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\mode<presentation>

\title{Confirmatory Factor Analysis Part One\footnote{See last slide for copyright information.}}
\subtitle{STA 2101 Fall/Winter 2019}
\date{} % To suppress date

% Earlier versions preserved standardized observed variables, left over from exploratory factor analysis. In this edition, just drop that feature. 

\begin{document}

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

\begin{frame}
\frametitle{A confirmatory factor analysis model}
\framesubtitle{One Factor: Starting simply} \pause
% Put model equations and path diagram side by side.
\begin{tabular}{cc}
\raisebox{.45in}{\parbox{2in} {
\begin{eqnarray*}
    D_1 &=& \lambda_1 F + e_1 \\
    D_2 &=& \lambda_2 F + e_2 \\
    D_3 &=& \lambda_3 F + e_3 
\end{eqnarray*}    
}} % End parbox and then raisebox
& \includegraphics[width=1.5in]{OneFactor}
\end{tabular} \pause
  \begin{itemize}
    \item $Var(F)=1$ \pause
    \item $Var(e_j)=\omega_j$ \pause
    \item $F,e_1,e_2,e_3$ all independent.
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Calculate $\boldsymbol{\Sigma}$}
%\framesubtitle{} 
\begin{columns}  
\column{0.3\textwidth}
\begin{eqnarray*}
    D_1 &=& \lambda_1 F + e_1 \\
    D_2 &=& \lambda_2 F + e_2 \\
    D_3 &=& \lambda_3 F + e_3 
\end{eqnarray*} 
\column{0.7\textwidth}
\begin{displaymath}
    {\Large \boldsymbol{\Sigma} } ~~~=~~~
\begin{array}{c|ccc}
    & D_1 &       D_2         &         D_3        \\  \hline
D_1 & \lambda_1^2 + \omega_1  &\lambda_1\lambda_2      & \lambda_1\lambda_3 \\
D_2 &                         & \lambda_2^2 + \omega_2 & \lambda_2\lambda_3 \\
D_3 &                         &                        & \lambda_3^2 + \omega_3
\end{array}
\end{displaymath} \pause
\end{columns}

\vspace{10mm}

Are the parameters identifiable?  \pause What if just one $\lambda$ is zero?
\end{frame}

\begin{frame}
\frametitle{Suppose no factor loadings equal zero} \pause
%\framesubtitle{} 
{\footnotesize
\begin{displaymath}
    \boldsymbol{\Sigma} =
\left(\begin{array}{ccc}
\sigma_{11}  & \sigma_{12} & \sigma_{13} \\
   &        \sigma_{22}    & \sigma_{23} \\
   &             &      \sigma_{33}  
\end{array}\right)
= 
\left(\begin{array}{ccc}
\lambda_1^2 + \omega_1  &\lambda_1\lambda_2 & \lambda_1\lambda_3 \\
     & \lambda_2^2 + \omega_2 & \lambda_2\lambda_3 \\
    &                   & \lambda_3^2 + \omega_3  
\end{array}\right)
\end{displaymath} \pause
} % End size
\begin{eqnarray*}
    \lambda_1^2 & = & \frac{\sigma_{12}\sigma_{13}}{\sigma_{23}}  \pause
                  =   \frac{\lambda_1\lambda_2 \, \lambda_1\lambda_3}{\lambda_2\lambda_3} \\ \pause
&&\\
    \lambda_2^2 & = & \frac{\sigma_{12}\sigma_{23}}{\sigma_{13}} \\
&&\\
    \lambda_3^2 & = & \frac{\sigma_{13}\sigma_{23}}{\sigma_{12}}
\end{eqnarray*}  \pause
\begin{itemize}
     \item Squared factor loadings are identifiable, but not the loadings. \pause
     \item Replace all $\lambda_j$ with $-\lambda_j$, get same $\boldsymbol{\Sigma}$ \pause
     \item Likelihood function will have two maxima, same height. \pause
     \item Which one you find depends on where you start.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Solution: Decide on the sign of one loading}
\framesubtitle{Based on \emph{meaning}}  \pause
\begin{center}
\includegraphics[width=2.5in]{OneFactor}
\end{center} \pause
\begin{itemize}
     \item Is $F$ math ability or math \emph{inability}? You decide. \pause
     \item It's just a matter of naming the factors.
\end{itemize}

\end{frame}

\begin{frame}
\frametitle{If $\lambda_1>0$} \pause
%\framesubtitle{} 
%{\footnotesize
\begin{displaymath}
    \boldsymbol{\Sigma} =
\left(\begin{array}{ccc}
\sigma_{11}  & \sigma_{12} & \sigma_{13} \\
   &        \sigma_{22}    & \sigma_{23} \\
   &             &      \sigma_{33}  
\end{array}\right)
= 
\left(\begin{array}{ccc}
\lambda_1^2 + \omega_1  &\lambda_1\lambda_2 & \lambda_1\lambda_3 \\
     & \lambda_2^2 + \omega_2 & \lambda_2\lambda_3 \\
    &                   & \lambda_3^2 + \omega_3  
\end{array}\right)
\end{displaymath} \pause 
%} % End size

\vspace{10mm}

  \begin{itemize}
    \item Signs of $\lambda_2$ and $\lambda_3$ can be recovered right away from $\boldsymbol{\Sigma}$. \pause 
    \item And all the parameters are identified.
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Add another variable: $D_4 = \lambda_4 F + e_4$} \pause
%\framesubtitle{} 
{\footnotesize
\begin{displaymath}
    \boldsymbol{\Sigma} =
\left(\begin{array}{cccc}
\lambda_1^2 + \omega_1  &\lambda_1\lambda_2 & \lambda_1\lambda_3 & \lambda_1\lambda_4 \\
     & \lambda_2^2 + \omega_2 & \lambda_2\lambda_3      & \lambda_2\lambda_4 \\
     &                        & \lambda_3^2 + \omega_3  & \lambda_3\lambda_4 \\
     & & & \lambda_4^2 + \omega_4 
\end{array}\right)
\end{displaymath} \pause
} % End size


  \begin{itemize}
    \item Parameters will all be identifiable as long as 3 out of 4 loadings are non-zero, and one sign is known. \pause
    \item For example, if only $\lambda_1=0$ then the top row = 0,  \pause and you can get $\lambda_2, \lambda_3, \lambda_4$ as before. \pause
    \item For 5 variables, two loadings can be zero, etc. \pause
    \item How many equality restrictions? $6-4=2$.
%    \item Inequality restrictions? It's like an Easter egg hunt. 
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Now add another factor}
\framesubtitle{$Var(F_1)=Var(F_2) = 1$} 
\begin{center}
\includegraphics[width=3in]{TwoFactors}
\end{center}
%{\footnotesize
\begin{eqnarray*}
    D_1 &    =   & \lambda_1 F_1 + e_1 \\
        & \vdots &                     \\
    D_6 &=& \lambda_6 F_2 + e_6 
\end{eqnarray*}
%} % End size
\end{frame}

\begin{frame}
\frametitle{Covariance matrix of observable variables}
%\framesubtitle{} 
\begin{displaymath}
\boldsymbol{\Sigma} = 
\left(\begin{array}{rrrrrr}
\lambda_1^2 + \omega_1 & \lambda_{1} \lambda_{2}
  & \lambda_{1} \lambda_{3}   & \lambda_{1}
\lambda_{4} \phi_{12} & \lambda_{1} \lambda_{5} \phi_{12} &
\lambda_{1} \lambda_{6} \phi_{12} \\
    & \lambda_2^2 + \omega_2 & \lambda_{2} \lambda_{3}   & \lambda_{2}
\lambda_{4} \phi_{12} & \lambda_{2} \lambda_{5} \phi_{12} &
\lambda_{2} \lambda_{6} \phi_{12} \\
    & 
  & \lambda_3^2 + \omega_3 & \lambda_{3}
\lambda_{4} \phi_{12} & \lambda_{3} \lambda_{5} \phi_{12} &
\lambda_{3} \lambda_{6} \phi_{12} \\
  &   &   & \lambda_4^2 + \omega_4 & \lambda_{4} \lambda_{5}   &
\lambda_{4} \lambda_{6}   \\
  &   &  &     & \lambda_5^2 + \omega_5 &
\lambda_{5} \lambda_{6}   \\
  &   &   &     &   & \lambda_6^2 + \omega_6
\end{array}\right) 
\end{displaymath}  \pause
  \begin{itemize}
    \item Identify $\lambda_1, \lambda_2, \lambda_3$ from set One \pause (assuming one sign is known). \pause
    \item Identify $\lambda_4, \lambda_5, \lambda_6$ from set Two (lower right). \pause
    \item Identify $\phi_{12}$ from any unused covariance. \pause
    \item What if you added more variables? \pause
    \item What if you added more factors? 
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Three-variable identification rule}
\framesubtitle{For standardized factors}  \pause
For a factor analysis model, the parameters will be identifiable provided \pause
  \begin{itemize}
    \item Errors are independent of one another and of the factors. \pause
    \item Variances of all factors equal one. \pause
    \item Each observed variable is a function of only one factor. \pause
    \item There are at least three observable variables with non-zero loadings per factor. \pause
    \item The sign of one non-zero loading is known for each factor.
  \end{itemize}
\end{frame}


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


\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/2101f19} 
{\small\texttt{http://www.utstat.toronto.edu/$^\sim$brunner/oldclass/2101f19}}

\end{frame}


\end{document}

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

\begin{frame}
\frametitle{Frame Title}
%\framesubtitle{} 

  \begin{itemize}
    \item 
    \item 
    \item 
  \end{itemize}
\end{frame}

{\LARGE
\begin{displaymath}
    
\end{displaymath} 
} % End size


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

\begin{frame}
\frametitle{Inequality constraints}
\framesubtitle{Slippery devils} 
{\footnotesize
\begin{displaymath}
    \boldsymbol{\Sigma} =
\left(\begin{array}{ccc}
1  & \sigma_{12} & \sigma_{13} \\
   &        1    & \sigma_{23} \\
   &             &      1  
\end{array}\right)
= 
\left(\begin{array}{ccc}
1  &\lambda_1\lambda_2 & \lambda_1\lambda_3 \\
   &        1          & \lambda_2\lambda_3 \\
   &                   &          1  
\end{array}\right)
\end{displaymath}
} % End size
  \begin{itemize}
    \item $\sigma_{12}\sigma_{13}\sigma_{23} = \lambda_1^2\lambda_2^2\lambda_3^2$, so
          $0 \leq \sigma_{12}\sigma_{13}\sigma_{23} < 1$.
    \item But $\sigma_{12}\sigma_{13}\sigma_{23} < 1$ is true of \emph{any} correlation matrix, so that's one inequality, not two.
    \item How about $\lambda_1^2 = \frac{\sigma_{12}\sigma_{13}}{\sigma_{23}} $ so $0 \leq \frac{\sigma_{12}\sigma_{13}}{\sigma_{23}} <1$? 
   \end{itemize}
\begin{eqnarray*}
     &  & \frac{\sigma_{12}\sigma_{13}}{\sigma_{23}} \frac{\sigma_{23}}{\sigma_{23}} < 1 \\
     & \Rightarrow & 0 \leq \sigma_{12}\sigma_{13}\sigma_{23} < \sigma_{23}^2
\end{eqnarray*}
% Do this with the other two, get something geometrically complicated ...
\end{frame}



Sage work for inequality constraints -- Basically came up empty.


sem = 'http://www.utstat.toronto.edu/~brunner/openSEM/sage/sem.sage' 
load(sem)
var('lambda1 lambda2 lambda3')
S = SymmetricMatrix(3,'sigma',corr=True); S
S.eigenvalues() # Terrifying. Sage does not know sigma_ij is even real.
d = S.determinant(); d
factor(expand(d)) # Nothing good
Sigma = S(sigma12=lambda1*lambda2, sigma13=lambda1*lambda3, sigma23=lambda2*lambda3)
Sigma
Sigma.determinant()
expand(_)