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Let $X,Y$ be probability measures on $\{1,2,\dots,n\}$, and set $K=\sum_i\sqrt{X(i)Y(i)}$ so that $Z:=\frac{1}{K}\sqrt{XY}$ is also a probability measure on $\{1,2,\dots,n\}$. How can we prove the ...

Short version Gibbs's inequality is a simple inequality for real numbers, usually understood information-theoretically. In the jargon, it states that for two probability measures on a finite set, ...

Let $ p, q $ be two probability densities on $ [0,1] $, strictly positive over $ (0,1) $. Let $ P $ be the cumulative function of $ p $, i.e., $ P(x) = \int_0^x p(x') \, \mathrm{d}x' $, $ x \in [0,1] $...

Gibbs' inequality is equivalent to: \begin{equation} \sum_{i} \ln q_i^{p_i}-\ln p_i^{p_i} \leq 0 \end{equation} where $p_i,q_i \in [0,1]$ and $\sum_i p_i = \sum_i q_i=1$. Now, a friend of mine ...