# An Inequality Involving a Product of Mutual Informations

Many inequalities in information theory have entropies or mutual informations appearing as linear additive terms (e.g. Shannon-type inequalities). It’s also not uncommon to see entropies appearing as exponents (e.g. entropy power inequality). But perhaps it’s not immediately seen how products of mutual informations make sense.

Recently I have encountered an inequality involving a product of mutual informations, which I cannot find a good way to prove (or disprove, though some numerics and asymptotic analysis seem to suggest its validity). I would much appreciate it if someone could be smart and gracious enough to provide a proof, or counter example, or generalization.

The formulation is quite simple: suppose $X,Y$ are binary random variables with  $\displaystyle P(X=0)=P(Y=0)=1/2$, and $Z$ is another random variable such that $X-Y-Z$ forms a Markov chain. The claim is that $I(X;Z)\ge I(X;Y)I(Y;Z)$. (Note that the left side is also upper bounded by either one of the two factors on the right side, by the well-known data processing  inequality.

At first glance this inequality seems absurd because different units appear on the two sides; but this may be resolved by considering that $Y$ has only one bit of information.

For a given joint distribution of $X$ and $Y$, the equality can be achieved when $Z$ is an injective function of $Y$.

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Update: the inequality has been proved, thanks very much for the help by Sudeep Kamath (see comments). An alternative proof (which avoids computing the monotonicity of  $f(r) = D(r*\epsilon||1/2)/D(r||1/2) = (1-h(r*\epsilon))/(1-h(r)))$ is sketched as follows: by convexity of the function $h(\epsilon*h^{-1}(x))$ (see for example, proof of Mrs Gerbers Lemma), we have $H(X|Z=z)=h(\epsilon*h^{-1}(H(Y|Z=z)))\le H(Y|Z=z)+(1-H(Y|Z=z))h(\epsilon)$ for any $z$, where $\epsilon:=Prob(X\neq Y)$. Averaging over $z$ under $P_Z$ gives $H(X|Z)\le H(Y|Z)+(1-H(Y|Z))h(\epsilon)$, and the claim follows after rearrangements.

# Update (4/3/2014): I believe I have solved the conjecture, and proven it to be correct.  I will make a preprint available shortly(see link below). The original blog post remains available below. — Tom

Preprint available here:An extremal inequality for long Markov chains“.

I have an extremal conjecture that I have been working on intermittently with some colleagues (including Jiantao Jiao, Tsachy Weissman, Chandra Nair, and Kartik Venkat). Despite our efforts, we have not been able to prove it. Hence, I thought I would experiment with online collaboration by offering it to the broader IT community.

In order to make things interesting, we are offering a \$1000 prize for the first correct proof or counterexample! Feel free to post your thoughts in the public comments. You can also email me if you have questions or want to bounce some ideas off me.

Although I have no way of enforcing them, please abide by the following ground rules:

1. If you decide to work on this conjecture, please send me an email to let me know that you are doing so. As part of this experiment with online collaboration, I want to gauge how many people become involved at various degrees.
2. If you solve the conjecture or make significant progress, please keep me informed.
3. If you repost this conjecture, or publish any results, please cite this blog post appropriately.

One final disclaimer: this post is meant to be a brief introduction to the conjecture, with a few partial results to get the conversation started; it is not an exhaustive account of the approaches we have tried.

1. The Conjecture

Conjecture 1. Suppose ${X,Y}$ are jointly Gaussian, each with unit variance and correlation ${\rho}$. Then, for any ${U,V}$ satisfying ${U-X-Y-V}$, the following inequality holds:

$\displaystyle 2^{-2I(Y;U)} 2^{-2I(X;V|U)} \geq (1-\rho^2)+ \rho^2 2^{-2I(X;U)} 2^{-2I(Y;V|U)} . \ \ \ \ \ (1)$

2. Partial Results

There are several partial results which suggest the validity of Conjecture 1. Moreover, numerical experiments have not produced a counterexample.

Conjecture 1 extends the following well-known consequence of the conditional entropy power inequality to include long Markov chains.

Lemma 1 (Oohama, 1997). Suppose ${X,Y}$ are jointly Gaussian, each with unit variance and correlation ${\rho}$. Then, for any ${U}$ satisfying ${U-X-Y}$, the following inequality holds:

$\displaystyle 2^{-2 I(Y;U)} \geq 1-\rho^2+\rho^2 2^{-2I(X;U)}. \ \ \ \ \ (2)$

Proof: Consider any ${U}$ satisfying ${U-X-Y}$. Let ${Y_u, X_u}$ denote the random variables ${X,Y}$ conditioned on ${U=u}$. By Markovity and definition of ${X,Y}$, we have that ${Y_u = \rho X_u + Z}$, where ${Z\sim N(0,1-\rho^2)}$ is independent of ${X_u}$. Hence, the conditional entropy power inequality implies that

$\displaystyle 2^{2h(Y|U)} \geq \rho^2 2^{2h(X|U)} + 2 \pi e(1-\rho^2) = 2 \pi e \rho^2 2^{-2I(X;U)} + 2 \pi e(1-\rho^2). \ \ \ \ \ (3)$

From here, the lemma easily follows. $\Box$

Lemma 1 can be applied to prove the following special case of Conjecture 1. This result subsumes most of the special cases I can think of analyzing analytically.

Proposition 1. Suppose ${X,Y}$ are jointly Gaussian, each with unit variance and correlation ${\rho}$. If ${U-X-Y}$ are jointly Gaussian and ${U-X-Y-V}$, then

$\displaystyle 2^{-2I(Y;U)} 2^{-2I(X;V|U)} \geq (1-\rho^2)+ \rho^2 2^{-2I(X;U)} 2^{-2I(Y;V|U)}. \ \ \ \ \ (4)$

Proof: Without loss of generality, we can assume that ${U}$ has zero mean and unit variance. Define ${\rho_u = E[XU]}$. Since ${U-X-Y}$ are jointly Gaussian, we have

$\displaystyle I(X;U) =\frac{1}{2}\log\frac{1}{1-\rho_u^2} \ \ \ \ \ (5)$

$\displaystyle I(Y;U) =\frac{1}{2}\log\frac{1}{1-\rho^2\rho_u^2}. \ \ \ \ \ (6)$

Let ${X_u,Y_u,V_u}$ denote the random variables ${X,Y,V}$ conditioned on ${U=u}$, respectively. Define ${\rho_{XY|u}}$ to be the correlation coefficient between ${X_u}$ and ${Y_u}$. It is readily verified that

$\displaystyle \rho_{XY|u} = \frac{\rho\sqrt{1-\rho_u^2}}{\sqrt{1-\rho^2\rho_u^2}}, \ \ \ \ \ (7)$

which does not depend on the particular value of ${u}$. By plugging (5)-(7) into (4), we see that (4) is equivalent to

$\displaystyle 2^{-2I(X;V|U)} \geq (1-\rho_{XY|u}^2)+ \rho_{XY|u}^2 2^{-2I(Y;V|U)}. \ \ \ \ \ (8)$

For every ${u}$, the variables ${X_u,Y_u}$ are jointly Gaussian with correlation coefficient ${\rho_{XY|u}}$ and ${X_u-Y_u-V_u}$ form a Markov chain, hence Lemma 1 implies

$\displaystyle 2^{-2I(X_u;V_u)} \geq (1-\rho_{XY|u}^2)+ \rho_{XY|u}^2 2^{-2I(Y_u;V_u)}. \ \ \ \ \ (9)$

The desired inequality (8) follows by convexity of

$\displaystyle \log\left[(1-\rho_{XY|u}^2)+ \rho_{XY|u}^2 2^{-2z}\right] \ \ \ \ \ (10)$

as a function of ${z}$. $\Box$

3. Equivalent Forms

There are many equivalent forms of Conjecture 1. For example, (1) can be replaced by the symmetric inequality

$\displaystyle 2^{-2(I(X;V)+I(Y;U))} \geq (1-\rho^2)2^{-2I(U;V)}+ \rho^2 2^{-2(I(X;U)+I(Y;V))}. \ \ \ \ \ (11)$

Alternatively, we can consider dual forms of Conjecture 1. For instance, one such form is stated as follows:

Conjecture 1′. Suppose ${X,Y}$ are jointly Gaussian, each with unit variance and correlation ${\rho}$. For ${\lambda\in [{1}/({1+\rho^2}),1]}$, the infimum of

$\displaystyle I(X,Y;U,V)-\lambda\Big(I(X;UV)+I(Y;UV)\Big), \ \ \ \ \ (12)$

taken over all ${U,V}$ satisfying ${U-X-Y-V}$ is attained when ${U,X,Y,V}$ are jointly Gaussian.