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I was trying to solve the following problem. But I dont know how to proceed. I would be really grateful if anybody would point me in the right direction.

Let $P = (p_1,p_2, \cdots, p_n)$ be a probability vector (That is $\sum p_i = 1$ and $p_i \geq 0$). Let $Q = (q_1,q_2, \cdots, q_n)$ be a permutation of the vector $P$.

If $I = \frac14 \left(\sum p_i \log \frac{p_i}{q_i}\right) + \frac14 \left(\sum q_i \log \frac{q_i}{p_i}\right)$ (involves the Kullback - Leibler divergence)

And $Z = \sum \sqrt{p_iq_i}$ (called the Bhattacharyya distance)

Prove that $I^2 + Z^2 \leq 1$

Thank you

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Order? Then no answer. – Did Sep 29 '11 at 5:54
@DidierPiau: What does that mean? What is 'Order'? – Isomorphism Sep 29 '11 at 6:47
It is not homework. It is just an interesting problem. Let me put the appropriate thank yous and pleases. thank you – Isomorphism Sep 29 '11 at 8:09
Oh brother. Mathematical problems are often naturally framed as commands, so when discussing them politely one has to frame them as hypothetical commands using appropriate delimiters or risk distancing oneself from the intended audience. Frequently we get newcomers that aren't aware of this community sensitivity, and frequently we have regulars that don't respond nicely when newcomers aren't conscious of how they come off. I've put the mathematical content of the problem in a blockquote to set an example. – anon Sep 29 '11 at 8:27
@Iso, The point is not what I want (I want nothing). I mentioned that the first formulation of your question prevented me (as well as others, I believe) to answer it. The new formulation is better, even though the description of what you tried should be included in the question, and not buried in a comment. – Did Sep 29 '11 at 19:42

1 Answer 1

up vote 2 down vote accepted

Not sure the inequality always holds. Let us have a quick look at the $n=2$ case, that is, let us choose $p=(x,1-x)$ an $q=(1-x,x)$. Unless I am mistaken, one gets $Z=2\sqrt{x(1-x)}$ and $$ 2I=x\log(x/(1-x))+(1-x)\log((1-x)/x)=(1-2x)\ell(x), $$ with $\ell(x)=\log((1-x)/x)$, hence $$ I^2+Z^2-1=\frac14(1-2x)^2\ell(x)^2+4x(1-x)-1=\frac14(1-2x)^2(\ell(x)^2-4). $$ If $\ell(x)>2$ (for example, for every $x\le\frac19$), $I^2+Z^2>1$.

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Whoops! That is bad :( Thank you anyway! – Isomorphism Sep 29 '11 at 23:56

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