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\title{Answer Key for Ch14 Exercise2: Show $\hat{\beta}$ is unbiased \vspace{0.0in}}
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Show that the OLS estimate $\hat{\beta}_1$ is unbiased for the model $Y_i = \beta_0 + \beta_1X_i + \epsilon_i$.
    \begin{enumerate}
    \item We first substitute the ``true'' equation for $Y_i$ into our estimate of $\hat{\beta}_1$.  Use the fact that $\overline{Y} = \beta_0 + \beta_1\overline{X} + \overline{\epsilon}$ and that $\overline{\epsilon}= 0$ to generate the following.  Note that the difference between $\beta_1$ and $\hat{\beta}_1$ is very important here.
        \begin{eqnarray*}
        \hat{\beta}_1   &=& \frac{\sum (\beta_0 + \beta_1X_i + \epsilon_i -  \beta_0 - \beta_1\overline{X})(X_i-\overline{X})}                                      {\sum  (X_i - \overline{X})^2}  \\
                        &=& \frac{(\beta_1 \sum  (X_i - \overline{X})  + \epsilon_i)(X_i-\overline{X})}                                                             {\sum  (X_i - \overline{X})^2}  \\
                        &=& \frac{ \beta_1 \sum  (X_i - \overline{X})(X_i-\overline{X})}{\sum  (X_i - \overline{X})^2}  + \frac{\sum \epsilon_i(X_i-\overline{X})}  {\sum  (X_i - \overline{X})^2}  \\
                        &=& \beta_1 + \frac{\sum \epsilon_i(X_i-\overline{X})}                                                                                      {\sum  (X_i - \overline{X})^2}  \\
                        &=& \beta_1 + \frac{\sum \epsilon_iX_i}{\sum  (X_i - \overline{X})^2} -\frac{\overline{X}\sum \epsilon_i}                                  {\sum  (X_i - \overline{X})^2}
        \end{eqnarray*}
    The expectation of $\sum \epsilon_i = 0$ by assumption of OLS model.  If $\epsilon_i$ is uncorrelated with $X_i$ then the expectation of $\sum \epsilon_iX_i$ is also zero, meaning that
        \begin{eqnarray*}
        E[\hat{\beta}_1 ] & = & \beta_1
        \end{eqnarray*}
        which means that the OLS estimate of $\beta_1$ is unbiased as long as the error is uncorrelated with the independent variable.
    \end{enumerate}
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