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I am confused by the definitions of ergodicity in wikipedia, see formal definition here which says that a measure-preserving transformation $T$ is ergodic if for every event $E$, $T^{-1}(E) = E$ implies that $P(E)=0$ or $P(E)=1$. Is this definition in anyway related to the definition of ergodic process here which talks about the statistics of a process being captured by a long trajectory sample? If so, can anyone demonstrate the relation to me?

Also, what property must a stochastic process $X(t)$ possess such that a sample of the process with a very long time trajectory can be used to infer statistical properties of $X(t)$ for any time $t$? Is it ergodicity and stationary in the strict sense?

Tomas Jorovic
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1 Answers1

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There is a theorem in dynamical systems known as the pointwise ergodic theorem. What it says (in part) is that if $T$ is a measure theoretic transformation of some probability space, and if $f$ is a random variable with finite expectation $\int f$, i.e. if $f$ is integrable, then the time average $\hat f(x) = \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n f(T^n(x))$ exists for almost all $x$, the function $\hat f$ is integrable, and its integral $\int\!\hat f$ is equal to the expectation of $f$, namely to $\int\!f$.

Ergodicity of $T$, combined with the pointwise ergodic theorem, implies that $\hat f=\int\!f$ almost everywhere.

If you now translate this conclusion into the language of stochastic processes by taking $X(n) = f \circ T^n$, you'll see that ergodicity of $T$ implies ergodicity of $X(n)$ as defined in this part of the wikipedia page you linked.

Lee Mosher
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  • This is very helpful. Now it makes sense. But also, the Birkhoff–Khinchin theorem which is what you're referring to mentions that $T$ has to be a measure preserving map, which I take to mean that the process is stationary in the strict sense. I know that not every stationary process is ergodic but are ergodic processes stationary? – Tomas Jorovic Feb 18 '19 at 22:50
  • I don't really know the answer to your last question, but it seems quite reasonable to me that the concept of an ergodic stochastic process could be generalized to nonstationary processes. – Lee Mosher Feb 19 '19 at 18:25
  • Also, it's true that every stochastic process that can be written in the form $X(n) = f \circ T^n$ is stationary, but I see no reason why every stationary process can (or should) be written in that form (for some $T$). – Lee Mosher Feb 19 '19 at 18:27