2
$\begingroup$

I would like to understand the proof that mutual information $I(X;Y)$ is concave in $p(x)$ - as presented in Elements of Information Theory by Cover & Thomas, theorem 2.7.4.

Here's the proof from the textbook:

enter image description here

I understand that $p(y) = \sum_x p(x,y) = \sum_x \color{blue}{p(x)} \, p(y \mid x)$, i.e. $p(y)$ is a linear combination of $p(x)$. But why from the fact that $H(Y)$ concave in $p(y)$ it follows that it's concave in $p(x)$?

And also, why fixing $p(y \mid x)$ is important? It's never negative anyway - it's a probability distribution.

$\endgroup$
3
$\begingroup$

In general, if $f:\mathbb{R}^n\to \mathbb{R}$ is concave, then for any matrix $A \in\mathbb{R}^{n\times n}$, $f(Ax)$ is also concave in $x$, and the proof uses this fact, i.e., the fact that the concave function of a linear combination is still a concave function.

Proof: For any $0\leq \lambda\leq 1$, and any $x_1,x_2 \in \mathbb{R}^n$, $$ f\left(A\left( \lambda x_1 + (1-\lambda)x_2\right)\right) = f\left(\lambda Ax_1 + (1-\lambda)Ax_2\right) \geq \lambda f(Ax_1) + (1-\lambda)f(Ax_2) , $$ where the inequality follows from the fact that $f$ is concave, and by the definition of concavity. $\square$

Again, in general, it is possible for a function $g: \mathbb{R}^n\to \mathbb{R}$ to be concave in a subset of its arguments when the remaining arguments are fixed, but still not be concave in the entire set of arguments, e.g., a function with two inputs, $g(x,y)$, can be concave in $x$ for every fixed $y$, but not concave in the pair $(x,y)$. In fact, this is exactly the case for mutual information: For a fixed $p(y|x)$, it is concave in $p(x)$; for a fixed $p(x)$, it is convex in $p(y|x)$; and it is neither convex nor concave in the pair $\left( p(x), p(y|x)\right)$.

$\endgroup$
  • $\begingroup$ Thanks! Maybe you can recommend a reference to the proof of the first fact, i.e. that the concave function of a linear combination is a concave function? $\endgroup$ – Alexey Grigorev Jan 14 '15 at 9:17
  • $\begingroup$ I added a proof of that in between. $\endgroup$ – karakusc Jan 14 '15 at 9:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.