lnE(X) and E(lnX)
Since \(lnx\) is the concave function, by Jensen's inequality, we know that \[ ln[E(X)]>E[lnX] \] Another proof can be displayed as follows:
From Simplest or nicest proof that \(1+x \le e^x\), we can prove that
\[ e^x\geq 1+x \]
Expectation of \(e^x\) is \[ \begin{gather*} E(e^Y)=e^{E(Y)}E(e^{Y-E(Y)})\\ \geq e^{EY}E(1+Y-EY)\\ =e^{EY} \end{gather*} \] Therefore, we have \[ e^{EY}\leq E[e^Y] \] Denote \(Y=lnX\), we have \[ e^{ElnX}\leq E(e^{lnX})=EX\\ \Rightarrow E(lnX)\geq ln(EX) \] The equality holds if and only if \(X\) is almost surely constant.