Any ideas on how I could prove the veracity or falseness of the following inequality?

Let $X:\Omega \to \mathbb{R}$ a random variable such that the expressions under are well-defined. Then

$$E[e^X] \leq 1 + e^{E[|X|]}.$$

I have the feeling that this is true but i do not know how to show it. I was thinking Jensen's inequality but it goes the wrong way.

One more question, if the above inequality is false, is there a way to upperbound $E[e^X] \leq f(E|X|)$ ?

Thanks you very much for your help.


The inequality is not true: take $X$ with $P\{X=n\}=\frac 1n$ and $P\{X=0\}=1-\frac 1n$. Then the LHS is $\frac{e^n}n+1-\frac 1n$ and the RHS is $1+e$ (or $f(1)$ in the general version).

  • $\begingroup$ But if I apply Jensen's inequality to $t\mapsto e^t$ (since it is convex) I obtain $e^{E[X]} \leq E[e^X]$ and I would like some how, the opposite bound. Nevertheless, it seems that $E[e^X]\leq e^{E[|X|]}$ is true? $\endgroup$ – marc Aug 27 '13 at 15:32
  • $\begingroup$ I've edited, since everything I wrote was misleading. $\endgroup$ – Davide Giraudo Aug 27 '13 at 15:49
  • $\begingroup$ @DavideGiraudo: Just a nit-pick, but since $\mathrm{E}[X]=1$, isn't $1+e^{\mathrm{E}[X]}=1+e$, not $2$? $\endgroup$ – robjohn Aug 27 '13 at 15:56
  • $\begingroup$ Oh thanks! Yes I understood, I also mix Jensen's inequality often. Do you think there is any chance to bound $E[e^X]$ by means of $e^{E[|X|]}$?, some function :) $\endgroup$ – marc Aug 27 '13 at 15:58
  • 1
    $\begingroup$ Sure I see the point.. Thanks a lot :) $\endgroup$ – marc Aug 27 '13 at 16:09

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