I have a question regarding the ergodicity of a transformation.

Given a probability space,



$({T_{N})}_{n\geqslant 0}$

And T is some ordered transformation through time. Then for ergodicity to hold we require,

$\frac{1}{n} \sum_{i=0}^{n-1}f(T^{n}(x)) = \int f(x)$

Which implies that $\mu(T^{-1}(A)) = \mu(A) $, where $ A \subseteq \Sigma $. In consequence, if we can demonstrate that $\mu(T^{-1}(A)) \neq \mu(A) $, then $T_N$ is presumably not ergodic. Question one: is this true?

Question two is based on the following reservation. If what I say above is true, a legitimate criticism (in my view) could be the following. That is, we don't often actually seen $T_N$ but rather, say, $T_j$, whereby $0 < j < N$. In which case (hypothetically at least) one could argue that the process in question could still be ergodic if it were left to 'run' longer, given that the process may then converge as required. Because all we need for ergodicity to apply is that,

$\frac{1}{N} \sum_{i=0}^{n-1}f(T^{n}(x)) = \int f(x)$


$\frac{1}{j} \sum_{i=0}^{j-1}f(T^{j}(x)) = \int f(x)$

Is this valid reasoning?

In advance, many thanks.

  • $\begingroup$ Huh? The definition of ergodicity is $\mu(T^{-1}(A))=\mu(A)$ implies $\mu(A)=0$ or $\mu(A)=1$. $\endgroup$ – Alex R. Oct 19 '16 at 18:46
  • $\begingroup$ Hi Alex, thanks for your reply. Would you mind expanding on your comment somewhat. $\endgroup$ – Kier Oct 19 '16 at 20:30

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