1. Now take any n > n and suppose that ? (x, ? n?1 (x, z)) is known for all (x, z) ? Y (n ? 1);We proceed by induction, starting with n = n. By the normalization (1), ? (x 0 , 0) = 0, where for some z we have (x 0 , z) ? Y (n)
2. the arguments used to show Lemmas 1 and 2 will often imply several forms of overidentification. For example, Lemma 1 implies overidentification of ? (x , ? m (x , z )) ? ? (x, ? m (x, z)) for any m which is both smaller than min {n (x , z ) , n (x, z)} and larger than max {n (x , z ) , n (x, z)}. And while Assumption 7 ensures only that there exist one (? x, ? z) ? Y (n) that is also in Y (n ? 1), when there is more than one such pair the proof of Lemma 2 will provide multiple ways of constructing the same value of a given difference ? (x , ? n (x , z )) ? ? (x , ? n?1 (x , z )). Finally, in practice there may often be more than one value of n * satisfying Assumption 8;The Market for Lemons: Quality Uncertainty and the Market Mechanism,1970