Question 1 Affiliated Private Values

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1 Econ 85 Fall 7 Problem Set Solutions Professor: Dan Quint Question Affiliated Private alues a - proving the ranking lemma Suppose g(y) < h(y) for some y > x. Let z = max {x : x < y, g(x) h(x)} (Since g and h are continuous, the set of points where g(x) h(x) is closed. Since it excludes y, the intersection of it with [x, y) is still closed. Since the set includes x, it is nonempty, and therefore the maximum exists.) By continuity, g(z) = h(z), and by definition, g(x) < h(x) for all x (z, y]. But then by assumption, g (x) h (x) for all x [z, y], so g(y) g(z) h(y) h(z), or g(y) h(y), a contradiction. b - value function in the first-price auction If bidder j bids according to β F, and bidder i, given type t i, bids β F (t), he wins whenever β F (t) > β F (t j ), which is when t j < t, which occurs with probability F j (t t i ). The object is worth t i to him, and he pays β F (t) when he wins. (If I hadn t told you to assume β F was an equilibrium, you d still have to check separately for deviations outside the range of β F, but by the usual arguments, they wouldn t be profitable.) By the envelope theorem, then, ( ) F P [ i = Fj (t t i ) ( t i β F (t) )] t i since β F (t i ) is i s equilibrium strategy at t i. Calculating the partial derivative gives t=ti [ Fj (t t i ) ( t i β F (t) )] = F j (t t i ) + Fj (t t i ) ( t i β F (t) ) t i where Fj is the partial derivative of F j with respect to its second argument; plugging in t = t i gives ( ) F P i = Fj (t i t i ) + Fj (t i t i ) ( t i β F (t i ) ) c - bid function in the second-price auction In the second-price auction, if I bid t, I win when b j (t j ) < t, in which case I pay b j (t j ). In the dominant-strategy equilibrium, b j (t j ) = t j ; and since types are correlated, its distribution is conditioned on my true type. So if I have a true type s, bid t, and win, my expected payment is E (t j t i = s, t j < t)

2 d - value function in the second-price auction In the second price auction, if I have a true type t i and bid t, I win whenever t j < t, which happens with probability F j (t t i ). When I win, the object is worth t i to me, and by part c above, I pay on average β S (t i, t). My value function is then the solution to the maximization problem over my possible bids. [ Fj (t t i ) ( t i β S (t i, t) )] = F j (t t i ) F j (t t i )β S (t i, t) + Fj (t t i ) ( t i β S (t i, t) ) t i where β S is the partial of βs with respect to its first argument; plugging in t = t i, the envelope theorem gives us ( ) SP i (ti ) = F j (t i t i ) F j (t i t i )β S (t i, t i ) + Fj (t i t i ) ( t i β S (t i, t i ) ) e - β S increasing in t i This is really the crux of the proof. One of the affiliation results we mentioned in class (without proof) was that if (x,..., x n ) are affiliated random variables and g is any increasing function, E (g(x,..., x n ) x [a, b ], x [a, b ],..., x n [a n, b n ]) is increasing in a, b,..., a n, b n. (This is Theorem 5 in Milgrom and Weber (98); it s proved in the appendix, but using some abstract terms and lattice theory.) Direct application of that result would imply that E (t j t j [, t], t i [s, s]) is increasing in s, since g(t i, t j ) = t j is nondecreasing. To see a clean proof for this case, though, rewrite this expectation as β S (s, t) = wf j(w s)dw F j (t s) Integrating the numerator by parts, with u = w and dv = f j (w s)dw, we find that so wf j (w s)dw = wf j (w s) w=t w= F j (w s)dw = tf j (t s) β S F j (w s) (s, t) = t F j (t s) dw F j (w s)dw Recall one of our earlier affiliation lemmas (one that we did prove in class see the notes from lecture ): if x and x are affiliated, then F (x x ) is log-supermodular. This means that for x > x and x > x, F (x x )F (x x ) F (x x )F (x x )

3 or F (x x ) F (x x ) F (x x ) F (x x ) that is, for x > x, F (x x ) F (x x ) is decreasing in x. Since t j and t i are affiliated and w t everywhere within the integral, F j(w s) F j (t s) decreasing in s; so is increasing in s as claimed. is decreasing in t i, so its integral, β S F j (w s) (s, t) = t F j (t s) dw F j (w s) F j (t s) dw, is f - value functions and bids Knowing that t = t i solves the maximization problem posed in both value functions, and and so SP i F P i (t i ) = F j (t i t i ) ( t i β F (t i ) ) SP i (t i ) = F j (t i t i ) ( t i β S (t i, t i ) ) (t i ) i F P (t i ) if and only if β S (t i, t i ) β F (t i ) as claimed. g - applying the ranking lemma By symmetry, the lowest types never win, so i SP () = = i F P (). To apply the ranking lemma, we need to show that for t i, i F P (t i ) SP From parts b and d above, ( F P i i (t i ) ( i F P ) (ti ) ( i SP ) (ti ) ) (ti ) ( i SP ) (ti ) = F j (t i t i ) + Fj (t i t i ) ( t i β F (t i ) ) F j (t i t i ) + F j (t i t i )β S(t i, t i ) Fj (t i t i ) ( t i β S (t i, t i ) ) = F j (t i t i )β S (t i, t i ) + F j (t i t i ) ( β S (t i, t i ) β F (t i ) ) We showed in part e that β S is positive, so the first term is always positive. F j is negative, since one of our lemmas in class was that under affiliation, F (x x ) is decreasing in x (the distribution of x increases in a first-order stochastic dominance sense when x rises, so F (x ) decreases). We showed in part f that i F P (t i ) i SP (t i ) implies β F (t i ) β S (t i, t i ), and so both pieces of the second term are negative; so the whole expression is positive. 3

4 h - revenue dominance By part f, knowing i F P (t i ) i SP (t i ) everywhere implies that β F (t i ) β S (t i, t i ) everywhere. Since both auctions are efficient, the winner is the same in either auction at any signal pair, and pays weakly more in expectation in the second-price auction, so the second-price auction earns weakly higher expected revenue. Question Extra Credit I wasn t kidding when I said I didn t know the answer. When t i and t j are affiliated, the expression for ( i F P ) (ti ) ( i SP ) (ti ) is the same as before, but now β S is negative and Fj is positive. But that means that ( β S (t i, t i ) β F (t i ) ) is related in the wrong direction to ( i F P ) (ti ) ( i SP ) (ti ), so we can t apply the ranking lemma. I m curious to see what you guys come up with. 4

5 Question 3 Common alues and Information Aggregation 3a Conditional on =, the joint density of (X i, Y ) at (x, x) is 3b Conditional on =, the joint density is 3c ( ) ( ) ( + x (n ) + x x + ) n ( ) ( ) ( x (n ) x x ) n Note that E( whatever) = Pr( = whatever). Using Bayes Law, letting Z denote the event that X i = Y = x, and abusing notation a little bit by allowing Pr(Z) to refer to the probability density at that event. Pr( = Z) = Pr( =, Z) Pr(Z) = Pr( = ) Pr(Z = ) Pr( = ) Pr(Z = ) + Pr( = ) Pr(Z = ) Since Pr( = ) = Pr( = ) =, this is exactly 3d A A + B = A A + B b i (x) = E( X i = Y = x) is exactly the usual equilibrium we ve been looking at in common-value second-price auctions. (Even if you learned Y and b j (Y ), you don t expect to gain by changing your bid.) The simplest way to disprove information aggregation is to show that when =, the winning bid is bounded above by b i (), which is bounded away from. To see this, note that at x =, and and so A = (n ) B = (n ) b i () = ( ) 3 ( ) A A + B = 9 and does not change as n. So even as n grows arbitrarily large, b i () =.9, and so the winning bid cannot converge to in probability when =. 5

6 3e To agree with the notation in Pesendorfer and Swinkels, we ll first do this for n bidders and k objects, then plug in n = r + and k = r +. 3f When =, the joint density of (X i, Y k ) at (x, x) is E = ( ) ( ) ( + x (n ) + x n k = (r + ) ( r r ) ( x ) k ( x + ) n k ) ( ) ( + x x ) r ( x + ) r x Similarly, when =, the joint density of (X i, Y k ) at (x, x) is 3g F = ( ) ( ) ( 3 3 x (n ) x n k = (r + ) ( r By the same Bayes -Law logic as before, r ) ( 3 x + ) k ( 3 x ) n k ) (3 ) ( x 3 x + ) r ( 3 x ) r x E( X i = Y k = x) = E E + F Note the (r + ) r C r appears in both E and F, and therefore drops out of both the numerator and denominator. Similarly, to get rid of all the one-halfs everywhere, multiply numerator and denominator by r. This leaves us with b r i (x) = ( + x) [( x x ) ( x + x )] r ( + x) [( x x ) (x + x )] r + ( 3 x) [( 3x + x ) (3x x )] r 3h We can rewrite the last expression as b r i (x) = + Q(r, x) where Q(r, x) = ( 3 x ) [( 3x + x )(3x x ] r ) + x ( x x )(x + x ) = ( 3 x ) [( ] x)( x)(3 x)x r + x = ( + x)( x)( + x)x 6 ( 3 x + x ) [( ] x)(3 x) r ( + x)( + x)

7 ( ) Note that the first term in this expression, 3/ x, /+x is bounded below by 9 by 9; specifically, it never approaches either or. If ( x)(3 x) > ( + x)( + x) and above then the term in square brackets is greater than, and so Q(r, x) grows unboundedly as r, and so b i (r). Conversely, if ( x)(3 x) < ( + x)( + x) then the fraction in square brackets is less than, so Q(r, x) goes to as r, and so b i (r). Let s look at the latter case. Note that ( x)(3 x) < ( + x)( + x) 6 5x + x < + 3x + x 4 8x < < x So for x >, Q and so br i (x) as r. On the other hand, for x <, Q and b r i (x). Now, consider what happens when r. When =, the distribution of the r + nd -highest type (the one whose bid sets the price) becomes a very narrow distribution around the median of the conditional distribution f(x = ), which is around.68; so the probability that it lies within, say, the interval [.55,.65] goes to. But since b r i (x) is increasing in x and b r i (.55), when =, the price converges in probability to. By the same argument, when =, the r + nd -highest type converges in probability to the median of f(x = ), which is around.38; since b r i goes to around there, the price converges in probability to when =. So information aggregates. 7