Let $f,g$ be two pdfs, and suppose $X$ is a random variable that has pdf $f$. Is it necessarily true that $E[f(X)] \ge E[g(X)]$?
Although I doubt this will help, but I got this problem from studying the Kullback-Leibler divergence, which is defined as $D(f,g) = E[ln (f(X)/g(X))]$ (with $X$ having pdf $f$ like above). It can be shown using Jensen's inequality that $D(f,g) \ge 0$, which is equivalent to saying that $E[ln(f(X))] \ge E[ln(g(X))]$. But my question removes the $ln$ and I am wondering whether it will still be true.
In fact, I hypothesize that not only is it true for the $ln$ function, but it will be true for any function that is increasing (which includes the identity function, which is my question here). So as a bonus, maybe we could ask if $E[h(f(X))] \ge E[h(g(X))]$ where $h$ is any strictly increasing function on positive reals. It seems like a continuous form of the rearrangement inequality to me.
No. Take $f(x)=(2\pi)^{-1/2}\exp(-x^2/2)$, $x\in\mathbb{R}$, i.e., the standard gaussian (normal) density, and $g(x)=(2/\pi)^{-1/2}\exp(-2x^2)$, $x\in\mathbb{R}$, i.e., the gaussian (normal) density with variance $1/4$. Then, with $X$ denoting a standard gaussian (normal) random variable, \begin{equation*} E[f(X)]=\int_{\mathbb{R}}[f(x)]^2\, dx=\frac{1}{2\sqrt{\pi}}<\sqrt{\frac{2}{5\pi}}= \int_{\mathbb{R}}g(x)f(x)\, dx=E[g(X)],\end{equation*} if I am not mistaken with my calculations.
Incidentally, the middle inequality follows from the inequality $5/2<4$, whence $\sqrt{5/2}<2$, and hence $\sqrt{5\pi}<2\sqrt{2\pi}$.