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.

Nope. let Y be the value showing when a 6 sided die is rolled. X=-ln(7-Y).
Then $e^{-X}=7-Y$ so $\operatorname E[e^{-X}Y]=E[Y(7-Y)]=(6+10+12+12+10+6)/6=28/3$
So let c=10. 28/3<10.
$E[e^{X}]=E[e^{-ln(7-Y)}]=E[1/(7-Y)]$=(6+10+12+12+10+6)/6=49/120
10*49/120=49/12
E[Y]=3.5<49/12
