0
$\begingroup$

For (discrete, finite) probability distributions $P,Q$, the Bhattacharyya coefficient is $B(P,Q) := \sum_x \sqrt{P_x Q_x}$. It can be shown that this is jointly concave in $P$ and $Q$. My question is, is there a monotone increasing function of the Bhattacharyya coefficient that is instead jointly convex in $P$ and $Q$? For instance, is it possible that $B(P,Q)^k$ might be jointly convex for some sufficiently large $k$? (As a starting point, it can be shown that $k=2$ is not sufficient.)

As a follow-up, does the joint convexity of that variant also hold if its domain is extended to include subnormalized probability distributions as well?

(The context of the question is that I wish to minimize $B(p,q)$ for subnormalized distributions $p$, $q$ belonging to convex sets $\mathcal{S}$ and $\mathcal{T}$ respectively. If I can transform the objective (in a monotone increasing fashion) to a convex function, then I would be able to apply standard convex-optimization algorithms.)

$\endgroup$
1
  • $\begingroup$ (The proof of concavity that I'm familiar with is based on Uhlmann's theorem in quantum information theory, so I'm not sure if this question might be better placed on quantumcomputing.stackexchange.com.) $\endgroup$
    – helloworld
    Commented Jun 24, 2020 at 18:28

1 Answer 1

0
$\begingroup$

In hindsight, this question has an obvious negative answer, apart from the trivial case where the transformed function is simply a constant function. Suppose there exists an increasing function $f: [0,1] \to \mathbb{R}$ such that the function $\tilde{B}(P,Q) := f(B(P,Q))$ is jointly convex with respect to $(P,Q)$. Pick any distribution $R$ in the interior of the set of distributions being considered, in which case the point $(P,Q)=(R,R)$ lies in the interior of the domain of $\tilde{B}$. Now note that we have $\tilde{B}(R,R) = f(B(R,R)) = f(1)$, and since $f$ is increasing on $[0,1]$, this means $\tilde{B}$ attains its maximum possible value (over its domain) at an interior point of its domain. But the only convex functions that can do so are constant functions, implying that $\tilde{B}$ can only be a constant function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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