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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?

ColorStatistics
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  • 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 Jan 03 '22 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. – ColorStatistics Jan 03 '22 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. – Dilip Sarwate Jan 04 '22 at 03:38
  • @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 $f_{X_1,X_2}=f_{X_3,X_4}$? – ColorStatistics Jan 04 '22 at 12:11
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    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? – Dilip Sarwate Jan 04 '22 at 22:24
  • Made an edit to that end. I mentioned the latter condition in my initial post. – ColorStatistics Jan 04 '22 at 22:37
  • @whuber: I am trying to understand your comment in light of Ben's answer. I imagine that you agree with Ben's answer. Did you misunderstand the question? If not, please elaborate a bit on the point you were making so that I can grasp it. Thank you. – ColorStatistics Jan 06 '22 at 00:58
  • The case of a "time series" of four points is of no interest: stationarity is irrelevant. I therefore understood your question as asking about a standard time series process indexed by *all natural numbers.* All you are really asking is how to express the equality of any two multivariate distributions in terms of conditional distributions. There are many ways to do that, but they all come down to the basic step of writing the distribution in terms of a marginal (also potentially multivariate) distribution and the complementary conditional (also multivariate) distribution. – whuber Jan 06 '22 at 04:23
  • Thank you for clarifying that. This is really helpful. Yes, I was indeed really asking about how to express the equality of multivariate distributions in terms of conditional distributions. I kept it at 4 random variables just to keep it manageable but I think one gets what is going on quite nicely with just these 4 variables and can extrapolate to longer time series processes. – ColorStatistics Jan 06 '22 at 11:25

1 Answers1

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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. $\blacksquare$

Ben
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  • 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? – ColorStatistics Jan 06 '22 at 00:48
  • No, haven't seen it in a book before. The conditions woudl become cumbersome for large $n$ though. – Ben Jan 06 '22 at 05:32