The differential Renyi entropy for a probability distribution is given by $H_q(P(X))=\frac{1}{1-q}\log\int p^q(x)dx$. In the limit of $q\to 1$, it reduces to the usual Shannon entropy. We can write down the mutual information between two variables X and Y simply by $I(X;Y)=H_q(P(X))+H_q(P(Y))-H_q(P(X,Y))$. Is this always a non-negative quantity? Again, in the case $q=1$ it is very easy to show it, but what about in general?

  • $\begingroup$ Hint: In the Shannon case you need to consider the Kullback Leibner divergence $D_{\pm 1}$ to prove that the mutual information is non negative. In the Renyi case you need to consider the $f$-divergence $D_f$ associated to it. Can you do it? $\endgroup$
    – Avitus
    Jun 27, 2013 at 17:25
  • $\begingroup$ Thanks, but I really don't see it :-(. In the Shannon case I can prove it with a simple application of Jensen's to xln(x), but in Renyi's case I seem to come up short... $\endgroup$
    – Ivan
    Jun 27, 2013 at 17:46
  • $\begingroup$ Ok, I wrote an answer, but I will add some more details, if you need them $\endgroup$
    – Avitus
    Jun 27, 2013 at 17:57

3 Answers 3


EDIT. I justify the positivity of the Renyi mutual information using its interpretation as Renyi divergence. I follow the expositions in

T. Cover, J. A. Thomas "Elements of Information Theory" (chapter 2)


D. Xu, D. Erdogmuns "Renyi's Entropy, Divergence and their Nonparametric Estimators"

  • Shannon entropy and mutual information

In the setting of "classical" information theory the mutual information $I(X,Y)$ of the random variables $X$ and $Y$ is defined as


where $D_{KL}(p_{XY}||p_Xq_Y),$ denotes the Kullback Leibler divergence (KL divergence) between the joint probability $p_{XY}$ and the product $p_Xq_Y$ of the prob. distribution of $X$ and $Y$.

Using the Jensen inequality on the KL divergence it follows that $I(X,Y)$ is always non negative. I refer to the first reference for the computation in the discrete case.

Introducing the Shannon entropies $H(X)$ $H(Y)$ of $X$ resp. $Y$ and the conditional entropy $H(X|Y)$ we arrive at the equivalent formulation


  • Renyi Entropy and mutual information

Let us consider the Renyi $\alpha$-setting , now. With

$$H_{\alpha}(X)=\frac{1}{1-\alpha}\log\int p^{\alpha}_X(x)dx$$

we denote the Renyi entropy of the r.v. $X$. The Renyi divergence of the distribution $g(x)$ from the distribution $f(x)$ is

$$D_{\alpha}(f||g):=\frac{1}{\alpha-1}\log\int f(x)\left(\frac{f(x)}{g(x)}\right)^{\alpha-1}dx.$$

It can be proved that (please see the second reference at pag.81)

$$D_{\alpha}(f||g)\geq 0 \forall ~f, g, \text{and}~\alpha>0,~~(*)$$ $$\lim_{\alpha\rightarrow 1}D_{\alpha}(f||g)=D_{1}(f||g)=D_{KL}(f||g).~~(*)$$

The Renyi mutual information $I_{\alpha}(X,Y)$ is defined naturally as the Renyi divergence between the joint distribution $p_{XY}$ of $X$ and $Y$ and the product of the marginal distributions $p_X$, $q_Y$, i.e.


This is a definition; you can find it, for example, at pag. 83 in the second reference. You can justify it through the overall $\alpha$-setting and the limit

$$\lim_{\alpha\rightarrow 1}I_{\alpha}(X,Y)=I(X,Y),$$

which follows from property $(**)$ of the Renyi divergence. This limit is parallel to the fundamental $\lim_{\alpha\rightarrow 1}H_{\alpha}(X)=H(X):$

From property $(*)$ one derives nonnegativity of the Renyi mutual information.

For these reasons, I would prove non negativity of the Renyi mutual information through the above definition. At the present stage I haven't been able to prove that


or to find such characterization in the literature. Even in the discrete case I got blocked because of the coefficient $\frac{1}{1-\alpha}$ in front of the entropies. The cases $0<\alpha<1$ and $\alpha>1$ must be studied separately and it seems that a straightforward application of Jensen's inequality is not possible.

  • $\begingroup$ I don't think that the Renyi mutual information is given by the integral on the R.H.S. In Renyi's case, the logarithm is outside the integral. $\endgroup$
    – Ivan
    Jun 27, 2013 at 18:05
  • $\begingroup$ @Ivan you are right, that is a typo. To apply Jensen one starts with the $ln$ outside. I edit it now. $\endgroup$
    – Avitus
    Jun 27, 2013 at 18:10
  • $\begingroup$ Sorry, but I really don't see how you end up with a single integral and not a ratio of integrals. If I follow the definition that I have in the question, I get $I(X;Y)=1/(q-1)log(\int p^q(x,y)/(\int p^q(x) \int p^q(y))$ which cannot be reduced as far as I can see $\endgroup$
    – Ivan
    Jun 27, 2013 at 18:18
  • $\begingroup$ I seem to have found a literature reference that suggests that Renyi mutual informations don't have to be positive definite: maths.nottingham.ac.uk/personal/ga/papers/12PRL_109_190502.pdf $\endgroup$
    – Ivan
    Jun 27, 2013 at 18:24
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    $\begingroup$ That is how the Renyi entropy is defined. Without the logarithm (and with a -1), it is called the Tsallis entropy. They are generally easier to calculate than Shannon, but don't satisfy the same properties. I was wondering about Renyi in particular. $\endgroup$
    – Ivan
    Jun 27, 2013 at 18:50

Unfortunately the quantity $I_q(X,Y) = H_q(X) + H_q(Y) - H_q(X,Y)$ is not always positive. This is particularly annoying given that this quantity has the useful property of being $0$ when $X$ and $Y$ are independent.

This is due to the fact that this definition implicitly states that the conditional Renyi entropy is: $$H_q(X|Y) = H_q(X,Y) - H_q(Y),$$ and with this definition it is not always true that conditioning a random variable decreases its Renyi entropy. That is: $$ H_q(X|Y) \leq H_q(X) \text{ is false}$$

From Conditional Renyi Entropies [Teixeira2012], we can find a joint probability distribution $(X,Y)$ for which the inequality above does not hold: $$P(X = x_1, Y = y_1) = 0.2, P(X = x_2, Y = y_1) = 0.4$$ $$P(X = x_1, Y = y_2) = 0, P(X = x_2, Y = y_2) = 0.4$$ In this case, $$I_q(X,Y) = \frac{1}{1-q} \log{\Big(\frac{\int \int p(x)^q p(y)^q}{\int \int p(x,y)^q}\Big)}$$ for $q = 3$ is equal to: $$ \frac{1}{1-3}\log(\frac{(0.2 \cdot 0.4)^3 + (0.2 \cdot 0.6)^3 + (0.6\ \cdot 0.8)^3 + (0.4 \cdot 0.8)^3}{0.2^3 + 0.4^3 + 0.4^3}) = -\frac{1}{2} \log(\frac{0.1456}{0.136}) = - 0.015 $$ which is negative.

  • $\begingroup$ so this is an argument to just use classical mutual information, as opposed to Renyi mutual information, because the first is always positive? $\endgroup$
    – develarist
    Aug 14, 2020 at 17:46
  • 1
    $\begingroup$ I guess it is if you always want positive values. There might be cases where you do not need this though. $\endgroup$
    – Simone
    Aug 14, 2020 at 17:53

The following paper elaborates on a few different possible definitions of "Rényi mutual information". It points out the negativity issue raised on Simone's answer and observes the nonnegativity of the proposal in Avitus' answer.

α-Mutual Information

Sergio Verdú


Verdú comes down in favor of a slight variant of Avitus' definition (attributed to Sibson): $$I_\alpha(X;Y) = \min_{q_Y} D_\alpha(p_{XY} \| p_X q_Y)$$ where $p_X$ is the marginal of $X$ under $p_XY$ and $q_Y$ ranges over all possible distributions for $Y$ (not necessarily the marginal corresponding to $p_{XY}$).

Perhaps the simplest takeaway here is that there is no single clear "Rényi mutual information". Which version you need will vary based on context.


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