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The simplest form of the information theoretic CLT is the following:

Let $X_1, X_2,\dots$ be iid with mean $0$ and variance $1$. Let $f_n$ be the density of the normalized sum $\frac{\sum_{i=1}^n X_i}{\sqrt{n}}$ and $\phi$ be the standard Gaussian density. Then the information theoretic CLT states that, if $D(f_n\|\phi)=\int f_n \log(f_n/\phi) dx$ is finite for some $n$, then $D(f_n\|\phi)\to 0$ as $n\to \infty$.

Certainly this convergence, in a sense, is "stronger" than the well establised convergences in the literature, convergence in distribution and convergence in $L_1$-metric, thanks to Pinsker's inequality $\left(\int |f_n-\phi|\right)^2\le 2\cdot \int f_n \log(f_n/\phi)$. That is, convergence in KL-divergence implies convergence in distribution and convergence in $L_1$ distance.

I would like to know two things.

  1. What is so great about the result $D(f_n\|\phi)\to 0$?

  2. Is it just because of the reason stated in the third paragraph we say convergence in KL-divergence (i.e., $D(f_n\|\phi)\to 0$) is stronger?

NB: I asked this question sometime ago in math.stackexchange where I didn't get any answer.

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    $\begingroup$ Your statement seems to implicitly assume the existence of a density (with respect to Lebesgue measure). You may be interested in this short and delightful paper: A. R. Barron (1986), Entropy and the Central Limit Theorem Ann. Probab., vol 14, no. 1, 336-342. (open access). $\endgroup$
    – cardinal
    Commented Jun 28, 2012 at 13:36
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    $\begingroup$ I had looked at that paper already. He has given a motivation in information theoretic perspective in the second paragraph of page 1. It was not all that clear to me at that time. Now it looks ok. Still, if one can explain the following clearly and post as an answer, it would be great. "From information theory, the relative entropy $D_n$ is the least upper bound to the redundancy (excess average description length) of the Shannon code based on the normal distribution when describing quantizations of samples from $f_n$." I have deleted that question in math.SE as it didn't attract anybody there $\endgroup$
    – Ashok
    Commented Jun 28, 2012 at 14:02

3 Answers 3

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One thing which is great with this theorem is that it suggests limit theorems in some settings where the usual central limit theorem do not apply. For instance, in situations where the maximum entropy-distribution is some nonnormal distribution, such as for distributions on the circle, it suggests convergence to a uniform distribution.

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  • $\begingroup$ I don't understand. As I already mentioned, convergence in KL divergence implies convergence in distribution, know? So whereever information theoretic CLT applies, usual CLT also applies. Moreover, information theoretic CLT also assumes finite variance. Or am I missing something? $\endgroup$
    – Ashok
    Commented Oct 1, 2012 at 8:02
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    $\begingroup$ What I meant is that the entropy method suggests what the limit could be in situations where the limit is not a normal distribution. The limit is then a distribution which maximizes entropy. $\endgroup$ Commented Nov 19, 2014 at 10:44
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After looking around, I could not find any example of convergence in distribution without convergence in relative entropy, so this is hard to measure the "greatness" of that result.

To me, it looks like this result simply describes the relative entropy of convolution products. It is often viewed as an alternative interpretation and proof framework of the Central Limit Theorem, and I am not sure it has a direct implication in probability theory (even though it does in information theory).

From Information Theory and the Central Limit Theorem (page 19).

The Second Law of Thermodynamics states that the thermodynamic entropy always increases with time, implying some kind of convergence to the Gibbs state. Conservation of energy means that $E$ remains constant during this time evolution, so we can tell from the start which Gibbs state will be the limit.
We will regard the Central Limit Theorem in the same way, by showing that the information-theoretic entropy increases to its maximum as we take convolutions, implying convergence to the Gaussian. Normalizing appropriately means that the variance remains constant during convolutions so we can tell from the start which Gaussian will be the limit.

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    $\begingroup$ There are lots of examples of convergence in distribution without convergence in relative entropy -- any time the $X_i$ have a discrete distribution and the CLT applies. $\endgroup$ Commented Aug 27, 2012 at 19:36
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$D(f_n\|\phi)\rightarrow 0$ assures that there's no "distance" between the distribution of the sum of random variables and the gaussian density as $n\rightarrow\infty$ just because of the definition of K-L divergence, so it's the proof itself. Perhaps I misunderstood your question.

About the second point as you appointed, it's responded in your paragraph.

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    $\begingroup$ The normal (Lindberg) CLT states that the sample mean converges in distribution to a normal RV. That means that the CDF converges pointwise to $\Phi$. There's a subtle measure theoretic difference between that and the OP's result that is not reflected in your answer here. $\endgroup$
    – AdamO
    Commented Dec 24, 2017 at 19:38

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