Consider the following hipothetical data. $A_t$ is a time-series tested to be I(0) with one known structural break and $B_t$ is another time-series on the same data set also tested to be I(0) with a structural break in another period (not the same as $A's$).

If I regress the following model:

$$A_t=\beta_0+\beta_1B_t+\epsilon_t \ \ (1) $$

Using the theoritical model in $(1)$ I obtain an estimated residual ($\hat{\epsilon_t}$) that is actually tested to be I(1). In such situation, if my intention is to obtain a stable relationship between the variables, I should insert the dummies of the known structural breaks? As in the following new model, with $DA_t$ representing the dummy for $A's$ structural break and $DB_t$ for $B's$:

$$A_t=\alpha_0+\alpha_1B_t+\gamma_1DA_t+\gamma_2DB_t+u_t \ \ (2)$$

Edit: The comment bellow (by @ChrisHaug) mentions the way I am testing the existence of unit root in the residuals. If I have intuited adequately what @ChrisHaug was trying to say, I should also run a test considering structural breaks in the residuals. Say, if I obtain I(0) residuals in that situation, there will be no bias in the coefficients estimated by running OLS with $(1)$ as reference?

My intuitive guess still is that I should include the dummies for the structural breaks (of each variable) in the equation, as I included in $(2)$, to solve a possible bias. But what theory does say?

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    What test did you use to check if the residuals are integrated or not? If you don't include the structural breaks in your model, their effect will be passed on to the residuals. Tests like ADF are very sensitive to structural breaks, which would explain why your residuals appear to be I(1). – Chris Haug Feb 9 '17 at 17:26
  • I tested the residuals with Stata's dfuller. But this is merely a question on the necessity of including the breaks on your final estimation. So, if I do not not include the structural breaks - as in the second example of estimation -, my residuals do not become stationary? – John Doe Feb 9 '17 at 17:48
  • @ChrisHaug, you meant that I should test the residuals with the same known structural breaks, right? If I do find that the residuals are I(0) with strucutural breaks, I won't be incurring in problems while testing the significance of the coeficients? – John Doe Feb 10 '17 at 17:33

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