167hw7 - Game Theory Steven Heilman Please provide complete...

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Game Theory Steven Heilman Please provide complete and well-written solutions to the following exercises. Due March 1st, in the discussion section. Homework 7 Exercise 1. Let n be a positive integer. Let v : 2 { 1 ,...,n } → { 0 , 1 } be a characteristic function that only takes values 0 and 1. Assume also that v is monotonic. That is, if S, T ⊆ { 1 , . . . , n } with S T , then v ( S ) v ( T ). The Shapley-Shubik power index of each player is defined to be their Shapley value. By monotonicity of v , we have v ( S ∪ { i } ) v ( S ) for all S ⊆ { 1 , . . . , n } and for all i { 1 , . . . , n } . Also, since v only takes values 0 and 1, we have v ( S ∪ { i } ) - v ( S ) = ( 1 when v ( S ∪ { i } ) > v ( S ) 0 when v ( S ∪ { i } ) = v ( S ) . Consequently, we have the following simplified formula for the Shapley-Shubik power index of player i ∈ { 1 , . . . , n } : φ i ( v ) = X S ⊆{ 1 ,...,n } : v ( S ∪{ i } )=1 and v ( S )=0 | S | !( n - | S | - 1)! n ! . Compute the Shapley-Shubik power indices for all players on the UN security council, with pre-1965 and post-1965 structure. Which structure is better for nonpermanent members? In pre-1965 rules, the UN security council had five permanent members, and six nonperma- nent members. A resolution passes only if all five permanent members want it to pass, and at least two nonpermanent members want it to pass. So, we can model this voting method, by letting { 1 , 2 , . . . , 11 } denote the council, and letting { 1 , 2 , 3 , 4 , 5 } denote the permanent members. Then we use the characteristic function v : 2 { 1 ,..., 11 } → { 0 , 1 } so that, for any S ⊆ { 1 , . . . , 11 } , v ( S ) = 1 if { 1 , 2 , 3 , 4 , 5 } ⊆ S and if | S | ≥ 7. And v ( S ) = 0 otherwise. This voting method was called unfair, so it was restructured in 1965. After the restructuring, the council had the following form (which is still used today). The UN security council has five permanent members, and now ten nonpermanent members. A resolution passes only if
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