Course 4, November 2000, #12.

[J-Man]

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]

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]

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]

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]

Wow was I dumb. Thanks!!