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\mode<presentation>

\title{Instrumental Variables\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{One Explanatory Variable}

\begin{frame}
\frametitle{Seeking identifiability}
%\framesubtitle{} 
We have seen that in simple regression, parameters of a model with measurement error are not identifiable.
\begin{eqnarray*}
    Y_i &=& \alpha_1 + \beta_1 X_i + \epsilon_i  \\
    W_i &=& \nu +  X_i + e_i,  
\end{eqnarray*}

\vspace{5mm}

  \begin{itemize}
    \item For example, $X$ might be income and $Y$ might be credit card debt.
    \item Include another response variable $Y_2$, like value of automobile.
  \end{itemize}


\end{frame}

\begin{frame}
\frametitle{Include a second response variable}
  \begin{itemize}
    \item Response variable of primary interest is now called $Y_{i,1}$ 
    \item The second response variable $Y_{i,2}$ is called an \emph{instrumental variable}.
    \item It's just a tool.
  \end{itemize}

\begin{eqnarray*} 
   W_{i\mbox{~}} & = & \nu +  X_i + e_i            \\  \nonumber
   Y_{i,1}          & = & \alpha_1 + \beta_1 X_i + \epsilon_{i,1}  \\  \nonumber
   Y_{i,2}          & = & \alpha_2 + \beta_2 X_i + \epsilon_{i,2}  \\  \nonumber
\end{eqnarray*}
where $X_i$, $e_i$,  $\epsilon_{i,1}$ and $\epsilon_{i,2}$ are all independent,
$Var(X_i)=\phi$, $Var(e_i)=\omega$, $Var(\epsilon_{i,1})=\psi_1$, $Var(\epsilon_{i,2})=\psi_2$, $E(X_i)=\mu_x$ and the expected values of all error terms are zero. The regression coefficients $\alpha_j$ and $\beta_j$ are unknown constants.
\end{frame}

\begin{frame}
\frametitle{Are the parameters identifiable?}
{\scriptsize
\begin{eqnarray*} 
   W_{i\mbox{~}} & = & \nu +  X_i + e_i            \\  \nonumber
   Y_{i,1}          & = & \alpha_1 + \beta_1 X_i + \epsilon_{i,1}  \\  \nonumber
   Y_{i,2}          & = & \alpha_2 + \beta_2 X_i + \epsilon_{i,2}  \\  \nonumber
\end{eqnarray*}
} % End size

  \begin{itemize}
    \item Assume everything is normal: 
    $\mathbf{D}_i \sim N(\boldsymbol{\mu}, \boldsymbol{\Sigma})$.
    \item $\boldsymbol{\theta} = (\nu,\alpha_1 ,\alpha_2 ,\beta_1, \beta_2, \mu_x, \phi,\omega, \psi_1, \psi_2)$: Ten parameters.
    \item $\boldsymbol{\mu}$ is $3 \times 1$.
    \item $\boldsymbol{\Sigma}$ is $3 \times 3$.
    \item Nine moment structure equations in ten unknowns.
  \end{itemize}
\end{frame}


\begin{frame}
\frametitle{Look at the covariance structure equations}
\framesubtitle{We are pessimistic about the expected values} 
{\scriptsize
\begin{eqnarray*} 
   W_{i\mbox{~}} & = & \nu +  X_i + e_i            \\  \nonumber
   Y_{i,1}          & = & \alpha_1 + \beta_1 X_i + \epsilon_{i,1}  \\  \nonumber
   Y_{i,2}          & = & \alpha_2 + \beta_2 X_i + \epsilon_{i,2}  \\  \nonumber
\end{eqnarray*}
} % End size

\vspace{5mm}

\begin{displaymath} 
    \boldsymbol{\Sigma} = 
    V\left( \begin{array}{l}
    W_i \\ Y_{i,1} \\ Y_{i,2} 
    \end{array} \right) = 
        \left[ \begin{array}{c c  c }
    \phi+\omega & \beta_1\phi & \beta_2\phi  \\ 
     &  \beta_1^2 \phi + \psi_1  & \beta_1\beta_2\phi  \\
     &      &  \beta_2^2 \phi + \psi_2   \\
        \end{array} \right].
\end{displaymath}
\end{frame}

\begin{frame}
\frametitle{Six equations in six unknowns}
\framesubtitle{A unique solution is possible but not guaranteed}
\begin{displaymath} 
    \left[ \begin{array}{c c  c }
    \sigma_{11} & \sigma_{12} & \sigma_{13}  \\ 
     &  \sigma_{22}  & \sigma_{23}  \\
     &      &  \sigma_{33}  \\
        \end{array} \right] =
    \left[ \begin{array}{c c  c }
    \phi+\omega & \beta_1\phi & \beta_2\phi  \\ 
     &  \beta_1^2 \phi + \psi_1  & \beta_1\beta_2\phi  \\
     &      &  \beta_2^2 \phi + \psi_2   \\
        \end{array} \right]
\end{displaymath}
Identifiability depends on where you are in the parameter space. Consider 
  \begin{itemize}
    \item $\beta_1=0$ and $\beta_2=0$
    \item $\beta_1=0$ and $\beta_2 \neq 0$
    \item $\beta_1 \neq 0$ and $\beta_2=0$
    \item $\beta_1 \neq 0$ and $\beta_2 \neq 0$
  \end{itemize}
The parameter $\beta_1$ is identifiable if $\beta_2 \neq 0$: 
$\beta_1 = \frac{\sigma_{23}}{\sigma_{13}}$.
\end{frame}

% This simple example is quite interesting because the parameter is not identifiable when H_0: \beta_1=0 is true. Is the likelihood ratio test still okay? The null hypothesis model imposes TWO constraints on the covariance matrix, provided beta_2 \neq 0. This is pretty important, since H_0 beta_1=0 is the most interesting null hypothesis. Maybe primary test is okay but standard errors of beta2-hat and other parameter estimates go through the ceiling. No I bet primary test has wrong df, inflate Type I error rate.

\begin{frame}
\frametitle{Suppose both $\beta_1 \neq 0$ and $\beta_2 \neq 0$}
{\scriptsize
\begin{displaymath} 
    \left[ \begin{array}{c c  c }
    \sigma_{11} & \sigma_{12} & \sigma_{13}  \\ 
     &  \sigma_{22}  & \sigma_{23}  \\
     &      &  \sigma_{33}  \\
        \end{array} \right] =
    \left[ \begin{array}{c c  c }
    \phi+\omega & \beta_1\phi & \beta_2\phi  \\ 
     &  \beta_1^2 \phi + \psi_1  & \beta_1\beta_2\phi  \\
     &      &  \beta_2^2 \phi + \psi_2   \\
        \end{array} \right]
\end{displaymath}
} % End size
  \begin{itemize}
    \item[] $\beta_1 = \frac{\sigma_{23}}{\sigma_{13}}$
    \item[] $\beta_2 = \frac{\sigma_{23}}{\sigma_{12}}$
    \item[] $\phi = \frac{\sigma_{12}\sigma_{13}}{\sigma_{23}}$
  \end{itemize}
Solve for $\omega$, $\psi_1$ and $\psi_2$ by subtraction. Can write
  \begin{itemize}
    \item[] $\omega = \sigma_{11}-\phi$ 
    \item[] $\psi_1 = \sigma_{22}-\beta_1^2 \phi$
    \item[] $\psi_2 = \sigma_{33}-\beta_2^2 \phi$
  \end{itemize}
Without substituting for parameter that have already been identified.
Don't need to give complete explicit solution. This shows it can be done.
\end{frame}

\begin{frame}
\frametitle{What about the expected values?}
{\scriptsize
\begin{eqnarray*} 
   W_{i\mbox{~}} & = & \nu +  X_i + e_i            \\  \nonumber
   Y_{i,1}          & = & \alpha_1 + \beta_1 X_i + \epsilon_{i,1}  \\  \nonumber
   Y_{i,2}          & = & \alpha_2 + \beta_2 X_i + \epsilon_{i,2}  \\  \nonumber
\end{eqnarray*}
} % End size
\begin{eqnarray*} 
    \mu_1 & = & \nu+\mu_x \\
    \mu_2 & = & \alpha_1+\beta_1\mu_x \nonumber \\
    \mu_3 & = & \alpha_2+\beta_2\mu_x \nonumber \\
\end{eqnarray*}
  \begin{itemize}
    \item Three equations in four unknowns, even assuming $\beta_1$ and $\beta_2$ known.
    \item Re-parameterize.
  \end{itemize}
\end{frame}
\begin{frame}

\frametitle{Re-parameterize}
{\scriptsize
\begin{eqnarray*} 
    \mu_1 & = & \nu+\mu_x \\
    \mu_2 & = & \alpha_1+\beta_1\mu_x \nonumber \\
    \mu_3 & = & \alpha_2+\beta_2\mu_x \nonumber \\
\end{eqnarray*}
} % End size
  \begin{itemize}
    \item Absorb $\nu, \mu_x, \alpha_1, \alpha_2$ into $\boldsymbol{\mu}$.
    \item Parameter was $\boldsymbol{\theta} = (\nu, \mu_x, \alpha_1 ,\alpha_2 ,\beta_1, \beta_2, \phi,\omega, \psi_1, \psi_2)$
    \item Now it's $\boldsymbol{\theta} = (\mu_1, \mu_2, \mu_3, \beta_1, \beta_2, \phi, \omega, \psi_1, \psi_2)$
    \item Dimension of the parameter space is now one less.
    \item We haven't lost much.
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{We haven't lost much especially because the model was already re-parameterized}
Model is
{\scriptsize
\begin{eqnarray*} 
   W_{i\mbox{~}} & = & \nu +  X_i + e_i            \\  \nonumber
   Y_{i,1}          & = & \alpha_1 + \beta_1 X_i + \epsilon_{i,1}  \\  \nonumber
   Y_{i,2}          & = & \alpha_2 + \beta_2 X_i + \epsilon_{i,2}  \\  \nonumber
\end{eqnarray*}
} % End size
But of course there is measurement error in $Y_1$ and $Y_2$. Recall
\begin{eqnarray*}
Y &=& \alpha + \beta X  + \epsilon \\
V &=& \nu_0 +  Y +  e \\
  &=& \nu_0 + (\alpha + \beta X  + \epsilon) +  e \\
  &=& (\nu_0 + \alpha) + \beta X  
                                + (\epsilon +  e) \\
  &=& \alpha^\prime +  \beta X  + \epsilon^\prime  
\end{eqnarray*} 

\end{frame}


\begin{frame}
\frametitle{Summary}
%\framesubtitle{} 
  \begin{itemize}
    \item Adding the instrumental variable didn't help identify the expected values and intercepts. That's  hopeless.
    \item But we did identify $\beta_1$, which is the most interesting parameter.
    \item Re-parameterizing, absorbed the intercepts and expected values into $\boldsymbol{\mu}$.
    \item Where $\beta_1$ and $\beta_2$ are both non-zero, the entire parameter vector is identifiable.  
    \item For maximum likelihood estimation, it helps to have the \emph{entire} parameter vector identifiable at the true parameter value.
    \item This is definitely a success. 
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Testing $H_0: \beta_1=0$}
\framesubtitle{The most interesting null hypothesis} 
  \begin{itemize}
    \item The parameter $\beta_1$ is identifiable, so a valid test is possible.
    \item But the whole parameter \emph{vector} is not identifiable when $\beta_1=0$.  
    \item Technical conditions of the likelihood ratio test are not satisfied.
    \item It becomes quite ``interesting."
    \item Likelihood ratio statistic actually has 2 $df$ even though $H_0$ appears to impose only one restriction on the parameter.
    \item Too interesting.
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{It's better with two (or more) instrumental variables.}
\begin{eqnarray*} 
   W_{i\mbox{~}} & = & \nu +  X_i + e_i            \\  \nonumber
   Y_{i,1}          & = & \alpha_1 + \beta_1 X_i + \epsilon_{i,1}  \\  \nonumber
   Y_{i,2}          & = & \alpha_2 + \beta_2 X_i + \epsilon_{i,2}  \\  \nonumber
   Y_{i,3}          & = & \alpha_3 + \beta_3 X_i + \epsilon_{i,3},
\end{eqnarray*}

\begin{displaymath} 
    \boldsymbol{\Sigma} = 
        \left( \begin{array}{c c  c c}
    \phi+\omega & \beta_1\phi & \beta_2\phi & \beta_3\phi \\ 
     &  \beta_1^2 \phi + \psi_1  & \beta_1\beta_2\phi & \beta_1\beta_3\phi \\
     &      &  \beta_2^2 \phi + \psi_2  & \beta_2\beta_3\phi \\
     &      &      & \beta_3^2 \phi + \psi_3 \\
        \end{array} \right).
\end{displaymath}
\end{frame}

\begin{frame}
\frametitle{With two instrumental variables}
%\framesubtitle{} 
  \begin{itemize}
    \item Again, identification of the expected values and intercepts is out of the question.
    \item So we re-parameterize,
    \item Absorbing the expected values and intercepts into $\boldsymbol{\mu} = E(\mathbf{D}_i)$
    \item And look at the covariance structure equations.
  \end{itemize}
\end{frame}


\begin{frame}
\frametitle{Covariance structure equations}
\begin{eqnarray*}
    V\left( \begin{array}{c}
W_{i\mbox{~}} \\ Y_{i,1} \\ Y_{i,2} \\ Y_{i,3}
\end{array} \right) 
& = & \left( \begin{array}{c c  c c}
    \sigma_{11} & \sigma_{12} & \sigma_{13} & \sigma_{14} \\ 
     &  \sigma_{22}  & \sigma_{23} & \sigma_{24} \\
     &      &  \sigma_{33}  & \sigma_{3,4}\\
     &      &               & \sigma_{4,4}
        \end{array} \right) \\
& = &   \left( \begin{array}{c c  c c}
        \phi+\omega & \beta_1\phi & \beta_2\phi & \beta_3\phi \\ 
        &  \beta_1^2 \phi + \psi_1  & \beta_1\beta_2\phi & \beta_1\beta_3\phi \\
        &      &  \beta_2^2 \phi + \psi_2  & \beta_2\beta_3\phi \\
        &      &      & \beta_3^2 \phi + \psi_3 \\
        \end{array} \right) 
\end{eqnarray*}
\begin{itemize}
     \item Ten equations in eight unknowns.
     \item Unique solution possible but not guaranteed.
     \item Primary interest is still in $\beta_1$.
     \item Assume $\beta_2 \neq 0$ and $\beta_3 \neq 0$, meaning only that $Y_2$ and $Y_3$ are well chosen.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Solve for $\phi$}
\framesubtitle{Assuming $\beta_2 \neq 0$ and $\beta_3 \neq 0$} 
{\scriptsize
\begin{displaymath}
   \left( \begin{array}{c c  c c}
    \sigma_{11} & \sigma_{12} & \sigma_{13} & \sigma_{14} \\ 
     &  \sigma_{22}  & \sigma_{23} & \sigma_{24} \\
     &      &  \sigma_{33}  & \sigma_{3,4}\\
     &      &               & \sigma_{4,4}
        \end{array} \right) 
=       \left( \begin{array}{c c  c c}
        \phi+\omega & \beta_1\phi & \beta_2\phi & \beta_3\phi \\ 
        &  \beta_1^2 \phi + \psi_1  & \beta_1\beta_2\phi & \beta_1\beta_3\phi \\
        &      &  \beta_2^2 \phi + \psi_2  & \beta_2\beta_3\phi \\
        &      &      & \beta_3^2 \phi + \psi_3 \\
        \end{array} \right) 
\end{displaymath}
} % End size
{\LARGE
\begin{displaymath}
        \frac{\sigma_{13} \sigma_{14}} {\sigma_{34}} 
    = \frac{\beta_2\beta_3\phi^2}{\beta_2\beta_3\phi} = \phi
\end{displaymath} 
} % End size
\end{frame}

\begin{frame}
\frametitle{Then all you have to write is}
%\framesubtitle{} 
\begin{eqnarray*} 
    \omega & = & \sigma_{11} - \phi              \\ \nonumber
    \beta_1      & = & \frac{\sigma_{12}}{\phi}        \\ \nonumber            
    \beta_2      & = & \frac{\sigma_{13}}{\phi}        \\ \nonumber
    \beta_3      & = & \frac{\sigma_{14}}{\phi}        \\ \nonumber
    \psi_1        & = & \sigma_{22} - \beta_1^2 \phi   \\ \nonumber
    \psi_2        & = & \sigma_{33} - \beta_2^2 \phi   \\ \nonumber
    \psi_3        & = & \sigma_{44} - \beta_3^2 \phi   \\ \nonumber
\end{eqnarray*} 
Notice again how once we have solved for a model parameter, we may use it to solve for other parameters without explicitly substituting in terms of $\sigma_{ij}$. 
\end{frame}

\begin{frame}[fragile]
\frametitle{Parameters can be recovered from the covariance matrix}
%\framesubtitle{} 

\begin{tabular}{ll}
\begin{minipage}{2in}
\begin{eqnarray*}
    \phi   & = & \frac{\sigma_{13} \sigma_{14}} {\sigma_{34}}   \\ 
    \omega & = & \sigma_{11} - \phi                             \\ 
    \beta_1      & = & \frac{\sigma_{12}}{\phi}                 \\             
    \beta_2      & = & \frac{\sigma_{13}}{\phi}                 \\ 
\end{eqnarray*} 
\end{minipage} 
&
\begin{minipage}{2in}
\begin{eqnarray*} 
    \beta_3       & = & \frac{\sigma_{14}}{\phi}        \\ 
    \psi_1        & = & \sigma_{22} - \beta_1^2 \phi   \\ 
    \psi_2        & = & \sigma_{33} - \beta_2^2 \phi   \\ 
    \psi_3        & = & \sigma_{44} - \beta_3^2 \phi   \\ 
\end{eqnarray*} 
\end{minipage}
\end{tabular}

  \begin{itemize}
    \item Parameter vector is identifiable almost everywhere in the parameter space.
    \item Everywhere $\beta_2$ and $\beta_3$ are both non-zero
    \item $\beta_1 = 0 \Leftrightarrow \sigma_{12}=0$ presents no problem.
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{But there is more than one way to recover the parameter values from $\boldsymbol{\Sigma}$}
{\scriptsize
\begin{displaymath}
   \left( \begin{array}{c c  c c}
    \sigma_{11} & \sigma_{12} & \sigma_{13} & \sigma_{14} \\ 
     &  \sigma_{22}  & \sigma_{23} & \sigma_{24} \\
     &      &  \sigma_{33}  & \sigma_{3,4}\\
     &      &               & \sigma_{4,4}
        \end{array} \right) 
=       \left( \begin{array}{c c  c c}
        \phi+\omega & \beta_1\phi & \beta_2\phi & \beta_3\phi \\ 
        &  \beta_1^2 \phi + \psi_1  & \beta_1\beta_2\phi & \beta_1\beta_3\phi \\
        &      &  \beta_2^2 \phi + \psi_2  & \beta_2\beta_3\phi \\
        &      &      & \beta_3^2 \phi + \psi_3 \\
        \end{array} \right) 
\end{displaymath}
} % End size

\begin{eqnarray*}
    \beta_1 & = & \frac{\sigma_{12}\sigma_{34}} {\sigma_{13}\sigma_{14}}    \\
&&\\
    \beta_1  & = & \frac{\sigma_{23}}{\sigma_{13}}      \\
&&\\
    \beta_1  & = & \frac{\sigma_{24}}{\sigma_{14}}
\end{eqnarray*} 
\end{frame}

\begin{frame}
\frametitle{Is there a problem?}
%\framesubtitle{} 
\begin{displaymath}
    \beta_1 = \frac{\sigma_{12}\sigma_{34}} {\sigma_{13}\sigma_{14}}
            = \frac{\sigma_{23}}{\sigma_{13}} = \frac{\sigma_{24}}{\sigma_{14}}
\end{displaymath} 

Does this mean the solution for $\beta_1$ is not ``unique?"
  \begin{itemize}
    \item No -- everything is okay.
    \item If the parameters can be recovered from the covariances in any way at all, they are identifiable.
    \item If the model is correct, all the seemingly different ways must be the same.
    \item That is,
        \begin{displaymath}
            \frac{\sigma_{12}\sigma_{34}} {\sigma_{13}\sigma_{14}}
                    = \frac{\sigma_{23}}{\sigma_{13}} \mbox{ and }
            \frac{\sigma_{12}\sigma_{34}} {\sigma_{13}\sigma_{14}}
                    = \frac{\sigma_{24}}{\sigma_{14}}
        \end{displaymath} 
    \item Simplifying a bit,
        \begin{displaymath}
            \sigma_{12}\sigma_{34} = \sigma_{14}\sigma_{23} 
                                   = \sigma_{13}\sigma_{24} 
        \end{displaymath} 
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Model implies two constraints on the covariance matrix}
%\framesubtitle{} 
{\LARGE
\begin{displaymath}
    \sigma_{12}\sigma_{34} = \sigma_{14}\sigma_{23} 
           = \sigma_{13}\sigma_{24}
\end{displaymath} 
} % End size
  \begin{itemize}
    \item All three products equal $\beta_1\beta_2\beta_3\phi^2$
    \item True even when some $\beta_j=0$
  \end{itemize}

\vspace{5mm}

{\scriptsize
\begin{displaymath}
   \left( \begin{array}{c c  c c}
    \sigma_{11} & \sigma_{12} & \sigma_{13} & \sigma_{14} \\ 
     &  \sigma_{22}  & \sigma_{23} & \sigma_{24} \\
     &      &  \sigma_{33}  & \sigma_{3,4}\\
     &      &               & \sigma_{4,4}
        \end{array} \right) 
=       \left( \begin{array}{c c  c c}
        \phi+\omega & \beta_1\phi & \beta_2\phi & \beta_3\phi \\ 
        &  \beta_1^2 \phi + \psi_1  & \beta_1\beta_2\phi & \beta_1\beta_3\phi \\
        &      &  \beta_2^2 \phi + \psi_2  & \beta_2\beta_3\phi \\
        &      &      & \beta_3^2 \phi + \psi_3 \\
        \end{array} \right) 
\end{displaymath}
} % End size
\end{frame}

\begin{frame}
\frametitle{Model implies $\sigma_{12}\sigma_{34} = \sigma_{14}\sigma_{23} 
           = \sigma_{13}\sigma_{24}$}

  \begin{itemize}
    \item Parameter is identifiable.
    \item Ten equations in eight unknowns.
    \item Call the parameter \emph{over-identifiable}.
    \item If there were the same number of equations as unknowns, it would be \emph{just identifiable}.
    \item Model imposes two equality constraints (restrictions) on the covariance matrix: $10-8=2$
    \item Sometimes called \emph{over-identifying restrictions}.
    \item These are the constraints that are tested in the likelihood ratio test for goodness of fit. 
    \item More instrumental variables can't hurt.
  \end{itemize}
\end{frame}


\section{Multiple Explanatory Variables}

\begin{frame}
\frametitle{Multiple Explanatory Variables}
\framesubtitle{An example with just two explanatory variables (and two instrumental variables)}
Independently for $i=1, \ldots, n$,
\begin{eqnarray*} 
   W_{i,1} & = & \nu_1 +  X_{i,1} + e_{i,1}            \\  
   Y_{i,1} & = & \alpha_1 + \beta_1 X_{i,1} + \epsilon_{i,1}  \\  
   Y_{i,2} & = & \alpha_2 + \beta_2 X_{i,1} + \epsilon_{i,2}  \\  
   W_{i,2} & = & \nu_2 +  X_{i,2} + e_{i,2}            \\  
   Y_{i,3} & = & \alpha_3 + \beta_3 X_{i,2} + \epsilon_{i,3}  \\  
   Y_{i,4} & = & \alpha_4 + \beta_4 X_{i,2} + \epsilon_{i,4}  \\  
\end{eqnarray*}
where $E(X_{i,j})=\mu_j$, $e_{i,j}$ and  $\epsilon_{i,j}$ are independent of one another and of $X_{i,j}$, $Var(e_{i,j})=\omega_j$, $Var(\epsilon_{i,j})=\psi_j$, and
\begin{displaymath}
V\left(
\begin{array}{c} X_{i,1} \\ X_{i,1} \end{array}
 \right) = \left(
\begin{array}{c c} 
\phi_{11} & \phi_{12} \\
\phi_{12} & \phi_{22}
\end{array} \right)
\end{displaymath} 
\end{frame}

\begin{frame}
\frametitle{As usual, intercepts and expected values can't be recovered individually}
%\framesubtitle{} 
  \begin{itemize}
    \item Eight intercepts and expected values of latent variables.
    \item Six expected values of observable variables.
    \item Re-parameterize, absorbing them into $\mu_1, \ldots, \mu_6$.
    \item Estimate with the vector of 6 sample means and set them aside, forever.
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Covariance matrix of $(W_{i,1}, Y_{i,1}, Y_{i,2}, W_{i,2}, Y_{i,3}, Y_{i,4})^\prime$}
%\framesubtitle{} 
$ [\sigma_{ij}] = $

\vspace{5mm}

{\scriptsize
$
    \left( \begin{array}{c c c c c c}
        \phi_{11}+\omega_1 & \beta_1\phi_{11} & \beta_2\phi_{11} & \phi_{12} & \beta_3\phi_{12} & \beta_4\phi_{12} \\ 
        &  \beta_1^2 \phi_{11} + \psi_1  & \beta_1\beta_2\phi_{11} & \beta_1\phi_{12} & \beta_1\beta_3\phi_{12} & \beta_1\beta_4\phi_{12} \\
        &      &  \beta_2^2 \phi_{11} + \psi_2  & \beta_2\phi_{12} & \beta_2\beta_3\phi_{12} &  \beta_2\beta_4\phi_{12} \\
        &      &      & \phi_{22} + \omega_2 & \beta_3\phi_{22} & \beta_4\phi_{22} \\
        &      &      &     &  \beta_3^2 \phi_{22} + \psi_3  &  \beta_3\beta_4\phi_{22} \\
        &      &      &     &     &   \beta_4^2 \phi_{22} + \psi_4   \\
        \end{array} \right)
$ 
} % End size

\vspace{5mm}

$\boldsymbol{\theta} = (\beta_1, \beta_2, \beta_3, \beta_4, \phi_{11}, \phi_{12}, \phi_{22}, \omega_1, \omega_2, \psi_1, \psi_2, \psi_3, \psi_4)$

{\small
  \begin{itemize}
    \item Does this model pass the test of the parameter count rule?
    \item Where the parameter vector is identifiable, how many over-identifying restrictions are there? % 6*7/2 - 13 = 7
    \item How many degrees of freedom in the likelihood ratio test for model fit?
  \end{itemize}
} % End size
\end{frame}

\begin{frame}
\frametitle{Where is the entire parmeter vector identifiable?}
%\framesubtitle{} 
{\scriptsize
$
    \left( \begin{array}{c c c c c c}
        \phi_{11}+\omega_1 & \beta_1\phi_{11} & \beta_2\phi_{11} & \phi_{12} & \beta_3\phi_{12} & \beta_4\phi_{12} \\ 
        &  \beta_1^2 \phi_{11} + \psi_1  & \beta_1\beta_2\phi_{11} & \beta_1\phi_{12} & \beta_1\beta_3\phi_{12} & \beta_1\beta_4\phi_{12} \\
        &      &  \beta_2^2 \phi_{11} + \psi_2  & \beta_2\phi_{12} & \beta_2\beta_3\phi_{12} &  \beta_2\beta_4\phi_{12} \\
        &      &      & \phi_{22} + \omega_2 & \beta_3\phi_{22} & \beta_4\phi_{22} \\
        &      &      &     &  \beta_3^2 \phi_{22} + \psi_3  &  \beta_3\beta_4\phi_{22} \\
        &      &      &     &     &   \beta_4^2 \phi_{22} + \psi_4   \\
        \end{array} \right)
$ 
} % End size

\vspace{5mm}

  \begin{itemize}
    \item What happens if $\beta_1=\beta_2=0$?
    \item Why is it reasonable to assume $\beta_2 \neq 0$ and $\beta_4 \neq 0$?
    \item In that case, what else do you need?
    \item Would any other condition identify the whole parameter vector?
  \end{itemize}
\end{frame}


\begin{frame}
\frametitle{My answer}
%\framesubtitle{To ``Where is the entire parmeter vector identifiable?"} 
{\scriptsize
$
    \left( \begin{array}{c c c c c c}
        \phi_{11}+\omega_1 & \beta_1\phi_{11} & \beta_2\phi_{11} & \phi_{12} & \beta_3\phi_{12} & \beta_4\phi_{12} \\ 
        &  \beta_1^2 \phi_{11} + \psi_1  & \beta_1\beta_2\phi_{11} & \beta_1\phi_{12} & \beta_1\beta_3\phi_{12} & \beta_1\beta_4\phi_{12} \\
        &      &  \beta_2^2 \phi_{11} + \psi_2  & \beta_2\phi_{12} & \beta_2\beta_3\phi_{12} &  \beta_2\beta_4\phi_{12} \\
        &      &      & \phi_{22} + \omega_2 & \beta_3\phi_{22} & \beta_4\phi_{22} \\
        &      &      &     &  \beta_3^2 \phi_{22} + \psi_3  &  \beta_3\beta_4\phi_{22} \\
        &      &      &     &     &   \beta_4^2 \phi_{22} + \psi_4   \\
        \end{array} \right)
$ 
} % End size

\vspace{5mm}

EITHER
  \begin{itemize}
    \item One of $\beta_1$ and $\beta_2$ non-zero, and
    \item One of $\beta_3$ and $\beta_4$ non-zero, and
    \item $\phi_{12} \neq 0$
  \end{itemize}
OR, all of $\beta_1, \ldots, \beta_4$ non-zero.
\end{frame}

\begin{frame}
\frametitle{Could we get by with less information?}
\framesubtitle{If we wanted to identify just some interesting parameters?} 
{\scriptsize
$
    \left( \begin{array}{c c c c c c}
        \phi_{11}+\omega_1 & \beta_1\phi_{11} & \beta_2\phi_{11} & \phi_{12} & \beta_3\phi_{12} & \beta_4\phi_{12} \\ 
        &  \beta_1^2 \phi_{11} + \psi_1  & \beta_1\beta_2\phi_{11} & \beta_1\phi_{12} & \beta_1\beta_3\phi_{12} & \beta_1\beta_4\phi_{12} \\
        &      &  \beta_2^2 \phi_{11} + \psi_2  & \beta_2\phi_{12} & \beta_2\beta_3\phi_{12} &  \beta_2\beta_4\phi_{12} \\
        &      &      & \phi_{22} + \omega_2 & \beta_3\phi_{22} & \beta_4\phi_{22} \\
        &      &      &     &  \beta_3^2 \phi_{22} + \psi_3  &  \beta_3\beta_4\phi_{22} \\
        &      &      &     &     &   \beta_4^2 \phi_{22} + \psi_4   \\
        \end{array} \right)
$ 
} % End size

\vspace{5mm}

  \begin{itemize}
    \item Usual rule in Econometrics is at least one instrumental variable for each explanatory variable.
    \item What if no instrumental variable for $X_2$?
    \item What if no response variables at all for $X_2$?
  \end{itemize}
\end{frame}


\begin{frame}
\frametitle{Observations}
%\framesubtitle{} 
  \begin{itemize}
    \item Instrumental variables can solve some of the terrible problems with measurement error in regression.
    \item General rules like ``At least one instrumental variable for each explanatory variable" are useful  even if they are over-simplifications.
    \item Awareness of parameter identifiability is vital in the \emph{planning} of data collection.
    \item[] 
    \item Most observational data sets are collected without the right kind of planning. 
  \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}}

\end{frame}


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