<p>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.</p>

<p>Build up: </p>

<p>You are essentially asking that </p>

<p><span class="math-container">\begin{equation}
   E_v \left(u(a - v) \right) \leq E_c \left(u(a-c) |a\leq c \leq b\right)
\end{equation}</span></p>

<p>for a function <span class="math-container">$u$</span> increasing and concave and <span class="math-container">$[a, b] \subset Supp_v = [x,y]$</span>.</p>

<p>In the language of measures, the above condition can be rewritten as</p>

<p><span class="math-container">\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}</span> </p>

<p>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: <span class="math-container">$u(a-c)$</span> is negative in the integration interval, but <span class="math-container">$\frac{f(c)}{F(b)-F(a)}$</span> cannot be uniformly lower than <span class="math-container">$g(c)$</span> on <span class="math-container">$[a,b]$</span>, as the former on this interval must sum up to 1 and the latter to <span class="math-container">$G(b)-G(a) \in (0,1)$</span>.</p>

CC BY-SA 4.0
2019-10-31T20:12:57.503Z
2019-10-31T20:12:57.503Z
Konstantin
<p>Consider two random variables (costs and valuations) distributed <span class="math-container">$v\backsim G(.)$</span> and
<span class="math-container">$c \backsim F(.)$</span> with pdfs <span class="math-container">$g(.)$</span> and <span class="math-container">$f(.)$</span>. Let the supports of <span class="math-container">$c$</span> and
<span class="math-container">$v$</span> be <span class="math-container">$[x,y]$</span>. Let <span class="math-container">$x&lt;a=E(v)&lt;b&lt;y$</span>, so <span class="math-container">$[a,b]\subset\lbrack x,y]$</span>. Now
consider a strictly concave (twice differentiable and continuous) utility function
<span class="math-container">$u(.)$</span>, with <span class="math-container">$u^{\prime}(.)&gt;0$</span>, <span class="math-container">$u^{^{\prime\prime}}(.)&lt;0$</span>, and <span class="math-container">$u(0)=0$</span>
(passes through the origin). Establish sufficient conditions such that the
expression 
<span class="math-container">$$\int_{a}^{b}u(E(v)-c)f(c)dc-\int_{a}^{b}\int_{x}^{y}%
u(E(v)-v)g(v)f(c)dvdc\geq0,$$</span> where <span class="math-container">$E(v)=\int_{x}^{y}vg(v)dv.$</span></p>

<p>Things I've tried:</p>

<ol>
<li><p><span class="math-container">$\int_{0}^{\bar{v}}u(E(v)-v)g(v)dv\leq0$</span> by Jensen's inequality. To see
this, let <span class="math-container">$E(v)-v=t$</span>. But <span class="math-container">$E(t)=E_{v}[E(v)-v]=0$</span>, and so <span class="math-container">$E(u(t))\leq
u(E(t))=0$</span>, since <span class="math-container">$u(0)=0$</span> by assumption.</p></li>
<li><p>Clearly, <span class="math-container">$\int_{a}^{b}u(E(v)-c)f(c)dc\leq0$</span>, since we are integrating the
integrand <span class="math-container">$(E(v)-c)$</span> from <span class="math-container">$a=E(v)$</span> to <span class="math-container">$b$</span>.</p></li>
<li><p>Intuitively, a variant of Jensen's inequality should apply if <span class="math-container">$c$</span> and <span class="math-container">$v$</span>
are i.i.d. Let <span class="math-container">$c$</span> and <span class="math-container">$v$</span> be i.i.d. with identical supports. Then the
integrands are the same, and we have the expression
<span class="math-container">$$\int_{a}^{b}%
u(E(v)-v)f(c)dc-\int_{a}^{b}\int_{x}^{y}u(E(v)-v)g(v)f(c)dvdc.$$</span> 
However, we
can't apply Jensen's inequality directly since <span class="math-container">$\int_{a}^{b}u(E(v)-v)f(c)dc$</span>
is not <span class="math-container">$u(E(x))$</span>, even if we "factor out" the outer integrals. <span class="math-container">$\int_{x}%
^{y}u(.)g(v)dv$</span> seems to be a form of <span class="math-container">$E(u(x))$</span>.</p></li>
</ol>

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

CC BY-SA 4.0
2019-12-03T20:31:35.333Z
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
<game-theory><expected-utility><probability><contract-theory>
Expected Utility and Jensen's Inequality
carlogambino
# Expected Utility and Jensen's Inequality

(Asked by: username_0 on 2019-10-28)

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!

**Tags:** game-theory, expected-utility, probability, contract-theory

> **Comments:**
> **username_1**: Note that  $$\int_{a}^{b}\int_{x}^{y}u(E(v)-v)g(v)f(c)dvdc =\int_{a}^{b}u(E(v)-v)g(v)dv\int_{x}^{y}f(c)dc =\int_{a}^{b}u(E(v)-v)g(v)dv$$  

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## 1 answers

### Answer by username_2 on 2019-10-31. Score: 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 $G(b)-G(a) \in (0,1)$.


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