Education Signaling

Two types with productivity $\theta_H = 2$ , $\theta_L = 1$ , and education cost $c(e, \theta) = e/\theta$ .

When firms believe $e \ge 1$ signals high (wage $2$ ) and $e < 1$ signals low (wage $1$ ), verify with both types' incentive constraints that $e^* = 1$ is the least-cost separating equilibrium.

U_H(e) = w(e) - \frac{e}{2}, \qquad U_L(e) = w(e) - e

w(e) = \begin{cases} 2 & e \ge 1 \\ 1 & e < 1 \end{cases}

e^* = 1

— the least-cost separating equilibrium, where the low type's incentive constraint binds exactly.

Show proof

At $e^* = 1$ the high type's payoff is $U_H(1) = 2 - 1/2 = 1.5$ . Deviating to $e = 0$ earns wage $1$ for a payoff of $1$ , so with $1.5 > 1$ there is no deviation — the high type's participation constraint ( $e^* \le 2$ ) holds.
The low type's payoff is $U_L(1) = 2 - 1 = 1$ .
Choosing $e = 0$ also gives $1 - 0 = 1$ , so there is no gain from mimicking — the low type's incentive constraint ( $e^* \ge 1$ ) holds with equality.
That is why this is the least-cost equilibrium: any lower $e$ would bring the low type along.
The signaling cost borne by the high type is $e^*/\theta_H = 0.5$ — pure waste, burned only to earn the wage $2$ while contributing nothing to productivity.
$\text{signaling cost} = \frac{1}{2} = 0.5, \qquad U_H = 1.5 > 1 = U_H(0), \quad U_L = 1 = U_L(0)$

In the same model, explain through the low type's mimicking incentive why $e = 0.5$ cannot be a separating-equilibrium education level.

U_L(0.5) = 1 - 0.5 = 0.5, \qquad U_L(1) = 2 - 1 = 1

e = 0.5

— the low type would mimic

e = 1

for the wage

2

, so types do not separate.

Show proof

At $e = 0.5$ the wage is $1$ , so the low type's payoff is $U_L(0.5) = 1 - 0.5 = 0.5$ . But choosing $e = 1$ earns the wage $2$ for a payoff of $U_L(1) = 2 - 1 = 1 > 0.5$ .
So the low type strictly gains from mimicking, and $e = 0.5$ cannot separate the types. Separation requires the low type's incentive constraint $2 - e^* \le 1$ , i.e. $e^* \ge \theta_L(\theta_H - \theta_L) = 1$ .

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