8
$\begingroup$

I'm reading a paper that refers to the limit as n goes to infinity of Rényi entropy. It defines it as ${{H}_{n}}\left( X \right)=\dfrac{1}{1-n} \log_2 \left( \sum\limits_{i=1}^{N}{p_{i}^{n}} \right)$. It then says that the limit as $n\to \infty $ is $-\log_2 \left( p_1 \right)$. I saw another article that uses the maximum of the ${{p}_{i}}'s$ instead of ${{p}_{1}}$. I think that this works out fairly easily if all of the ${{p}_{i}}'s$ are equal (a uniform distribution). I have no idea how to prove this for anything other than a uniform distribution. Can anyone show me how it's done?

$\endgroup$
7
$\begingroup$

Suppose $p_1 = \max_i p_i$. We have $$ p_1^n \leq \sum_{i=1}^N p_i^n \leq N p_1^n. $$ Therefore $$ \frac{n \log p_1 + \log N}{1-n} \leq H_n(X) \leq \frac{n \log p_1}{1-n}. $$ As $n \rightarrow \infty$, $\log N/(1-n) \rightarrow 0$ while $n/(1-n) \rightarrow -1$.

$\endgroup$

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.