Riemann surface
matlab

Consider the optimization problem of minimizing the variance of the weighted estimator. If the estimate is to be unbiased, it must be of the form $c_1 \hat{\theta}1+c_2 \hat{\theta}_2$ where $c_1$ and $c_2$ sum to 1 . Thus, $c_2=1-c_1$. The function to minimize is $\operatorname{Min}_c L*=c_1^2 v_1+\left(1-c_1\right)^2 v_2$. The necessary condition is $\partial L_{\mathrm{s}} / \partial c_1=2 c_1 v_1-2\left(1-c_1\right) v_2=0$ which implies $c_1=v_2 /\left(v_1+v_2\right)$. A more intuitively appealing form is obtained by dividing numerator and denominator by $v_1 v_2$ to obtain $c_1=\left(1 / v_1\right) /\left[1 / v_1+1 / v_2\right]$. Thus, the weight is proportional to the inverse of the variance. The estimator with the smaller variance gets the larger weight.

a. The sample means are (1/100) times the elements in the first column of $\mathbf{X}^{\prime} \mathbf{X}$. The sample covariance matrix for the three regressors is obtained as $(1 / 99)\left[\left(\mathbf{X}^{\prime} \mathbf{X}\right)_{i j}-100 \bar{x}_i \bar{x}_j\right]$.
\begin{aligned} & \text { Sample Var[ } \mathbf{x}]=\left[\begin{array}{ccc} 1.0127 & 0.069899 & 0.555489 \ 0.069899 & 0.755960 & 0.417778 \ 0.555489 & 0.417778 & 0.496969 \end{array}\right] \text { The simple correlation matrix is } \ & {\left[\begin{array}{ccc} 1 & .07971 & .78043 \ .07971 & 1 & .68167 \ .78043 & .68167 & 1 \end{array}\right]} \ & \end{aligned}
b. The vector of slopes is $\left(\mathbf{X}^{\boldsymbol{\prime}} \mathbf{X}\right)^{-1} \mathbf{X}^{\boldsymbol{t}} \mathbf{y}=[-.4022,6.123,5.910,-7.525]^{\boldsymbol{t}}$.
c. For the three short regressions, the coefficient vectors are
(1) one, $x_1$, and $x_2:[-.223,2.28,2.11]^{\prime}$
(2) one, $x_1$, and $x_3[-.0696, .229,4.025]^{\prime}$
(3) one, $x_2$, and $x_3:[-.0627,-.0918,4.358]^{\prime}$
d. The magnification factors are
\begin{aligned} & \text { for } x_1:\left[(1 /(99(1.01727)) / 1.129]^2=.094\right. \ & \text { for } x_2:[(1 / 99(.75596)) / 1.11]^2=.109 \ & \text { for } x_3:[(1 / 99(.496969)) / 4.292]^2=.068 \end{aligned}
e. The problem variable appears to be $x_3$ since it has the lowest magnification factor. In fact, all three are highly intercorrelated. Although the simple correlations are not excessively high, the three multiple correlations are .9912 for $x_1$ on $x_2$ and $x_3, .9881$ for $x_2$ on $x_1$ and $x_3$, and .9912 for $x_3$ on $x_1$ and $x_2$.

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