This question was previously posted on MSE at [About the monotonicity of the exponential entropy][1].

In the paper [The Unifying Frameworks of Information Measures][2] the conditional exponential entropy (see equation 29) is defined as:

$\mathcal{E}_{\alpha}(X|Y) = E_y\left(\int_{\mathbb{R}} f^{\alpha}(x|y)\,dx\right)^{\frac{1}{1-\alpha}}$

while the exponential entropy (see equation 9) is:

$\mathcal{E}_{\alpha}(X) = \left(\int_{\mathbb{R}} f^{\alpha}(x)\,dx\right)^{\frac{1}{1-\alpha}}$

$f(x)$ is the density of the random variable $X$, which is a non-negative real-valued Borel measurable function on $\mathbb{R}$ (i.e. is absolutely continuous).

I would like to prove the monotonicity of exponential entropy with order $\alpha$ by showing that the following expression holds for the continuous random variables X, Y, Z:
 
$\mathcal{E}_{\alpha}(X|Y Z) \leq \mathcal{E}_{\alpha}(X|Z) $.

**Why this question**

The condition of monotonicity is crucial and fairly obvious: the amount of information required to determine a particular choice cannot decrease as the number of available alternatives increases. The traditional framework for formalizing uncertainty theories can be extended by using a monotonic measure. In fact, entropy generalizations can be achieved within the framework of classical set theory by replacing the additivity requirement of probability measures with the weaker monotonicity.

Exponential entropy is related to Rényi entropy, but the situation is more complex when exponential conditional entropy is considered. The problem is that there is no single definition of conditional Renyi entropy. For example, the following definitions:
\begin{equation}
H_{\alpha}(X|Y)=\sum_{y}P_{Y}(y)H_{\alpha}(X|Y=y)
\end{equation}
\begin{equation}
H_{\alpha}(X|Y)=H_{\alpha}(X,Y)-H_{\alpha}(Y)
\end{equation}
And
\begin{equation}
H_{\alpha}(X|Y)=\frac{1}{1-\alpha} \log(\sum_{y}P_{Y}(y)\sum_{x}P_{X|Y}(x|y)^{\alpha})
\end{equation}
have been proposed in the literature. Regardless of the fact that there is no single definition, the three definitions mentioned above do not even have the same properties. For example, the first two definitions do not respect monotonicity, while the third definition does not respect the chain rule. The definition introduced 
 in the paper [The Unifying Frameworks of Information Measures][2] I think respects both monotonicity and the chain rule, but I would like to prove it rigorously.In conclusion, Kodlu, I fully agree with your comment, but I do not find so simple to adapt previous results I have seen in the literature, probably because I am not a mathematician, just a curious person.

**My attempt to prove the above inequality**

I am not sure how to proceed and would like to get your opinion. This is my attempt. Using Jensen's inequality, for $0 < \alpha < 1$, since $t^{\frac{1}{1-\alpha}}$ is convex for $t > 0$, we have:

$\mathcal{E}_{\alpha}(X|Z) = E_{z} \left(\int_{\mathbb{R}} f^{\alpha}(x|z)\,dx\right)^{\frac{1}{1-\alpha}}$
$\geq \left(E_{yz}\left(\int_{\mathbb{R}} f^{\alpha}(x|y,z)\,dx\right)\right)^{\frac{1}{1-\alpha}} \geq E_{z} \left(E_{y}\left(\int_{\mathbb{R}} f^{\alpha}(x|y,z)\,dx\right)\right)^{\frac{1}{1-\alpha}}
= \mathcal{E}_{\alpha}(X|Y Z)$

For $\alpha > 1$, since $t^{\frac{1}{1-\alpha}}$ is concave and decreasing for $t > 0$, we similarly obtain:

$\mathcal{E}_{\alpha}(X|Z) = E_{z} \left(\int_{\mathbb{R}} f^{\alpha}(x|z)\,dx\right)^{\frac{1}{1-\alpha}}$
$\geq \left(E_{yz}\left(\int_{\mathbb{R}} f^{\alpha}(x|y,z)\,dx\right)\right)^{\frac{1}{1-\alpha}} \geq E_{z} \left(E_{y}\left(\int_{\mathbb{R}} f^{\alpha}(x|y,z)\,dx\right)\right)^{\frac{1}{1-\alpha}}
= \mathcal{E}_{\alpha}(X|Y Z)$

Therefore, $\mathcal{E}_{\alpha}(X|Y Z) \leq \mathcal{E}_{\alpha}(X|Z)$ holds for all $\alpha > 0$ and $\alpha \neq 1$. Is this proof correct?


  [1]: https://math.stackexchange.com/questions/4907023/about-the-monotonicity-of-the-exponential-entropy
  [2]: https://www.hindawi.com/journals/mpe/2018/1791954/