I trying to prove that something is finite. I have given that $\mathbb{E}\left[ e^{X}\right]$ is finite and I am supposed to show that: $\int\limits_{\mathbb{R}}e^zP(X>z)dz$ is finite. I tried applying Markov's inequality which led me to: $\int\limits_{\mathbb{R}}e^zP(X>z)dz=\int\limits_{\mathbb{R}}e^zP(e^X>e^z)dz\leq \int\limits_{\mathbb{R}}\mathbb{E}\left[ e^{X}\right]dz$ but that doesn't work out. Someone has a clue how to do it?
Thank you to everyone that contributes.
Best regards, Chris
This answer is quite heuristic, but with a little work it could be made rigorous. Let $F(x)=P(X\le x)$ and $G(x)=1-F(x)$. $E(e^X)=\int_Re^xdF(x)$. Integrate by parts to get $E(e^X)=e^xF(x)]_R-\int_R e^xF(x)dx=e^xF(x)]_R+\int_Re^x(G(x)-1)dx=e^x(F(x)-1)]_R+\int_Re^xG(x)dx$. Since $F(\infty)=1$ and $e^{-\infty}=0$ we have $E(e^X)=\int_Re^xG(x)dx$.
Obviously using $F(\infty)=1$ may not be enough. We need $\lim_{x\to \infty} e^x(1-F(x))$ finite. .This probably can be derived form $E(e^X)$ being finite