# 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?

• The setting is unnecessarily complicated. Your question comes down to how to express equality of two bivariate distributions in terms of conditional distributions. Do you know of any way to do that?
– whuber
Commented Jan 3, 2022 at 21:26
• So it sounds like what I've written above is not true. If someone could help me understand why that is case, what is missing - that might just do it for me. Commented Jan 3, 2022 at 22:24
• What you have written is correct; if the process is stationary, the three bulleted statements are correct; they are necessarily true given the premise that the process is strictly stationary. But, the three bulleted statements are not sufficient to assert strict stationarity of the process. For example, the three bulleted statements don't have anything to account for the requirement $f_{X_1,X_2} = f_{X_3,X_4}$ that a strictly stationary process must satisfy. Commented Jan 4, 2022 at 3:38
• @DilipSarwate: thank you the guidance. We've included the condition $f_{X_2|X_1}=f_{X_4|X_3}$. What else do we need to include to have fully accounted for $f_{X_1,X_2}=f_{X_3,X_4}$? Commented Jan 4, 2022 at 12:11
• Why not start with writing down every condition that the joint distributions must satisfy? For example, the joint distribution of $X_1$ and $X_3$ must be the same as the joint distribution of $X_2$ and $X_4$? Similarly, the joint trivariate distribution of $X_1, X_2, X_3$ is the same the joint trivariate distribution of $X_2, X_3, X_4$? You have lots ot things that you are missing, and then you have to worry about expressing everything in terms of conditional distributions instead of joint distributions. Do you have to use only conditional distributions to express everything? Commented Jan 4, 2022 at 22:24

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. $$\blacksquare$$

• Thank you for the confirmation, Ben. I am surprised that I haven't seen this stated in any textbook. I imagine this is probably filed under self-evident, but for me it was not so self-evident. Have you seen it presented in any book? Commented Jan 6, 2022 at 0:48
• No, haven't seen it in a book before. The conditions woudl become cumbersome for large $n$ though.
– Ben
Commented Jan 6, 2022 at 5:32