TITLE: Lower bound for product of expectation?
QUESTION [0 upvotes]: Is it possible to find a positive number $\alpha>0$ such that
$$E(F(X)e^{-X})\ge \alpha E(e^{-X})$$
for any positive-valued random variable $X$ with CDF $F$? ($E(\cdot)$ is the expectation operator, as usual.)
I'm wondering whether we can use the fact that $F(X)$ is uniform to prove it, but I've not managed to do it.
I would like to use this result for my research in operations research.
Thanks!

REPLY [0 votes]: A first thought: because $F$ is increasing and $x \mapsto e^{-x}$ is decreasing, the random variables $F(X)$ and $e^{-X}$ must be negatively correlated. Thus $E[F(X)e^{-X}] \leq E[F(X)]E[e^{-X}]$, and it will be by quite a bit if we make the negative correlation stronger.
The negative correlation is stronger if $F(X)$ is much less than $E[F(X)]$ when $e^{-X}$ is much greater than $E[e^{-X}]$. So we can achieve this by making $X$ be skewed to the right. For instance we can take $X=3\log(b)$ with probability $b$ and $X=0$ with probability $1-b$. Then $E[e^{-X}]=\frac{1}{b^2}+1-b$ and $E[F(X)e^{-X}]=\frac{1}{b}+1-b$. Sending $b \to 0^+$ gives the desired contradiction.
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