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\begin{centering}
\large
Intro to Econometric Theory \linebreak
Heinz School, Carnegie Mellon University \linebreak
90-906, Spring 2000 \linebreak
\vspace{0.2in} \linebreak
Solution \#3
\normalsize

\end{centering}
\vspace{0.3in}

\begin{enumerate}
\item \textbf{
      Consider the generalized regression model.  Suppose we try to run
      GLS using a matrix $\Omega_1$ which we believe is equal to 
      (or a good estimator of) $\Omega$.  Suppose further that we are 
      wrong, and that $\Omega_1 \ne \Omega$.  Let's call the estimator
      we calculate using $\Omega_1$, $\hat{\beta}_{WGLS}$ (W for wrong).
      }
      \begin{enumerate}
        \item \textbf{
              Is $\hat{\beta}_{WGLS}$ unbiased?
              }

              The formula is 
              $\hat{\beta}_{WGLS}=
                \left(X'\Omega_1^{-1}X\right)^{-1}X'\Omega_1^{-1}Y$.
              We take expectations conditional upon $X$ as follows:

              \begin{eqnarray}
                E\left\{\left.\hat{\beta}_{WGLS}\right|X\right\} 
                  &amp;=&amp; \left(X'\Omega_1^{-1}X\right)^{-1}X'\Omega_1^{-1}E\{Y|X\} 
                      \nonumber\\
                  &amp;=&amp; \left(X'\Omega_1^{-1}X\right)^{-1}
                      X'\Omega_1^{-1}X\beta 
                      \nonumber \\
                  &amp;=&amp; \beta \nonumber
              \end{eqnarray}            

              Therefore, $\hat{\beta}_{WGLS}$ is unbiased.

        \item \textbf{
              What is its variance?
              }

              Notice, throughout the answers I will omit the $\sigma^2$
              from the expressions for variance --- including them would
              change little and would be tedious.

              \begin{eqnarray}
                V\left\{\left.\hat{\beta}_{WGLS}\right|X\right\} 
                  &amp;=&amp; V\left\{\left.
                     \left(X'\Omega_1^{-1}X\right)^{-1}
                      X'\Omega_1^{-1}Y\right|X\right\} \nonumber\\
                  &amp;=&amp; V\left\{\left.
                     \beta + 
                     \left(X'\Omega_1^{-1}X\right)^{-1}
                      X'\Omega_1^{-1}\epsilon\right|X\right\} \nonumber\\
                  &amp;=&amp; \left(X'\Omega_1^{-1}X\right)^{-1}X'\Omega_1^{-1}
                      V\left\{\epsilon|X\right\}
                      \Omega_1^{-1}X\left(X'\Omega_1^{-1}X\right)\nonumber\\
                  &amp;=&amp; \left(X'\Omega_1^{-1}X\right)^{-1}X'\Omega_1^{-1}
                      \Omega
                      \Omega_1^{-1}X\left(X'\Omega_1^{-1}X\right)\nonumber
              \end{eqnarray}

        \item \textbf{
              Can you think of a reasonable estimator for this variance?
              }

              In general, no.  We would need an estimator of $\Omega$, which
              has $N(N-1)$ parameters.  However, in the case of heteroskedasticity,
              we could substitute in $\hat{\Omega}$, a matrix with $e^2_i$ on the
              diagonal, similar to White's estimator for robust standard errors.

        \item \textbf{
              Is this variance bigger or smaller than the GLS variance?
              }

              It is larger than the GLS variance.  Recall from class that
              we can show 
              $V\left(\hat{\beta}_{GLS}\le\right)V\left(\hat{\beta}_{WGLS}\right)$
              by showing
              $\left(V\left(\hat{\beta}_{GLS}\right)\right)^{-1} \ge
               \left(V\left(\hat{\beta}_{WGLS}\right)\right)^{-1}$
              which is the same as showing that
              $\left(V\left(\hat{\beta}_{GLS}\right)\right)^{-1} -
               \left(V\left(\hat{\beta}_{WGLS}\right)\right)^{-1}$
               is positive definite.

               \begin{align}
                  \left(V\left(\hat{\beta}_{GLS}\right)\right) &amp; ^{-1} -
                    \left(V\left(\hat{\beta}_{WGLS}\right)\right)^{-1}
                 \nonumber\\
                    &amp;= \left(X'\Omega^{-1}X\right) - 
                       \left(X'\Omega_1^{-1}X\right)
                       \left(X'\Omega_1^{-1}\Omega\Omega_1^{-1}X\right)^{-1}
                       \left(X'\Omega_1^{-1}X\right) 
                 \nonumber\\
                    &amp;= X'\Omega^{-\frac{1}{2}}
                       \left(I - \Omega^{\frac{1}{2}}
                             \Omega_1^{-1}X
                             \left(X'\Omega_1^{-1}\Omega\Omega_1^{-1}X\right)^{-1}
                             X'\Omega_1^{-1}
                             \Omega^{\frac{1}{2}} \right) \Omega^{-\frac{1}{2}}X
                 \nonumber
               \end{align}

              This difference is positive semi-definite since the matrix in
              the middle is idempotent (hence positive semi-definite) and it
              is then pre and post multiplied by a matrix and its transpose.

        \item \textbf{
              What should we conclude about the dangers of trying to 
              correct for a non-scalar covariance matrix, but screwing it
              up?
              }

              If we try and fail to correct for heteroskedasticity, we are
              not obviously any worse off than if we did not try at all.  Our
              estimator is still unbiased (and CAN).  Our estimator still has
              a covariance matrix we can calculate and estimate, as long as 
              we could have calculated and estimated the OLS estimator's.  
              Our estimator is still
              worse than GLS.  Finally, our estimator may be worse or 
              better than OLS.  There don't seem to be any really bad dangers
              to trying to correct for heteroskedasticity --- if we get it 
              wrong we are in a similar situation to where we started with OLS,
              and if we get it right, we are in a better situation than with
              OLS.

       \end{enumerate}

\item \textbf{
      Consider the conditional expectation assumption necessary 
      for consistency of OLS in the random
      $X$ version of the regression model.  Think about what this 
      assumption means in the context of equation~\ref{eqn:levels} below.
      Tell me in natural language what the assumption means.  Does this
      assumption seem reasonable to you?
      }

      The conditional expectation assumption in the context of 
      equation~\ref{eqn:levels} below says that  hospitals with higher
      (unobservable) costs have no more or fewer days or visits than 
      do hospitals with lower costs.  This does not seem to be a very
      good assumption.  There are several reasons we might expect that
      a hospital's (unobservable) costs would be correlated with its 
      level of output.  If higher quality hospitals are more costly than
      low quality hospitals and if patients (or more likely their physicians)
      have some information about this quality, then we should expect high
      (unobserved) cost hospitals to have more patients.  Conversely,
      low cost hospitals are likely to charge lower rates for their services
      than are high cost hospitals --- this is likely to lead to low cost
      hospitals having more patients as insurance companies and HMOs direct
      their patients to low cost providers.  It's hard to know which of these
      forces (or others you can probably name) will prevail, but it would
      be an amazing stroke of luck if they precisely cancelled. 

      Notice two things about the above stories.  First, they depend on
      \emph{unobservables} --- it would be wrong to say that costs and
      output might be related because of scale economies, since we observe
      output and therefore can measure scale economies (this would be
      a story of correlation between $C$ and $D^2$ or $V^2$ not a story
      of correlation between $C$ and $\epsilon$).  Second, we could ``fix''
      the problem with quality discussed above if we had a reliable
      measure of hospital quality which we could drop into the regression.

\item \textbf{
      With respect to the hospital cost dataset from the web site, please
      estimate the following two models:
      }

      \begin{eqnarray}
         lnC &amp;=&amp; \beta_1 + \beta_2lnD + \beta_3lnV + \beta_4time + \epsilon
               \label{eqn:logs}\\
         C   &amp;=&amp; \alpha_1 + \alpha_2D + \alpha_3V + \alpha_4time + \epsilon
               \label{eqn:levels}
      \end{eqnarray}

      \textbf{
      In the above, $D$ is inpatient days, $V$ is outpatient visits, and 
      $time$ is the number of quarters since the first quarter of 1991.
      }

      \begin{enumerate}
        \item \textbf{
              Discuss how the estimates that these two equations provide
              for marginal costs differ.
              }

              In the linear case, marginal costs may just be read off as
              parameter estimates.  Our best estimate is that the marginal
              cost of a day is \$202 and the marginal cost of a visit is
              \$1492.  In the logarithmic case, the expected marginal cost 
              of a visit is
 
              \begin{displaymath}
              exp\left(\beta_1 + \beta_2lnD + \beta_3lnV 
                        + \beta_4time\right)
               E\left\{exp(\epsilon)\right\}
               \frac{\beta_3}{visits}
               \end{displaymath}
               
               and the expected marginal cost of a day is

               \begin{displaymath}
               exp\left(\beta_1 + \beta_2lnD + \beta_3lnV 
                        + \beta_4time\right)
               E\left\{exp(\epsilon)\right\}
               \frac{\beta_2}{days}
               \end{displaymath}

               Obviously, these differ for different observations.  We can
               estimate the two marginal costs by replacing $\beta$ with
               $\hat{\beta}$ and by replacing $Eexp(\epsilon)$ by
               $\frac{1}{N}\sum exp(e_i)$.  But the answer still depends upon
               the values of $D,V,time$.  To get a single number, I just 
               average over the points in the dataset (see sas program).

               After all this, our estimate of the average of the marginal
               costs of a day are \$360 and the average of the estimated 
               marginal costs of a visit are \$1154.

        \item \textbf{
              Test for non-linearity in equation~\ref{eqn:levels}.  Discuss
              the meaning of your findings.
              }

              We can do this with an F-test comparing the regressions on
              pages 1 and 2 of the output:

              \begin{eqnarray}
                F-stat &amp;=&amp; \frac{\left(3.79 \times 10^{17} 
                                 - 3.34 \times 10^{17}\right)/6}
                               {3.34 \times 10^{17}/7525} \nonumber\\
                       &amp;=&amp; 169 \nonumber
              \end{eqnarray}

              So, we reject the null hypothesis of linearity.  This is
              some kind of evidence in favor of the log model, since it
              permits some non-linearity.

        \item \textbf{
              Think carefully about the variable $C$.  Think carefully 
              about equation~\ref{eqn:levels} (not the estimated 
              version, but the theoretical version).  Are there 
              circumstances in which equation~\ref{eqn:levels} can give 
              obviously wrong predictions for$C$?  Does 
              equation~\ref{eqn:logs} suffer from a similar problem?
              }

              Equation~\ref{eqn:levels} can give negative predicted values
              for costs; therefore, it is suspect.  Equation~\ref{eqn:logs}
              does not have this defect.

        \item \textbf{
              Calculate predicted values for $C$ from both models.  
              Calculate the mean squared error for both models: 
              $\sum(C-\hat{C})$.  Which model is better by this criterion?
              }  

              The mean squared error for the linear model can be read off
              of the regression output on page 1.  It is 
              $5.033 \times 10^{13}$.  The mean squared error for the log
              model must be calculated (see SAS program).  Its value appears
              on page 5 and is $5.947 \times 10^{13}$.

        \item \textbf{
              Would it make sense to compare the $R^2$ from the two 
              equations in order to decide which is best?  Why or why not?
              }

              Absolutely not.  The $R^2$ from the linear model tells you
              how much of the variance in costs the RHS variables explain;
              whereas, the $R^2$ from the log model tells you how much of the
              variance in the log of costs the RHS variables explain.  Since
              different LHS variables are being explained, different 
              $R^2$ are to be expected and these differences tell us nothing
              about the relative merits of the two models.  If your goal is
              to predict costs, then the MSE in predicted costs is the 
              right measure (or at least a much better measure)
              to use to compare models.

              It is often the case that, when the LHS is distributed with
              a ``thick right tail'' (lots of really big values) that 
              a log model looks better on its $R^2$ than does a linear
              model.  This is because logarithms really shrink big numbers
              a lot.

        \item \textbf{
              If we care about getting the right answer, which equation should
              we use and why? (You may do more analysis if you like to answer
              this question)
              }

              I probably would prefer the logarithmic model on theoretical
              grounds --- it can't give nonsense predictions, marginal 
              cost can be different at different levels of the RHS variables,
              elasticities can be read off easily, and the coefficient on time
              represents a growth rate.

              On empirical grounds, however, the linear model looks better,
              with its lower MSE (although the test for non-linearity did
              show that the linear model was inadequate).

              At this point, I would probably proceed by making both models
              more flexible --- put in squares and interactions for both
              models and then compare MSE again.

       \end{enumerate}

\item \textbf{
      Redo question 3a from assignment 2.  This time address heteroskedasticity.
      }

      First, we will want to test for heteroskedasticity using White's test.
      Notice, there are a lot of duplicative terms in the RHS variables of
      White's test.  For example $V^2$ should appear on the RHS both because
      $V$ appears on the RHS in the original model and because $V^2$ appears
      on the RHS in the original model.  On page 7 of the output is the regression
      for White's test.  $NR^2=7535(0.1114)=839$.  Thus the null hypothesis of
      no heteroskedasticity will be rejected at any reasonable significance 
      level.  

      The parts needing to be redone are ii, iii, and iv.  First, we are to
      test for scale economies in visits and in days.  To do the test, I just
      use robust standard errors, since I don't have any theory of the
      heteroskedasticity.

      First, testing for constant returns separately for each of days and 
      visits:

      \begin{eqnarray}
        t-stat &amp;=&amp; \frac{-0.000075}{\sqrt{3.10 \times 10^{-8}}}
                  = -0.4 \nonumber\\
        t-stat &amp;=&amp; \frac{0.008}{\sqrt{7.3 \time 10^{-6}}} = 3 \nonumber
      \end{eqnarray}

      So, at reasonable significance levels, we will reject the null of
      constant returns in visits and accept the null of constant returns
      in days.

      Now, we test jointly:

      \begin{eqnarray}
        W &amp;=&amp; \left(\begin{array}{r r}
                     8.0 \times 10^{-3} &amp; -7.5 \times 10^{-5}
                    \end{array}
              \right)
              \left[\begin{array}{r r} 
                     7.3 \times 10^{-6} &amp; 3.3 \times 10^{-7} \\
                     3.3 \times 10^{-7} &amp; 3.1 \times 10^{-8}
                    \end{array}
              \right]^{-1}
              \left(\begin{array}{r}
                     8.0 \times 10^{-3} \\ -7.5 \times 10^{-5}
                    \end{array}
              \right) \nonumber\\
          &amp;=&amp; 20.6
      \end{eqnarray}

      Looking at the $\chi^2_2$ table,
      at the 1\% significance level I get 9.21.  Thus, I reject the
      null of no scale economies.

      A one standard deviation increase in days is an increase of 
      24,167 (see page 4).  My best estimate of its effect on the 
      marginal cost of a day is 
      $2(-7.5 \times 10^{-5})24,167=-3.63$.  So, my best guess is that
      the marginal cost of a day will fall by \$3.63.  A 95\% confidence
      interval will be 
      $3.63 \pm 1.96(\sqrt{3.1 \times 10^{-8}})(24,167)$ or
      $3.63 \pm 8.34$. 

\end{enumerate}

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