Is this trend series weakly stationary?

Question

A measured signal is modeled as Xt = 5 + 0.3\,t + \varepsilont, where \varepsilont is white noise with mean 0 and variance 4, and t = 1, 2, 3, \dots Is \{Xt\} weakly (covariance) stationary? Answer yes or no, and justify.

Accepted Answer

Weak stationarity asks for three things: 1. the mean E[Xt] is the same constant at every t, 2. the variance is finite and constant, 3. the covariance between Xt and X{t+k} depends only on the gap k, not on where you are in time. Here the noise averages to zero, so E[Xt] = 5 + 0.3\,t. That mean climbs with time: it is 5.3 at t=1, 5.6 at t=2, 5.9 at t=3, and so on. Condition 1 already fails. The other pieces actually look fine: \operatorname{Var}(Xt)=\operatorname{Var}(\varepsilont)=4 for every t, and distinct observations are uncorrelated. But a mean that drifts upward is by itself enough to r…

Is this trend series weakly stationary?

Your answer

Discussion

Learn the concepts

Related questions