My friend introduced me to association inequalities for expectation. Namely:
If $f$ is monotonically increasing and $g$ is monotonically decreasing, then for any random variable $X$ $$E[f(X)g(X)] \leq E[f(X)]E[g(X)]$$
provided the expectations are well defined. As an example, we can show for $k\geq 1$ $$E[X^ke^{-X}] \leq E[X^k]E[e^{-X}]$$
The Cauchy Schwarz inequality would give a looser bound in this case. Now for the above example, I wondered if it was possible to prove the following stronger claim: Does there exist a universal constant $C_k$ such that for $X$ having ANY distribution, $$E[X^ke^{-X}] \leq C_kE[e^{-X}]$$ ?
My understanding as of now is that it does exist but not universally (i.e. not distribution independent). But I couldn't come up with a counter for this? I'd appreciate it if someone could throw some light on this. If necessary we may assume that $X$ is non negative.
The reason I'm interested in such a bound is that there are cases where the kth moment of $X$ is infinity but $E[X^ke^{-X}]<\infty$.
Update: The question has been successfully answered. However I wanted to make a note that $E[X^ke^{-X}] \leq k^ke^{-k}$ for $X\geq 0$.
A trivial counterexample would be a $X$ which is a constant. In this case, $$ \frac{\mathsf{E}[X^ke^{-X}]}{\mathsf{E}[e^{-X}]} = X^k $$ Therefore, there does not exist a universal constant $C_k$.