Application of the Jensen's inequality

Question

For a convex function $\phi(x)$, Jensen's inequality states: \begin{equation} E_{x}(\phi(x)) \geq \phi(E_{x}(X)) \end{equation} where $E_x$ is the expectation. When applying the Jensen's inequality to the following expression \begin{equation} \left(E_{z} \left \{E_{y|z} \left\{\int_{\mathbb{R}} f^{r}(x|y,z)\,dx\right\}\right\}\right)^{\frac{1}{1-r}} \end{equation} assuming $0<r<1$ the function $\frac{1}{1-r} >1$ so \begin{equation} \left(E_{z} \left \{E_{y|z} \left\{\int_{\mathbb{R}} f^{r}(x|y,z)\,dx\right\}\right\}\right)^{\frac{1}{1-r}} \leq E_{z} \left\{ \left(E_{y|z} \left\{\int_{\mathbb{R}} f^{r}(x|y,z)\,dx\right\}\right)^{\frac{1}{1-r}}\right\} \end{equation} I think the result should be just the opposite (i.e. $\leq$ should be replaced by $\geq$), but I don't understand why. Could you please help me? When looking at the expression above we realize that \begin{equation} E_{z} \left\{ \left(E_{y|z} \left\{\int_{\mathbb{R}} f^{r}(x|y,z)\,dx\right\}\right)^{\frac{1}{1-r}}\right\} =\mathcal{E}_{r}(X|Y,Z) \end{equation} So if the relation I wrote is correct \begin{equation} \mathcal{E}_{r}(X|Y,Z) \geq \left(E_{z} \left \{E_{y|z} \left\{\int_{\mathbb{R}} f^{r}(x|y,z)\,dx\right\}\right\}\right)^{\frac{1}{1-r}} = \left(E_{z} \left\{\int_{\mathbb{R}} f^{r}(x|z)\,dx\right\}\right)^{\frac{1}{1-r}} \end{equation} By applying the Jensen's inequality to the rhs we have \begin{equation} \mathcal{E}_{r}(X|Z) = E_{z} \left \{\left(\int_{\mathbb{R}} f^{r}(x|z)\,dx\right)^{\frac{1}{1-r}} \right \} \geq \left(E_{z} \left \{ \int_{\mathbb{R}} f^{r}(x|z)\,dx\right\}\right)^{\frac{1}{1-r}} \end{equation} I expect to get \begin{equation} \mathcal{E}_{r}(X|Y,Z) \leq \mathcal{E}_{r}(X|Z) \end{equation} which is possible only if the sign is reversed.As a matter of fact conditioning on more variables (Y and Z) can only reduce the entropy compared to conditioning on fewer variables (only Z).

Answer 1

I think you are looking at the wrong function. \begin{equation} \mathcal{E}_r(X|Z) = E_Z \left\{ \left( \int_\mathbb{R} f^r(x|z) \, dx \right)^{\frac{1}{1-r}} \right\} = E_Z \left\{ \left( \int_\mathbb{R} \left( E_{Y|Z} \left[ f(x|y,z) \right] \right)^r \, dx \right)^{\frac{1}{1-r}} \right\}. \end{equation} Assuming $0<r<1$ the function $\phi(x) = x^r$ is concave, so \begin{equation} \left( E_{Y|Z} \left[ f(x|y,z) \right] \right)^r \ge E_{Y|Z} \left[ f^r(x|y,z) \right] \end{equation} \begin{equation} \mathcal{E}_r(X|Z) \ge E_Z \left\{ \left( \int_\mathbb{R} E_{Y|Z} \left[ f^r(x|y,z) \right] \, dx \right)^{\frac{1}{1-r}} \right\} = \mathcal{E}_r(X|Y,Z). \end{equation}