# Expected Utility and Jensen's Inequality

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 $$x, so $$[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!

• Note that $$\int_{a}^{b}\int_{x}^{y}u(E(v)-v)g(v)f(c)dvdc =\int_{a}^{b}u(E(v)-v)g(v)dv\int_{x}^{y}f(c)dc =\int_{a}^{b}u(E(v)-v)g(v)dv$$ – Bertrand Dec 3 '19 at 20:31

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:

$$$$E_v \left(u(a - v) \right) \leq E_c \left(u(a-c) |a\leq c \leq b\right)$$$$
for a function $$u$$ increasing and concave and $$[a, b] \subset Supp_v = [x,y]$$.
$$$$\int_x^a u(a-c) dG(c) + \int_b^y u(a-c)dG(c) \leq \int_a^b u(a-c) \cdot \left[\frac{f(c)}{F(b)-F(a)} - g(c) \right]dc.$$$$
First term on the left is certainly positive, second term on the left is certainly negative. The term on the right is of undetermined sign: $$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)$$.