$\def\anonfunc#1{#1(\cdot)}$Consider two random variables distributed $v\sim \anonfunc G$ and $c \sim \anonfunc F$ with pdfs $\anonfunc g$ and $\anonfunc f$. Let the supports of $c$ and $v$ be $[x,y]$. Let $x<a=E(v)<b<y$, so $[a,b]\subset [x,y]$. Now consider a strictly concave (twice differentiable and continuous) function $\anonfunc u$, with $\anonfunc{u'}>0$, $\anonfunc{u''}<0$, and $u(0)=0$ (passes through the origin). Establish sufficient conditions such that the expression $$ \int_{a}^{b}u(E(v)-c)f(c)dc-\int_{a}^{b}\int_{x}^{y}% u(E(v)-v)g(v)f(c)dvdc\geq0\quad\forall v,c, $$ where $E(v)=\int_{x}^{y}vg(v)dv.$

Things I've tried:

$\int_{0}^{\bar{v}}u(E(v)-v)g(v)dv\leq0$ by Jensen's inequality. To see this, let $E(v)-v=t$. But $E(t)=E_{v}[E(v)-v]=0$, and so $E(u(t))\leq u(E(t))=0$, since $u(0)=0$ by assumption.

Clearly, $\int_{a}^{b}u(E(v)-c)f(c)dc\leq0$, since we are integrating the integrand $(E(v)-c)$ from $a=E(v)$ to $b$.

Intuitively, a variant of Jensen's inequality should apply if $c$ and $v$ are i.i.d. Let $c$ and $v$ be i.i.d. with identical supports. Then the integrands are the same, and we have the expression $\int_{a}^{b} u(E(v)-v)f(c)dc-\int_{a}^{b}\int_{x}^{y}u(E(v)-v)g(v)f(c)dvdc$. However, we can't apply Jensen's inequality directly since $\int_{a}^{b}u(E(v)-v)f(c)dc$ is not $u(E(x))$, even if we "factor out" the outer integrals. $\int_{x}% ^{y}\anonfunc u g(v)dv$ seems to be a form of $E(u(x))$.

`$u^{^{\prime\prime}}$`

for the second derivative, which should be $u''$`$u''$`

or $u^{\prime\prime}$`$u^{\prime\prime}$`

, and an unexpected`\lbrack`

where`[`

would work fine). I have edited to try to fix everything that seemed suspicious, TeX-wise. I also deleted the "Thank you", since the consensus is to omit such pleasantries. $\endgroup$ – LSpice Oct 29 '19 at 14:15