Joint distributions of a strictly stationary time series model

Question

A strictly stationary time series model $\{X_t\}$ is by definition a stochastic process where $$(X_1,\ldots,X_n) \overset{d}{=} (X_{1+h}, \ldots, X_{n+h}).$$

Why does this imply that $(X_t,X_{t+h}) \overset{d}{=} (X_1,X_{1+h})$ and $\text{Cov}(X_t, X_{t+h}) = \text{Cov}(X_1, X_{1+h})$?

Answer 1

It looks like all indexes are intended to be whole numbers $1,2,3,\ldots$ and that the lag $h$ is intended to be a natural number $0,1,2,\ldots.$ (If not, an obvious modification of the following explanation will work.)

1. Apply the definition to the case $h=t-1$ and (any) $n\ge t+h=t + (t-1),$ where it asserts $$(X_1,\ldots,X_{n}) \overset{d}{=} (X_{1+(t-1)}, \ldots, X_{n+(t-1)}).$$ Because $n\ge t + (t-1),$ $t\ge 1,$ and $h\ge 0$ imply $t \le t+h \le n+t-1,$ two of the components on the right hand side are indexed by $t$ and $t+h.$ The corresponding components on the left hand side are indexed by $1$ and $1+h\le n+(t-1).$ In other words, we can write this distributional equality more specifically as $$(X_1,\ldots,X_{1+h}, \ldots, X_{n}) \overset{d}{=} (X_{t}, \ldots, X_{t+h}, \ldots, X_{n+(t-1)}).$$Since equality of the joint distribution implies equality of the marginal distributions it is immediate that $$(X_1, X_{1+h}) \overset{d}{=} (X_t, X_{t+h}).$$

2. Now that we know $(X_1, X_{1+h})$ and $(X_t, X_{t+h})$ have the same distribution, they will have the same moments. Since the covariance can be defined in terms of the first two moments, the covariances must be equal.

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With a modification of this technique (extending it to sets of three, four, ..., etc. components) you can derive the usual definition of (strict) stationarity. Conversely, the Brockwell & Davis definition we are discussing here obviously is a special case of the usual definition. Thus there is no controversy concerning what "stationary" means.