# Joint distributions of a strictly stationary time series model

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 $$\text{Cov}(X_t, X_{t+h}) = \text{Cov}(X_1, X_{1+h})$$?

• Plug in $n=2$ and $t=1:$ done.
– whuber
Jun 4, 2021 at 22:35
• For $n=2$ we get that $(X_1, X_2) \stackrel{d}{=} (X_{1+h}, X_{2+h})$. Why does this imply that $(X_1, X_{1+h}) \stackrel{d}{=} (X_{t}, X_{t+h})$? Jun 5, 2021 at 12:35
• Thanks @whuber! The definition I quoted is the one in the textbook Introduction to Time Series and Forecasting by Brockwell and Davis. It all makes sense with the definition that is stated on Wikipedia. Jun 5, 2021 at 14:13
• Actually, Brockwell & Davis are not in error -- they are just a little too parsimonious and elegant. I'll post a brief explanation.
– whuber
Jun 5, 2021 at 14:16

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.