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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 \#1
\normalsize

\end{centering}
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\begin{enumerate}
\item Consider a sample of people who were asked to report their
      sex (Everyone answers, and everyone answers either male or
      female).  From their answers, we construct the following 
      matrix:

      \begin{equation}
      X = \left[ \begin{array}{c c c} \mathbf{1} & M & F \end{array} \right]
      \end{equation}

      In the matrix, each row corresponds to one person's response;
      $\mathbf{1}$ is a column of 1s, $M$ is a column with a 1 in
      the $ith$ row if the $ith$ person answered male and a zero 
      otherwise, and $F$ is a column with a 1 in the $ith$ row if
      person $i$ answered female and a zero otherwise.

      Is $X'X$ invertible.  Why or why not?

      \textbf{
      No, $X'X$ is not invertible.  Notice that $M+F=\mathbf{1}$.  This
      is a linear dependency in the columns of $X$.  Thus, the rank of
      $X$ must be less than three.  Since the rank of $X'X$ (a 
      $3 \times 3$ matrix) is the same as the rank of $X$, its rank 
      must be less than 3.  Thus, it is not full rank.  Thus it is not
      invertible.
      }

\item We discussed in class that variance matrixes are always 
      positive semi-definite but not always positive definite.
      In the case of scalar random variables, a singular 
      variance matrix means a zero variance --- that is, that
      the random variable is not really random at all, but fixed.

      Now, let's think about what a positive semi-definite but
      not positive definite (hereafter PSDNPD) 
      variance matrix means for the random
      vector it corresponds to.  Let x be a random vector with
      PSDNPD variance matrix V.

      \begin{enumerate}
      \item Must there be a fixed non-zero vector $\alpha$, such that
            $\alpha'V\alpha$=0?  Why or why not?
      
       \textbf{
       Yes.  If there were no such vector, then $V$ would be
       positive definite, and we are told that it is not.
       }


      \item Suppose there is such an $\alpha$, what would the variance
            of $\alpha'x$ be?

      \textbf{
      Zero.  $V(\alpha'x)=\alpha 'V \alpha=0$.
      }

      \item What does this tell us about $\alpha'x$?

      \textbf{
      It is fixed, non-random.
      }

      \item Does this mean that $\alpha'x=0$?

      \textbf{
      No, it could just as easily be 8 or 0.73 or any fixed number.
      The variance of 0 is zero, but the variance of 8 or 0.73 are
      is zero.
      }

      \item So, what does it mean for a random vector to have a
            PSDNPD variance matrix?

      \textbf{
      When a random vector has a PSDNDP variance matrix, this means
      that there are, in some sense, fewer random variables in the 
      vector than appear to be.  Consider a scalar random variable 
      $z$ and a random vector:
      }
 
      $x=\left( \begin{array}{c} z \\ -z \end{array} \right)$

      \textbf{
      As you can see, although $x$ has two elements it does not 
      \emph{really} contain two random variables, just $z$.
      }

      \textbf{
      The variance matrix of $x$ is:
      }

      $V(x) = \left[ \begin{array}{c c} 
                     \sigma^2_z & -\sigma^2_z \\
                     -\sigma^2_z & \sigma^2_z
                     \end{array}
              \right]$

      \textbf{
      This matrix is singular, and:
      }

      $\left( \begin{array}{c c} 1 & 1 \end{array} \right)
       \left[ \begin{array}{c c} 
              \sigma^2_z & -\sigma^2_z \\
              -\sigma^2_z & \sigma^2_z
             \end{array}
       \right]
      \left( \begin{array}{c} 1 \\ 1 \end{array} \right) = 0$

      and:

      $V\left( (1 \qquad 1)x \right) = 0$

      \textbf{
      Another way of saying the same thing is that if V is 
      PSDNPD, then you can always write one of the elements of 
      $x$ as a linear function of the other elements plus a 
      fixed number (this follows from $V(\alpha'x)=0)$.
      }

      \end{enumerate}

\item We want to estimate the proportion of US households with 
      broadband Internet access.  We randomly and independently call
      up 1,000 households and ask them whether or not they do.  We
      make a $1,000 \times 1$ vector $x$ which has a 1 in the $ith$ row
      if the $ith$ household answered yes and a 0 if no.  Assume
      that all households in the US have a phone, all answer our
      call, and all truthfully answer the question.

      Consider the following two estimators for $p$, the proportion
      of US households who, in fact, have broadband:

      \begin{enumerate}
      \item $\overline{x}$, also known as the sample proportion.
      \item 0.73 for surveys with sample size $<$ ten million, and 
            $\overline{x}$ for surveys with sample size $>$ ten million.
      \end{enumerate}

      Which is unbiased or are both?  Which is consistent or are both?
      What is my point?

      \textbf{
      Clearly the first is unbiased.  
      $E \left\{ \overline{x} \right\} = \mu_x =p$.  The second is not.  
      Unbiasedness requires that the expected value of the estimator 
      equal the true value of the parameter no matter what the true
      value of the parameter is.  When the true value of the parameter
      is not 0.73, then, at least sometimes, the second estimator 
      has an expectation different from the true value.
      }

      \textbf{
      The first is consistent.  This is because it is unbiased and 
      its variance goes to zero in n.  It variance is $p(1-P)/n$.
      Thus it converges in mean square to p.  Thus it converges in
      probability to p.
      }

      \textbf{
      The second is also consistent.  Notice that, for $n$ bigger than
      ten million, the two estimators are the same.  Consistency is
      about \emph{limiting} behavior --- behavior for $n$ ``near''
      infinity.  Since the two estimators are the same for large $n$,
      if one is consistent, then so must the other be.
      }

      \textbf{
      My point:  Consistency is a property which says that an estimator
      is good in \emph{large} samples and large samples only.  A 
      consistent estimator may be very bad, or even very silly, in
      small samples.  Notice that little in my answer would change if
      the second estimator was -57 for samples less than ten million.
      Whether a consistent estimator performs well in sample sizes we
      are likely to encounter in reality is an important question,
      though not one we are going to treat extensively in the 
      course.
      }

\item Suppose we have an estimator of per-capita income in the US,
      $\hat{I}$, and an estimator of the share of income saved
      (NOT the average over people of share of income saved but
      rather the average over dollars of share of income saved!),
      $\hat{s}$.  The true values of these parameters are
      $\mu_I$ and $\mu_s$. 
      These two estimators are derived somehow from a 
      survey of size $n$, and we have proved the following result for
      these estimators:

      \begin{equation}
      \sqrt{n} \left( \left( \begin{array}{c} 
                             \hat{I}\\ \hat{s}
                             \end{array} 
                      \right) -
                      \left( \begin{array}{c}
                              \mu_I\\ \mu_s 
                             \end{array} 
                      \right)
               \right)
       \overset{A}{\displaystyle{\sim}}
       N \left( \left( \begin{array}{c} 0\\ 0 \end{array} \right) ,
         \left[ \begin{array}{r r}
                400 & 20\\
                 20 &  9
                \end{array}
         \right]
         \right)
      \end{equation}

      Please construct an approximate (why approximate?) 
      95\% confidence interval for the per-capita savings, $\mu_I \mu_s$.

      \textbf{
      Approximate because we are using asymptotic (or approximate) 
      distributions, not the true distributions of the estimators to
      construct the CI --- this is true even though the problem I gave 
      did not contain a statement about asymptotic distribution since 
      the theorem we will use has only a conclusion for asymptotic
      distribution.  The CI is right if the sample is very large,
      and (we hope!) pretty close in sample sizes we are likely to
      encounter.
      }

      \textbf{
      Anyway, we have a result from class which says:
      }

      \textbf{
      If 
      $\sqrt{n}\left( x-\mu_x \right) 
       \overset{A}{\displaystyle{\sim}} N(0,\Sigma)$ then 
      $\sqrt{n} \left( g(x)-g(\mu_x) \right) 
       \overset{A}{\displaystyle{\sim}}
       N \left( 0, \triangledown _g' \Sigma \triangledown _g \right)$
      In this formula, let's set 
      $x=\left( \begin{array}{c} \hat{I} \\ \hat{s} \end{array} \right)$
      and
      $g(x) = x_1x_2$.  That means that 
      $\triangledown_g = \left( \begin{array}{c} x_2 \\ x_1 
                                \end{array} \right)$
      or
      $\triangledown_g = \left( \begin{array}{c} \hat{s} \\ \hat{I} 
                                \end{array} \right)$
      }

      \textbf{
      Now, applying the theorem, we get:
      }

      \begin{eqnarray}
      \sqrt{n} \left( \left( \begin{array}{c} \hat{I}\hat{s} 
                             \end{array} 
                      \right)
                      -
                      \left(\left( \begin{array}{c} \mu_I\mu_s 
                                   \end{array}
                            \right) 
                      \right)
                \right)
       & = &
      \sqrt{n} \left( \left( g(x) 
                      \right)
                      -
                      \left( g(\mu)
                      \right)
                \right)
       \nonumber\\
       & \overset{A}{\displaystyle{\sim}} &
       N \left( 0,
               \left( \begin{array}{c c} \hat{s} & \hat{I} \end{array} \right)
               \left[ \begin{array}{l l} 400 & 20\\
                                          20 &  9
                      \end{array}
               \right]
               \left( \begin{array}{c} \hat{s} \\ \hat{I} \end{array} \right)
          \right)
       \nonumber\\
       &=&
       N \left( 0, 400 \hat{s}^2 + 40 \hat{s} \hat{I} + 9 \hat{I}^2
         \right)
       \nonumber
      \end{eqnarray}

      \textbf{
      Let's call the rather messy quadratic $V$.  Had we been given the
      values of $\hat{I}$ and $\hat{s}$, we could have calculated $V$.
      Then, our 95\% confidence interval would be:
      }

      $\hat{I}\hat{s} \pm 1.96 V$

\end{enumerate}

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