# Stationary Process Ergodicity

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

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.

• 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? – Andrea May 14 at 18:35