2
$\begingroup$

I have this linear regression: $y_{i}=\beta_{0} + \beta_{1}x_{i} + u_{i}$ with $i=\{1..n\}$.

Say $\hat{\beta}_{1}$ is the OLS estimator of $\beta_{1}$. $\hat{\beta}_{1}$ is BLUE since the Gauss Markov assumptions hold.

Now I am asked to show that $\tilde{\beta}_{1}=\frac{y_{n}-y_{1}}{x_{n}-x_{1}}$ is less efficient than the OLS estimator.

Say $u_{i}\sim iid $ then $\text{Var}(\tilde{\beta}_{1})=(\frac{1}{x_{n}-x_{1}})^2\times(\text{Var}(u_{n})-\text{Var}(u_{1}))=0$.

So $\text{Var}(\tilde{\beta}_{1})<\text{Var}(\hat{\beta}_{1})$ which contradict the fact that $\hat{\beta}_{1}$ is BLUE. So obviously the assumption that $u_{i}\sim iid$ is wrong. Why?

$\endgroup$
  • 5
    $\begingroup$ Is the variance of the difference between two random variables the difference of their variances? $\endgroup$ – Scortchi Feb 24 '17 at 15:01
  • $\begingroup$ Oh I agree that if the two random variables were dependent then you would have to add the 2*cov(un,u1). But here I am asking about the case when the error terms are independent. Or maybe that's impossible? $\endgroup$ – jeake Feb 24 '17 at 16:00
  • 2
    $\begingroup$ Whether the variables are dependent or independent, variance of the difference between them is not the difference of their variances. $\endgroup$ – Richard Hardy Feb 24 '17 at 16:02
  • 1
    $\begingroup$ Is the variance of the difference between two independent random variables the difference of their variances? You should be able to answer that without looking anything up. $\endgroup$ – Scortchi Feb 24 '17 at 16:03
  • $\begingroup$ How/Should I mark this as answered? $\endgroup$ – jeake Feb 24 '17 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.