5
$\begingroup$

I'm reading a paper in which it is stated that

The entropy of an ergodic measure is defined as $$\lim_{n \to \infty} -\frac{1}{n} \sum_{|w|=n} \mu[w] \log \mu[w].\tag{1} \label{eq:1}$$

Here the underlying space is $(\{0,1\}^{\mathbb{N}}, \mathscr{B}, \mu)$, where $\mathscr{B}$ is the $\sigma$-algebra generated by cylinders. So for a fixed $n$, the sum is over all binary strings of lenght $n$.

I've not seen the entropy of a measure defined anywhere, so I need to make sure I know what that means. Since the definition at (1) above is actually of the entropy of an ergodic measure, and a measure can only be ergodic with respect to some underlying measure-preserving transformation, it seems that the transformation must be the shift since the shift, $T$, is already mentioned. With that assumption, I conclude then, that the entropy of $\mu$ must be what most books (that I've seen) call the entropy of $T$.

Question 1: Is my interpretation of the definition at (1) correct?

If the answer to question 1 is yes, then the definition of the entropy of $T$ (or $\mu$) differs from the one I've seen, which is that $$h(T) = \sup_{\mathcal{A}} \, h(T,\mathcal{A}),\tag{2} \label{eq:2}$$ where $\mathcal{A}$ ranges over all finite (measurable) partitions and for any fixed finite partition $\mathcal{A}$, $$h(T,\mathcal{A}) = \limsup_{n\to \infty} \frac{1}{n} H \left( \bigvee_{k=0}^{n-1} T^{-k} \mathcal{A} \right),$$ where $\bigvee$ denotes the common refinement of partition (i.e. the join).

Question 2: Are the definitions at (1) and (2) equivalent in this scenario?

$\endgroup$
3

1 Answer 1

3
$\begingroup$
  • Consider the context of your second question. On the underlying probability space, let ${(Z_n)}_{n \geq 0}$ be the stationary (shift-invariant) process defined by $Z_n(x)=P(T^n(x))$ where $P(x)$ denotes the element of the partition $P$ to which $x$ belongs. Then $h(T,P)=\lim \frac{H(Z_1,\ldots,Z_n)}{n}$ where $H(\cdot)$ denotes the entropy of a discrete random variable.

  • Now consider the context of your first question. As you feel, the transformation $T$ coming into play is the shift. The partition $P$ of $(\{0,1\}^{\mathbb{N}}$ defined by the equivalence relation $(x_n) \sim (y_n)$ $\iff$ $x_0=y_0$ is a generator of $T$ and then Kolmogorov & Sinai's theorem mentioned by yourself says that $h(T)=h(T,P)$.

That shows that "yes" is the answer to your both questions, because here $H(Z_1,\ldots,Z_n)= - \sum_{|w|=n} \mu[w] \log \mu[w]$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .