Integral inequality of transformed integrand with second order stochastic dominance flavor

93 Views Asked by At

Let $f,g : [0,1] \rightarrow [0,1]$ be two functions such that for all $x \in [0,1]$

$\int_0^x f(t) dt \geq \int_0^x g(t) dt$

and

$\int_0^1 f(t) dt = \int_0^1 g(t) dt.$

Can I conclude that

$\int_0^x f(t)(\Lambda (f(t) -1)-1) dt \geq \int_0^x g(t)(\Lambda (g(t) -1)-1) dt$

for all $x\in [0,1]$ where $\Lambda \in [0,1]$ is some parameter? Or is the inequality reversed?

There is a flavor of second order stochastic dominance and it seems that a statement should be possible, but I've been stuck on this one for a while.