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Caillaud Jullien (2003) - Chicken And Egg (view source)
Revision as of 20:26, 21 June 2010
, 20:26, 21 June 2010no edit summary
*Each agent has a unique match in the opposite population
**Probability of finding a match without an intermediary = 0
**Probability of finding a match without an intermediary = <math>\lambda \le 1 \,</math> if both parties are registered with that intermediary
*Two matchmakers will compete: <math>k \in \{I,E\}\,</math> (Incumbent and Entrant)
**Each matchmaker uses the same technology <math>\lamdbalambda\,</math> (at least to start with)
*Matchmakers earn revenue:
**Registration fees <math>p\,</math>
**Transaction fees <math>t\,</math> (charged only if trade takes place)
*and have costs <math>c\,</math> per registration
*A type 2 agent seeking a type 1 agent on a matchmaker with <math>n_1\,</math> type 1 agents has a match probability of <math>n_1 \lamdalambda\,</math>
*It is assumed that intermediation is efficient: <math>\lambda > c = c_1 +c_2\,</math>
E then conquers j users through the externality:
:<math>p_j^E + \lambda u_j t^E < \lambda u_j + \inf\{p_j^I,0\} \quad \mbox(as presented in the paper}\,</math>
:<math>p_j^E + \lambda u_j(1-t^E) < \lambda u_j(1-t^I) + \inf\{p_j^I,0\} \quad \mbox(as understood by Ed}\,</math>
And sets <math>t^E = 1\,</math>.
Note that both sides are negative as <math>\lambda > c\,</math>, so the intermediary pays the customers.
The the minimal surplus for using I as a second source is:
:<math>z^I \equiv \min_h \{\frac{ \lambda(1-\lambda)u_h + \lambda^2 u_h \t^I - r_h^I}{\lambda^2 u_h} \}\,</math>
:<math>\pi^F > \pi^SS \,</math> iff <math>t^E > \frac{\lambda(1-\lambda)u_2}{\lambda(u_1 + \lambda u_2)}\,</math>.
#<math>t^I\,</math>
#<math>t^I-z^I\,</math>
#<math>\frac{\lambda(1-\lambda)u_2}{\lambda(u_1 + \lambda u_2)}\,</math>
