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(1) For all agents i1, i2, , im N, Ki1Ki2 Kim(A)Hence, K*N(A) iff (1) is the case for each m 1.
Proof.
Note first that
( 2 ) |
i 1 N |
K i 1 ( |
i 2 N |
K i 2 ( . . . ( |
i m - 1 N |
K i m - 1 ( |
i m N |
K i m( A ) ) ) ) ) |
= |
i 1 N |
K i 1 ( |
i 2 N |
K i 2 ( . . . ( |
i m - 1 N |
K i m - 1 (K 1N ( A ) ) ) ) ) |
= |
i 1 N |
K i 1 ( |
i 2 N |
K i 2 . . . ( |
i m - 2 N |
K i m - 2 ( K 2N ( A ) ) ) ) |
= . . . |
= |
i 1 N |
K i 1( K m - 1N ( A ) ) |
= | K mN ( A ) |
By ( 2 ), K mN ( A ) K i 1K i 2 . . . K i m( A ) for i 1, i 2, . . ., i m N so if K mN ( A ) then condition ( 1 ) is satisfied. Condition ( 1 ) is equivalent to
i 1 NK i 1 (
i 2 NK i 2 ( . . . (
i m - 1 NK i m - 1 (
i m NK i m ( A ) ) ) ) )
so by ( 2 ), if ( 1 ) is satisfied then K mN ( A ).
First published: August 27, 2001
Content last modified: August 27, 2001