(Solved): 10. Suppose that we want to fit the no-intercept model yi=xi+i, where i 's are in ...

10. Suppose that we want to fit the no-intercept model $y_i=?x_i+{?}_i$, where ${?}_i$ 's are independent normal random variables with mean zero, but they have unequal variances. This means that the variance of $y_i$ is not constant as well, i.e., depends on the value of $x_i$. In order to handle the violation of the identical distribution assumption, a weighted least squares estimation is applied. That is, the estimated slope, $\widehat{?}$, that minimizes the weighted sum of squares function, $WSSE={?}_{i=1}^n{w}_i{\left(y_i?{\hat{y}}_i\right)}^2$, was found as $\widehat{?}=\frac{{?}_{i=1}^n{w}_i{x}_i{y}_i}{{?}_{i=1}^n{w}_i{x}_i^2}$ Suppose the variance of $y_i$ is directly proportional to the corresponding $x_i$. Then the best weighing scheme will be $w_i={?}^2/x_i$, where ${?}^2$ is a "baseline" variance. What would $\widehat{?}$ become? 11. Consider the model in Question 10, i.e., suppose the variance of $y_i$ is proportional to the corresponding $x_i$, say, $\mathrm{Var}\left(y_i\right)={?}_i^2=c{x}_i$ where $c$ is a constant number, and $\mathrm{Cov}\left(y_i,y_j\right)=0$ for all $i?=j$ since they are independent. Considering the $\widehat{?}$ found in part (a), find $\mathrm{Var}(\widehat{?})$.