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View Full Version : Course 4, November 2000, #12.

J-Man
05-10-2002, 11:45 AM
Here is the problem. You are given the following linear regression results:

Actual(1) = 77.0, Fitted(1) = 77.6
Actual(2) = 69.9, Fitted(2) = 70.6
Actual(3) = 73.2, Fitted(3) = 77.6
Actual(4) = 72.7, Fitted(4) = 72.7
Actual(5) = 66.1, Fitted(5) = 67.1

Determine the estimated lag 1 serial correlation coefficient after one iteration of the Cochrane-Orcutt procedure.

What I did was: Compute the errors. Run the regression
eps(t)=rho*eps(t-1) + v(t) to find rho. I thought this was how the Cochrane-Orcutt procedure begins. It seems that they just computed the "sample correlation coefficient" to obtain their answer. Both, when rounded, gave the correct answer. But is their method correct?

Gordo
05-10-2002, 01:59 PM
The key, I think, is the wording "estimated lag 1 serial coefficient". Under exam conditions, I would have calculated it the same way as yourself.

J-Man
05-10-2002, 02:38 PM
It also says "after one iteration of the Cochrane Orcutt procedure"... which I thought means that you'd have to perform the procedure or at least start it.

phdmom
05-10-2002, 11:04 PM
Their solution is the result of the regression. The sample autocorrelation coefficient would have one more term in the denominator, e(n)^2.

From the regression y = rho*x + v, rho = sum (xy)/ sum(x^2). Put in e(t) for y and e(t-1) for x, and you get their solution.

J-Man
05-11-2002, 12:35 PM
Wow was I dumb. Thanks!!