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\begin{centering}
\large
Intro to Econometric Theory \linebreak
Heinz School, Carnegie Mellon University \linebreak
90-906, Spring 2003-4 \linebreak
\vspace{0.2in} \linebreak
Solution \#2
\normalsize

\end{centering}
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\begin{enumerate}
\item You have collected a dataset on consumers' purchases of wine
      and cars.  In particular, for each of 100 consumers, you know
      the price of their most recent bottle of wine and the 
      price of their most recent car.

      \begin{eqnarray}
      \overline{P_{\mathrm{wine}}} &=& 6 \mathrm{in \$}         \nonumber\\
      \hat{\sigma}^2_{\mathrm{wine}} &=& 10                     \nonumber\\
                                                                \nonumber\\
      \overline{P_{\mathrm{car}}} &=& 18 \mathrm{in \$}         \nonumber\\
      \hat{\sigma}^2_{\mathrm{car}} &=& 10                      \nonumber\\
                                                                \nonumber\\
      \widehat{\mathrm{Cov}}(P_w,P_c) &=& 5                     \nonumber
      \end{eqnarray}

      Now, suppose we wish to model the relationship between these two 
      prices as:

      \begin{equation}
      P_c = \beta_0 + \beta_1 P_w + \epsilon
      \end{equation}

      What are the OLS estimates for $\beta_0$ and $\beta_1$?

      \textbf{
      The OLS estimators are:
      }

      \begin{eqnarray}
      \hat{\beta}_{1,OLS} &=& \frac{\widehat{Cov}(P_c,P_w)}{\hat{V}(P_w)}
                              \nonumber\\
                          &=& \frac{5}{10} \nonumber\\
                          &=& 0.5 \nonumber\\
                          & & \nonumber\\
      \hat{\beta}_{0,OLS} &=& \overline{P_c} 
                              - \hat{\beta}_{1,OLS}\overline{P_w} \nonumber\\
                          &=& 18 - 0.5(6) \nonumber\\
                          &=& 15 \nonumber
      \end{eqnarray}


\item Should the clever people at General Motors start distributing
      expensive wine?

      \textbf{
      No.  Although the setup of the CLRM essentially assumes a causation
      from the RHS to the LHS variable, this causation seems quite 
      implausible here.  \emph{If} it were to be the case that drinking
      expensive wine causes people to buy expensive cars, then perhaps
      car-sellers would be wise to distribute expensive wine. 
      }

      \textbf{
      However, it seems likely that, in fact, income and/or taste for
      luxury goods explains both a tendency to buy expensive wine and 
      a tendency to buy expensive cars.  So, since handing out wine
      is unlikely to change a persons' income or tastes much, we should
      expect little or no effect of handing out expensive wine on 
      spending for cars.
      }

\item Suppose instead we model:

      \begin{equation}
      P_w = \beta_2 + \beta_3 P_c + \epsilon
      \end{equation}

      What are the OLS estimates for $\beta_0$ and $\beta_1$?

      \textbf{
      The OLS estimators are:
      }

      \begin{eqnarray}
      \hat{\beta}_{3,OLS} &=& \frac{\widehat{Cov}(P_c,P_w)}{\hat{V}(P_c)}
                              \nonumber\\
                          &=& \frac{5}{10} \nonumber\\
                          &=& 0.5 \nonumber\\
                          & & \nonumber\\
      \hat{\beta}_{2,OLS} &=& \overline{P_w} 
                              - \hat{\beta}_{1,OLS}\overline{P_c} \nonumber\\
                          &=& 6 - 0.5(18) \nonumber\\
                          &=& -3 \nonumber
      \end{eqnarray}


\item One might expect that $\beta_{1,OLS}=\frac{1}{\beta_{3,OLS}}$.
      True?  Why or why not?

      \textbf{
      False, both are 0.5.  We would expect 
      $\hat{\beta}_{1,OLS}=\frac{1}{\hat{\beta}_{3,OLS}}$ 
      if the two ways of
      running OLS produced the \emph{same line}.  But they should
      not, in general, produce the same line.  The reason for this
      is that the two ways of running OLS embody different ideas of 
      what a ``mistake'' is.  In the first case, a mistake is the
      distance between a data point and the line in $P_c$.  In the
      second case, a mistake is the distance between a data point
      and the line in $P_w$.  So, since the two ways of running OLS
      minimize different mistake criteria, they should be expected
      to reach different answers.
      }

\item Exercise 6 on page 63 of Greene.

      \textbf{
      It is pretty easy to show that the profit-maximizing level of 
      output is:}

      \begin{equation}
      q^* = \frac{1}{2} \alpha + \frac{1}{2} \beta 10 + \frac{1}{2} \epsilon
      \nonumber
      \end{equation}

      \textbf{
      The expected value of this is going to be:
      }

      \begin{equation}
      E(q^*) = \frac{1}{2} \alpha + \frac{1}{2} \beta 10 \nonumber
      \end{equation}

      \textbf{
      Or, $\alpha/2 + 5 \beta$.  We will use OLS to find estimates of
      $\alpha$ and $\beta$.  Refer to the SAS program and output for
      the calculations.  Using the output of the corr procedure, we 
      can calculate the OLS coefficients as follows:
      }

      \begin{eqnarray}
      \hat{\beta}_{OLS} &=& \widehat{Corr}(P,Q)
                            \frac{\sqrt{\hat{V}(Q)}}
                                 {\sqrt{\hat{V}(P)}}
                            \nonumber\\
                        &=& -0.70 \frac{5.357}{4.464} \nonumber\\
                        &=& -0.840 \nonumber\\
                        & & \nonumber\\
      \hat{\alpha}_{OLS} &=& \overline{Q} 
                              - \hat{\beta}_{OLS}\overline{P} \nonumber\\
                          &=& 11.467 - (-0.84)(11.067) \nonumber\\
                          &=& 20.763 \nonumber
      \end{eqnarray}

      \textbf{
      Then, we can calculate fitted values for $Q$ and residuals $e$.
      This is also done in the SAS program and the results are in the
      SAS output.  The sum of squared residuals is 204.65, so that
      $\hat{\sigma}^2_{OLS} = 204.65/(15-2)=15.74$.
      }

      \textbf{
      Using the formulae for variance in the bivariate CLRM:
      }

      \begin{eqnarray}
      \hat{V}(\hat{\beta}) &=& \frac{\hat{\sigma}^2}
                                    {\sum (P_i-\overline{P})^2} \nonumber\\
                           &=& \frac{\hat{\sigma}^2}
                                    {(N-1)\hat{V}(P)} \nonumber\\
                           &=& \frac{15.74}{14(4.464)^2} \nonumber\\
                           &=& 0.0564 \nonumber
      \end{eqnarray}

      \begin{eqnarray}
      \hat{V}(\hat{\alpha}) &=& \frac{\hat{\sigma}^2}{N}
                                \frac{\sum P_i^2}{\sum (P_i - \overline{P})^2}
                                \nonumber\\
                            &=& \frac{\hat{\sigma}^2}{N}
                                \frac{N \overline{P_i^2}}{(N-1) \hat{V}(P)}
                                \nonumber\\
                            &=& \frac{15.74}{15}
                                \frac{15(141.07)}{14(4.464)^2} \nonumber\\
                                \nonumber\\
                            &=& 7.959 \nonumber
      \end{eqnarray}

      \begin{eqnarray}
      \widehat{Cov}(\hat{\alpha},\hat{\beta}) 
               &=& -\hat{\sigma}^2
                   \frac{\overline{P}}{\sum (P_i-\overline{P})^2}
                   \nonumber\\
               &=& -15.74 \frac{11.07}{14(4.464)^2} \nonumber\\
               &=& -0.625 \nonumber
      \end{eqnarray}

      \textbf{
      Now, our estimate of the expected profit-maximizing quantity
      is $\hat{q}^* = 0.5(20.77)-5(0.62) = 7.29$.  We need to know
      the variance of $0.5\hat{\alpha}+5\hat{\beta}$ in order to
      calculate the confidence interval.  We have a formula for
      the variance of a sum:
      }

      \begin{eqnarray}
      \hat{V}(0.5\hat{\alpha}+5\hat{\beta})
             &=& (0.5)^2 \hat{V}(\hat{\alpha}) + (5)^2 \hat{V}(\hat{\beta})
                 + 2(0.5)(5) \widehat{Cov}(\hat{\alpha},\hat{\beta}) 
                 \nonumber\\
             &=& 0.25(7.961) + 25(0.056) + 2(0.5)(5)(-0.625) \nonumber\\
             &=& 0.27 \nonumber
      \end{eqnarray}

      \textbf{
      Assuming normality for the errors and therefore for the OLS 
      estimators, a 95\% CI is then:
      }

      \begin{equation}
      7.29 \pm 1.96 \sqrt{0.27}
      \end{equation}

      \begin{equation}
      7.29 \pm 1.02
      \end{equation}

      \textbf{
      Notice, I was able to check many of my calculations using the
      proc reg command at the end of the SAS program.
      }


\end{enumerate}

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