Strict stationarity in terms of conditional distributions Let's start with the definition of a strictly stationary process: The process $\{X_t\}=\{X_1,X_2,X_3,X_4\}$ is strictly stationary if the joint distribution of the vector $(X_1,...,X_n)$ and the time shifted vector $(X_{1+h},...,X_{n+h})$ is the same for every integer h and positive integer n.
I am realizing that my understanding of joint distributions and their relationship to conditional distributions is limited. I'd like to flush out what it means for the process $\{X_t\}=\{X_1,X_2,X_3,X_4\}$ to be strictly stationary and I'd like a definition in terms of conditional distributions instead. Suppose that $\{X_t\}$ is strictly stationary. Therefore it must be that the following holds true:

*

*$f_{X_1}=f_{X_2}=f_{X_3}=f_{X_4}$

*$f_{X_2|X_1}=f_{X_3|X_2}=f_{X_4|X_3}$

*$f_{X_3|X_1}=f_{X_4|X_2}$

*$f_{X_3|X_1,X_2}=f_{X_4|X_2,X_3}$
Did I miss anything? Did I wrongly include anything? How do I express the fact that $(X_1,X_2,X_3)$ and $(X_2,X_3,X_4)$ have the same joint distribution in terms of conditional distributions?
 A: No, you haven't missed anything (after your edit) --- those four statements are necessary and sufficient for the condition of strict stationarity (for a time-series vector with four elements) and are therefore equivalent to the strict stationarity condition.  Here it is formalised in a theorem if you prefer.

Theorem: Consider a time-series vector $(X_1,X_2,X_3,X_4)$ with four elements, where all marginal and conditional densities are defined.  This time-series vector is strictly stationary if and only if the following equivalences hold between the marginal and conditional densities:

*

*$f_{X_1}=f_{X_2}=f_{X_3}=f_{X_4}$

*$f_{X_2|X_1}=f_{X_3|X_2}=f_{X_4|X_3}$

*$f_{X_3|X_1}=f_{X_4|X_2}$

*$f_{X_3|X_1,X_2}=f_{X_4|X_2,X_3}$
Proof: In order for $(X_1,X_2,X_3,X_4)$ to be strictly stationary, the list of all required distributional equivalences is:
$$\begin{matrix}
X_1 & \sim & X_2 & \sim & X_3 & \sim & X_4, \\[6pt]
(X_1,X_2) & \sim & (X_2,X_3) & \sim & (X_3,X_4), \\[6pt]
(X_1,X_3) & \sim & (X_2,X_4), \\[6pt]
(X_1,X_2,X_3) & \sim & (X_2,X_3,X_4). \\[6pt]
\end{matrix}$$
We can demonstrate equivalence by showing that the conditions in the theorem are both necessary and sufficient for these distributional equivalences.  ($\implies$): The conditions above follow trivially from the conditions in the theorem by application of the law of total probability.  ($\impliedby$): The conditions in the theorem follow immediately from the above conditions by taking ratios of the relevant joint densities to obtain equivalence of the relevant marginal/conditional densities.  $\blacksquare$

