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This question is coming from the fact that all the counter examples for which second order stochastical domination holds but first oder stochastical domination fails do not accept increasing likelihood ratio condition.

From this, a natural question is: If second order stochastical domination together with increasing likelihood ratio implies first order stochastical domination.

$(1)$ Second order stochastical domination: $\rightarrow$ $\int_{t=-\infty}^x (G(t)-F(t))\mbox{d}t\geq 0\quad\forall x.$

$(2)$ Increasing likelihood ratio: $\rightarrow$ $\frac{\mbox{d}F(t)}{\mbox{d}G(t)}=\frac{f(t)}{g(t)}$ increasing function in $t$.

Do $(1)$ and $(2)$ imply $G(x)\geq F(x)\forall x$?

Any ideas? Thanks.

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Increasing likelihood ratio alone implies first-order domination due to the following (very) special case of the FKG inequality:

For any random variable $X$ and any two positive increasing functions $\phi,\psi$ we have $\mathsf{E} \, \phi(X) \psi(X) \ge \mathsf{E} \, \phi(X) \, \mathsf{E} \, \psi(X)$.

Apply that to $X$ distributed according to $G$, and functions $\phi := \mathsf{1} [c,+\infty)$, $\psi := dF/dG$, and you will get exactly $G(c) \ge F(c)$.

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