Let $X\sim \mathcal{N}(\mu,\sigma^2)$ be a Gaussian random variable with random mean $\mu\sim {\sf Bernoulli}(p)$, i.e., $\mu=1$ with probability $p$ and $\mu=0$ with probability $1-p$. In other words, to draw a realization of $X$, first one flips a coin to decide whether $\mu=0$ or $\mu=1$ and then, one samples from the Gaussian distribution with chosen mean.

Consider the family $\mathcal{M}$ of Bernoulli random variables taking on the values $0$ or $1$.

Let $\mathcal{I}\left(Y,X\right)$ be the mutual information between $Y$ and $X$ (a natural definition can be found, e.g., in this mathoverflow post).

My intuition tells that a global solution for the optimization problem

$$\max_{Y\in\mathcal{M}} \mathcal{I}(Y,X)$$

is given by $\mu$ itself, i.e.,

$$\mathcal{I}(\mu,X) = \max_{Y\in\mathcal{M}} \mathcal{I}(Y,X).$$

Any reference or ideas on how to prove or disprove this claim would be appreciated.

  • $\begingroup$ @IosifPinelis: (i) Maybe I am missing something, but I think the definition is symmetric (in any case, I have edited my post to reflect your comment); (ii) mutual information is defined for any two random variables -- these random variables will have a joint distribution whether we know it explicitly or not (in the optimization problem, they are not given explicitly). $\endgroup$ Jan 29, 2023 at 5:40
  • $\begingroup$ I see. Still, you cannot say "a global solution for the optimization problem [...] is given by $\mu$ itself" -- because any such solution is a joint distribution. $\endgroup$ Jan 29, 2023 at 12:53
  • $\begingroup$ Iosif, I understand that when we break down the mutual information, the most natural landscape of optimization is the space of (joint) distributions (that is actually one of my difficulties in handling this problem). We can still formulate the problem in the space of measurable functions (or random variables). "$\mu$ is a global solution" reads in the sense of the last identity in my post. $\mu$ has a distribution and there is a joint distribution between $X$ and $\mu$. In other words, I believe this (distribution) is a global optimum for the optimization problem. $\endgroup$ Jan 29, 2023 at 13:12

1 Answer 1


$\newcommand{\de}{\delta}\newcommand{\M}{\mathcal M} \newcommand{\ep}{\varepsilon} \newcommand{\thh}{\theta}\newcommand\I{\mathcal I}\newcommand{\Si}{\Sigma}$Here is the (slightly edited) definition of the mutual information given in the answer linked in the OP:

Let $D$ be any discrete random variable (r.v.) with distinct values $d_i$ taken with probabilities $p_i=P(D=d_i)>0$ for $i\in I$, where $I$ is a denumerable (that is, at most countable) set. Let $X$ be any r.v. (defined on the same probability space as $D$), with values in any nonempty set $S$ (given also some sigma-algebra $\Sigma$ over $S$, to make $S$ a measurable space). Let $\nu$ be the probability distribution of $X$, so that $\nu(B)=P(X\in B)$ for all $B\in\Sigma$. For each $i\in I$ and each $B\in\Sigma$, let \begin{equation*} \nu_i(B):=P(D=d_i,X\in B). \end{equation*} Then $\nu_i$ is a (sub-probability) measure absolutely continuous with respect to $\nu$, so that we can consider a Radon--Nikodym density \begin{equation*} \rho_i:=\frac{d\nu_i}{d\nu} \end{equation*} of the measure $\nu_i$ with respect to $\nu$, so that the values of $\rho_i$ are in $[0,1]$.

Then the mutual information between $D$ and $X$ is defined as follows: \begin{equation*} \I(D,X):=\sum_{i\in I}\int_S d\nu\;\rho_i\ln\frac{\rho_i}{p_i}. \end{equation*}

Let us now answer the current question.

Much more generally than in the OP, let $X$ be any r.v. as in the highlighted text above. Let $\nu$ denote the distribution of $X$.

Let $\M$ denote the set of all Bernoulli r.v.'s $Y$ (defined on the same probability space as $X$) with $P(Y=1)\in(0,1)$; we suppose that the probability space is rich enough so that for any coupling $\gamma$ of any Bernoulli distribution and the distribution $\nu$ of $X$ there is a r.v. $Y$ such that the joint distribution of the pair $(Y,X)$ is $\gamma$.

In view of the highlighted definition, for any $Y\in\M$, \begin{equation*} \I(Y,X)=\int_S d\nu\; \Big(\rho\ln\frac{\rho}{p}+(1-\rho)\ln\frac{1-\rho}{1-p}\Big), \end{equation*} where $p:=P(Y=1)$, $\rho:=\frac{d\nu_1}{d\nu}$, and $\nu_1(B):=P(Y=1,X\in B)$ for all $B\in\Si$; we are assuming the convention that $0\times\text{anything}:=0$.

We want to maximize $\I(Y,X)$ over all $Y\in\M$. This is very easy to do, noting that $u\ln\frac{u}{p}+(1-u)\ln\frac{1-u}{1-p}$ is convex in $u\in[0,1]$. So, $\I(Y,X)$ is maximized when $\rho=1_A$ for some $A\in\Si$, and then $p=\nu(A)$ and $\I(Y,X)=\int_A d\nu\; \ln\frac1{p}=p\ln\frac1p$. Maximizing the latter expression in $p\in(0,1)$, we see that, contrary to the conjecture in the OP,
\begin{equation} \max_{Y\in\M}\I(Y,X)=\frac1e, \end{equation} for any r.v. $X$ whatsoever provided that $X$ is defined on a rich enough probability space.


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