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Gilligan Krehbiel (1987) - Collective Decision Making And Standing Committees (view source)
Revision as of 19:05, 30 September 2011
, 19:05, 30 September 2011→Open rule, specialization
<b>Proof</b>:
* <b><math>p^{\ast}(b)</math></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>\omega\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.
<b>Proposition 3</b>: Compare the expected utilities for the committee in proposition 1 (no specialization) and proposition 2 (specialization) -- both with open amendment rules. With open rules, the committee will specialize iff <math>k<\tilde{k}^{u}</math> where <math>\tilde{k}^{u}=\frac{N^{2}-1}{N^{2}}\sigma^{2}_{\omega}-\frac{N^{2}-1}{3}x_{c}^{2}</math>.
===Closed rule, no specialization===
