Econ 805 Advanced Micro Theory I Dan Quint Fall 2009 Lecture 8 Last week we laid a bunch of technical groundwork on affiliation In particular we defined affiliation for variables with a density equivalent to log f being supermodular which is equivalent to log f having pairwise increasing differences showed affiliation is robust to order preserving transformations if g1 gn are strictly increasing functions from to then x1 xn are affiliated if and only if g1 x1 g2 x2 gn xn are affiliated and a bunch of lemmas leading up to the punchline if x1 and x2 are affiliated then F x1 x2 is nonincreasing in x2 that is the distribution of x1 conditional on a high realization of x2 first order stochastically dominates the distribution of x1 conditional on a lower realization of x2 Next we show that if a set of random variables are affiliated then any subset of them are affiliated as well Lemma 1 If f x1 xn is a log supermodular probability density then Z g x1 xn 1 f x1 xn 1 s ds is a log supermodular probability density as well We skipped the proof in lecture but here it is for your viewing pleasure We ll show it for the case where f is twice differentiable so that log supermodularity is the same as positive mixed partials The result holds more generally log g is increasing in xj Taking the derivative g is log supermodular if x i log g x1 xn 1 xi R R R f x1 xn 1 s ds f x1 xn 1 s ds R fi x1 xn 1 s ds R f x1 xn 1 s ds fi x1 xn 1 s ds f x1 xn 1 s ds f x1 xn 1 s ds R E xi f x1 xn 1 s ds fi x1 xn 1 s x1 xn 1 f x1 xn 1 s where the expectation is taken over s conditional on the first n 1 variables We need to show this is increasing in xj 1 An increase in xj has two effects First since x1 xn 1 s are affiliated an increase in xj increases the ratio ffi directly When we fix the values of all variables but xj and xn it s clear that these remaining two are affiliated since log f has positive mixed partials when other variables are held fixed That means that an increase in xj leads to an increase in the conditional distribution of s in the firstorder stochastic dominance sense by the last result But by affiliation ffi is increasing in xn and therefore its expectation over xn is increasing in xj So an increase in xj increases the whole expression which implies that g is log supermodular The next result is really the punchline of affiliation that if a bunch of variables are affiliated the expected value of any increasing function of all of them is increasing in the realization of any of them That is Lemma 2 Suppose x1 xn are affiliated For any function g n which is bounded and isotone increasing in all its arguments h x E g x1 xn x1 x is increasing in x The proof is by induction on n and iterated expectations First suppose n 2 Then for x x0 h x E g x x2 x1 x E g x0 x2 x1 x E g x0 x2 x1 x0 h x0 The first inequality is because x x0 and g is increasing in both its arguments The second is because the distribution of x2 conditional on x1 x first order stochastically dominates the distribution of x2 conditional on x1 x0 and we just showed the expected value of any increasing function is higher over a stochastically dominant distribution Now suppose the lemma is known to hold for functions of n 1 affiliated variables Define j x y E g x1 x2 xn x1 x x2 y and rewrite h x as h x Ex2 x1 x j x x2 By the inductive assumption for a fixed x j x y is increasing in y and the distribution of x2 conditional on x1 x is increasing in x by FOSD So by the same logic for x x0 h x E j x x2 x1 x E j x0 x2 x1 x E j x0 x2 x1 x0 h x0 2 The Milgrom and Weber paper actually give a different formulation of this result if x1 xn are affiliated and g is any increasing function then the expectation E g x1 xn x1 a1 b1 x2 a2 b2 xn an bn is increasing in all 2n of its arguments ai and bi The last result we need is that a bunch of symmetric random variables are affiliated if and only if their order statistics are affiliated Symmetry has bite here if the variables are affiliated but ex ante asymmetric the result does not hold Lemma 3 Suppose x1 xn y have a joint density f which is symmetric in the first n arguments Then x1 xn y are affiliated if and only if x 1 x 2 x n y are affiliated where x i is the ith highest of x1 xn The density of the n 1 random variables x 1 x 2 x n y is f x1 x2 xn y n f x1 x2 xn y 1x1 x2 xn P This is because the density is 0 unless x1 x2 xn and f x y when this does hold where the sum is taken over the permutations of the elements of x but there are n of these permutations and by assumption f takes the same value at all of them We need to show that f satisfies the affiliation inequality everywhere if and only if f does We ll show the reverse First suppose f violates the affiliation inequality somewhere that is suppose there exist x y and x0 y 0 such that f x y x0 y 0 f x y x0 y 0 f x y f x0 y 0 There are two possible cases either one of the indicator functions 1 is zero on the left hand side or not Since the right hand side is strictly positive the indicator functions must both be 1 on the right hand side If not then f n f at all four points where it is being evaluated and so the affiliation inequality is also violated by f If so this means that although x1 x2 xn and x01 x02 x0n either the meet or the join are ordered incorrectly It is easy to show that this cannot occur so if f violates the affiliation inequality so does f On the other hand suppose f violates the affiliation inequality somewhere Then there is some point where f xi x0j x i j f xi xj x i j f x0i xj x i j f x0i x0j x i j By continuity we can find such points like this such that xj and x0j are very close together and similarly we can find such points where xi and x0i are very …