0
$\begingroup$

Consider the following two simple linear model specifications:

$log(y) = \beta_0 + \beta_1log(x) + u$ $(1)$

$log(y/x) = \alpha_0 + \alpha_1log(x) + v$ $(2)$

where y and x are two random variables for which we have a random sample size n and both u and v are error terms.

(a) Show that $\hat\alpha_1 = \hat\beta_1 - 1$ where both $\hat\alpha_1$ and $\hat\beta_1$ are the OLS estimates of the population parameters.

(b) Show that $se(\hat\beta_1) = se(\hat\alpha_1)$

This is what I have completed thus far but now I am having issues with Part (b) of this problem

$$ln(y) = \beta_0 + \beta_1ln(x) +u$$

$$ln(\frac{y}{x}) = ln(y) - ln(x) = \alpha_0 + \alpha_1ln(x) +v$$ $$\Rightarrow $$ $$ln(y) = \alpha_0 + \alpha_1 ln(x) +v \Rightarrow ln(y) = \alpha_0 +(x_1+1)ln(x)$$

And since we know that the ln(x) must equal we can conclude:

$$1 + \hat\alpha_1 = \hat\beta_1\Rightarrow \hat\alpha_1=\hat\beta_1 - 1$$

I will now work on Part (b) which I will add later on today hopefully. (Also, studying for an exam tomorrow) If anyone catches an error that be appreciated if you can bring it to my attention.

(b)

$$se(\hat\alpha_1)=\frac{\frac{\hat\alpha_1-\alpha_1}{sd(\hat\beta_1)}}{\sqrt(\frac{\hat\sigma^2}{\sigma^2})} = \frac{\hat\alpha_1-\alpha_1}{sd(\hat\alpha_1)} \sim N(0,1)$$

we know that:

$$tratio = \frac{\hat\alpha_1-\alpha_1}{se(\hat\alpha)}\sim t_{n-k-1}$$

If anyone can help me proceed with this question and put me on the right track that would be appreciated.

$\endgroup$
6
  • 1
    $\begingroup$ Since it looks like a homework question, it is encouraged that you include what you have done so far to solve the problem. $\endgroup$
    – hakanc
    Dec 14 '20 at 8:12
  • $\begingroup$ Ok, I will try to add it onto my question above and edit my question. Would that be okay with you? $\endgroup$ Dec 14 '20 at 8:29
  • $\begingroup$ You should try to use actual OLS estimator formula. It may be different depending on course, because few ways may lead to correct solution. Try to use the one, that is used on your course, however most universal is $b=(X'X)^{-1}X'y$ if you had it. $\endgroup$
    – cure
    Dec 14 '20 at 9:47
  • 1
    $\begingroup$ You have a typo in (b) as there is no $\beta_2$ $\endgroup$
    – Henry
    Dec 14 '20 at 10:45
  • $\begingroup$ Thank you for the comment, I have edited the question to make sure it is correct. $\endgroup$ Dec 14 '20 at 23:39
0
$\begingroup$

For (a) I would have thought $\log(y/x)= \log(y)-\log(x)$ made the answer almost immediate in terms of minimising the sum of the squares of the residuals

In the first model you want to find $\beta_0,\beta_1$ to minimise $\sum (\log(y_i)-\beta_0-\beta_1 \log(x_i))^2$ while in the second you want to find $\alpha_0,\alpha_1$ to minimise $\sum (\log(y_i/x_i)-\alpha_0-\alpha_1\log(x_i))^2$ which is $\sum (\log(y_i)-\alpha_0-(\alpha_1+1)\log(x_i))^2$.

So optimal $\hat\beta_0,\hat\beta_1$ in the first is equivalent to optimal $\hat\alpha_0=\hat\beta_0$ and $\hat\alpha_1+1=\hat\beta_1$ in the second

$\endgroup$
1
  • $\begingroup$ Hi, thank you fro commenting and providing advice. Prior to me seeing your response I tried my best to answer the problem. If you have any advice on where to go from here, that would be appreciated. $\endgroup$ Dec 14 '20 at 23:00

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.