Does $\operatorname E[e^{-X}Y]\le c$ imply $\operatorname E[Y]\le c\operatorname E[e^X]$? Quick question: Let $X,Y$ be real-valued random variables, $Y$ being nonnegative, such that $\operatorname E[e^{-X}Y]\le c$ for some $c\ge0$. Can we somehow show that $\operatorname E[Y]\le c\operatorname E[e^X]$? It's clearly true when $X$ is nonrandom.
In the situation I've got in mind, $X$ is $\sigma(Y)$ measurable, which might be useful.

This is equivalent to asking for the inequality $E[ZW]\le E[Z]E[W]$ to hold for all $Z\ge0$ and $W> 0$ (set $W=e^X$ and $Z=e^{-X}Y$). In other words, is it true that if $Z,W>0$, then $\operatorname{Cov}(Z,W)\le 0$?
This is not the case, see for instance $U$ uniformly distributed on $(0,1]$, $Z=U^3$ and $W=U^5+1$. We have $$E[ZW]-E[Z]E[W]=\frac19+\frac14-\frac14\cdot\left(\frac16+1\right)=\frac5{72}$$
Also notice that if we recover the original random variables via the relations $Z=e^{-X}Y$ and $W=e^X$, we obtain $Y=U^8+U^3$ and $X=\ln(U^5+1)$, and since the map $t^8+t^3$ is strictly increasing for $t\ge0$, we have that $U=h(Y)$ for some continuous function $h:[0,\infty)\to [0,\infty)$. Therefore $\sigma(Y)=\sigma(U)=\sigma(X)$.
