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Let $X$ be a random variable such that $X \geq 0$ and $\exists \varepsilon > 0: E[e^{X \varepsilon}] < \infty$. Is it true that $\exists$ a real number $B < \infty$ such that $E[X]E[e^{-X}] \leq B$ for all such $X$? If so, please provide a constructive proof that derives a real value for $B$.

The question is motivated by the observation that as $X$ increases point-wise, $e^{-X}$ decreases point-wise. If $X$ is non-random then $Xe^{-X} \leq \frac{1}{e}$. The question asks whether the fact that $Xe^{-X}$ is a bounded function over $X \geq 0$ may be generalized to $X$ being a non-negative random variable with new goal being to bound $E[X]E[e^{-X}]$.

I provide no detail regarding the specific distribution of $X$ because I am asking whether a hard real-valued bound exists for all random variables satisfying the given criteria. Otherwise, $B = E[X]$ suffices for a particular random variable.

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    $\begingroup$ What about it is it that you are having trouble with? Please add your thoughts. $\endgroup$
    – user126540
    Jul 21 '16 at 0:01
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If $X$ is $0$ or $N$ with equal probability, then $\mathbb E[X] = \frac N2$ and $\mathbb E[e^{-X}] = \frac{1 + e^{-N}}{2} > \frac12$, so $\mathbb E[X] \mathbb E[e^{-X}] > \frac N4$ and no upper bound can exist.

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    $\begingroup$ (Not sure why someone decided to reopen this question four years after it was asked, but as long as it's active I might as well answer it??) $\endgroup$ Nov 8 '20 at 4:38
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There is no such $B$, this is just wrong.
For example taking $X_n$ such that :$\mathcal{L}(X_n)= \frac{1}{2} \delta_0 +\frac{1}{2}\delta_n$

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