Economics Asked by carlogambino on October 5, 2021
Consider two random variables (costs and valuations) distributed $vbacksim G(.)$ 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 $[a,b]subsetlbrack x,y]$. Now
consider a strictly concave (twice differentiable and continuous) utility function
$u(.)$, with $u^{prime}(.)>0$, $u^{^{primeprime}}(.)<0$, and $u(0)=0$
(passes through the origin). Establish sufficient conditions such that the
expression
$$int_{a}^{b}u(E(v)-c)f(c)dc-int_{a}^{b}int_{x}^{y}%
u(E(v)-v)g(v)f(c)dvdcgeq0,$$ where $E(v)=int_{x}^{y}vg(v)dv.$
Things I’ve tried:
$int_{0}^{bar{v}}u(E(v)-v)g(v)dvleq0$ 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, $int_{a}^{b}u(E(v)-c)f(c)dcleq0$, 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. $int_{x}%
^{y}u(.)g(v)dv$ 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!
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) |aleq c leq bright) 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 $G(b)-G(a) in (0,1)$.
Answered by Konstantin on October 5, 2021
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