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?

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

1 Answer 1


I cross-posted on Math Overflow and quickly obtained a very nice answer. See https://mathoverflow.net/questions/377913/probability-of-a-deviation-when-jensen-s-inequality-is-almost-tight


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