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\title{Two-stage Proofs of Identifiability\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{Overview}
\tableofcontents
\end{frame}

\section{The Two-stage Idea}

\begin{frame}
\frametitle{The two-stage model: $V(\mathbf{D}_i)=\boldsymbol{\Sigma}$}
%\framesubtitle{} 
{\LARGE
\begin{eqnarray*}
    \mathbf{Y}_i &=& \boldsymbol{\beta} \mathbf{Y}_i + \boldsymbol{\Gamma} \mathbf{X}_i + \boldsymbol{\epsilon}_i \\
\mathbf{F}_i &=& \left( \begin{array}{c}
         \mathbf{X}_i  \\ \mathbf{Y}_i 
             \end{array} \right)  \\
\mathbf{D}_i &=& \boldsymbol{\Lambda}\mathbf{F}_i + \mathbf{e}_i 
\end{eqnarray*}
} % End size
\begin{itemize}
     \item  $V(\mathbf{X}_i)=\boldsymbol{\Phi}_{11}$, 
            $V(\boldsymbol{\epsilon}_i)=\boldsymbol{\Psi}$
     \item $V(\mathbf{F}_i) = V\left( \begin{array}{c}
         \mathbf{X}_i  \\ \mathbf{Y}_i 
             \end{array} \right)
     =\boldsymbol{\Phi} 
= \left( \begin{array}{c c}
         \boldsymbol{\Phi}_{11}  & \boldsymbol{\Phi}_{12} \\
         \boldsymbol{\Phi}_{12}^\prime  & \boldsymbol{\Phi}_{22} \\
             \end{array} \right)$
     \item $V(\mathbf{e}_i)=\boldsymbol{\Omega}$
\end{itemize}
\end{frame}


\begin{frame}
\frametitle{Identify parameter matrices in two steps}
\framesubtitle{It does not matter which one you do first.} 
{\Large
\begin{enumerate}
     \item \emph{Measurement model}: Show $\boldsymbol{\Phi}$ and $\boldsymbol{\Omega}$ can be recovered from $\boldsymbol{\Sigma}=V(\mathbf{D}_i)$.
     \item \emph{Latent model}: Show $\boldsymbol{\beta}$, $\boldsymbol{\Gamma}$, $\boldsymbol{\Phi}_{11}$ and $\boldsymbol{\Psi}$ can be recovered from $\boldsymbol{\Phi}$.
\end{enumerate}

This means the parameters of the latent model are recovered from  $\boldsymbol{\Sigma}$ as well.
} % End size

\vspace{3mm}
\hrule 
\vspace{3mm}

%{\footnotesize
    \begin{itemize}
        \item $\mathbf{D}_i = \boldsymbol{\Lambda}\mathbf{F}_i + \mathbf{e}_i $
        \begin{itemize}
            \item[] $V(\mathbf{F}_i)=\boldsymbol{\Phi}$, 
                    $V(\boldsymbol{e}_i)=\boldsymbol{\Omega}$
        \end{itemize}
        \item $\mathbf{Y}_i = \boldsymbol{\beta} \mathbf{Y}_i + \boldsymbol{\Gamma} \mathbf{X}_i + \boldsymbol{\epsilon}_i$
        \begin{itemize}
            \item[] $V(\mathbf{X}_i)=\boldsymbol{\Phi}_{11}$, 
                    $V(\boldsymbol{\epsilon}_i)=\boldsymbol{\Psi}$
        \end{itemize}
    \end{itemize}
%} % End size
\end{frame}

\begin{frame}
\frametitle{Parameter count rule}
\framesubtitle{A necessary condition} 
If a model has more parameters than covariance structure equations,
the parameter vector can be identifiable on at most a set of volume zero in the
parameter space. This applies to all models.
\end{frame}

\begin{frame}
\frametitle{All the following rules}
%\framesubtitle{} 
  \begin{itemize}
    \item Are sufficient conditions for identifiability, not necessary.
    \item Assume that errors are independent of exogenous variables that are not errors.
    \item Assume  all variables have expected value zero. 
  \end{itemize}
\end{frame}

\section{Measurement Model Rules}

\begin{frame}
\frametitle{Measurement Model Rules}
\framesubtitle{Factor Analysis} 

In these rules, latent variables that are not error terms are described as ``factors."

\end{frame}

\begin{frame}
\frametitle{Double Measurement Rule}
%\framesubtitle{} 
  \begin{itemize}
    \item Each factor is measured twice. 
    \item  All factor loadings equal one.
    \item There are two sets of measurements, set one and set two.
    \item Correlated measurement errors are allowed within sets of measurements, but not between sets.
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Three-Variable Rule for Standardized Factors}
%\framesubtitle{} 
            \begin{itemize}
                \item Errors are independent of one another.
                \item Each observed variable is caused by only one factor.
                \item The variance of each factor equals one.
                \item There are at least 3 variables with non-zero loadings per factor.
                \item The sign of one non-zero loading is known for each factor.
            \end{itemize}
Factors may be correlated.
\end{frame}

\begin{frame}
\frametitle{Three-Variable Rule for Unstandardized Factors}
%\framesubtitle{} 
The parameters of a measurement model will be identifiable if
            \begin{itemize}
                \item Errors are independent of one another.
                \item Each observed variable is caused by only one factor.
                \item For each factor, at least one factor loading equals one.
                \item There are at least 2 additional variables with non-zero loadings per factor.
            \end{itemize}
Factors may be correlated.
\end{frame}

\begin{frame}
\frametitle{Two-Variable Rule for Standardized Factors}
%\framesubtitle{} 
A factor with just two variables may be added to a measurement model whose parameters are identifiable, and the parameters of the combined model will be identifiable provided
            \begin{itemize}
                \item The errors for the two additional variables are independent of one another and of those already in the model.
                \item The two new variables are caused only by the new factor.
                \item The variance of the new factor equals one.
                \item Both new factor loadings are non-zero.
                \item The sign of one new loading is known.
                \item The new factor has a non-zero covariance with at least one factor already in the model.
            \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Two-Variable Rule for Unstandardized Factors}
%\framesubtitle{} 
A factor with just two variables may be added to a measurement model whose parameters are identifiable, and the parameters of the combined model will be identifiable provided
            \begin{itemize}
                \item The errors for the two additional variables are independent of one another and of those already in the model.
                \item The two new variables are caused only by the new factor.
                \item At least one new factor loading equals one.
                \item The other new factor loading is non-zero.
                \item The sign of one new loading is known.
                \item The new factor has a non-zero covariance with at least one factor already in the model.
            \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Four-variable Two-factor Rule}
%\framesubtitle{} 
The parameters of a measurement model with two factors and four observed variables will be identifiable provided
             \begin{itemize}
                \item All errors are independent of one another.
                \item All factor loadings are non-zero.
                \item For each factor, either the variance of the factor equals and the sign of one new loading is known, or at least one factor loading equals one.
                \item The covariance of the two factors does not equal zero.
            \end{itemize}
\end{frame}



\begin{frame}
\frametitle{Proof of the Four-variable Two-factor Rule}
\framesubtitle{With standardized factors} 
The model equations are
        \begin{eqnarray*}
        D_1 & = & \lambda_1 F_1 + e_1 \\
        D_2 & = & \lambda_2 F_1 + e_2 \\
        D_3 & = & \lambda_3 F_2 + e_4 \\
        D_4 & = & \lambda_4 F_2 + e_5, 
        \end{eqnarray*}
where all expected values are zero, $V(e_j)=\omega_j$ for $j=1, \ldots, 4$, and
\begin{displaymath}  
\begin{array}{ccc}    % Array of Arrays: Nice display of matrices.
     V\left[ \begin{array}{c}
                 F_1 \\
                 F_2
    \end{array} \right]         & = &  
     \left[ \begin{array}{c c}
                 1 & \phi_{12} \\
                 \phi_{12} & 1
    \end{array} \right].                                  
\end{array}
\end{displaymath}  
Also suppose no loadings = 0 and $\lambda_1>0$, $\lambda_3>0$.
\end{frame}

\begin{frame}
\frametitle{Covariance matrix}
\framesubtitle{For the 4-variable 2-factor problem} 
{\footnotesize
        \begin{eqnarray*}
        D_1 & = & \lambda_1 F_1 + e_1 \\
        D_2 & = & \lambda_2 F_1 + e_2 \\
        D_3 & = & \lambda_3 F_2 + e_4 \\
        D_4 & = & \lambda_4 F_2 + e_5, 
        \end{eqnarray*}
} % End size
\begin{displaymath}
    \boldsymbol{\Sigma} ~~~=~~~
\begin{array}{c|cccc}
    & D_1 &       D_2         &         D_3      & D_4  \\  \hline
D_1 & \lambda_1^2+\omega_1  &\lambda_1\lambda_2 & 
      \lambda_1\lambda_3\phi_{12}  & \lambda_1\lambda_4 \phi_{12}  \\
D_2 &     & \lambda_2^2+\omega_2  & 
    \lambda_2\lambda_3 \phi_{12} & \lambda_2\lambda_4 \phi_{12} \\
D_3 &     &                   & \lambda_3^2+\omega_3 &  \lambda_3\lambda_4   \\
D_4 &  &  &  &  \lambda_4^2+\omega_4
\end{array}
\end{displaymath}

\end{frame}


\begin{frame}
\frametitle{Using the assumption that $\lambda_1>0$ and $\lambda_3>0$}
%\framesubtitle{} 
{\footnotesize
\begin{displaymath}
    \boldsymbol{\Sigma} ~~~=~~~
\begin{array}{c|cccc}
    & D_1 &       D_2         &         D_3      & D_4  \\  \hline
D_1 & \lambda_1^2+\omega_1  &\lambda_1\lambda_2 & 
      \lambda_1\lambda_3\phi_{12}  & \lambda_1\lambda_4 \phi_{12}  \\
D_2 &     & \lambda_2^2+\omega_2  & 
    \lambda_2\lambda_3 \phi_{12} & \lambda_2\lambda_4 \phi_{12} \\
D_3 &     &                   & \lambda_3^2+\omega_3 &  \lambda_3\lambda_4   \\
D_4 &  &  &  &  \lambda_4^2+\omega_4
\end{array}
\end{displaymath}} % End size
\begin{eqnarray*}
     & &   \frac{\sigma_{12}\sigma_{13}}{\sigma_{23}} 
         = \frac{\lambda_1^2\lambda_2\lambda_3\phi_{12}}{\lambda_2\lambda_3 \phi_{12}} = \lambda_1^2 \\
&& \\
     & \Rightarrow & \lambda_1 = \sqrt{\frac{\sigma_{12}\sigma_{13}}{\sigma_{23}}}
\end{eqnarray*} 
Similarly, $ \lambda_3 = \sqrt{\frac{\sigma_{34}\sigma_{23}}{\sigma_{24}}}$, and the rest is easy.
\end{frame}

\begin{frame}
\frametitle{Please don't do both!}
\framesubtitle{Don't set the variance \emph{and} a factor loading to one!} 
\begin{itemize}
    \item Setting the variance of factors to one looks arbitrary, but it's really a smart re-parameterization.
    \item Setting one loading per factor to one also is a smart re-parameterization.
    \item It's smart because the resulting models impose the \emph{same restrictions on the covariance that the original model does.}
    \item And, the \emph{meanings} of the parameters have a clear connection to the meanings of the parameters of the original model.
    \item But if you do both, it's a mess. Most or all of the meaning is lost.
    \item And you put an \emph{extra} restriction on $\boldsymbol{\Sigma}$ that is not implied by the original model.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Combination Rule}
%\framesubtitle{} 
Suppose that the parameters of two measurement models are identifiable by any of the rules above. The two models may be combined into a single model provided that the error terms of the first model are independent of the error terms in the second model. The additional parameters of the combined model are the covariances between the two sets of factors, and these are all identifiable.
\end{frame}

\begin{frame}
\frametitle{Cross-over Rule}
%\framesubtitle{} 
Suppose that 
            \begin{itemize}
                \item The parameters of a measurement models are identifiable, and
                \item For each factor there is at least one observable variable that is caused only by that factor (with a non-zero factor loading).
            \end{itemize}
Then any number of new observable variables may be added to the model and the result is a model whose parameters are all identifiable, provided that 
            \begin{itemize}
                \item The error terms associated with the new variables are independent of the error terms in the existing model.
            \end{itemize}
 Each new variable may be caused by any or all of the factors, potentially resulting in a cross-over pattern in the path diagram. The error terms associated with the new set of variables may be correlated with one another.
\end{frame}

\begin{frame}
\frametitle{Error-Free Rule}
%\framesubtitle{} 
 A vector of observable variables may be added to the factors of a measurement model whose parameters are identifiable. Suppose that 
            \begin{itemize}
                \item The new observable variables are independent of the errors in the measurement model, and
                \item For each factor in the measurement model there is at least one observable variable that is caused only by that factor (with a non-zero factor loading).
            \end{itemize}
Then the parameters of a new measurement model, where some of the variables are assumed to be measured without error, are identifiable. The practical consequence is that variables assumed to be measured without error may be included in the latent component of a structural equation model, provided that the measurement model for the other variables has identifiable parameters. 
\end{frame}

\section{Latent Model Rules}

\begin{frame}
\frametitle{Latent Model Rules}
%\framesubtitle{} 
  \begin{itemize}
    \item $\mathbf{Y}_i = \boldsymbol{\beta} \mathbf{Y}_i + \boldsymbol{\Gamma} \mathbf{X}_i + \boldsymbol{\epsilon}_i$
    \item  Here, identifiability means that the parameters involved are functions of $V(\mathbf{F})=\boldsymbol{\Phi}$.
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Regression Rule}
\framesubtitle{Someimes called the Null Beta Rule} 
Suppose 
  \begin{itemize}
    \item No endogenous variables cause other endogenous variables.
    \item[]
    \item $\mathbf{Y}_i = \boldsymbol{\Gamma} \mathbf{X}_i + \boldsymbol{\epsilon}_i$
    \item Of course $C(\mathbf{X}_i, \boldsymbol{\epsilon}_i) = \mathbf{0}$, always.
    \item $\boldsymbol{\Psi} = V(\boldsymbol{\epsilon}_i)$ need not be diagonal.
  \end{itemize}
\vspace{5mm}
Then $\boldsymbol{\Gamma}$ and $\boldsymbol{\Psi}$ are identifiable.
\end{frame}

\begin{frame}
\frametitle{Acyclic Rule}
%\framesubtitle{} 
Parameters of the Latent Variable Model are identifiable if the model is acyclic (no feedback loops through straight arrows) and $V(\boldsymbol{\epsilon}) = \boldsymbol{\Psi}$ has the following block diagonal structure.
  \begin{itemize}
    \item Organize the variables into sets. Set 0 consists of the exogenous variables. 
    \item For $j=1,\ldots ,k$, each variable in set $j$ is caused by at least one variable in set $j-1$, and also possibly by variables in earlier sets.
    \item Error terms for the variables in a set may (or may not) have non-zero covariances.
    \item All other covariances between error terms are zero.
  \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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