# Probability of a deviation when Jensen’s inequality is almost tight

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?

• I couldn't follow why $\epsilon=0$ in (1) gives equality in Jensen's inequality. Nov 26, 2020 at 3:45
• Jensen implies that $\log E X \geq E \log X$ which combined with (1) and $\epsilon = 0$ gives the equality. Nov 27, 2020 at 2:29
• I did not notice the opposite sign. Thanks. Nov 27, 2020 at 3:55