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Gilligan Krehbiel (1987) - Collective Decision Making And Standing Committees (view source)
Revision as of 18:57, 30 September 2011
, 18:57, 30 September 2011→Open rule, specialization
* <b>p^{\ast}(b)</b>: Start by fixing the committee's strategy. Suppose <math>b\in[x_{c}-a_{i+1},x_{c}-a_{i}]</math>. This implies that <math>\omega\in (a_{i},a_{i+1})</math>. The floor's problem is now <math>\max_{p}-\int_{a_{i}}^{a_{i+1}}(p+\omega)^{2}f(\omega)d(\omega) \implies p=\frac{-(a_{i+1}+a_{i})}{2}</math>. SOC: <math>-2<0 \implies p^{\ast}(b)=\frac{-(a_{i+1}+a_{i})}{2}</math>.
* <b>b^{\ast}(\omega)</b>: Note that at the cutpoint, the committee is indifferent between which choice the committee makes. <math>[-\frac{a_{i}+a_{1}}{2}]^{2}=[-\frac{a_{i-1}+a_{i}}{2}+a_{i}-x_{c}]^{2}</math>. With <math>a_{i+1}=2a_{i}-a_{i-1}-4x_{c} \implies</math> definitions of <math>a_{i}</math> above.
* Beliefs: Straight forward since when <math>\omgeaomega\in[a_{i},a_{i-1}]\implies b^{\ast}(\omega)\in[x_{c}-a_{i+1},x_{c}-a_{i}]\implies g^{\ast}(b)=\{\omega|\omega\in[a_{i},a_{i+1}]\}</math>. Which is consistant because ...
* See paper for proofs of expected utilities.
