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.

**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/