Educational supply policies: distortions and labor market performance

“Now, as a nation, we don’t promise equal outcomes, but we were founded on the idea everybody should have an equal opportunity to succeed. No matter who you are, what you look like, where you come from, you can make it. Where you start should not determine where you end up.”

Barack Obama.

Abstract

Is it always worth implementing an open enrollment policy? And implementing policies that pursue equity in school supply? What is the impact of these two policies on the labor market? Do they produce efficient aggregate outcomes in both the labor and educational markets? This paper theoretically provides answers to these questions by studying the link between distortionary school supply policies and labor market performance. We build a two-sector labor market matching model, where the skilled segment of the economy is composed of workers who differ in the quality of the school they attended. We show the impact of government interventions to eliminate educational supply policy distortions within this theory. We demonstrate that both open enrollment and school equity policies have ambiguous effects on the labor market. Whenever their impact on the measure of workers choosing to become better educated is stronger than the additional school quality gains generated by the policy, the effects on the economy are negative. We also study the central planner solution, emphasizing the existing inefficiencies.

Appendices

Appendix 1

Proof

For each $\epsilon$, let ${\tilde{I}}(\epsilon )$ be the solution of $Q_o({\tilde{I}}(\epsilon )) = Q_d({\tilde{I}}(\epsilon ))$. Now, consider an increase from $\epsilon$ to $\epsilon$’ and let:

$$\begin{aligned} {Q'_o(I)} = q_H - (q_H - q_L) \left( \frac{I - I_L}{I_H - I_L}\right) ^{\epsilon '}. \end{aligned}$$

It follows that $Q'_o(I) > Q_o(I)$, for all $I \in (I_L,I_H)$. Since ${\tilde{I}} \in (I_L, I_H)$ for any $\epsilon \ge 1$ we have that:

$$\begin{aligned} Q'_o({\tilde{I}}(\epsilon ))>Q_d({\tilde{I}}(\epsilon )). \end{aligned}$$

However, from the definition of ${\tilde{I}}(\epsilon ')$, we have that:

$$\begin{aligned} Q'_o({\tilde{I}}(\epsilon '))=Q_d({\tilde{I}}(\epsilon ')). \end{aligned}$$

Adding the last two expressions and multiplying by $\frac{1}{2}$ yields:

$$\begin{aligned} \frac{1}{2}Q'_o({\tilde{I}}(\epsilon )) + \frac{1}{2}Q'_o({\tilde{I}}(\epsilon '))>\frac{1}{2}Q_d({\tilde{I}}(\epsilon )) + \frac{1}{2}Q_d({\tilde{I}}(\epsilon ')). \end{aligned}$$

By the concavity of $Q'_o$ and the strict convexity of $Q_d$, it follows that:

$$\begin{aligned} Q'_o\left( \frac{{\tilde{I}}(\epsilon ) + {\tilde{I}}(\epsilon ')}{2}\right) >Q_d\left( \frac{{\tilde{I}}(\epsilon ) + {\tilde{I}}(\epsilon ')}{2}\right) . \end{aligned}$$

From these two last expressions, we have that:

$$\begin{aligned} \frac{{\tilde{I}}(\epsilon ) + {\tilde{I}}(\epsilon ')}{2}<{\tilde{I}}(\epsilon ') \end{aligned}$$

then, ${\tilde{I}}(\epsilon )<{\tilde{I}}(\epsilon ')$.

Now, we show that $Q_d$ is decreasing in ${\tilde{I}}$. Consider that:

$$\begin{aligned} Q_d=M\nu _d, \end{aligned}$$

for:

$$\begin{aligned} M = \frac{\frac{1}{\rho }\left[ \left( b_S+\frac{\beta _S}{1-\beta _S} k_S \theta _S\right) -\left( b_N+\frac{\beta _N}{1-\beta _N} k_N \theta _N\right) \right] }{\left( 1- e^{- \rho T}\right) }; \end{aligned}$$

and

$$\begin{aligned} \nu _d = {\mathbb {E}}\left[ \frac{1}{I} \bigm \vert I \le {\tilde{I}} \right] . \end{aligned}$$

Notice that changes in $Q_d$, due to changes in ${\tilde{I}}$, occur through the term $\nu _d$. We may rewrite $\nu _d$ as:

$$\begin{aligned} \nu _d = {\mathbb {E}}\left[ x | x \ge {\tilde{x}} \right] , \end{aligned}$$

where $x = \frac{1}{I}$ and ${\tilde{x}} = \frac{1}{{\tilde{I}}}$.

Then, increases in ${\tilde{I}}$ are equivalent to reductions of ${\tilde{x}}$ and $\nu _d$. This guarantees that the higher $\epsilon$ is, the lower $Q_d$ will be.

Now we verify the impact of $\epsilon$ on $Q_o$. Consider that:

$$\begin{aligned} Q_o = q_H -(q_H -q_L) \nu _o \end{aligned}$$

for

$$\begin{aligned} \nu _o = {\mathbb {E}} \left[ \left( \frac{I - I_L}{I_H - I_L}\right) ^{\epsilon } | I \le {\tilde{I}}\right] . \end{aligned}$$

As previously, consider that:

$$\begin{aligned} x = \frac{I - I_L}{I_H - I_L} \in [0,1] \end{aligned}$$

and

$$\begin{aligned} {\tilde{x}} = \frac{{\tilde{I}} - I_L}{I_H - I_L} \in [0,1] \end{aligned}$$

for all $I \in [I_L,I_H]$. Then,

$$\begin{aligned} \nu _o = \int _{0}^{{\tilde{x}}} x^\epsilon \frac{d{\tilde{H}}(x)}{{\tilde{H}}({\tilde{x}})}, \end{aligned}$$

where ${\tilde{H}}(x)$ is the distribution of transformation $x = (I-I_L)/(I_H-I_L)$.

Differentiating this last expression with respect to $\epsilon$, we get

$$\begin{aligned} \frac{d \nu _o}{d \epsilon } = \Big [( {\tilde{x}}^{\epsilon } - \nu _o) {\tilde{h}}({\tilde{x}}) + \int _{0}^{{\tilde{x}}} x^\epsilon ln(x) \frac{d{\tilde{H}}(x)}{{\tilde{H}}({\tilde{x}})}\Big ] \frac{1}{{\tilde{H}}({\tilde{x}})} \frac{d {\tilde{x}}}{d \epsilon } < 0. \end{aligned}$$

The sign of the previous expression is derived from: ${\tilde{x}}^{\epsilon } - \nu _o < 0$, $0< x < 1$ and $d {\tilde{x}}/d \epsilon \propto d {\tilde{I}}/d \epsilon$.

$\square$

Appendix 2

Proof

To simplify the proof, consider that $q^{e}(Q_{d},Q_{o})=q^{e}$. We have that:

$$\begin{aligned} \frac{dq^{e}}{d\varepsilon }=\frac{\partial q^{e}}{\partial Q_{d}}\frac{ dQ_{d}}{d\varepsilon }+\frac{\partial q^{e}}{\partial Q_{o}}\frac{dQ_{o}}{ d\varepsilon }. \end{aligned}$$

By using the fact that $\frac{\partial q^e(Q_d, Q_o)}{\partial Q_d} > 0$, $\frac{\partial q^e(Q_d, Q_o)}{\partial Q_o} > 0$ and Proposition 1, we have that all previous derivatives are positive except $dQ_{d}/d\varepsilon$.

We now rewrite the previous expression as:

$$\begin{aligned} \frac{dq^{e}}{d\varepsilon }=-\frac{[q^{e}-Q_{d}]g(Q_{d})}{G(Q_{o})-G(Q_{d})} \left| \frac{dQ_{d}}{d\varepsilon }\right| +\frac{ [Q_{o}-q^{e}]g(Q_{o})}{G(Q_{o})-G(Q_{d})}\frac{dQ_{o}}{d\varepsilon }. \end{aligned}$$

Therefore, we have that $dq^{e}/d\varepsilon >0$ if and only if:

$$\begin{aligned} \frac{[Q_{o}-q^{e}]g(Q_{o})}{[q^{e}-Q_{d}]g(Q_{d})}>\frac{\left| \frac{dQ_{d}}{d\varepsilon }\right| }{\frac{dQ_{o}}{d\varepsilon }}. \end{aligned}$$

To complete the demonstration we need the following result.

Lemma

For any values of $Q_{d}$ and $Q_{o}$ in $[q_{L},q_{H}]$ and $\varepsilon \in [1,\infty )$, we have:

$$\begin{aligned} \left| \frac{dQ_{d}}{d\varepsilon }\right| <\frac{dQ_{o}}{ d\varepsilon }. \end{aligned}$$

Proof

By definition we have:

$$\begin{aligned} Q_{d}=\frac{U_{S}-U_{N}}{(1-e^{-\rho T})}\int \limits _{I_{L}}^{I(\varepsilon )}\frac{1}{s}\frac{h(s)}{H(I(\varepsilon ))}ds \end{aligned}$$

$\square$

where $I(\varepsilon )$ is such that $Q_{d}(I(\varepsilon ))=Q_{o}(I(\varepsilon ))$. Differentiating the previous expression with respect to $\varepsilon$ yields:

$$\begin{aligned} \frac{dQ_{d}}{d\varepsilon }= & \, \frac{U_{S}-U_{N}}{(1-e^{-\rho T})}\left[ \frac{1}{I(\varepsilon )}\frac{h(I(\varepsilon ))}{H(I(\varepsilon ))}\frac{ dI}{d\varepsilon }-\int \limits _{I_{L}}^{I(\varepsilon )}\frac{1}{s}\frac{h(s) }{H(I(\varepsilon ))}\frac{h(I(\varepsilon ))}{H(I(\varepsilon ))}\frac{dI}{ d\varepsilon }ds\right] \\= & \, \frac{U_{S}-U_{N}}{(1-e^{-\rho T})}\left[ \frac{1}{I(\varepsilon )} -\int \limits _{I_{L}}^{I(\varepsilon )}\frac{1}{s}\frac{h(s)}{H(I(\varepsilon ))}ds\right] \frac{h(I(\varepsilon ))}{H(I(\varepsilon ))}\frac{dI}{ d\varepsilon } \\= & \, \left[ Q_{d}(I(\varepsilon ))-Q_{d}\right] \frac{h(I(\varepsilon ))}{ H(I(\varepsilon ))}\frac{dI}{d\varepsilon }. \end{aligned}$$

It is easy to shown that for any $\varepsilon \in [1,\infty )$, $Q_{d}(I(\varepsilon ))<Q_{d}$. In addition, from the proof of Proposition 1 in “Appendix 1”, it is shown that $dI/d\varepsilon >0$. Thus we have:

$$\begin{aligned} \left| \frac{dQ_{d}}{d\varepsilon }\right| =\left[ Q_{d}-Q_{d}(I(\varepsilon ))\right] \frac{h(I(\varepsilon ))}{ H(I(\varepsilon ))}\frac{dI}{d\varepsilon }. \end{aligned}$$

(27)

By using the definition of $Q_{o}$, we have:

$$\begin{aligned} Q_{o}=q_{H}-(q_{H}-q_{L})\int \limits _{I_{L}}^{I(\varepsilon )}\left( \frac{ s-I_{L}}{I_{H}-I_{L}}\right) ^{\varepsilon }\frac{h(s)}{H(I(\varepsilon ))}ds. \end{aligned}$$

Differentiating this previous expression with respect to $\varepsilon$ gives us:

$$\begin{aligned} \frac{dQ_{o}}{d\varepsilon }= & \, -(q_{H}-q_{L})\left\{ \left( \frac{ I(\varepsilon )-I_{L}}{I_{H}-I_{L}}\right) ^{\varepsilon }\frac{ h(I(\varepsilon ))}{H(I(\varepsilon ))}\frac{dI}{d\varepsilon }\right. \nonumber \\&+\int \limits _{I_{L}}^{I(\varepsilon )}\left( \frac{s-I_{L}}{I_{H}-I_{L}} \right) ^{\varepsilon }\ln \left( \frac{s-I_{L}}{I_{H}-I_{L}}\right) \frac{ h(s)}{H(I(\varepsilon ))}ds \nonumber \\&\left. -\int \limits _{I_{L}}^{I(\varepsilon )}\left( \frac{s-I_{L}}{ I_{H}-I_{L}}\right) ^{\varepsilon }\frac{h(s)}{H(I(\varepsilon ))}\frac{ h(I(\varepsilon ))}{H(I(\varepsilon ))}ds\right\} . \end{aligned}$$

(28)

Manipulating the previous expression:

$$\begin{aligned} \frac{dQ_{o}}{d\varepsilon }= & \, \left[ -(q_{H}-q_{L})\left( \frac{ I(\varepsilon )-I_{L}}{I_{H}-I_{L}}\right) ^{\varepsilon }\right. \nonumber \\&+\left. (q_{H}-q_{L})\int \limits _{I_{L}}^{I(\varepsilon )}\left( \frac{ s-I_{L}}{I_{H}-I_{L}}\right) ^{\varepsilon }\frac{h(s)}{H(I(\varepsilon ))}ds \right] \frac{h(I(\varepsilon ))}{H(I(\varepsilon ))}\frac{dI}{d\varepsilon } \nonumber \\&-(q_{H}-q_{L})\int \limits _{I_{L}}^{I(\varepsilon )}\left( \frac{s-I_{L}}{ I_{H}-I_{L}}\right) ^{\varepsilon }\ln \left( \frac{s-I_{L}}{I_{H}-I_{L}} \right) \frac{h(s)}{H(I(\varepsilon ))}ds. \end{aligned}$$

(29)

By Adding and subtracting $q_H$ in the term between brackets, we have:

$$\begin{aligned} \frac{dQ_{o}}{d\varepsilon }= & \, \left[ Q_{o}(I(\varepsilon ))-Q_{o}\right] \frac{h(I(\varepsilon ))}{H(I(\varepsilon ))}\frac{dI}{d\varepsilon } \\&-(q_{H}-q_{L})\int \limits _{I_{L}}^{I(\varepsilon )}\left( \frac{s-I_{L}}{ I_{H}-I_{L}}\right) ^{\varepsilon }\ln \left( \frac{s-I_{L}}{I_{H}-I_{L}} \right) \frac{h(s)}{H(I(\varepsilon ))}ds. \end{aligned}$$

Note that the above expression is positive, since $Q_{o}(I(\varepsilon ))>Q_{o}$ for any $\varepsilon \in [1,\infty )$ and $\ln [(s-I_{L})/(I_{H}-I_{L})]<0$ for any $s<I_{H}$. Finally, using (27) and (29), we have:

$$\begin{aligned} \left| \frac{dQ_{d}}{d\varepsilon }\right| -\frac{dQ_{o}}{ d\varepsilon }= & \, \left[ Q_{d}-Q_{d}(I(\varepsilon ))\right] \frac{ h(I(\varepsilon ))}{H(I(\varepsilon ))}\frac{dI}{d\varepsilon } \\&-\left[ Q_{o}(I(\varepsilon ))-Q_{o}\right] \frac{h(I(\varepsilon ))}{ H(I(\varepsilon ))}\frac{dI}{d\varepsilon } \\&+(q_{H}-q_{L})\int \limits _{I_{L}}^{I(\varepsilon )}\left( \frac{s-I_{L}}{ I_{H}-I_{L}}\right) ^{\varepsilon }\ln \left( \frac{s-I_{L}}{I_{H}-I_{L}} \right) \frac{h(s)}{H(I(\varepsilon ))}ds \\= & \, \left[ \left( Q_{o}+Q_{d}\right) -\left( Q_{o}(I(\varepsilon ))+Q_{d}(I(\varepsilon )\right) )\right] \frac{h(I(\varepsilon ))}{ H(I(\varepsilon ))}\frac{dI}{d\varepsilon } \\&+(q_{H}-q_{L})\int \limits _{I_{L}}^{I(\varepsilon )}\left( \frac{s-I_{L}}{ I_{H}-I_{L}}\right) ^{\varepsilon }\ln \left( \frac{s-I_{L}}{I_{H}-I_{L}} \right) \frac{h(s)}{H(I(\varepsilon ))}ds. \end{aligned}$$

By definition $Q_{o}(I(\varepsilon ))=Q_{d}(I(\varepsilon ))$ such that $Q_{o}(I(\varepsilon ))+Q_{d}(I(\varepsilon ))=2Q_{o}(I(\varepsilon ))$. Thus, we have $Q_{o}(I(\varepsilon ))+Q_{d}(I(\varepsilon ))=2Q_{o}(I(\varepsilon ))>2Q_{o}>Q_{o}+Q_{d}$. Therefore:

$$\begin{aligned} \left| \frac{dQ_{d}}{d\varepsilon }\right| <\frac{dQ_{o}}{ d\varepsilon }. \end{aligned}$$

$\square$

Using the previous result in the proposition, we have that:

$$\begin{aligned} \frac{[Q_{o}-q^{e}]g(Q_{o})}{[q^{e}-Q_{d}]g(Q_{d})}>1>\frac{ \left| \frac{dQ_{d}}{d\varepsilon }\right| }{\frac{dQ_{o}}{ d\varepsilon }} \end{aligned}$$

or equivalently:

$$\begin{aligned} \frac{d}{d\varepsilon }q^{e}(Q_{d},Q_{o})>0 . \end{aligned}$$

$\square$

Appendix 3

Proof

The expressions that characterize $l_{S}$ and $l_{N}$ are respectively given by:

$$\begin{aligned} l_{S}= & \, \chi _{S}q^{e}(Q_{D},Q_{O})^{-\frac{1}{\alpha _{S}}} \\ l_{N}= & \, \chi _{N}q_{L}^{-\frac{1}{\alpha _{N}}} \end{aligned}$$

where

$$\begin{aligned} \chi _{j}=\left\{ \frac{\left[ \alpha _{j}\beta _{j}+(1-\beta _{j})\right] (1+\rho )C_{j}}{(1-\alpha _{j})(1-\beta _{j})}\right\} ^{\frac{1}{\alpha _{j} }}{\text {, for}}j=N,S. \end{aligned}$$

Therefore we have that

$$\begin{aligned} \frac{\partial l_{S}}{\partial \varepsilon }= & \, -\frac{1}{\alpha _{S}}\chi _{S}q^{e}(Q_{D},Q_{O})^{-\frac{1}{\alpha _{S}}-1}\frac{d}{d\varepsilon } q^{e}(Q_{D},Q_{O})<0; \end{aligned}$$

(30)

$$\begin{aligned} \frac{\partial l_{N}}{\partial \varepsilon }= & \, 0. \end{aligned}$$

(31)

By definition we have:

$$\begin{aligned} u_{S}= & \, F({\tilde{I}})-l_{S}; \end{aligned}$$

(32)

$$\begin{aligned} u_{N}= & \, 1-F({\tilde{I}})-l_{N}. \end{aligned}$$

(33)

Using (30) and (31) in (32) and (33) we concludes that:⁴¹

$$\begin{aligned} \frac{\partial u_{S}}{\partial \varepsilon }= & \, f({\tilde{I}})\frac{d{\tilde{I}} }{d\varepsilon }-\frac{\partial l_{S}}{\partial \varepsilon }>0; \\ \frac{\partial u_{N}}{\partial \varepsilon }= & \, -f({\tilde{I}})\frac{d{\tilde{I}} }{d\varepsilon }<0. \end{aligned}$$

This shows the first part of the corollary.

From Eqs. (20) and (21) we have:

$$\begin{aligned} \frac{\partial w_{N}}{\partial \varepsilon }= & \, 0;\\ \frac{\partial w_{S}}{\partial \varepsilon }= & \, \frac{\beta _{S}\alpha _{S}}{ \beta _{S}\alpha _{S}+(1-\beta _{S})}\left[ l_{S}^{\alpha _{S}-1}\frac{dq^{e} }{d\varepsilon }+(\alpha _{S}-1)l_{S}^{\alpha _{S}-2}q^{e}\frac{\partial l_{S}}{\partial \varepsilon }\right] >0. \end{aligned}$$

Notice that the previous results depend on the assumption of decreasing returns to scale. $\square$

Appendix 4

Proof

Initially, it should be mentioned that:

$$\begin{aligned} Q_{o}\ge q^{e}(Q_{d},Q_{o}) \end{aligned}$$

Thus, it suffices to show that ${\bar{q}}>Q_{o}$. By definition, we have that:

$$\begin{aligned} Q_{o}=q_{H}-(q_{H}-q_{L})\int \limits _{I_{L}}^{{\tilde{I}}}\left( \frac{I-I_{L} }{I_{H}-I_{L}}\right) ^{\varepsilon }\frac{h(I)}{H({\tilde{I}})}dI. \end{aligned}$$

As $h(I)\ge 0$ for any $I\in [I_{L},{\tilde{I}}]$, the mean value theorem for integrals guarantees that there exists $\delta \in [I_{L},{\tilde{I}}]$ such that:

$$\begin{aligned} \int \limits _{I_{L}}^{{\tilde{I}}}\left( \frac{I-I_{L}}{I_{H}-I_{L}}\right) ^{\varepsilon }\frac{h(I)}{H({\tilde{I}})}dI=\left( \frac{\delta -I_{L}}{ I_{H}-I_{L}}\right) ^{\varepsilon }\int \limits _{I_{L}}^{{\tilde{I}}}\frac{h(I) }{H({\tilde{I}})}dI=\left( \frac{\delta -I_{L}}{I_{H}-I_{L}}\right) ^{\varepsilon }. \end{aligned}$$

Therefore, if ${\bar{I}}<\delta$ then:

$$\begin{aligned} \left( \frac{{\bar{I}}-I_{L}}{I_{H}-I_{L}}\right) ^{\varepsilon }<\left( \frac{ \delta -I_{L}}{I_{H}-I_{L}}\right) ^{\varepsilon }=\int \limits _{I_{L}}^{ {\tilde{I}}}\left( \frac{I-I_{L}}{I_{H}-I_{L}}\right) ^{\varepsilon }\frac{h(I) }{H({\tilde{I}})}dI \end{aligned}$$

which implies the desired result. $\square$

Appendix 5

Proof

The condition ${\bar{q}}>q^{e}(Q_{d},Q_{o})$ implies that ${\bar{l}}_{S}<l_{S}$. On other hand we get $F({\bar{I}})<F({\tilde{I}})$. Therefore by usin the definition of unemployment rate we see that

$$\begin{aligned} {\bar{u}}_{S}-u_{S}=\left[ l_{S}-{\bar{l}}_{S}\right] -\left[ F({\tilde{I}})-F( {\bar{I}})\right] \end{aligned}$$

(34)

The components on the right side of (1) are both positives. The first one represents the loss of employment due to productivity gains and the other is the reduction of the mass of students required for reach that gains. The final effect on the unemployment rate depends on which the effect is dominant. Finally in the unskilled sector there is no effect on the employment but, due to the decreasing of the mass of students the unemployment rate rises. These arguments shows item a).

From the Eqs. (20) and (21), we have:

$$\begin{aligned} {\bar{w}}_{S}-w_{S}=\frac{\beta _{S}\alpha _{S}}{\beta _{S}\alpha _{S}+(1-\beta _{S})}\left[ {\bar{q}}{\bar{l}}_{S}^{\alpha _{S}-1}-q^{e}(Q_{d},Q_{o})l_{S}^{\alpha _{S}-1}\right] \end{aligned}$$

(35)

Using the facts shown in the item a) and decreasing returns to scale we have ${\bar{w}}_{S}-w_{S}>0$. The wage in the unskilled segment do not depends of the changes in the productivity such that ${\bar{w}}_{N}=w_{N}$. That finish the proof.

$\square$

Appendix 6

Proof

We initially solve the centralized problem. Consider again the problem presented in (26). Manipulating the previous constraints, the central planner problem can be redefined as:

$$\begin{aligned}&\max _{Q, \theta _S, \theta _N} {\mathcal {W}}(Q, v_S, v_N, l_S, l_N)\\&\quad = G(Q) \left\{ q_L [\frac{z(\theta _N)G(Q)}{\lambda _N + z(\theta _N)}]^{\alpha _N}\frac{1}{G(Q)} +\frac{\lambda _N b_N - k_N \lambda _N \theta _N}{\lambda _N + z(\theta _N)} - \frac{C_N}{G(Q)} \right\} \\&\qquad +[1-G(Q)] \left\{ q^e(Q) [\frac{z(\theta _S)[1-G(Q)]}{\lambda _S + z(\theta _S)}]^{\alpha _S}\frac{1}{[1-G(Q)]}\nonumber \right. \\&\qquad \left. +\frac{\lambda _S b_S - k_S \lambda _S \theta _S}{\lambda _S + z(\theta _S)} - \frac{C_S}{[1-G(Q)]} \right\} \\&\qquad -[1-G(Q)] \frac{1}{\rho } [(b_S + \frac{\beta _S}{1-\beta _S} k_S \theta _S) - (b_N + \frac{\beta _N}{1-\beta _N} k_N \theta _N)]. \end{aligned}$$

Assume that the Hosios Condition is met.⁴² The set of expressions that characterize the social optimum is given by:

$$\begin{aligned} \frac{k_N( \lambda _N + \beta _N z(\theta ^{P}_N))}{p(\theta ^{P}_N))}= & \, \alpha _N q_L (1-\beta _N)[\frac{z(\theta ^{P}_N) G(Q^P) }{\lambda _N + z(\theta ^{P}_N)}]^{(\alpha _N - 1)} \nonumber \\&- \frac{1}{\rho }[\frac{1- G(Q^P)}{G(Q^P)}] \frac{\beta _N k_N}{(1-\beta _N)p(\theta ^{P}_N)} - (1-\beta _N)b_N; \end{aligned}$$

(36)

$$\begin{aligned} \frac{k_S( \lambda _S + \beta _S z(\theta ^{P}_S))}{p(\theta ^{P}_S))}= & \, \alpha _S q^e (Q^P) (1-\beta _S)[\frac{z(\theta ^{P}_S)( 1- G(Q^P)) }{\lambda _S + z(\theta ^{P}_S)}]^{(\alpha _S - 1)} \nonumber \\&+ \frac{1}{\rho } [\frac{1- G(Q^P)}{G(Q^P)}] \frac{\beta _S k_S}{(1-\beta _S)p(\theta ^{P}_S)} - (1-\beta _S)b_S; \end{aligned}$$

(37)

$$\begin{aligned}&{\mathcal {A}}(\theta ^{P}_N) G(Q^P)^{\alpha _N - 1} q_L + [1-G(Q^P)]^{\alpha _S - 1} [{\mathcal {B}}(\theta ^{P}_S) q^e (Q^P) \nonumber \\&\qquad - {\mathcal {C}}(\theta ^{P}_S)Q^P] \nonumber \\&\quad = {\mathcal {D}}(\theta ^{P}_N) - {\mathcal {E}}(\theta ^{P}_S) - {\mathcal {F}}(\theta ^{P}_N, \theta ^{P}_S); \end{aligned}$$

(38)

where:

$$\begin{aligned} {\mathcal {A}}(\theta ^{P}_N)= & \, \alpha _N [ \frac{ z(\theta ^{P}_N)}{\lambda _N + z(\theta ^{P}_N)}]^{\alpha _N}; \quad {\mathcal {B}}(\theta ^{P}_S) = (1- \alpha _S)[\frac{ z(\theta ^{P}_S)}{\lambda _S+ z(\theta ^{P}_S))}]^{\alpha _S};\\ {\mathcal {C}}(\theta ^{P}_S)= & \, \alpha _S[\frac{ z(\theta ^{P}_S)}{\lambda _S+ z(\theta ^{P}_S))}]^{\alpha _S} ; \quad {\mathcal {D}}(\theta ^{P}_N) = \frac{k_N \lambda _N \theta ^{P}_N - \lambda _N b_N}{\lambda _N + z(\theta ^{P}_N)};\\ {\mathcal {E}}(\theta ^{P}_S)= & \, \frac{k_S \lambda _S \theta ^{P}_S - \lambda _S b_S}{\lambda _S + z(\theta ^{P}_S)};\\ {\mathcal {F}}(\theta ^{P}_N, \theta ^{P}_S)= & \, \frac{1}{\rho } [(b_S + \frac{\beta _S}{1- \beta _S}k_S \theta ^{P}_S)- (b_N + \frac{\beta _N}{1- \beta _N}k_N \theta ^{P}_N)]; \end{aligned}$$

and $\theta ^{P}_S$, $\theta ^{P}_N$ and $Q^P$ represent the efficient values of the market tightness in the skilled and unskilled sectors and the efficient mass of skilled individuals, respectively.

Notice from (30) that the cost of opening a new vacancy in the unskilled sector equalizes the social surplus generated by this sector. This expression defines the optimal steady-state value of $\theta ^{P}_N$. The following equation characterizes the efficient value of $\theta ^{P}_S$. Finally, expression (32) determines the optimal mass of the educated workforce that generates the efficient outcome. $\square$

Appendix 7

Let

$$\begin{aligned} H_N(\theta ; \rho )= & \, \frac{(\rho + \lambda _N + \beta _N q(\theta ))k_N}{p(\theta )};\\ H_S(\theta ; \rho )= & \, \frac{(\rho + \lambda _S + \beta _S q(\theta ))k_S}{p(\theta )};\\ H^P_N(\theta )= & \, \frac{(\lambda _N + \beta _N q(\theta ))k_N}{p(\theta )};\\ H^P_S(\theta )= & \, \frac{(\lambda _S + \beta _S q(\theta ))k_S}{p(\theta )}. \end{aligned}$$

It can be seen that all previous expressions are increasing in $\theta$. Furthermore, note that for any $\theta$:

$$\begin{aligned} H^P_j(\theta ) = \lim _{\rho \rightarrow 0} H_j (\theta ; \rho ), \quad for \quad j={N,S}. \end{aligned}$$

(39)

Consider that $\alpha _N = \alpha _S = 1$, $C_N = C_S = \frac{1}{(1+ \rho )}$ and the Hosios Condition are satisfied. In this case, the equilibrium expressions of $\theta _N$ and $\theta ^P_N$ satisfy, respectively:

$$\begin{aligned} H_N(\theta _N; \rho ) = (1-\beta _N)(q_L - b_N); \end{aligned}$$

(40)

$$\begin{aligned} H^P_N(\theta ^P_N) = (1-\beta _N)(q_L - b_N) - \left[ \frac{1-G(Q^P)}{\rho G(Q^P)}\right] \left[ \frac{\beta _N k_N}{(1-\beta _N) p(\theta _N)}\right] . \end{aligned}$$

(41)

By using (34) and (35) we have that:

$$\begin{aligned} H^P_N(\theta ^P_N) < H_N(\theta _N; \rho ). \end{aligned}$$

At the limit, when $\rho \rightarrow 0$ and using (33), we conclude that:

$$\begin{aligned} H^P_N(\theta ^P_N) < H^P_N(\theta _N). \end{aligned}$$

Then $\theta ^P_N < \theta _N.$

By using the same reasoning for the expressions that define $\theta _S$ and $\theta ^P_S$ we have that:

$$\begin{aligned} H^P_S(\theta ^P_S) > H_S(\theta _S; \rho ) + (1- \beta _S)[q^e (Q^P) - q^e(Q_d, Q^o)]. \end{aligned}$$

(42)

By definition, $q^e (Q^P) = q^e (Q^P, q_H).$ Thus:

$$\begin{aligned} q^e (Q^P) \ge q^e (Q^P, Q_o), \quad for \quad any \quad Q_o \in [q_L, q_H]. \end{aligned}$$

Therefore, if $Q^P > Q_d$ then $q^e (Q^P) > q^e(Q_d, Q^o)$. By using (36) we have that:

$$\begin{aligned} H^P_S(\theta ^P_S) > H_S(\theta _S; \rho ). \end{aligned}$$

Finally, taking the limit when $\rho \rightarrow 0$, we have:

$$\begin{aligned} H^P_S(\theta ^P_S) > H^P_S(\theta _S). \end{aligned}$$

Then $\theta ^P_N < \theta _N.$$\square$