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Directed Search Lecture 2: Matching Patterns and Inequality October 2012 Shouyong Shi c Main sources of this lecture: Becker, G., 1973. A Theory of Marriage: Part I, JPE 81, 813846. Shi, S., 2001, Frictional Assignment I:


  1. Directed Search Lecture 2: Matching Patterns and Inequality October 2012 ° Shouyong Shi c

  2. Main sources of this lecture: • Becker, G., 1973. “A Theory of Marriage: Part I,” JPE 81, 813—846. • Shi, S., 2001, “Frictional Assignment I: E ffi ciency,” JET 98, 232-260. • Shi, S., 2005, “Frictional Assignment, Part II: In fi nite Horizon and Inequality,” RED 8, 106-137. • Shimer, R. and L. Smith, 2000, “Assortative Matching and Search,” ECMA 68, 343—370 • Shi, S., 2002, “A Directed Search Model of Inequality with Heterogeneous Skills and Skill-Biased Technology,” RES 69, 467-491. 2

  3. 1. Motivation and Issues • many markets have heterogeneity on both sides: — labor market: workers di ff er in skills, fi rms in capital and size — loan market: borrowers di ff er in project quality, lenders in funds — marriage market: men and women di ff er in income, beauty, etc. 3

  4. • positive assortative matching (PAM): individuals are matched according to their ranking: — workers with higher skills match with better fi rms; — projects with higher quality match with better loans; — rich people marry rich people; handsome men marry beautiful women, etc. • two questions about the matching pattern: — positive: is PAM an equilibrium? — normative: is PAM socially e ffi cient? 4

  5. • answer by Gary Becker (73, JPE) and Tinbergen (51): — PAM is an equilibrium and it is socially e ffi cient when markets are frictionless — necessary and su ffi cient condition for this result: joint surplus of a match is complementary (supermodular) in the two sides’ attributes • think again: — most matching markets are frictional — not all observed matching patterns are PAM 5

  6. Main questions: when there are search frictions, • does the e ffi cient allocation have PAM? • how to decentralize the e ffi cient allocation? • how does matching a ff ect inequality? With undirected search, Shimer and Smith (00) fi nd that complementarity is not enough for PAM to arise in eqm • but their equilibrium is ine ffi cient, generically; is this ine ffi ciency responsible for non-PAM? • still need to answer the normative questions above 6

  7. Directed search: • makes sense with homogeneous individuals • makes even more sense with heterogeneity: observable heterogeneity helps directing search — job ads typically specify worker quali fi cations; workers can observe fi rms’ attributes — di ff erentiated loan terms target di ff erent borrowers — people may date selectively 7

  8. Roadmap: • analyze a market with TU and matching between workers who di ff er + machines that di ff er in skill levels in qualities • eqm and e ffi cient allocation with no friction • with search friction and directed search, characterize: e ffi cient allocation decentralization, inequality • extend to in fi nite horizon; dynamics • calibrate to examine e ff ects of skill-biased technology 8

  9. 2. Frictionless Economy and Assignment One-period environment • risk-neutral workers: exogenous supply; observable skill  ∈  ⊂ R + : number =  (  ); • machine quality  ∈ K ⊂ R + : costs  (  ); endogenous supply determined by free entry • one worker operates one machine; • output of the pair (   ):  (   ) 9

  10. Assumptions on  : • complementarity (supermodularity):    0 • both inputs are necessary:  (0   ) =  (  0) = 0 • every skill is employable with some machine quality:  (    ) −  (  )  0 for some  ∈ K • regularity conditions:    0;   ≥ 0, (     −     )   (   −   )  2  10

  11. Frictionless assignment • no frictions: all pairs are matched instantaneously • e ffi cient assignment   :  → K max [  (   ) −  (  )] , for each  ∈  ,    (   (  )   ) =   (   (  )) i.e., •   (  ) exists and is unique for each  • PAM:     0 (  ) =  0 i ff    0   −   11

  12. Decentralization: • wage function: ½  (   ) −  (  )  if  (   ) −  (  ) ≥ 0  (   ) = 0  otherwise • a fi rm chooses  for each  to attract workers: ⇒ solution:  =   (  ) max  ∈ K  (   ) =   (  ) =  (   (  )   ) −  (   (  )) • equilibrium wage: • Does PAM increase wages for higher  ? (relevant for identifying sorting through wage or value added) 12

  13. Answer: Not marginally. • assignment pattern has NO fi rst-order e ff ect on wage:   (  ) =  (   (  )   ) −  (   (  ))   0 (  ) =   (   (  )   ) e ff ect of  by itself +   0 (  )[   (   (  )   ) −   (  )] e ff ect of a better match (but this is = 0) 13

  14. 3. E ffi cient Assignment with Frictions Frictional economy qualities in skill  unit subset  (  ) workers # :  (  1   ) → # :  (  1   )  (  1   )  1 (  1   )  : workers/machines . . . . . . . . .   (     ) # :  (     ) → # :  (     )  (     ) Matching probability in a unit (   ): for a worker: 1 −  −  (  ) for a machine: 1 −  −  (  ) ;  (  ) 14

  15. E ffi cient allocation : The planner chooses •   (  ) ⊆ K : machine qualities assigned to  ∈  •   (   ): # of machines created for the unit (   ) •   (   ): worker/machine ratio in the unit (   ) h³ 1 −  −  (  ) ´ i X X max  (   )  (   ) −  (  ) | {z } (  )  ∈   ∈  (  ) expected surplus of a match (   ) X s.t.  (   )  (   ) =  (  )  ∈  (  ) | {z } # of skill  workers assigned to  15

  16. Component problem of the e ffi cient allocation: For each  ∈  , the e ffi cient allocation (   (  )    (   )) solves: (   ) (  )  −  (  )  (   ) max social value of a worker  h 1 − (1 +  (   ))  −  (  ) i s.t.  (   ) =  (  ) | {z } social value of a machine in unit (   ) • FOC of   (   ) leads to the constraint in (   ) • FOC of   coincides with that of (   ) • if  1 ∈   (  ) does not solve (   ), welfare can be increased 16

  17. Why can the planner’s problem be decomposed so? • The planner chooses machines for each  separately; there is no direct interaction between di ff erent  • For each  , the planner should — maximize the worker’s social marginal value, which is the objective function in (   ) — create as many machines in each unit (   ) as to equate: social marginal value of a machine = the cost; (this is the constraint in (   )) 17

  18. B : w o rk er/m ach in e ratio zero n et p ro fit Z N P (k ) d irectio n o f h ig h er so cial m arg in al n et v alu e o f a m ach in e IN D (k ) in d ifferen ce cu rv e b o E d irectio n o f h ig h er so cial m arg in al v alu e o f a w o rk er Ф ο m ach in e q u ality k E ffi cient allocation 18

  19. E ffi cient allocation: solution • Assignment is distinct:   (  1 ) ∩   (  2 ) = ∅ if  1 6 =  2 — suppose  1 and  2 are both assigned to  , with  2   1 . Let   =  (    ) and   =  (    ). Then,  −  1  1 =  −  2  2 = ⇒  2   1 | {z } social value of  1 and  2 — contradiction: net value of using skill  2 is higher: h i h i 1 − (1 +  2 )  −  2 1 − (1 +  1 )  −  1  2 −  (  )   1 −  (  ) • assignment is one-to-one:   (  ) is unique for each  if (     −     )   (   −   )  2  19

  20. E ffi cient allocation: solution (continued) • e ffi cient choice of  for  (where   (  ) =   (   (  )   )): h 1 −  −   (  ) i   (   (  )   )   (   (  )) = | {z } | {z } expected marginal product of  marginal cost recall: frictionless assignment   (   (  )   ) =   (   (  )) = ⇒   (  )    (  ) • e ffi cient choice of  for  : h 1 − (1 +   (  ))  −   (  ) i  (   (  )   ) =  (   (  )) | {z } social value of a machine   (  ) 20

  21. E ffi cient allocation: solution (continued) Write these conditions more explicitly: ∙ ¸ 1 −   (   (  ))   (  ) = − ln   (   (  )   )  (   (  )) ∙ ¸  (   (  )  )   (   (  )   ) −   (   (  )) 1 −   (   (  )) ln =   (   (  )   )   (   (  )   ) −   (   (  )) 21

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