Adverse Selection: Positive Selection of Worker Types (Mas-Collel)

Question

I'm reviewing some question from Mas-Collel and I am stuck on a chapter 13 question related to adverse selection.

Consider a model of positive selection in which there are workers of two possible productivity types, $\theta_H$ and $\theta_L$, with $\infty > \theta_H > \theta_L > 0$ and $\lambda = Prob(\theta=\theta_H) \in (0,1)$. A worker of type $\theta_i$ can produce $\theta_i$ for a firm in exchange for a wage $w$ or work at home and gain $r(\theta_i)$. r(.) is strictly decreasing and r$(\theta_H) < \theta_H$ and that $r(\theta_L)>\theta_L$. Show that the highest-wage competitive equilibrium need not be a constrained Pareto optimum. [13.B.9]

If the equilibrium is not a constrained Pareto optimal, then it could be a Pareto optimal, this occurs where all types are full employed when $w = \theta$, i.e. the condition that $\Theta^* = \{\theta:r(\theta) \leq w^*\}$.

Answer 1

In PO, you want all types with opportunity cost $r(\theta)\leq \theta$ to trade, because the firm gets more productivity than the worker has to give up on home productivity ("opportunity cost").

By your assumptions $r(\theta_H) < \theta_H$ and $r(\theta_L)>\theta_L$ such that in PO, L shall stay at home and H shall work. However, you need to offer H a wage of at least $w\geq r(\theta_H)$. The highest wage a firm would offer is the expected $\theta$ of a worker who is willing to accept this wage. Suppose that $\mathbb E[\theta]=\lambda \theta_H + (1-\lambda)\theta_L >r(\theta_H)$ and, by definition, $\theta_L<\mathbb E[\theta]<\theta_H$.

Now, it could be that $w=E[\theta] > r(\theta_L)$ such that L would also want to accept this wage, but this is inefficient as $$w > r(\theta_L) > \theta_L.$$ Low types should not work, but because there are so few of them ($\lambda$ is large), the firms would still profit if they also accepted $w$.

An uniformed third party could Pareto improve the allocation by changing the wage to $w' \in (r(\theta_H),r(\theta_L))$ such that L types would reject $w'$.