_3 &=& F_3 + e_3 
\end{eqnarray*}  \pause  
{\footnotesize
\begin{columns}  
\column{0.4\textwidth}
\begin{displaymath}
    \boldsymbol{\Phi} = \left(
\begin{array}{ccc}
\phi_{11}             & \phi_{12}            & \phi_{13} \\
\phi_{12}             & \phi_{22}            & \phi_{23} \\
\phi_{13}             & \phi_{23}            & \phi_{33}           
\end{array} \right)~~
\end{displaymath} \pause 
\column{0.6\textwidth}
\begin{displaymath}
    \boldsymbol{\Sigma}  = \left(
\begin{array}{ccc}
\phi_{11} + \omega_1  & \phi_{12}            & \phi_{13} \\
\phi_{12}             & \phi_{22} + \omega_2 & \phi_{23} \\
\phi_{13}             & \phi_{23}            & \phi_{33} + \omega_3
\end{array} \right)
\end{displaymath} \pause
\end{columns}
} % End size
\vspace{3mm}
What if you knew the reliabilities? \pause
\vspace{3mm}
\begin{center}
\begin{tabular}{ccc}
$\rho^2_1 = \frac{\phi_{11}}{\phi_{11} + \omega_1}$ &
$\rho^2_2 = \frac{\phi_{22}}{\phi_{22} + \omega_2}$ &
$\rho^2_3 = \frac{\phi_{33}}{\phi_{33} + \omega_3}$
\end{tabular}
\end{center}
\end{frame}

\begin{frame}
\frametitle{Solve for $\boldsymbol{\Phi}$: Compute $\phi_{jj} = \sigma_{jj}\rho^2_j$}\pause 
%\framesubtitle{} 
{\footnotesize
\begin{columns}  
\column{0.4\textwidth}
\begin{displaymath}
    \boldsymbol{\Phi} = \left(
\begin{array}{ccc}
\phi_{11}             & \phi_{12}            & \phi_{13} \\
\phi_{12}             & \phi_{22}            & \phi_{23} \\
\phi_{13}             & \phi_{23}            & \phi_{33}           
\end{array} \right)~~
\end{displaymath} 
\column{0.6\textwidth}
\begin{displaymath}
    \boldsymbol{\Sigma}  = \left(
\begin{array}{ccc}
\phi_{11} + \omega_1  & \phi_{12}            & \phi_{13} \\
\phi_{12}             & \phi_{22} + \omega_2 & \phi_{23} \\
\phi_{13}             & \phi_{23}            & \phi_{33} + \omega_3
\end{array} \right)
\end{displaymath}
\end{columns}
} % End size
\begin{center}
\begin{tabular}{ccc}
$\rho^2_1 = \frac{\phi_{11}}{\phi_{11} + \omega_1}$ &
$\rho^2_2 = \frac{\phi_{22}}{\phi_{22} + \omega_2}$ &
$\rho^2_3 = \frac{\phi_{33}}{\phi_{33} + \omega_3}$
\end{tabular}
\end{center} \pause

  \begin{itemize}
    \item Do it using the estimates: $\widehat{\phi}_{jj} = \widehat{\sigma}_{jj} \widehat{\rho}^2_j$,  \pause
    \item Converting $\widehat{\boldsymbol{\Sigma}}$ to $\widehat{\boldsymbol{\Phi}}$. \pause
    \item Operate on $\widehat{\boldsymbol{\Phi}}$ as if it's a $\widehat{\boldsymbol{\Sigma}}$ and the variables are measured without error.
  \end{itemize}
\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/431s17} 
{\small\texttt{http://www.utstat.toronto.edu/$^\sim$brunner/oldclass/431s17}}
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