Suppose that $X$ is a $\mathcal{X}$-valued random variable and $Y$ is a $\mathcal{Y}$-valued random variable.
Assume that $X$ and $Y$ are independent of each other.
Suppose that $f: \mathcal{X} \times \mathcal{Y} \to [0,+\infty]$ is a measurable function.
Minkowski integral inequality implies that for any $p\ge 1$ it holds that $$\mathbb{E}\bigg[\Big(\mathbb{E}\big[f(X,Y)^p \mid Y\big]\Big)^{1/p}\bigg] \ge \Bigg(\mathbb{E}\bigg[\Big(\mathbb{E}\big[f(X,Y) \mid X\big]\Big)^{p}\bigg]\Bigg)^{1/p}$$
Here, defining $\varphi:z \mapsto z^p$ and $\psi:z \mapsto z^{1/p}$, we see that $\varphi$ is convex, $\psi$ is concave, $\varphi$ and $\psi$ are inverses of each other, and Minkowski integral inequality tells us that $$\mathbb{E}\Bigg[\psi\bigg(\mathbb{E}\Big[\varphi\big(f(X,Y)\big) \mid Y\Big]\bigg)\Bigg] \ge \psi\Bigg(\mathbb{E}\bigg[\varphi\Big(\mathbb{E}\big[f(X,Y) \mid X\big]\Big)\bigg]\Bigg)$$ I suspect the convexity of $\varphi$, the concavity of $\psi$, and being them inverses of each other are insufficient to guarantee that the previous inequality holds true in general... On the other hand, I'm interested in two specific functions: does the same inequality hold true replacing $\varphi$ and $\psi$ with $\bar{\varphi}:z \mapsto \exp(z)$ and $\bar{\psi}:z\mapsto \log(z)$, respectively? I.e., does it hold true that:
$$\mathbb{E}\Bigg[\log\bigg(\mathbb{E}\Big[\exp\big(f(X,Y)\big) \mid Y\Big]\bigg)\Bigg] \ge\log\Bigg(\mathbb{E}\bigg[\exp\Big(\mathbb{E}\big[f(X,Y) \mid X\big]\Big)\bigg]\Bigg) \;?$$
The proof I know along which Minkowski integral inequality is proven (i.e., relying on Holder inequality) doesn't seem to work (since I don't see any analogous inequality to rely on here).
Is the suggested inequality utterly wrong? Or if it seems plausible, does anyone have an idea on how to attack the problem?
