The MA term created by over-differencing

Question

You take the first difference of a trend-stationary series $X_t = a + b\,t + \varepsilon_t$ , where $\varepsilon_t$ is white noise. The differenced series turns out to be an MA(1).

What is the moving-average coefficient $\theta$ of that induced MA(1)?

Accepted Answer

Difference the series directly. The deterministic trend contributes a constant, and the noise contributes a difference of white-noise terms: \Delta Xt = (a + bt + \varepsilont) - (a + b(t-1) + \varepsilon{t-1}) = b + (\varepsilont - \varepsilon{t-1}). Ignoring the constant b, the stochastic part is \varepsilont - \varepsilon{t-1}. Compare with the MA(1) form \varepsilont + 	heta\varepsilon{t-1}: the coefficient on \varepsilon{t-1} is -1, so 	heta = -1. This value sits right on the boundary 	heta = 1, so the induced MA(1) is non-invertible (it has a unit root in its MA polynomial). That is …

The MA term created by over-differencing

Your answer

Discussion

Learn the concepts

Related questions