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.
 A: Let $Z\sim N(0,1)$, $X=Z^2/3$, and $Y=e^{2Z^2/3}$. Then
$$
\mathsf{E}e^{-X}Y=\mathsf{E}e^{Z^2/3}<\infty, \quad \mathsf{E}e^X<\infty, \quad\text{but}\quad \mathsf{E}Y=\infty.
$$
However, when $X$ and $Y$ are independent,
$$
\mathsf{E}Y\le \frac{c}{\mathsf{E}e^{-X}}\le c\mathsf{E}e^X
$$
by Jensen's inequality.
A: 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 $\sigma(Y)=\sigma(U)=\sigma(X)$.
A: 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
