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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.

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.

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 likelhood 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.

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.