Expected Utility and Jensen's Inequality

Question

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

Things I've tried:

1. $\int_{0}^{\bar{v}}u(E(v)-v)g(v)dv\leq0$ 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.

2. 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 $b$.

3. 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. $\int_{x}% ^{y}u(.)g(v)dv$ seems to be a form of $E(u(x))$.

At a loss as to what to do here. Any help would be greatly appreciated. Thank you!

Answer 1

Didn't manage to get to a definitve answer in one shot, but it seems to me that Jensen inequality is not going to help much.

Build up:

You are essentially asking that

\begin{equation} E_v \left(u(a - v) \right) \leq E_c \left(u(a-c) |a\leq c \leq b\right) \end{equation}

for a function $u$ increasing and concave and $[a, b] \subset Supp_v = [x,y]$.

In the language of measures, the above condition can be rewritten as

\begin{equation} \int_x^a u(a-c) dG(c) + \int_b^y u(a-c)dG(c) \leq \int_a^b u(a-c) \cdot \left[\frac{f(c)}{F(b)-F(a)} - g(c) \right]dc. \end{equation}

First term on the left is certainly positive, second term on the left is certainly negative. The term on the right is of undetermined sign: $u(a-c)$ is negative in the integration interval, but $\frac{f(c)}{F(b)-F(a)}$ cannot be uniformly lower than $g(c)$ on $[a,b]$, as the former on this interval must sum up to 1 and the latter to $G(b)-G(a) \in (0,1)$.