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?



1 Answer 1


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

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

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

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

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

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

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 $$\frac{\partial E(\theta|\theta<r^{-1}(w))}{\partial w}=\frac{w^2+\underline{\theta}^2}{2w^2},$$ so $E(\theta|\theta<r^{-1}(w))$ is non-linear.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.