TITLE: Difference between logarithm of an expectation value and expectation value of a logarithm
QUESTION [27 upvotes]: Assuming I have a always positive random variable $X$, $X \in \mathbb{R}$, $X > 0$. Then  I am now interested in the difference between the following two expectation values:

$E \left[ \ln X \right]$
$\ln E \left[ X \right]$

Is one maybe always a lower/upper bound of the other?
Many thanks in advance...

REPLY [10 votes]: To add on Didier's answer, it is instructive to note that the inequality ${\rm E}(\ln X) \le \ln {\rm E}(X)$
can be seen as a consequence of the AM-GM inequality combined with the strong law of large numbers, upon writing 
the AM-GM inequality
$$
\sqrt[n]{{X_1  \cdots X_n }} \le \frac{{X_1  +  \cdots  + X_n }}{n}
$$
as
$$
\exp \bigg(\frac{{\ln X_1  +  \cdots  + \ln X_n }}{n}\bigg) \le \frac{{X_1  +  \cdots  + X_n }}{n},
$$
and letting $n \to \infty$.
EDIT: For completeness, let me note that ${\rm E}[\ln X]$ might be equal to $-\infty$. For example, if $X$ has density function 
$$
f(x) = \frac{{\ln a}}{{x\ln ^2 x}},\;\;0 < x < \frac{1}{a},
$$
where $a>1$ (note that $\int f  = 1$), then 
$$
{\rm E}[\ln X] = 
\int_0^{1/a} {\frac{{\ln a}}{{x\ln x}}} \,{\rm d}x = -\infty.
$$