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Consider the estimator $b_1=\frac{\sum y_i}{\sum x_i}$. Suppose that $y_i = \beta x_i + \epsilon_i$, $E[\epsilon_i]=0$, $E[\epsilon_i \epsilon_j] (i \neq j)$ and $E[\epsilon_i^2]=\sigma_i^2$. Find a model for the variance of $b_1$ for which th estimator is BLUE.

The answer is supposed to be:

$v_i=x_i$ and $\sigma_i^2=\sigma^2 x_i$

However, I am not sure how they got to this answer. Could anyone please help?

Update:
I tried computing the variance of $b_1$. I get: $$ {\rm Var}(b_1)=\frac 1 {(∑x^2_i)^2}∑x^2_i\ {\rm Var}(\varepsilon_i) $$ I don't see however how I could proceed from here or whether this was the right thing to do.

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    $\begingroup$ Please see the self-study tag wiki. $\endgroup$
    – Nick Stauner
    Jul 4, 2014 at 10:26
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    $\begingroup$ Can you say a little about what you have tried / understand so far? $\endgroup$
    – gung - Reinstate Monica
    Jul 4, 2014 at 11:52
  • $\begingroup$ @gung I tried computing the variance of $b_1$. I get that $\operatorname{Var}(b_1)=\frac{1}{(\sum x_i^2)^2} \sum x_i^2 \operatorname{Var}(\epsilon_i)$. I don't see however how I could proceed from here or whether this was the right thing to do. $\endgroup$
    – student777
    Jul 5, 2014 at 8:35
  • $\begingroup$ Should the definition of $b_1$ be different from what you have in the first line (e.g. should it have summation over $i$ or something?) $\endgroup$
    – Glen_b
    Jul 5, 2014 at 14:58
  • $\begingroup$ @Glen_b Big sorry. That's really sloppy. I edited the OP. $\endgroup$
    – student777
    Jul 5, 2014 at 15:01

1 Answer 1

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For weighted regression through the origin, I presume you either know or can show that $b_1=\frac{\sum_i w_ix_iy_i}{\sum_i w_ix_i^2} = \frac{\sum_i (w_ix_i)y_i}{\sum_i (w_ix_i)x_i}$

Since in this case $b_1=\frac{\sum_i y_i}{\sum x_i}$, you can see by inspection that the weights must be such that $w_ix_i$ is a constant.

Can you do it now?

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  • $\begingroup$ Ah, I see it's that easy... I thought variances needed to be computed and compared with that of the OLS estimator. This makes it a lot easier though. Thanks for your help. $\endgroup$
    – student777
    Jul 5, 2014 at 15:34
  • $\begingroup$ I imagine it can be done in several more complicated ways. It's really a matter of what things you are able to take as given, and what things you're required to show. If you can take a few things as given (which it seems you probably can), then it's pretty easy. $\endgroup$
    – Glen_b
    Jul 5, 2014 at 15:41

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