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Can you give me an example of a stationary nonergodic stochastic process that is time continuous?

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Let $X_0$ have a Bernoulli$(p)$ distribution, $0\lt p\lt 1,$ and define $X_t=X_0$ for all $t.$

  • It is stationary because all finite-dimensional joint distributions of $(X_{t_1}, \ldots, X_{t_k})$ are time-invariant.

  • It is time continuous because all realizations, being constant, are (obviously) continuous.

  • It is not ergodic because any realization, being constant, does not display the full statistical properties of the process. E.g., if you were to estimate $p$ from any realization the estimate would either be $0$ or $1,$ neither of which will equal $p.$


OK, maybe this seems too trivial to be of interest. But it does capture something essential, as you can see by generalizing it. For instance, let $(X_t)$ and $(Y_t)$ be independent processes that are "time continuous" in any sense you like, but with different marginal distributions. Let $U$ be an independent Bernoulli$(p)$ variable. Use it to select which process is realized by defining

$$Z_t = UX_t + (1-U)Y_t.$$

The same reasoning as before shows this is not ergodic (no realization exhibits the statistical characteristics of the process) but it is stationary when both $(X_t)$ and $(Y_t)$ are and, because its realizations are either realizations of $(X_t)$ or $(Y_t),$ it is as continuous as both of these component processes.

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  • $\begingroup$ I don't understand this part about any realization being constant, so it is continuous? Do you mean that any two realizations can be either 0 or 1 and that makes it continuous? Also, about the last part, the estimate would basically be the realization I get, but the expected value is p? $\endgroup$ – Andrea May 14 at 18:35

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