<p>Suppose that $r'(\theta)&gt;0$. The following figure should make clear that saying $r(\theta)\leq w$ is equivalent to saying that $\theta\leq r^{-1}(w)$ (where $r^{-1}(\cdot)$ is the inverse of $r$):</p>

<p><a href="https://i.sstatic.net/RFcN8.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/RFcN8.png" alt="r as a function of theta"></a></p>

<p>So we can rephrase your question as 'why should $E(\theta|\theta&lt;r^{-1}(w))$ be non-linear?'</p>

<hr>

<p>Let's calculate this expectation (assuming that $\theta$ is distributed on support $[\underline{\theta},\overline{\theta}]$ according to CDF $F$):</p>

<p>$$E(\theta|\theta&lt;r^{-1}(w))=\frac{1}{F[r^{-1}(w)]}\int_\underline{\theta}^{r^{-1}(w)}\!\theta F'(\theta)d\theta.$$</p>

<p>For $E(\theta|\theta&lt;r^{-1}(w))$ to be linear in $w$ we would need $\partial E(\theta|\theta&lt;r^{-1}(w))/\partial w$ to be constant.</p>

<p>Differentiating with respect to $w$ we have
$$\frac{\partial E(\theta|\theta&lt;r^{-1}(w))}{\partial w}=\frac{F'[r^{-1}(w)]\frac{\partial r^{-1}(w)}{\partial w}\left[F[r^{-1}(w)]r^{-1}(w)-\int_\underline{\theta}^{r^{-1}(w)}\!\theta F'(\theta)d\theta\right]}{F[r^{-1}(w)]^2}$$</p>

<hr>

<p>The above expression isn't the prettiest thing to look at, but a quick glance should convince you that it will only be constant under very special circumstances. For example, even if we take a highly linear environment like $r(\theta)=\theta$, $F(\theta)=\theta$, we have
$$\frac{\partial E(\theta|\theta&lt;r^{-1}(w))}{\partial w}=\frac{w^2+\underline{\theta}^2}{2w^2},$$
so $E(\theta|\theta&lt;r^{-1}(w))$ is non-linear.</p>

CC BY-SA 3.0
2015-11-28T21:00:40.710Z
2015-11-28T21:15:23.710Z
2015-11-28T21:15:23.710Z
Ubiquitous
<p>I'm referring to the Figure 13.B.1 titled 'A competitive equilibrium with adverse selection' from 'Microeconomics Theory' by Mas Colel et al. It basically graphs the expected value of the workers' productivity ($E[\theta|r(\theta)\leq w])$ against the wage offered $w$ in an attempt to explain the wage equilibrium when the opportunity cost $r(\theta)$ is lower than $\theta$ for all $\theta$.. For the life of it, I cannot understand why this function is a curve (as opposed to straight line) in the graph. What is the distributional assumption for worker's type $\theta$ that is made in order to obtain the curve? </p>

<p>Thanks. </p>

<wages><adverse-selection>
Adverse selection in competitive markets- labor market
ForumWhiner
# Adverse selection in competitive markets- labor market

(Asked by: username_0 on 2015-11-28)

I'm referring to the Figure 13.B.1 titled 'A competitive equilibrium with adverse selection' from 'Microeconomics Theory' by Mas Colel et al. It basically graphs the expected value of the workers' productivity ($E[\theta|r(\theta)\leq w])$ against the wage offered $w$ in an attempt to explain the wage equilibrium when the opportunity cost $r(\theta)$ is lower than $\theta$ for all $\theta$.. For the life of it, I cannot understand why this function is a curve (as opposed to straight line) in the graph. What is the distributional assumption for worker's type $\theta$ that is made in order to obtain the curve?

Thanks.

**Tags:** wages, adverse-selection

---

## 1 answers

### Answer by username_1 on 2015-11-28. Score: 2

Suppose that $r'(\theta)>0$. The following figure should make clear that saying $r(\theta)\leq w$ is equivalent to saying that $\theta\leq r^{-1}(w)$ (where $r^{-1}(\cdot)$ is the inverse of $r$):

[![r as a function of theta](https://i.sstatic.net/RFcN8.png)](https://i.sstatic.net/RFcN8.png)

So we can rephrase your question as 'why should $E(\theta|\theta

---

Let's calculate this expectation (assuming that $\theta$ is distributed on support $[\underline{\theta},\overline{\theta}]$ according to CDF $F$):

$$E(\theta|\theta

For $E(\theta|\theta

Differentiating with respect to $w$ we have
$$\frac{\partial E(\theta|\theta

---

The above expression isn't the prettiest thing to look at, but a quick glance should convince you that it will only be constant under very special circumstances. For example, even if we take a highly linear environment like $r(\theta)=\theta$, $F(\theta)=\theta$, we have
$$\frac{\partial E(\theta|\theta


---

