Does ergodicity imply stationarity? In general, if a random process is ergodic, does it imply that it is also stationary in any sense?
 A: Yes, ergodicity implies stationarity.
Consider an ensemble of realizations generated by a random process. Ergodicity states that the time-average is equal to the ensemble average. The time-average is obtained by taking the average of a single realization, giving you a particular number. To obtain the ensemble average, you take the average across the realizations at a particular time-point.
If the process was not stationary in regard to the mean, the ensemble average would vary depending upon the time-point that you chose. It could not then be equal to the time-average of a single realization.
A: This man says that it is https://www.youtube.com/watch?v=k6y2kzayV6A&#t=1433. So, ergodicity implies stationarity.
A: In the theory of Dynamical Systems, one usually only defines ergodicity for invariant measures, so the short answer could be 'yes'.
But, consider the following theorem (taken from Norris' Markov Chains):
Theorem 1.10.2: let $P=(p_{ij}:\,i,j\in I)$ be an irreducible, positive recurrent transition matrix indexed by $I\times I$, where $I$ is a countable set, and let $\lambda = (\lambda_i:\,i\in I)$ be any distribution on $I$. If $(X_n:\,n\in\mathbb{Z}_+)$ is a Markov chain with initial distribution $\lambda$ and transition matrix $P$, then for any bounded function $f\colon I\to \mathbb{R}$ there is a set $\Omega_f$ with $\mathbb{P}(\Omega_f) = 1$ such that $$\frac{1}{n}\sum_{k=0}^{n-1} f\circ X_k(\omega) \to \sum_{i\in I} \pi_i f(i),\qquad \forall\omega\in \Omega_f,$$ where $\pi = (\pi_i:\,i\in I)$ is the unique invariant distribution of $P$.
By taking $\lambda\neq \pi$, clearly the corresponding chain is not stationary, but nevertheless the time averages converge to the "limiting space averages".
