Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

There appear to be two definitions of the word ergodic.

The dynamical systems definition says that a measure space $(X,\mathit B, \mu)$ and measure preserving transformation $T: X \mapsto X$ is ergodic if

the only $T$-invariant sets have measure 0 or 1.

However, a Markov chain is ergodic if

there exists $t$ such that for all $x,y \in \Omega, P^t(x,y) >0$

I've used the Markov chain notation and definition found here

I would like to know if these definitions are equivalent.

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.

share|improve this question

1 Answer 1

up vote 12 down vote accepted

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.

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 finite (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.

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.

share|improve this answer
3  
As you say, the more consistent definition (and I think the way I learned it) of ergodicity for Markov chains is to say that ergodic is a synonym for irreducible. –  Anthony Quas Sep 4 '11 at 22:28
    
R W: I have the same question. Thanks for the nice clarification. (1) What is the definition of "mixing" in ergodic theory used in your reply? (2) Is a Markov chain generated by a measure-preserving mapping? I know a measure-preserving mapping can generate a stationary process, but I don't know if it can also generate a (homogeneous?) Markov chain ? Thanks. –  Tim Jul 21 '13 at 1:42
    
A more common terminology for Markov processes is to call a process ergodic if it has a unique invariant measure that attracts all the other measure. This again is not consistent with the terminology of ergodic theory but consistent with Boltzmann's usage, don't you think? –  Algernon Mar 8 at 13:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.