Different uses of the word "ergodic" - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T06:03:32Z http://mathoverflow.net/feeds/question/74503 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74503/different-uses-of-the-word-ergodic Different uses of the word "ergodic" Daniel Mansfield 2011-09-04T09:10:52Z 2011-09-04T10:38:36Z <p>There appear to be two definitions of the word ergodic. </p> <p>The dynamical systems definition says that a measure space $(X,\mathit B, \mu)$ and measure preserving transformation $T: X \mapsto X$ is <em>ergodic</em> if</p> <blockquote> <p>the only $T$-invariant sets have measure 0 or 1.</p> </blockquote> <p>However, a Markov chain is ergodic if </p> <blockquote> <p>there exists $t$ such that for all $x,y \in \Omega, P^t(x,y) >0$</p> </blockquote> <p>I've used the Markov chain notation and definition found <a href="http://www.cc.gatech.edu/~vigoda/MCMC_Course/MC-basics.pdf" rel="nofollow">here</a></p> <p>I would like to know if these definitions are equivalent.</p> <p>Of course, I am asking here because it seems to me that they are not. For example, if $X=\{0,1\}$, $\mathit B = \{\emptyset, \{0\},\{1\},X\}$, $\mu(\{0\})=0,\mu(\{1\})=1$ and $T(x) = 1$ for all $x\in X$, then $(X,\mathit B, \mu, T)$ is ergodic as a dynamical system, but the equivalent Markov chain is not ergodic, since the probability of traveling from $0$ to $1$ is zero. </p> http://mathoverflow.net/questions/74503/different-uses-of-the-word-ergodic/74505#74505 Answer by R W for Different uses of the word "ergodic" R W 2011-09-04T09:36:04Z 2011-09-04T09:49:21Z <p>Unfortunately, the way the term "ergodic" is used in the theory of (finite) Markov chains is completely misleading from the point of view of general ergodic theory. To be consistent, one should have called "ergodic" the chains whose state space does not admit a decomposition into non-trivial non-communicating subsets. The notion of ergodicity you are referring to would rather correspond to what is called "mixing" in ergodic theory. </p> <p>More precisely, an initial distribution $m$ of a Markov chain on a state space $X$ determines the associated measure $\mathbf P_m$ on the space of sample paths $X^{\mathbb Z_+}$. The measure $\mathbf P_m$ is shift invariant iff the measure $m$ is stationary. Now, if $m$ is <b>finite</b> (this condition is important; otherwise the following claim is false), then ergodicity of the time shift is equivalent to absence of non-trivial partitions of $X$ into non-communicating subsets. </p> <p>By the way, your example is really too degenerate: the standard example for difference between ergodicity and mixing for Markov chains is presence of so-called periodic classes $A_1\to A_2\to\dots\to A_k\to A_1$ (the only allowed transitions are from $A_i$ to $A_{i+1}$ mod k). For finite chains this is actually the only reason for difference between ergodicity and mixing, but for general state spaces the situation is more complicated. </p>