Product and Labor Market Entry Costs, Underemployment and International Trade

Abstract

We develop a small, open economy, two-sector model with heterogeneous agents and endogenous participation in a labor matching market. There are two types of agents: workers and entrepreneurs. Both populations are heterogeneous. Workers are distinguished by their potential ability as skilled workers and entrepreneurs by their potential ability to manage a firm. To capture the notion of decentralized labor markets we assume random matching. Those agents on the long side of the market who are not matched find employment in the unskilled sector as do those agents who decided not to attempt to enter the matching market. The output of matched pairs is a function of the two partners’ abilities. We find that disparities in labor institutions become a source of comparative advantage. The exact patterns will depend not only on the costs of entering the skilled sector but also on the mechanism used for dividing the surplus. We analyze the implications of asymmetric market entry costs for the patterns of international trade and underemployment. We find that if labor market inefficiencies are sufficiently strong trade liberalization can lead to welfare losses. We also examine the robustness of our results when we allow for complementarities in the production function and for alternative matching mechanisms.

Appendix

In all proofs we have suppressed the *.

Proof of Lemma 1

The system of Eqs. (1), (2) and (3) can be rewritten as

$$\displaystyle{ \frac{1} {2}\left (\frac{1 + z} {2} +\alpha \right )-\gamma = P }$$

(13)

$$\displaystyle{ \frac{1} {2}\left (z + \frac{1+\alpha } {2} \right ) -\frac{1 - z} {1-\alpha } c = P }$$

(14)

$$\displaystyle{ P = \frac{(1-\alpha )\left (2 +\alpha +z\right )} {4\alpha } }$$

(15)

Differentiating (15) with respect to α we get

$$\displaystyle{\frac{\partial P} {\partial \alpha } = \frac{-2 -\alpha ^{2} - z} {4\alpha ^{2}} < 0}$$

Then the difference $\frac{1} {2}\left (\frac{1+z} {2} +\alpha \right ) - P = \frac{1} {2}\left (\frac{1+z} {2} +\alpha \right ) -\frac{(1-\alpha )\left (2+\alpha +z\right )} {4\alpha }$ is increasing in α. When the expression is evaluated at $\alpha = \frac{1} {2}$ we find that it is equal to $-\frac{1} {8} < 0$. Then, the Lemma follows from (13) and γ > 0.

Proof of Proposition 2

By substituting (15) into (13) and (14) we can reduce the above system into two equations in the two unknowns α and z. Totally differentiating the new system we get

$$\displaystyle{ \left (\frac{1} {2} -\frac{\partial P} {\partial \alpha } \right )d\alpha + \left (\frac{1} {4} -\frac{\partial P} {\partial z} \right )dz = d\gamma }$$

(16)

$$\displaystyle\begin{array}{rcl} & & \left (\frac{1} {4} - \frac{1 - z} {\left (1-\alpha \right )^{2}}c -\frac{\partial P} {\partial \alpha } \right )d\alpha + \left (\frac{1} {2} + \frac{1} {1-\alpha }c -\frac{\partial P} {\partial z} \right )dz \\ & =& \left (\frac{1 - z} {1-\alpha } \right )dc {}\end{array}$$

(17)

where

$$\displaystyle{\frac{\partial P} {\partial z} = \frac{1-\alpha } {4\alpha } > 0}$$

Next, we proceed to show that the determinant $\Delta$ is positive.

$$\displaystyle\begin{array}{rcl} \Delta & =& \left (\frac{1} {2} -\frac{\partial P} {\partial \alpha } \right )\left (\frac{1} {2} + \frac{1} {1-\alpha }c -\frac{\partial P} {\partial z} \right ) -\left (\frac{1} {4} -\frac{\partial P} {\partial z} \right )\left (\frac{1} {4} - \frac{1 - z} {\left (1-\alpha \right )^{2}}c -\frac{\partial P} {\partial \alpha } \right ) {}\\ & =& \left [ \frac{3} {16} -\frac{1} {4} \frac{\partial P} {\partial z} -\frac{1} {4} \frac{\partial P} {\partial \alpha } \right ] + \left [\frac{1} {2} \frac{1} {1-\alpha }c - \frac{1} {1-\alpha }c\frac{\partial P} {\partial \alpha } + \frac{1} {4} \frac{1 - z} {\left (1-\alpha \right )^{2}}c - \frac{1 - z} {\left (1-\alpha \right )^{2}}c\frac{\partial P} {\partial z} \right ] {}\\ \end{array}$$

Lemma 1 implies that $\frac{\partial P} {\partial z} < \frac{1} {4}$ and that $-\frac{\partial P} {\partial \alpha } > \frac{3} {4}$. The two inequalities imply that the expression in the first bracket is positive. The first inequality also implies that $\frac{1} {4} \frac{1-z} {\left (1-\alpha \right )^{2}} c - \frac{1-z} {\left (1-\alpha \right )^{2}} c\frac{\partial P} {\partial z} > 0$ which, in turn, implies that the expression in the second bracket is also positive.

Thus, $\Delta > 0$.

• (a) $\Delta > 0$ implies that

  $$\displaystyle{sign\left \{\frac{d\alpha ^{{\ast}}} {dc}\right \} = sign\left \{-\left (\frac{1 - z} {1-\alpha } \right )\left (\frac{1} {4} -\frac{\partial P} {\partial z} \right )\right \} = sign\left \{-\frac{1} {4}\left (\frac{1 - z} {1-\alpha } \right )\frac{2\alpha - 1} {\alpha } \right \}}$$

  where given that $\alpha > \frac{1} {2}$ is negative.

• (b) $\Delta > 0$ implies that

  $$\displaystyle{sign\left \{\frac{d\alpha ^{{\ast}}} {d\gamma } \right \} = sign\left \{\frac{1} {2} + \frac{1} {1-\alpha }c -\frac{\partial P} {\partial z} \right \} > 0}$$

  Given that $\frac{\partial P} {\partial z} < \frac{1} {4}$ the expression is positive.

• (c) $\Delta > 0$ implies that

  $$\displaystyle\begin{array}{rcl} sign\left \{\frac{dz^{{\ast}}} {d\gamma } \right \}& =& sign\left \{-\left (\frac{1} {4} - \frac{1 - z} {\left (1-\alpha \right )^{2}}c -\frac{\partial P} {\partial \alpha } \right )\right \} {}\\ & =& sign\left \{-\left (\frac{1} {4} - \frac{1 - z} {\left (1-\alpha \right )^{2}}c + \frac{2 +\alpha ^{2} + z} {4\alpha ^{2}} \right )\right \} {}\\ & =& sign\left \{-4\alpha ^{2}(1 - z)c + \left (1-\alpha \right )^{2}\left (2 + 2\alpha ^{2} + z\right )\right \} \gtrless 0 {}\\ \end{array}$$

  For sufficiently high values of γ (high α) the expression will be negative. For low values of γ the sign will depend on c, and given that an increase in c implies an increase in z (see below), for relative extreme values of c the expression will be positive.

• (d) $\Delta > 0$ implies that

  $$\displaystyle{sign\left \{\frac{dz^{{\ast}}} {dc} \right \} = sign\left \{\left (\frac{1} {2} -\frac{\partial P} {\partial \alpha } \right )\left (\frac{1 - z} {1-\alpha } \right )\right \} > 0}$$

  where the inequality follows from $\frac{\partial P} {\partial \alpha } < 0$.

Proof of Proposition 3

• (a) Totally differentiating (13) we get

  $$\displaystyle{\frac{dP} {dc} = \frac{1} {4} \frac{dz} {dc} + \frac{1} {2} \frac{d\alpha } {dc}}$$

  Given that $\Delta > 0$ the sign of the above expression is the same as the sign of

  $$\displaystyle\begin{array}{rcl} & & \frac{1} {4}\left (\frac{1} {2} -\frac{\partial P} {\partial \alpha } \right )\left (\frac{1 - z} {1-\alpha } \right ) -\frac{1} {2}\left (\frac{1 - z} {1-\alpha } \right )\left (\frac{1} {4} -\frac{\partial P} {\partial z} \right ) {}\\ & =& \frac{1} {4}\left (\frac{1 - z} {1-\alpha } \right )\left (\left (\frac{1} {2} + \frac{2 +\alpha ^{2} + z} {4\alpha ^{2}} \right ) -\frac{1} {2}\left (\frac{2\alpha - 1} {\alpha } \right )\right ) {}\\ & =& \frac{1} {4\alpha }\left (\frac{1 - z} {1-\alpha } \right )\left (2 -\alpha ^{2} + 2\alpha + z\right ) > 0 {}\\ \end{array}$$

• (b) Totally differentiating (13) we get

  $$\displaystyle{\frac{dP} {d\gamma } = \frac{1} {4} \frac{dz} {d\gamma } + \frac{1} {2} \frac{d\alpha } {d\gamma } - 1}$$

  The first two terms are equal to

  $$\displaystyle{\left \{-\frac{1} {4}\left (\frac{1} {4} - \frac{1 - z} {\left (1-\alpha \right )^{2}}c -\frac{\partial P} {\partial \alpha } \right ) + \frac{1} {2}\left (\frac{1} {2} + \frac{1} {1-\alpha }c -\frac{\partial P} {\partial z} \right )\right \}/\Delta }$$

  To complete the proof we need to show that the numerator is less than $\Delta$

  $$\displaystyle\begin{array}{rcl} & & -\frac{1} {4}\left (\frac{1} {4} - \frac{1 - z} {\left (1-\alpha \right )^{2}}c -\frac{\partial P} {\partial \alpha } \right ) + \frac{1} {2}\left (\frac{1} {2} + \frac{1} {1-\alpha }c -\frac{\partial P} {\partial z} \right ) - {}\\ & &\left [ \frac{3} {16} -\frac{1} {4} \frac{\partial P} {\partial z} -\frac{1} {4} \frac{\partial P} {\partial \alpha } \right ] -\left [\frac{1} {2} \frac{1} {1-\alpha }c - \frac{1} {1-\alpha }c\frac{\partial P} {\partial \alpha } + \frac{1} {4} \frac{1 - z} {\left (1-\alpha \right )^{2}}c - \frac{1 - z} {\left (1-\alpha \right )^{2}}c\frac{\partial P} {\partial z} \right ] {}\\ & =& \left (\frac{1} {2} + \frac{1} {1-\alpha }c\right )\frac{\partial P} {\partial \alpha } -\left (\frac{1} {4} - \frac{1 - z} {\left (1-\alpha \right )^{2}}c\right )\frac{\partial P} {\partial z} {}\\ & =& - \frac{1} {16\alpha ^{2}\left (1-\alpha \right )}\left (\left (2\left (1-\alpha \right ) + 4c\right )\left (2 +\alpha ^{2} + z\right ) +\alpha \left (1-\alpha \right )^{2} - 4\alpha \left (1 - z\right )c\right ) {}\\ & =& - \frac{1} {16\alpha ^{2}\left (1-\alpha \right )}\big(4\left (1-\alpha \right ) + 2\left (1-\alpha \right )\alpha ^{2} + 2\left (1-\alpha \right ) {}\\ & & \qquad \qquad \qquad \quad \times z + 8c + 4\alpha ^{2}c + 4zc +\alpha \left (1-\alpha \right )^{2} - 4\alpha \left (1 - z\right )c\big) {}\\ \end{array}$$

  The proof follows from $4\alpha \left (1 - z\right )c < 4c < 8c$.

Proof of Proposition 6

The function P(β; γ, c) is continuous but not differentiable at β = β ^∗.

For β > β ^∗ the equilibrium of the model is determined by

$$\displaystyle{ (1-\beta )\left (\frac{1 + z} {2} +\alpha \right )-\gamma = P }$$

(18)

$$\displaystyle{ \beta \left (z + \frac{1+\alpha } {2} \right ) -\frac{1 - z} {1-\alpha } c = P }$$

(19)

$$\displaystyle{ P = \frac{(1-\alpha )\left (2 +\alpha +z\right )} {4\alpha } }$$

(20)

Given that the definition of β ^∗ implies that α ^∗ > z ^∗ we have obtained the above system from the system (13), (14) and (15) after setting the entrepreneur’s share of surplus equal to β. Let P ⁺(β) denote the price function for β > β ^∗. We will sow that $\frac{dP^{+}(\beta )} {d\beta } _{(\beta =\beta ^{{\ast}})} = \left (\frac{\partial P} {\partial \alpha } \frac{d\alpha } {d\beta } + \frac{\partial P} {\partial z} \frac{dz} {d\beta } \right )_{(\beta =\beta ^{{\ast}})} < 0$. Notice that given that the market clearing condition (20) does not directly depend on β the expressions for $\frac{\partial P} {\partial \alpha }$ and $\frac{\partial P} {\partial z}$ are the same as those derived in the proof of Proposition 2. After substituting (20) in (18) and (19) and totally differentiating we get

$$\displaystyle{ \left (\left (1-\beta \right ) -\frac{\partial P} {\partial \alpha } \right )d\alpha + \left (\frac{1-\beta } {2} -\frac{\partial P} {\partial z} \right )dz = \left (\frac{1 + z} {2} +\alpha \right )d\beta }$$

(21)

$$\displaystyle{ \left ( \frac{\beta } {2} - \frac{1 - z} {\left (1-\alpha \right )^{2}}c -\frac{\partial P} {\partial \alpha } \right )d\alpha + \left (\beta + \frac{c} {1-\alpha }-\frac{\partial P} {\partial z} \right )dz = -\left (z + \frac{1+\alpha } {2} \right )d\beta }$$

(22)

Following the same steps as those used in the proof of Proposition 2 we can show that the determinant is positive for β close to $\frac{1} {2}$, and we have already shown that as γ and c come closer together β ^∗ approaches $\frac{1} {2}$. Then,

$$\displaystyle\begin{array}{rcl} & & sign\left \{\frac{\partial P} {\partial \alpha } \frac{d\alpha } {d\beta } + \frac{\partial P} {\partial z} \frac{dz} {d\beta } \right \}_{\beta =\beta ^{{\ast}}} {}\\ & =& sign\left \{\begin{array}{c} \frac{\partial P} {\partial \alpha } \left (\left (\frac{1+z} {2} +\alpha \right )\left (\beta + \frac{c} {1-\alpha } -\frac{\partial P} {\partial z} \right ) + \left (z + \frac{1+\alpha } {2} \right )\left (\frac{1-\beta } {2} -\frac{\partial P} {\partial z} \right )\right ) \\ -\frac{\partial P} {\partial z} \left (\left (\left (1-\beta \right ) -\frac{\partial P} {\partial \alpha } \right )\left (z + \frac{1+\alpha } {2} \right ) + \left ( \frac{\beta } {2} - \frac{1-z} {\left (1-\alpha \right )^{2}} c -\frac{\partial P} {\partial \alpha } \right )\left (\frac{1+z} {2} +\alpha \right )\right ) \end{array} \right \}_{\beta =\beta ^{{\ast}}}{}\\ \end{array}$$

Setting α = z = x and simplifying we get

$$\displaystyle{\frac{\partial P} {\partial \alpha } \left (\frac{1} {2} + \frac{\beta } {2} + \frac{c} {1 - x}\right ) -\frac{\partial P} {\partial z} \left (1 - \frac{\beta } {2} - \frac{c} {1 - x}\right ) < 0}$$

The inequality follows from $\frac{\partial P} {\partial z} < \frac{1} {4}$ and $-\frac{\partial P} {\partial \alpha } > \frac{3} {4}$.

To complete the proof we need to show that when β < β ^∗, $\frac{dP^{-}(\beta )} {d\beta } _{(\beta =\beta ^{{\ast}})} = \left (\frac{\partial P} {\partial \alpha } \frac{d\alpha } {d\beta } + \frac{\partial P} {\partial z} \frac{dz} {d\beta } \right )_{(\beta =\beta ^{{\ast}})} > 0$, where P ⁻(β) denotes the corresponding price function. In this case, the equilibrium is given by

$$\displaystyle{ (1-\beta )\left (\frac{1 + z} {2} +\alpha \right ) - \frac{1-\alpha } {1 - z}\gamma = P }$$

(23)

$$\displaystyle{ \beta \left (z + \frac{1+\alpha } {2} \right ) - c = P }$$

(24)

$$\displaystyle{ P = \frac{(1 - z)\left (2 +\alpha +z\right )} {4z} }$$

(25)

Notice that we can obtain (25) by substituting γ for c, c for γ, α for z and z for α in (18). Thus the partial derivatives $\frac{\partial P} {\partial \alpha }$ and $\frac{\partial P} {\partial z}$ are given by

$$\displaystyle{\frac{\partial P} {\partial \alpha } = \frac{1 - z} {4z} > 0}$$

and

$$\displaystyle{\frac{\partial P} {\partial z} = \frac{-2 - z^{2}-\alpha } {4z^{2}} < 0}$$

After substituting (25) in (23) and (24) and totally differentiating we get

$$\displaystyle{ \left (\left (1-\beta \right ) + \frac{\gamma } {1 - z} -\frac{\partial P} {\partial \alpha } \right )d\alpha + \left (\frac{1-\beta } {2} - \frac{1-\alpha } {\left (1 - z\right )^{2}}\gamma -\frac{\partial P} {\partial z} \right )dz = \left (\frac{1 + z} {2} +\alpha \right )d\beta }$$

(26)

$$\displaystyle{ \left ( \frac{\beta } {2} -\frac{\partial P} {\partial \alpha } \right )d\alpha + \left (\beta -\frac{\partial P} {\partial z} \right )dz = -\left (z + \frac{1+\alpha } {2} \right )d\beta }$$

(27)

Because of symmetry the determinant of this system is identical to the one for the previous case for $\beta = \frac{1} {2}$. (Let the old matrix be $\left [\begin{array}{*{10}c} A&B\\ C &D \end{array} \right ]$. Then the new matrix is $\left [\begin{array}{*{10}c} D&C\\ B & A \end{array} \right ]$. The determinant has not been affected by these changes, however, when the signs were different across a diagonal before now the positive sign has become negative and the other way around). Thus, at least for values of β close to $\frac{1} {2}$, the new determinant will also be positive. Then the sign of $\frac{dP^{-}(\beta )} {d\beta } _{(\beta =\beta ^{{\ast}})}$ is the same as

$$\displaystyle\begin{array}{rcl} & & sign\left \{\frac{\partial P} {\partial \alpha } \frac{d\alpha } {d\beta } + \frac{\partial P} {\partial z} \frac{dz} {d\beta } \right \}_{\beta =\beta ^{{\ast}}} {}\\ & =& sign\left \{\begin{array}{c} \frac{\partial P} {\partial \alpha } \left (\left (\frac{1+z} {2} +\alpha \right )\left (\beta -\frac{\partial P} {\partial z} \right ) + \left (z + \frac{1+\alpha } {2} \right )\left (\frac{1-\beta } {2} - \frac{1-\alpha } {\left (1-z\right )^{2}} \gamma -\frac{\partial P} {\partial z} \right )\right ) \\ -\frac{\partial P} {\partial z} \left (\left (\left (1-\beta \right ) + \frac{\gamma } {1-z} -\frac{\partial P} {\partial \alpha } \right )\left (z + \frac{1+\alpha } {2} \right ) + \left ( \frac{\beta } {2} -\frac{\partial P} {\partial \alpha } \right )\left (\frac{1+z} {2} +\alpha \right )\right ) \end{array} \right \}_{\beta =\beta ^{{\ast}}}{}\\ \end{array}$$

Setting α = z = x and simplifying we get

$$\displaystyle{\frac{\partial P} {\partial \alpha } \left (\frac{1} {2} + \frac{\beta } {2} - \frac{\gamma } {1 - x}\right ) -\frac{\partial P} {\partial z} \left (1 - \frac{\beta } {2} + \frac{\gamma } {1 - x}\right ) > 0}$$

The above inequality follows from using the same logic as the one used for the proof of Lemma 1 to show that $z > \frac{1} {2}$ and then using the last inequality to show that $\frac{\partial P} {\partial \alpha } < \frac{1} {4}$ and $-\frac{\partial P} {\partial z} > \frac{3} {4}$.

This completes the proof of the proposition.