3 ms·

Honest question: Aren't both equivalent? If the process will eventually traverse its entire state space, you should be able to infer the behavior of the process

• AstralStorm 7y ago
  Second version is much more loose, e.g. most semi-stationary processes fit the bill. In fact article makes a mistake by requiring stationary nature from ergodic process. You can have an ergodic process of the first kind where ML expectation is equal to the value of the process function, which nonetheless is not stationary in absolute sense. (Because definite integral does not match indefinite.) Just perfectly predictable and having the indefinite integral identical to integral over space state. Which implies strictly bounded but not exactly stationary in any defined term. Likewise you can prove an equivalence class of the process to be identical to another known ergodic process without actually calculating such difficult to obtain complete integrals. And ergodicity is not required for representative sampling either, just definite error bounds on predictive (statistical) power or likelihood estimates. Which can be estimated or directly calculated for many kinds of processes accurately. It is only required for certainty.
• Simon_says 7y ago
  Yes, that's the content of the Birkhoff Ergodic Theorem.