# Max mutual information with variance constraint

Let $$x_1$$ and $$x_2$$ be two independent random variables with the same probability distribution $$p(x)=p(x_1)=p(x_2)$$, and let $$z$$ be a normal random variable which is independent of $$x_1$$ and $$x_2$$. Consider the following problem: $$\tag{1} \label{1} \max_{p(x):\ \ \mathbb{E}[x^2]\le 1} I(x_1;x_1+x_2+z),$$ where $$I(\cdot;\cdot)$$ denote the mutual information between two random variables.

My question. Is it true that the optimal probability distribution $$p(x)$$ in \eqref{1} is Gaussian?

If we consider $$I(x_1;x_1+z)$$ instead of $$I(x_1;x_1+x_2+z)$$, then it is easy to show (using a maximum entropy argument) that the optimal distribution must be Gaussian. However it is not clear if a similar argument applies to my case. Any help is very appreciated. Thank you.

I don't have a very illuminating answer or anything, just numerical evidence that the Gaussian is not the maximiser above.

I'll work with natural logs for convenience.

Note that, by independence, $$I(X_1; X_1 + X_2 + Z) = h(X_1 + X_2 + Z) - h(X_2 + Z).$$

If $$X$$ is Gaussian of variance $$\sigma^2$$, then we can explicitly compute functional above to be $$1/2 \log (2 - 1/(1+\sigma^2))$$ This is increasing with $$\sigma$$, and under the constraints, the best any Gaussian can do is $$1/2 \log (3/2) \le 0.20274.$$

On the other hand, consider $$X$$ uniform on $$\{+ 1, -1\}$$.

Let $$\varphi$$ be the density of the standard Gaussian. Note then that the density of $$X + Z$$ is $$(\varphi(u-1) + \varphi(u+1))/2$$ and the density of $$X_1 + X_2 + Z$$ is $$(\varphi(u-2) + \varphi(u+2) + 2\varphi(u))/4.$$

At this point, I simply enter these expressions into wolfram alpha. By these computations,

1. $$h(X_1 + X_2 +Z) \ge 1.960$$ - see https://bit.ly/2RWQuyl
2. $$h(X_2 + Z) \le 1.756$$ - see https://bit.ly/2FWPo0A

Thus, for this $$X$$, the functional is $$\ge 0.204.$$

(In case you want to check the above - W|A helpfully displays the latex-ed out versions of the expressions keyed in).