let $X_1 \sim f_1(x)$ and $X_2 \sim f_2(x)$.
Suppose we know that $\mu_1=E(X_1)<E(X_2)=\mu_2$ and let $\nu_1=E(\log(X_1))$ and $\nu_2=E(\log(X_2))$.
Since $\log$ is monotonically increasing, my intuition tells me that $\nu_1 < \nu_2$ but I can't find a convincing result to prove it.
Jensen's inequality says that $\nu_1 < \log \mu_1$ and $\nu_2 < \log \mu_2$ but it doesn't seem to help with this question.
Are there any conditions under which $\nu_1 < \nu_2$ holds? Or does it always hold and why?
Help would be appreciated. Thank you.
It's not true. Let $X_1 = 1$ be a constant random variable, and let $\Pr(X_2 = s/2) = \Pr(X_2 = 2s) = 1/2$. Then $E(X_1) = 1$, $E(X_2) = 5s/4$, $E(\log(X_1)) = 0$, and $E(\log(X_2)) = \log(s)$. So this gives a counterexample if $4/5 < s < 1$. I know your distributions are continuous, not discrete, but a simple modification will work.