6
$\begingroup$

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

$\endgroup$
0

1 Answer 1

3
$\begingroup$

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 $x\le\frac19$), $I^2+Z^2>1$.

$\endgroup$
1
  • $\begingroup$ Whoops! That is bad :( Thank you anyway! $\endgroup$ Commented Sep 29, 2011 at 23:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .