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Baron, D. (1991), Bargaining Majoritarian Incentives, Pork Barrel Programs and Procedural Control (view source)
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, 19:14, 29 September 2020no edit summary
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|Has page=Baron, D. (1991), Bargaining Majoritarian Incentives, Pork Barrel Programs and Procedural Control
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|Has article title=Bargaining Majoritarian Incentives, Pork Barrel Programs and Procedural Control
|Has author=Baron, D.
|Has year=1991
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[http://www.edegan.com/wiki/images/3/3f/Baron_%281991%29_-_Bargaining_Majoritarian_Incentives_Pork_Barrel_Programs_and_Procedural_Control.pdf Full-text PDF]
Note similarity to Baron and Ferejohn (1989):
* Multi-lateral,
* Proposals are fully characterized by <math>b\in B</math> and net benefits are <math>z_{i}=b_{i}-T/n</math>.
* Payoffs are discounted: <math>\delta^{\tau}z_{i}=U_{i}(z,\tau)</math>. Extensive form is the same as before for closed rule.
Structure of game:
* P is drawn (which implies a ratio of B/T).
* A random legislator is chosen to distribute B. Note that per the above, all T are distributed equally no matter what.
* Legislators vote against the status quo, in which everyone gets nothing and is taxed nothing.
Stationarity implies members are paid their continuation value in equilibrium in exchange for their votes. <math>\delta v(g,t), \forall t\in\Tau</math>
Derivation of proposition 1:
* <math>z_{i}>\delta\bar{V}</math>. <math> b_{i}-T/n\geq\bar{V} \implies b_{i}\geq T/n+\delta\bar{V}</math>.
* Proposal will be accepted if <math>(n-1)/2</math> members vote yes, therefore proposals will be of the form of: Keep <math>B-\frac{n-1}{2}(\frac{T}{n+\delta\bar{V}}</math>. Give <math>T/n +\delta\bar{V}</math > to <math>(n-1)/2</math> others, and the rest zero. * <math>\bar{V}</math>=P(selected)E[Value of being selected|p^{\ast})+P(not selected)(value of not being selected)</math>.
* <math>\bar{V}=\frac{1}{n}(B-\frac{n-1}{2}(T/n+\delta\bar{V}))+\frac{n-1}{n}(\frac{1}{2}(T/n+\delta\bar{V}) +\frac{1}{2}(-T/n))</math>. Solve for <math>\bar{V}=\frac{B-T}{n}</math>.
* Offer is <math>T/n+\frac{\delta(B-T)}{n}=\frac{\delta B-(1-\delta)T}{n}</math>.
