Let $X>0$ be a random variable. Suppose that we knew that for some $\epsilon \geq 0$, \begin{eqnarray} \log(E[X]) \leq E[\log(X)] + \epsilon \tag{1} \label{eq:primary} \end{eqnarray} The question is: if $\epsilon$ is small, can we find a good bound for \begin{eqnarray*} P\left( \log(X) > E[\log(X)] + \eta \right) \end{eqnarray*} for a given $\eta > 0$. One bound can be obtained this way: \begin{eqnarray*} P\left( \log(X) > E[\log(X)] + \eta \right) &=& P\left( X > \exp(E[\log(X)] + \eta) \right) \\ & \leq & E[X] / \exp(E[\log(X)] + \eta) \\ & = & \exp( \log E[X] - E[\log(X)] - \eta ) \\ & \leq & \exp(\epsilon - \eta) \end{eqnarray*} where the first inequality follows from Markov’s inequality. This seems like a good bound due to the exponential decay with $\eta$, but upon closer examination it appears that it can be significantly improved. If we have $\epsilon = 0$, then this bounds gives \begin{eqnarray} P\left( \log(X) > E[\log(X)] + \eta \right) & \leq & \exp(-\eta) \tag{2} \label{eq:good_but_not_best} \end{eqnarray} However, from Jensen's inequality applied to (\ref{eq:primary}) with $\epsilon = 0$ we obtain $\log(E[X]) = E[\log(X)]$ and therefore $X$ is a constant almost everywhere. As a consequence, for any $\eta>0$, \begin{eqnarray*} P\left( \log(X) > E[\log(X)] + \eta \right) = 0. \end{eqnarray*} which is (of course) infinitely better than (\ref{eq:good_but_not_best}).
It would appear that a better bound should decay to zero as $\epsilon$ decays, and ideally preserve the exponential decay with $\eta$. Any suggestions?
