Stationarity

Study guides: Exam SRM, Exam MAS-II

A Time Series is (weakly) stationary when its mean, variance and autocovariances do not depend on time — only on the lag. Stationarity is the assumption every ARIMA result rests on, because a model whose behaviour drifts cannot be estimated from one realization.

E[Y_t] = \mu \quad \text{for all } t

\mathrm{Cov}(Y_t, Y_{t+k}) = \gamma_k \quad \text{depends on } k \text{ only}

• Weak (second-order) stationarity constrains the first two moments; strict stationarity requires the whole joint distribution to be shift-invariant and is rarely needed
• Symptoms of non-stationarity: a visible trend in the mean, a variance that grows with level, or an Autocorrelation Function that decays very slowly and nearly linearly
• Fixes: Differencing removes a stochastic trend, a log transform stabilizes a variance that scales with the mean, and seasonal differencing removes fixed Seasonality
• An AR(1) Y_t = \phi Y_{t-1} + \varepsilon_t is stationary exactly when \lvert\phi\rvert < 1; at \phi = 1 it is a Random Walk and there is no fixed mean to return to
• MA processes are stationary for any finite parameters — it is the AR side that carries the condition
• The "I" in ARIMA is the number of differences taken to reach stationarity

Is This AR(2) Stationary? Is Y_t = 0.5Y_{t-1} + 0.6Y_{t-2} + \varepsilon_t stationary?

Answer The characteristic equation is 1 - 0.5B - 0.6B^2 = 0; stationarity requires all roots outside the unit circle. B = \frac{0.5 \pm \sqrt{0.25 + 2.4}}{-1.2} \Rightarrow B = \frac{0.5 \pm 1.628}{-1.2} giving B = -1.774 and B = 0.940. The second root has \lvert B\rvert = 0.94 < 1 — inside the unit circle, so the series is not stationary. (Quick check for AR(2): stationarity needs \phi_1 + \phi_2 < 1. Here 0.5 + 0.6 = 1.1 > 1, which fails immediately.)