Let $b>0$ and $X$ be a (strictly) positive random variable such that $$ E(e^{bX})<\infty. $$ If $f(x)\ge0$ is a continuous decreasing function with $\lim_{x\to\infty}f(x)=b$, can I derive that $$ \limsup_{x\to\infty}E(e^{f(x)X})\le E(\limsup_{x\to\infty} e^{f(x)X})=E(e^{bX})? $$ Since $f$ is decreasing, I couldn't apply monotone convergence theorem. On the other hand, I couldn't find an integrable upper bound for $e^{f(x)X}$, so the dominated convergence theorem couldn't be used.
Any help, hints or references would be very much appreciated. Thank you.
There exist positive random variables $Y$ such that $EY<\infty$ but $EY^{a}=\infty$ for all $a >1$. Let $b=1$ and $X=\ln Y$. If $f$ is strictly decreasing then $f(x) >1$ for every $x$. Hence $Ee^{f(x)X}=EY^{f(x)}=\infty$ for every $x$. So RHS of your inequality is $\infty$ and RHS is finite.