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In this (quite elementary) paper, the scaled entropy of an unbounded measure $\mu$ on $\mathbb{N}$ is defined by $$ h_c(\mu) := \lim_{\epsilon \to 0} \limsup_{n \to \infty} \dfrac{H^\epsilon(X_n)}{c(n)}, $$ where $X_n \sim \mu(\cdot \mid 0:n)$ (the normalized truncation of $\mu$ to $\{0, \ldots, n\}$), the "scaling" $c$ is a given function $c \colon \mathbb{N} \to (0,\infty)$, and the $\epsilon$-entropy $H^\epsilon(Y)$ of a random variable $Y$ on a finite set is defined by
$$ H^\epsilon(Y) = \min\bigl\{H(F) \mid \Pr(F \neq Y) < \epsilon\bigr\}, $$ where the minimum is taken over $\sigma(Y)$-measurable random variables $F$ (in other words $F=f(Y)$). In fact $H^\epsilon(Y)$ is a quantity about the $\sigma$-field $\sigma(Y)$ rather than about the random variable $Y$.

The measure $\mu$ is assumed to be unbounded and to satisfy $\mu(n)>0 \; \forall n \geq 0$. I am particularly interested in the case when it also satisfies $$ (\ast)\colon\quad \sum \frac{\mu(n)}{\mu(0:n)} = \infty \quad \text{and } \quad \sum {\left(\frac{\mu(n)}{\mu(0:n)}\right)}^2 < \infty. $$

Say that a scaling $c$ is proper for $\mu$ when $\boxed{0 < h_c(\mu) < \infty}$. I am looking for:

  • an example of an unbounded measure $\mu$ satisfying $(\ast)$ for which there does not exist a proper scaling (an example not satisying $(\ast)$ would be welcome too),

  • or a proof that there always exists a proper scaling for every unbounded measure $\mu$ on $\mathbb{N}$ satisfying $(\ast)$.

Maybe the scaled entropy $h_c(\mu)$ is not very exciting but the motivation of this question is the fact that $h_c(\mu)$ coincides with the scaled entropy of a certain filtration when $(\ast)$ holds (as said in the paper), and then the answer to this question would be a progress for the same question about a general filtration, which is more important.

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  • $\begingroup$ What's to stop you defining $c(n)=H^\epsilon(X_n)$? $\endgroup$ Jun 30, 2016 at 19:55
  • $\begingroup$ @AnthonyQuas For a fixed $\epsilon$ ? That would give $h_c(\mu) \geq 1$ since $H^\epsilon$ increases as $\epsilon$ decreases. That does not guarantee $h_c(\mu) < \infty$. $\endgroup$ Jun 30, 2016 at 20:42

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