3.1 Strict stationarity
A process is first-order stationary if the distribution of \(X_t\) is the same for every \(t\): \(F(x_{t_1}) = F(x_{t_1+k})\) for all \(t_1\), \(k\). It is second-order stationary in distribution if \(F(x_{t_1}, x_{t_2}) = F(x_{t_1+k}, x_{t_2+k})\) for all \(t_1, t_2, k\), and \(n\)th-order stationary if \[ F(x_{t_1}, x_{t_2}, \dots , x_{t_n}) = F(x_{t_1+k}, x_{t_2+k}, \dots , x_{t_n+k}) \] for all \(t_1, \dots , t_n\) and \(k\).
Definition 3.1.1. A process is strictly stationary if it is \(n\)th-order stationary for every \(n = 1, 2, \dots \): the joint distribution of \(X_{t_1}, \dots , X_{t_n}\) is the same as that of \(X_{t_1+k}, \dots , X_{t_n+k}\) for every \(t_1, \dots , t_n\) and every shift \(k\).
In words, shifting the time origin by any amount \(k\) has no effect on the distributions.
Proposition 3.1.2. If \(\{X_t\}\) is strictly stationary with finite variance, then
- (i)
- \(\mu (t) = \mu \) and \(\sigma ^2(t) = \sigma ^2\) are constants;
- (ii)
- \(\gamma (t_1, t_2)\) depends only on the lag \(t_2 - t_1\).
Proof. (i) With \(n = 1\), every \(X_t\) has the same distribution, hence the same mean and variance. (ii) With \(n = 2\), \((X_{t_1}, X_{t_2})\) has the same joint distribution as \((X_{t_1+k}, X_{t_2+k})\) for every \(k\). Taking \(k = -t_1\), it is the same as that of \((X_0, X_{t_2-t_1})\), so its covariance is a function of \(t_2 - t_1\) alone. □
So for a stationary process we write \(\gamma (k) = \Cov (X_t, X_{t+k})\), the autocovariance at lag \(k\), and \[ \rho (k) = \frac {\gamma (k)}{\gamma (0)} , \] the autocorrelation function (ac.f.), which measures the correlation between \(X_t\) and \(X_{t+k}\). Here \(\gamma (0) = \sigma ^2\).
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.