Assortative Matching of Exporters and Importers

Table 16: Segment-Switching Hypothesis

DLog Imports US Importer Mexican Exporter
(1)
(2) (3) (4) (5)
Linear
-0.061 Linear Probit Linear Probit
(0.292)
Binding 0.022*** 0.62*** 0.056*** 0.132*** 0.156***
(0.006)
mvrank2004 -0.016** (0.024) (0.021) (0.038) (0.046)
(0.008)
rank binding 0.002*** 0.001*** -0.001* -0.001
Yes
Hs2-level Effects (0.001) (0.000) (0.001) (0.001)
R-Squared 0.018
Obs. 966 -0.002* -0.001* 0.000 -0.001

(0.001) (0.000) (0.001) (0.001)

Yes Yes Yes Yes

0.060 0.034

601 601 718 707

Note: In the first column, the dependent variable is the log difference of imports at the importer product level. In the second
and third columns, the dependent variable is a dummy variables indicating whether US importers upgrade the main Mexican
partners. .In the fourth and fifth columns, the dependent variable is a dummy variables indicating whether Mexican exporters
downgrade the main US partners. The table reports coefficients from firm-product level regressions of the dependent variables

on a dummy on whether the US imposed binding quotas on imports from China in 2004 (Binding), the initial rank, and

interaction between both of them together with chapter (HS 2 digit product) fixed effects. The probit results report the
marginal effects. Standard errors are clustered at the 6-digit product level and shown in parentheses. Significance: * 10
percent, ** 5 percent, *** 1 percent.

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Appendix

A.1 The Sign of Sorting

Consider two teams in a stable matching. Firms in team 1 have task quality (x, y) and firms in team
2 have (x0, y0). Let ⇧(x, y) be the joint profit for the team with a supplier of capability x and a final
producer of capability y. We first prove two claims.
Claim 1. The following inequality holds in a stable matching:

⇧ (x, y) + ⇧ x0, y0 ⇧ x0, y + ⇧ x, y0 (A.1)

Proof. From the maximization of suppliers, the following inequalities must hold

⇡x (x) = ⇧ (x, y) ⇡y (y) ⇧ x, y0 ⇡y y0
⇡x(x0) = ⇧ x0, y0 ⇡y y0 ⇧ x0, y ⇡y (y) .

Adding up the above two equations leads to (A.1).
Notice that

ˆ y0 ˆ x0 @2⇧ (x˜, y˜) dx˜dy˜ = ⇧ (x, y) + ⇧ x0, y0 ⇧ x0, y ⇧ x, y0 . (A.2)
@x˜@y˜
y x

The right hand side of (A.2) is always non-negative from (A.1).

In Case-C, @2⇧(x, y)/@x@y = ✓xy > 0. Therefore, if x0 > x, then y0 y must hold to make the
right hand side of (A.2) to be non-negative. This is positive assortative matching (PAM).

In Case-S, @2⇧(x, y)/@x@y = ✓xy < 0. Therefore, if x0 > x, then y0  y must hold to make the
right hand side of (A.2) to be non-negative. This is negative assortative matching (PAM).

In Case-I, @2⇧(x, y)/@x@y = ✓xy = 0. In this case, the right hand side of (A.2) is zero. The
inequality of (A.1) holds with equality, which implies firms are indifferent among all possible matches.

Therefore, the matching pattern is not determined.

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