Abstract
We develop a small, open economy, twosector 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.
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Notes
 1.
See McGuiness (2006) for a review of this literature.
 2.
 3.
A simplified version of the model with onesided uncertainty has been used by Bougheas and Riezman (2007) to examine the relationship between the distribution of endowments and the patterns of trade and by Davidson and Matusz (2006) and Davidson et al. (2006) to examine redistribution policy issues.
 4.
In Helpman et al. (2010) although both populations of firms and entrepreneurs are heterogeneous it is only the participation of the second group that is derived endogenously. Egger and Kreickemeier (2012) analyze a model with one heterogenous population and generalized endogenous participation where agents in addition to their level of skills also decide in which sector to be employed. In our model, both workers and entrepreneurs can choose whether or not to enter the matching market.
 5.
Let X denote the level of consumption of the hightech product, Y the level of consumption of the primary commodity and I the level of nominal income. By maximizing \(\sqrt{XY}\) subject to I = PX + Y, we obtain the solutions \(X = \frac{I} {4P}\) and \(Y = \frac{I} {4}\), which after substituting them back in the utility function and multiplying by 2 (because (a) the marginal utility of income is equal to 1, and (b) the measure of agents is equal to 2) we obtain the solution in the text.
 6.
In an earlier version of the paper, we had the costs denominated in numeraire units (units of the hightech good). As a result the relative price of the two goods depended on the size of these costs. By eliminating this bias we have simplified many derivations and we were able to derive some additional results.
 7.
Acemoglu (1996) also employs Nash bargaining in a random matching environment similar to the one in this paper.
 8.
For the derivation of the last term, given that the output of a matched pair is equal to the sum of the abilities of its members, it suffices to add individual abilities. Thus, we have that aggregate income of matched pairs equals
$$\displaystyle{\int \nolimits _{\alpha ^{{\ast}}}^{1}\alpha d\alpha + \frac{1 \alpha ^{{\ast}}} {1  z^{{\ast}}}\int \nolimits _{z^{{\ast}}}^{1}zdz}$$Notice that second term follows from random matching and z ^{∗} < α ^{∗}.
 9.
The * have been suppressed.
 10.
It is clear that symmetry implies that the result does not depend on which side of the market is short. What matters for our conclusions is the presence of underemployment and not on the type of underemployment.
 11.
Our solution for β ^{∗} corresponds to the Hosios (1990) condition derived from search models using the matching function.
 12.
Demonstrating the result for values of γ and c sufficiently apart has proven to be a very daunting task. However, calibrations of the model (Table 1 provides just a few examples), where we have allowed the two entry costs and the sharing rule to take values in the interval [0, 1], show that the result stated in Proposition 6 is not only valid when we allow the two values to differ considerably but also that the maximum is a global maximum.
 13.
This of course requires that this distribution is the same as the realized distribution resulting from random matching. AlósFerrer (2002) has shown that this is indeed the case.
 14.
The numerical results are provided in a separate Appendix that is available from the authors.
 15.
Our result that trade can potentially lead to welfare losses relies on comparisons between the competitive equilibrium under autarky and the corresponding equilibrium under trade and thus is not affected by this choice.
 16.
Keep in mind that the size of the population has measure 2.
 17.
See footnote 5.
 18.
In fact, this is a third source of inefficiency due to a participation externality which is common in many matching and search models.
 19.
Due to the choice of functional forms and parameter values the differences are small, however, they are robust in the sense that the qualitative results are obtained for a wide set of parameter values.
 20.
This is an application of the second welfare theorem. Suppose that the agents in the economy are allocated to sectors by the social planner (this step follows from the fact that the equilibrium allocation is inefficient) and then exchange goods in competitive markets. The equilibrium price would be the one that decentralizes the social planner’s optimal allocation under autarky.
 21.
We are indebted to Carl Davidson for suggesting this alternative mechanism.
 22.
Of course, if the measure of entrepreneurs who enter the matching market is more than twice the measure of corresponding workers then all workers will be matched with multiple entrepreneurs.
 23.
The numerical results are provided in a separate Appendix available from the Authors.
 24.
What matters is that they do not know how many others are trying to find a match in the same location.
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Acknowledgements
We would like to thank participants at the Society for the Advancement of Economic Theory Conference, Ischia 2009, the MidWest International Economics Meeting, Evanston, 2010, the GEP Annual Conference, University of Nottingham 2010, the CESifo Area Conference on Global Economy, Munich, 2011, the Conference on WorkerSpecific Effects of Globalization, Tubingen, 2011, the ETSG Conference, Copenhagen, 2011, the Conference on Globalization: Strategies and Effects, Kolding Denmark, 2011 and seminar participants at ECARES, Brussels, 2011, the University of Southern California, 2012 and the University of Tennessee, 2012 for helpful comments and suggestions. The usual disclaimer applies.
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Appendix
In all proofs we have suppressed the *.
Appendix
Proof of Lemma 1
The system of Eqs. (1), (2) and (3) can be rewritten as
Differentiating (15) with respect to α we get
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
where
Next, we proceed to show that the determinant \(\Delta \) is positive.
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 \(\frac{1} {4} \frac{1z} {\left (1\alpha \right )^{2}} c  \frac{1z} {\left (1\alpha \right )^{2}} c\frac{\partial P} {\partial z} > 0\) 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
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
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,
Setting α = z = x and simplifying we get
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
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
and
After substituting (25) in (23) and (24) and totally differentiating we get
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
Setting α = z = x and simplifying we get
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.
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Bougheas, S., Riezman, R. (2017). Product and Labor Market Entry Costs, Underemployment and International Trade. In: Christensen, B., Kowalczyk, C. (eds) Globalization. Springer, Berlin, Heidelberg. https://doi.org/10.1007/9783662495025_2
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