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HGR maximal correlation revisited : a corrected reverse inequality

Sudeep Kamath sent me a note about a recent result he posted on the ArXiV that relates to an earlier post of mine on the HGR maximal correlation and an inequality by Erkip and Cover for Markov chains Image may be NSFW.
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U -- X -- Y
which I had found interesting:
Image may be NSFW.
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I(U ; Y) \le \rho_m(X,Y)^2 I(U ; X)
.
Since learning about this inequality, I’ve seen a few talks which have used the inequality in their proofs, at Allerton in 2011 and at ITA this year. Unfortunately, the stated inequality is not correct!

On Maximal Correlation, Hypercontractivity, and the Data Processing Inequality studied by Erkip and Cover
Venkat Anantharam, Amin Gohari, Sudeep Kamath, Chandra Nair

What this paper shows is that the inequality is not satisfied with Image may be NSFW.
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\rho_m(X,Y)^2
, but by another quantity:
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I(U ; Y) \le s^*(X;Y) I(U ; X)

where Image may be NSFW.
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s^*(X;Y)
is given by the following definition.

Let Image may be NSFW.
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X
and Image may be NSFW.
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Y
be random variables with joint distribution Image may be NSFW.
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(X, Y) \sim p(x, y)
. We define
Image may be NSFW.
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s^*(X;Y) = \sup_{r(x) \ne p(x)} \frac{ D( r(y) \| p(y) ) }{ D( r(x) \| p(x) }
,
where Image may be NSFW.
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r(y)
denotes the Image may be NSFW.
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y
-marginal distribution of Image may be NSFW.
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r(x, y) := r(x)p(y|x)
and the supremum on the right hand side is over all probability distributions Image may be NSFW.
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r(x)
that are different from the probability distribution Image may be NSFW.
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p(x)
. If either Image may be NSFW.
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X
or Image may be NSFW.
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Y
is a constant, we define Image may be NSFW.
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s^*(X; Y)
to be 0.

Suppose Image may be NSFW.
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(X,Y)
have joint distribution Image may be NSFW.
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P_{XY}
(I know I am changing notation here but it’s easier to explain). The key to showing their result is through deriving variational characterizations of Image may be NSFW.
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\rho_m
and Image may be NSFW.
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s^*
in terms of the function
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t_{\lambda}( P_X ) := H( P_Y ) - \lambda H( P_X )

Rather than getting into that in the blog post, I recommend reading the paper.

The implication of this result is that the inequality of Erkip and Cover is not correct : not only is Image may be NSFW.
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\rho_m(X,Y)^2
not the supremum of the ratio, there are distributions for which it is not an upper bound. The counterexample in the paper is the following: Image may be NSFW.
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X \sim \mathsf{Bernoulli}(1/2)
, and Image may be NSFW.
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Y
is generated via this asymmetric erasure channel:

Image may be NSFW.
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Joint distribution counterexample

Joint distribution counterexample (Fig. 2 of the paper)


How can we calculate Image may be NSFW.
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\rho_m(X,Y)
? If either Image may be NSFW.
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X
or Image may be NSFW.
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Y
is binary-valued, then
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\rho_m(X,Y)^2 = -1 + \sum_{x,y} \frac{ p(x,y)^2 }{ p(x) p(y) }

So for this example Image may be NSFW.
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\rho_m(X,Y)^2 = 0.6
. However, Image may be NSFW.
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s^*( X,Y) = \frac{1}{2} \log_2(12/5) > 0.6
and there exists a sequence of variables Image may be NSFW.
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U_i
satisfying the Markov chain such that Image may be NSFW.
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U_i -- X -- Y
such that the ratio approaches Image may be NSFW.
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s^*
.

So where is the error in the original proof? Anantharam et al. point to an explanation that the Taylor series expansion used in the proof of the inequality with Image may be NSFW.
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\rho_m(X,Y)^2
may not be valid at all points.

This seems to just be the start of a longer story, which I look forward to reading in the future!


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