Can people explain how the Dickey Fuller root test is conducted? I feel a little shaky on this concept. Thanks, -J

Dickey fuller for unit root in AR models : If fail to reject NULL then unit root problem (NULL: g = 0 where g = b1-1) DF-EG test : For covariance stationarity and cointegration: If rejected NULL then cointegrated; can use OLS. Thats all I think you should care to know for the exam.

I’m confused though how it relates to first differencing though. Does the actual test use the difference in the dependent variable as a regression against the independent variable i.e. xt - xt-1 = a0 + (a1-1)xt-1 But when you actually first difference to resolve a unit root problem you use: xt - xt-1 = a0 + a1(xt-1 - xt-2) Saw a question like this on the CFA mock or Schwesser somewhere and it threw me off.

You regress xt - (xt-1) = b0 + g1(xt-1) + et where g1 = (b1 - 1) Perform a t-test for g1 = 0 which is the same as performing the test for b1 = 0.

not b1 =1 wyantjs?

Christ i hope this isnt on the exam

Dickey Fuller subtracts 1 from the slope as the test. If the test = 0 that means there is a unit root because 1 (the slope) - 1(the test) = 0 which proves there is a unit root.

Thanks everyone. Question though. Anyone understand why you use xt - (xt-1) for the dependent variable in conducting the test? This didn’t really make sense to me.

Dsylexic Wrote: ------------------------------------------------------- > not b1 =1 wyantjs? You’re right. Sorry.

joe, assume the AR(1) model. y(t)=phi(1)*y(t-1)+error term(t) theoretically you could test here for phi(1)=1, but the ADF test subtracts from the above equation y(t-1) y(t)-y(t-1)=phi(1)*y(t-1)-y(t-1) +error term(t) or delta y(t) = [phi(1)-1]*y(t-1) + error term(t) ADF test is then [phi(1)-1] / std (phi(1)) hope this makes sense

Oh. That does help a lot actually. Thanks very much. So you just are subtracting y(t-1) from both sides to do the test, not first differencing. You then use first differencing to solve a unit problem.