Suppose $X$ and $Y$ are positive random variables such that $P(X\geq x)\leq P(Y\geq x)$ and let $\lambda >0$
Then can we say $$E(e^{\lambda X})\leq \int_{0}^{\infty}e^{\lambda x}P(Y\geq x)\,dx $$
My attempt:-
Well, I'll assume everything to be finite to avoid complications.
I can say that $$E(e^{\lambda X})\leq E(e^{\lambda Y})=\int_{0}^{\infty}P(e^{\lambda Y}\geq t)\,dx$$ But I don't see how to get $e^{\lambda x}$ from the above.
Another way I tried is $$P(X\geq x)=P(e^{\lambda X}\geq e^{\lambda x})\leq e^{-\lambda x}E(e^{\lambda X}\mathbf{1}_{X\geq x})$$ but this alters the direction of the inequalities.
Well, I don't know if the bound that I said above is at all true or not.
I simply want to find the best possible bound. So I will get $\int_{0}^{\infty}\lambda e^{\lambda x}P(X\geq x)\,dx=E(e^{\lambda X})-1$. Which would lead me to the bound $1+\int_{0}^{\infty}\lambda e^{\lambda x}P(Y\geq x)\,dx$ . I simply want to know if this bound can be improved. Any help is appreciated.