This problem is about maximum likelihood estimation. Let $f\left(x;\theta\right)$ be a parametric density function with unknown parameter $\theta$, and denote $\theta_{0}$ the true value. By Jensen's inequality, we have $E_{0}\left\{ \log f\left(x;\theta\right)-\log f\left(x;\theta_{0}\right)\right\} \leq0$, where $E_{0}$ indicates that the expectation is under the true density function $f\left(x;\theta_{0}\right)$. In general, we do not have $E_0\left\{ f\left(x;\theta\right)-f\left(x;\theta_{0}\right)\right\} \leq0$ (c.f. the counter example below). This looks counter-intuitive for me. I thought the true density $f(x;\theta_0)$ should be the maximum in the sense that $E_0\left\{ f\left(x;\theta\right)-f\left(x;\theta_{0}\right)\right\} \leq0$. How to understand this? Why is $\log$ transformation indispensable?
Counter-example of $E_0\left\{ f\left(x;\theta\right)-f\left(x;\theta_{0}\right)\right\} \leq0$:
Consider a binary distribution. Let $\pi_{1}$ and $\pi_{2}=1-\pi_{1}$ be true distribution of $X\in\left\{ 0,1\right\} $, and denote $p_{1}$ and $p_{2}=1-p_{1}$ a generic distritbuion of $X$. Here $\pi_1=\Pr(X=0)$ and $p_1=\Pr(X=0)$. The expectation $E_{0}\left\{ f\left(x;\theta\right)-f\left(x;\theta_{0}\right)\right\} $ herein is \begin{eqnarray*} \pi_{1}\left(p_{1}-\pi_{1}\right)+\pi_{2}\left(p_{2}-\pi_{2}\right) & = & \pi_{1}\left(p_{1}-\pi_{1}\right)+\left(1-\pi_{1}\right)\left(\pi_{1}-p_{1}\right)\\ & = & \left(2\pi_{1}-1\right)\left(p_{1}-\pi_{1}\right). \end{eqnarray*} In order to force $E_{0}\left\{ f\left(x;\theta\right)-f\left(x;\theta_{0}\right)\right\} >0$, we simply need $\pi_{1}>1/2$ and $\pi_{1}<p_{1}\leq1$.