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\mode<presentation>

\title{Confirmatory Factor Analysis: The Measurement Model\footnote{See last slide for copyright information.}}
\subtitle{STA431 Winter/Spring 2013}
\date{} % To suppress date

\begin{document}

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


\begin{frame}
\frametitle{A confirmatory factor analysis model}
\framesubtitle{One Factor: Starting simply} 
\begin{eqnarray*}
    Z_1 &=& \lambda_1 F + e_1 \\
    Z_2 &=& \lambda_2 F + e_2 \\
    Z_3 &=& \lambda_3 F + e_3 
\end{eqnarray*}
\begin{itemize}
     \item[] $V(F)=V(e_1)=V(e_2)=V(e_3)=1$
     \item[] $F,e_1,e_2,e_3$ all independent
\end{itemize}

\begin{eqnarray*}
     &  & V(Z_1) = 1 = \lambda^2+V(e_1) \\
     & \Rightarrow & V(e_1) = 1-\lambda_1^2, \mbox{ etc.}
\end{eqnarray*}
\end{frame}

\begin{frame}
\frametitle{$\boldsymbol{\Sigma}$ is a correlation matrix}
\framesubtitle{$Corr(X,Y) = \frac{Cov(X,Y)}{SD(X)SD(Y)}$} 
{\footnotesize
\begin{eqnarray*}
    Z_1 &=& \lambda_1 F + e_1 \\
    Z_2 &=& \lambda_2 F + e_2 \\
    Z_3 &=& \lambda_3 F + e_3 
\end{eqnarray*}
} % End size

\begin{displaymath}
    \boldsymbol{\Sigma} ~~~=~~~
\begin{array}{c|ccc}
    & Z_1 &       Z_2         &         Z_3        \\  \hline
Z_1 &  1  &\lambda_1\lambda_2 & \lambda_1\lambda_3 \\
Z_2 &     &        1          & \lambda_2\lambda_3 \\
Z_3 &     &                   &          1  
\end{array}
\end{displaymath}

  \begin{itemize}
    \item $\boldsymbol{\theta} = (\lambda_1,\lambda_2,\lambda_3)$
    \item The parameter space is an open cube. 
    \item Are the parameters identifiable? What if just one is zero?
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Suppose no factor loadings equal zero}
%\framesubtitle{} 
{\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{eqnarray*}
    \lambda_1^2 & = & \frac{\sigma_{12}\sigma_{13}}{\sigma_{23}} 
                  =   \frac{\lambda_1\lambda_2 \, \lambda_1\lambda_3}{\lambda_2\lambda_3} \\
&&\\
    \lambda_2^2 & = & \frac{\sigma_{12}\sigma_{23}}{\sigma_{13}} \\
&&\\
    \lambda_3^2 & = & \frac{\sigma_{13}\sigma_{23}}{\sigma_{12}}
\end{eqnarray*} 
\begin{itemize}
     \item Squared factor loadings are identifiable, but not the loadings.
     \item Replace $\lambda_j$ with $-\lambda_j$, get same $\boldsymbol{\Sigma}$
     \item Likelihood function will have two maxima, same height.
     \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}} 
\begin{center}
\includegraphics[width=2.5in]{OneFactor}
\end{center}
\begin{itemize}
     \item Is $F_1$ math ability or math \emph{inability}? You decide.
     \item It's just a matter of naming the factors.
\end{itemize}

\end{frame}

\begin{frame}
\frametitle{If $\lambda_1>0$}
%\framesubtitle{} 
{\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 Signs of $\lambda_2$ and $\lambda_3$ can be recovered right away from $\boldsymbol{\Sigma}$. 
    \item And all the parameters are identified.
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Equality constraints}
%\framesubtitle{} 

  \begin{itemize}
    \item Three parameters minus three correlations (equations) = ZERO.
    \item The inequality constraints are more slippery.
  \end{itemize}
\end{frame}

\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} positive definite 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}

\begin{frame}
\frametitle{Add another variable: $Z_4 = \lambda_4 F + e_4$}
%\framesubtitle{} 
{\footnotesize
\begin{displaymath}
    \boldsymbol{\Sigma} =
\left(\begin{array}{cccc}
1  &\lambda_1\lambda_2 & \lambda_1\lambda_3  & \lambda_1\lambda_4 \\
   &        1          & \lambda_2\lambda_3  & \lambda_2\lambda_4 \\
   &                   &         1           & \lambda_3\lambda_4 \\
   &                   &                     &         1  
\end{array}\right)
\end{displaymath}
} % 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.
    \item For example, if $\lambda_1=0$ then the top row = 0 and you can get $\lambda_2, \lambda_3, \lambda_4$ as before.
    \item For 5 variables, two loadings can be zero, etc.
    \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{} 
\begin{center}
\includegraphics[width=3in]{TwoFactors}
\end{center}
%{\footnotesize
\begin{eqnarray*}
    Z_1 &    =   & \lambda_1 F_1 + e_1 \\
        & \vdots &                     \\
    Z_6 &=& \lambda_6 F_2 + e_6 
\end{eqnarray*}
%} % End size
\end{frame}


\begin{frame}
\frametitle{Correlation matrix of observable variables}
%\framesubtitle{} 
\begin{displaymath}
\boldsymbol{\Sigma} = 
\left(\begin{array}{rrrrrr}
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} \\
    & 1 & \lambda_{2} \lambda_{3}   & \lambda_{2}
\lambda_{4} \phi_{12} & \lambda_{2} \lambda_{5} \phi_{12} &
\lambda_{2} \lambda_{6} \phi_{12} \\
    & 
  & 1 & \lambda_{3}
\lambda_{4} \phi_{12} & \lambda_{3} \lambda_{5} \phi_{12} &
\lambda_{3} \lambda_{6} \phi_{12} \\
  &   &   & 1 & \lambda_{4} \lambda_{5}   &
\lambda_{4} \lambda_{6}   \\
  &   &  &     & 1 &
\lambda_{5} \lambda_{6}   \\
  &   &   &     &   & 1
\end{array}\right) 
\end{displaymath} 
  \begin{itemize}
    \item Identify $\lambda_1, \lambda_2, \lambda_3$ from set 1
    \item Identify $\lambda_4, \lambda_5, \lambda_6$ from set 2
    \item Identify $\phi_{12}$ from any unused correlation.
    \item What if you added more variables?
    \item What if you added more factors?
    \item What if observed variables were centered but not standardized?
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Three-variable identification rule}
% \framesubtitle{} 
For a factor analysis model, the parameters will be identifiable provided
  \begin{itemize}
    \item Errors are independent of one another and of the factors.
    \item Variances of factors equal one.
    \item Each observed variable is a function of only one factor.
    \item There are at least three observable variables with non-zero loadings per factor.
    \item The sign of one non-zero loading is known for each factor.
  \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/431s13} {\small\texttt{http://www.utstat.toronto.edu/$^\sim$brunner/oldclass/431s31}}

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