3
$\begingroup$

Suppose we have two distributions given by the vectors $p=(p_1,\dots,p_n)$ and $q=(q_1,\dots,q_n)$, with $p_i,q_i\geq 0$, and $\sum_i p_i = \sum_i q_i=1$.

Now suppose that for some $\alpha\in(0,\infty)$,

$$H_{\alpha}(p)=H_{\alpha}(q),$$

where $H_{\alpha}(p)=\frac{1}{1-\alpha}\log\sum_i p_i^{\alpha}$ is the Rényi entropy of $p$. What can we say about $p$ and $q$? Of course if $q$ is a permutation of $p$, then their entropies will be equal. But is the converse true?

$\endgroup$
  • $\begingroup$ The set of probability distributions $p$ is an $(n-1)$ dimensional region (a simplex). The set of distributions which satisfy $H_a(p)=C$ for some constant $C$ should be an $(n-2)$ dimensional set, since $H_a(p)=C$ is a single constraint. Therefore, there should be a large number of distributions which have the same Renyi entropy for any particular $\alpha$. However, if $H_\alpha(p)=H_\alpha(q)$ for all $\alpha>0$, you could probably conclude $p$ is a permutations of $q$. $\endgroup$ – Mike Earnest Jun 6 '18 at 13:51
1
$\begingroup$

As pointed out in the comment, all you can say is that the ${\alpha}-$ norm of the two vectors are equal: $$ \sum_{i=1}^n p_i^{\alpha}=\sum_{i=1}^n q_i^{\alpha} $$ if and only if $$ \lVert p\rVert_{\alpha}=\lVert q\rVert_{\alpha}. $$ Of course their $1-$norm is one (e.g., $\sum_{i=1}^n p_i=1$) and their entries are nonnegative as well. So they all lie in the positive orthant of $\mathbb{R}^n$ on the unit sphere $S_{n-1}.$

$\endgroup$

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.