2
$\begingroup$

Consider two firms are offering contracts to an employee who has either high or low productivity ($θ_H$ or $θ_L$, with $θ_H > θ_L$). The employee can acquire education $e$ which is costly but does not increase productivity; the cost of education is strictly increasing and convex in e but strictly decreasing in $θ$. The firms can observe $e$ but not $θ$ before deciding which wage $w$ to offer the employee. If the employee does not accept any contract, she gets wage $0$. Can there be a PBE in which both types do not get any education (i.e. $e = 0$ for both)?

My understanding is that we are in pooling equilibrium and the outcome of this scenario where neither type gets an education is the same as if there was no signal. Can there be an equilibrium? There could be if $\theta_H$ doesn't deviate by changing the education level. If we had to draw the figure would it be something similar to this but with exceptions were $\theta_L=E[\theta]$ and just remove the wage schedule since the education level is $e_H=e_L=0$

enter image description here

$\endgroup$
2
  • 1
    $\begingroup$ Looks like you already have the answer. Are you just seeking confirmation? $\endgroup$
    – Herr K.
    Sep 6 '21 at 17:31
  • $\begingroup$ Yes, I wasn't sure if my reasoning was correct as I'm not very familiar with these topics, so I just wanted to check if my answer is suitable or not $\endgroup$ Sep 6 '21 at 19:22
1
$\begingroup$

Can there be a PBE in which both types do not get any education (i.e. $e=0$ for both)?

Yes, workers' strategies $e^*(\theta_L)=e^*(\theta_H)=0$ and a wage function $w^*(e)$ satisfying: (i) $w^*(0)=E[\theta]$ and (ii) for all $e\ge0$, $w^*(e)$ is no higher than $u_H^0$ --- the high-type's indifference curve crossing point $(0,E[\theta])$ --- form such a PBE. For example, one possible such PBE is graphically depicted as follows:

enter image description here

$\endgroup$
1
  • $\begingroup$ Thank you for your answer, this is what I had in mind :) $\endgroup$ Sep 7 '21 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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