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For a stochastic process, does being Strict Sense Stationary (SSS) imply being Weak Sense Stationary (WSS) since WSS process is easier to fulfill?

S.Surace
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kubicwerke
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1 Answers1

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Strict/strong sense stationarity means (by definition) that the law of the process $(X_{t+a})_{t\in\mathbb{R}}$ is the same as the law of the process $(X_t)_{t\in\mathbb{R}}$ for all $a\in\mathbb{R}$. In particular, all finite dimensional distributions are invariant with respect to translations in time. This implies that the first moment $E[X_t]$ is constant and the second moment $E[X_tX_{t'}]$ only depends on the time difference $|t-t'|$, if they exist.

As you say and this argument shows, weak sense stationarity is weaker than strict/strong sense stationarity (as the name suggests). Examples of processes that are weak sense stationary but not strong sense stationary are processes where higher-order moments change over time or do not even exist.

S.Surace
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  • It is slightly pathological, but one exception could be a process that has no second, or perhaps even first, moment. For example, iid Cauchy. This is strictly stationary, but we can't really talk about weak stationarity. Nevertheless, if we rule out this (i.e., require that second moment exists and is finite, which implies that first exists and finite) than the statement is really true. – Tamas Ferenci Jan 21 '22 at 18:00