<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass{article}

\usepackage{LectureNotes}

\begin{document}
\pagestyle{empty}

\begin{centering}
\large
Intro to Econometric Theory \linebreak
Heinz School, Carnegie Mellon University \linebreak
90-906, Spring 2000 \linebreak
\vspace{0.2in} \linebreak
Solution \#1
\normalsize

\end{centering}
\vspace{0.3in}

\begin{enumerate}
\item  \label{q:econometrics tricks}
       \textbf{
       Please prove the following for scalar random variables.  I will call these 
       things ``econometrics tricks'' and use them throughout the
       course.
       }
       \begin{enumerate}
         \item \label{q:trick-E}
               $E\left\{(X-EX)(Y-EY)\right\} = 
                E\left\{(X-EX)(Y)\right\} = 
                E\left\{(X)(Y-EY)\right\}$

                \begin{align}
                  E\{(X-EX)(Y-EY)\} &amp;= E\{(X-EX)Y - (X-EX)EY\} \nonumber\\
                                    &amp;= E\{(X-EX)Y\} - E\{(X-EX)EY\} \nonumber\\
                                    &amp;= E\{(X-EX)Y\} - EY \dot E\{(X-EX)\} 
                                       \nonumber\\
                                    &amp;= E\{(X-EX)Y\} - EY \dot 0 \nonumber\\
                                    &amp;= E\{(X-EX)Y\} \nonumber
                \end{align}

                The third equality holds since $EY$ is not a random variable,
                it is a (constant) scalar.  The proof for 
                $E\left\{(X-EX)(Y-EY)\right\} = E\left\{(X)(Y-EY)\right\}$
                is the same, except that we expand $(X-EX)$ instead of 
                expanding $(Y-EY)$ as we did above.\vspace{\baselineskip}
                

         \item $\overline{\left(X-\overline{X}\right)
                         \left(Y-\overline{Y}\right)} = 
                \overline{\left(X-\overline{X}\right)
                         \left(Y\right)} = 
                \overline{\left(X\right)
                         \left(Y-\overline{Y}\right)}$

                \begin{align}
                  \overline{\left(X-\overline{X}\right)
                    \left(Y-\overline{Y}\right)} 
                  &amp;=  \overline{\left(X-\overline{X}\right)
                    Y - \left(X-\overline{X}\right)\overline{Y}} \nonumber\\ 
                  &amp;=  \overline{\left(X-\overline{X}\right)Y} -
                      \overline{\left(X-\overline{X}\right)\overline{Y}}
                      \nonumber\\
                  &amp;=  \overline{\left(X-\overline{X}\right)Y} -
                      \overline{Y}\overline{\left(X-\overline{X}\right)}
                      \nonumber\\
                  &amp;=  \overline{\left(X-\overline{X}\right)Y} -
                      \overline{Y}0
                      \nonumber\\
                  &amp;=  \overline{\left(X-\overline{X}\right)Y}
                      \nonumber
                \end{align}
      \end{enumerate}
\item \label{q:vector tricks}
      \textbf{
      Please formulate and prove vector versions of the tricks in 
      question \ref{q:econometrics tricks}.
      }

      A vector version of the trick (and its proof)
      in question~\ref{q:trick-E} would look like:

      \begin{align}
        E\{(X-EX)(Y-EY)'\} &amp;= E\{(X-EX)Y'-(X-EX)EY'\} \nonumber\\
                           &amp;= E\{(X-EX)Y'\}-E\{(X-EX)EY'\} \nonumber\\
                           &amp;= E\{(X-EX)Y'\}-E\{(X-EX)\}EY' \nonumber\\
                           &amp;= E\{(X-EX)Y'\}-0EY' \nonumber\\
                           &amp;= E\{(X-EX)Y'\} \nonumber
      \end{align}

      Again, the third equality follows since $EY'$ is a constant.  Notice,
      it \emph{is} important that $EY'$ is factored out to the right.  
      The other three proofs follow in an exactly analogous fashion.

\item \textbf{
      Suppose that $\left\{X_i\right\}_{i=1}^N$ is a sequence of iid
      random variables with mean $\mu$ and variance $\sigma^2$.  Please
      use the theorems discussed in class for convergence in probability
      and convergence in distribution to prove the following:
      }
      \begin{enumerate}
        \item ${\scriptstyle\sqrt{N}}\overline{X} \goesindist \N{\mu}{\sigma^2}$
              \vspace{0.5\baselineskip}

              This follows by the Lindberg-Levy Central Limit Theorem directly.
              \vspace{0.5\baselineskip}


        \item \label{q:z-stat}
              $\frac{
               {\scriptscriptstyle\sqrt{N}}\left(\overline{X}-\mu\right)
               }
               {\sqrt{\frac{1}{N}\sum\left(X-\overline{X}\right)^2}}
                \goesindist \N{0}{1}$
              \vspace{0.5\baselineskip}

               From the previous part, we have 
               ${\scriptstyle\sqrt{N}}\overline{X} \goesindist \N{\mu}{\sigma^2}$.
               By the Khinchine Law of Large Numbers (the one we discussed
               in class), 
               $\overline{X}   \goesinprob \mu$, 
               $\overline{X-\mu}   \goesinprob 0$, and
               $\overline{X^2} \goesinprob E\{X^2\} = \sigma^2 + \mu^2$.
               Now, since the sum of plims is the plim of sums and since
               the plim of a square is the square of the plim (plim of a 
               continuous function is function of the plim --- or plim
               ``goes through'' continuous functions):

               \begin{align}
                  \frac{1}{N}\sum\left(X-\overline{X}\right)^2
                    &amp;= \overline{X^2} - \left(\overline{X}\right)^2 \nonumber\\
                    &amp;\goesinprob \sigma^2 + \mu^2 - \mu^2 \nonumber\\
                    &amp;= \sigma^2 \nonumber\\
               \end{align}
               
               Now, since square root is a continuous function, we have 
               shown that the denominator of our ratio
               goes in probability to $\sigma$.  

               The numerator of our ratio, 
               $\scriptscriptstyle\sqrt{N}\left(\overline{X}-\mu\right)$
               goes in distribution to $N(0,\sigma^2)$ by the Lindberg Levy
               CLT. Finally, Slutsky's Theorem
               says that since the numerator of our ratio goes in distribution
               to $N(0,\sigma^2)$ and the denominator goes in probability to
               $\sigma$ that the ratio goes in distribution to 
               $N(0,\sigma^2)/\sigma = N(0,1)$.


        \item \textbf{
              Recall from your earlier statistics class, the t-statistic for
              testing hypotheses about the mean of a normal distribution.
              How would you use the result above in hypothesis testing?
              How are the assumptions you are making different from the
              conventional t-test framework?  How are the mechanics of
              the test itself different?
              }

              To test the hypothesis that $EX=\mu$, I would calculate the
              ratio in question~\ref{q:z-stat} 
              (which I will call a z-statistic).  Then I would choose a 
              significance level and go to the normal table to get the
              appropriate critical value.  If the z-stat is larger in
              absolute value than the normal table value, I reject the
              null hypothesis.  This test is approximately valid (that
              is, it gives a test with a type I error probability 
              \emph{approximately} equal to the significance level I
              chose.

              The assumptions I am making here are much weaker than in
              the conventional t-test framework.  I am not assuming that
              $X$ is distributed normal!

              The ratio above is roughly the ``t-statistic'' for the null
              hypothesis that the mean of $X$ is $\mu$.  The only difference
              between it and the typical t-stat is that the estimate of 
              the variance of $X$ in the denominator is divided by 
              $N$ rather than $N-1$.  Also, I look on the normal rather than
              the t table.
      \end{enumerate}
\item \textbf{
      You are interested in the effect of going to the emergency room on
      a person's life expectancy.  There is a sample of people collected
      on a particular day, some of whom went to an emergency room and
      some of whom did not.  They are followed over time until they die.
      From each person there are two items in the dataset: $T_i$ the
      length of time from the day of the survey until death and 
      $ER_i$ a variable which equals one if they went to the emergency room
      and zero if they did not.  You wish to run a regression, but the
      people who conducted the study will not let you have the data, only
      summary statistics.  What should you ask for and how will you use
      it? \label{q:er}
      }

      I can run the regression using only $\overline{T}$, $\overline{ER}$,
      $\widehat{\sigma^2_T}$, $\widehat{\sigma^2_{ER}}$, and 
      $\widehat{Cov}(T,ER)$.  Actually, if I only want to regress $T$ on
      $ER$, I could do without $\widehat{\sigma^2_{ER}}$.  Consider the 
      regression model $T=\beta_1+\beta_2ER+\epsilon$.  I will use the
      sample statistics to calculate the regression coefficients as
      follows:

      \begin{align}
        \hat{\beta}_{2,OLS}  &amp;= \frac{\widehat{Cov}(T,ER)}{\widehat{\sigma^2_{ER}}}
                             \nonumber\\
                             &amp;\\
        \hat{\beta}_{1,OLS}  &amp;= \overline{T}-\hat{\beta}_{2,OLS}\overline{ER}
                             \nonumber
      \end{align}

\item \textbf{
      Considering the example in problem \ref{q:er}, consider two 
      linear models:
      }
      \begin{enumerate}
        \item \label{q:reg-TonER}
              $T_i = \beta_1 + \beta_2ER_i + \epsilon_i$
        \item \label{q:reg-ERonT}
              $ER_i = \beta_3 + \beta_4T_i + \epsilon_i$
      \end{enumerate}
      \textbf{
      Let's call the OLS estimates you would get from regressions run
      on these equations $\widehat{\beta}_{1,OLS},\widehat{\beta}_{2,OLS},
      \ldots$.  What relationship will there be between 
      $\widehat{\beta}_{2,OLS},\widehat{\beta}_{4,OLS}$?  What sign ($\pm$)
      do you expect for $\widehat{\beta}_{2,OLS},\widehat{\beta}_{4,OLS}$?
      \label{q:er-models}  
      }

      Just by examining the formula for OLS estimates, we can see that
      the relationship will be:

      \begin{equation}
        \hat{\beta}_{2,OLS} = \hat{\beta}_{4,OLS}\frac{\widehat{\sigma^2_T}}
                                                      {\widehat{\sigma^2_{ER}}}
      \end{equation}

      Since people who go to the emergency room are probably sick, I expect
      $Cov(T,ER)&lt;0$ and therefore that both $\hat{\beta}_{2,OLS}&lt;0$ and 
      $\hat{\beta}_{4,OLS}&lt;0$.

\item \textbf{
      We usually interpret regression models as embodying some causal
      story (the RHS causing the LHS); however, OLS does not ``know''
      this: it produces estimates regardless of whether our implicit
      causal story is right.  Discuss this fact in the context of the
      models in question \ref{q:er-models} and your expectations about
      the signs of the relevant coefficients.
      }

      Well, the model in question~\ref{q:reg-TonER} would produce the result
      that emergency rooms kill people if we were to take its implicit 
      causal story
      seriously; whereas, the model in question~\ref{q:reg-ERonT} would 
      produce the result that sick people go to the emergency room if we 
      were to take its implicit causal story seriously.  Assuming for the
      moment that we don't believe that emergency rooms kill people, 
      obviously we can go astray if we don't think carefully about what
      causal model we are assuming when we write down and interpret 
      regression models.  That is, if we wrote down, estimated, and interpreted
      the
      model of question~\ref{q:reg-TonER} we would be led to the (false)
      conclusion that emergency rooms kill people; whereas, if we wrote
      down, estimated, and interpreted the model of 
      question~\ref{q:reg-ERonT} we would be led to the (true) conclusion that
      sick people go to emergency rooms.

      Regressions only \emph{use} correlation in making their estimates; 
      therefore, they can not \emph{tell} you about causation.  You must 
      already have a causal theory before you run your regression if you
      want to draw causal conclusions from your results.
\end{enumerate}
\end{document}</pre></body></html>