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I am trying to apply Cochrane-Orcutt procedure by using transformed equation derived from calculation with value of p (rho). In my actual case, I used some dummies as independent variables. My question is that, I made changes to the orginal data of dependent and some independent variables by using rho value, but i dont know whether I have to do the same with the dummies or not? Please give me an advice! Thank you. I am sorry if this topic is not relevant in this forum.

I am trying to apply Cochrane-Orcutt procedure by using transformed equation derived from calculation with value of p (rho). In my actual case, I used some dummies as independent variables. My question is that, I made changes to the orginal data of dependent and some independent variables by using rho value, but i dont know whether I have to do the same with the dummies or not? Please give me an advice! Thank you. I am sorry if this topic is not relevant in this forum.

Yes, you should do the same for the dummies. Mathematically, they are the same as any other independent variable.

Thank you for your answer. I have another question that if i have to use this procedure, thus how can i show the result in the research, the original estimation or the one used by cochorane-orcutt procedure? In other words, how should i explain it in my research?

Please help me one more time. If I would like to repeat the cochorane-orcutt procedure with new rho calculated from new DW, what data should i use, the original or the ones after the step-1 of cochorane-orcutt procedure?

Thank you for your answer. I have another question that if i have to use this procedure, thus how can i show the result in the research, the original estimation or the one used by cochorane-orcutt procedure? In other words, how should i explain it in my research?

Please help me one more time. If I would like to repeat the cochorane-orcutt procedure with new rho calculated from new DW, what data should i use, the original or the ones after the step-1 of cochorane-orcutt procedure?

Usually, one just shows the final result.

To do repeated Cochrane-Orcutt, you get a new rho from the new estimated residuals (not the residuals on the transformed model) and then do the transformation again.

But the real answer is that almost no one ever does Cochrane-Orcutt by hand anymore except in special circumstances (for example, a very short sample). In EViews, just do least squares and add AR(1) to the list of independent variables. You'll get results that are essentially the same as doing CO repeatedly in the way you describe. And it's all automatic.

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I did include the AR(1) in the estimation, its value in the results is new rho. Could you please explain me more about "just do least squares and add AR(1) to the list of independent variables. You'll get results that are essentially the same as doing CO repeatedly in the way you describe. And it's all automatic"? How does it give the same results as I do repeated CO?

I did include the AR(1) in the estimation, its value in the results is new rho. Could you please explain me more about "just do least squares and add AR(1) to the list of independent variables. You'll get results that are essentially the same as doing CO repeatedly in the way you describe. And it's all automatic"? How does it give the same results as I do repeated CO?

will give essentially the same results as

Code: Select all

ls y x ar(1)

will give essentially the same results as

Code: Select all

ls y-rho*y(-1) x-rho*x(-1)

Yes, you are right.
price1 c poil1 rtdricq sdvxb dths: this is the estimation of transformed model, it gave the DW stats of 1.45 (seems to be fine), but the R2 was 0.64, maynot be OK. That is why I would like to repeat CO one more time to see the difference. As you mentioned, in the above estimation of transformed model, I only need to add AR(1), so I can see the DW stats which will be the same as I repeat CO, is that right? I have tried, the new rho (if I add AR(1) is 0.27 and new DW is 2.01.

Dear Startz, I add AR(1) to the original equation, the estimate for AR is not the same with rho which I calculated from residuals (given by the original equation), and in the result window of original equation without AR(1), I click on View\residual tests\serial correlation LM test..., i select "1" 'in lag to include', the resid(-1) is very close to the above calculated value of rho. Could you explain me this?

Dear Startz, I add AR(1) to the original equation, the estimate for AR is not the same with rho which I calculated from residuals (given by the original equation), and in the result window of original equation without AR(1), I click on View\residual tests\serial correlation LM test..., i select "1" 'in lag to include', the resid(-1) is very close to the above calculated value of rho. Could you explain me this?

That's odd all right.

I really dont know what i have done wrongly. I used the equation for rho calculation as follows: rho=sum(t=2)(et*et-1)/sum(t=1)et^2. Anyway, can you tell me what exactly the test by "View\residual tests\serial correlation LM test..." means?

I really dont know what i have done wrongly. I used the equation for rho calculation as follows: rho=sum(t=2)(et*et-1)/sum(t=1)et^2. Anyway, can you tell me what exactly the test by "View\residual tests\serial correlation LM test..." means?

As a check, you can estimate rho by the regression My guess is that you've accidentally picked up the wrong residuals somewhere.

The LM test tests the hypothesis that there is no remaining serial correlation in the equation.

As a check, you can estimate rho by the regression

Code: Select all

ls e e(-1)

My guess is that you've accidentally picked up the wrong residuals somewhere.

The LM test tests the hypothesis that there is no remaining serial correlation in the equation.

I have done the regression e e(-1) as you suggested, and it gave the same result with value of rho I calculated before. I also double checked the residuals, I got the right ones.
How can I understand the result of LM test?

[quote="cuongnh]How can I understand the result of LM test?[/quote]

The LM test is an alternative to the Durbin-Watson test.

Thank you very much for your time, Startz.