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1 Auctioning Keywords in Online Search Jianqing Chen The Uniersity of Calgary De Liu Uniersity of Kentucky Andrew B. Whinston Uniersity of Texas at Austin Web Appendix Deriation of the Expected Reenue. Denote V ( as the equilibrium payoff of an adertiser of type, and P ( as the equilibrium probability for him to win the th share. So (W V ( = Q(s P ( E [equilibrium payment of type ] First notice that (W2 V ( V (ṽ + ( ṽ Q(s P (ṽ, where the righthand side is the payoff of a type bidder if he follows instead an equilibrium strategy of a typeṽ bidder. Therefore we hae (W3 (W4 V ( V ( + d + ( d Q(s P ( + d V ( + d V ( + d Q(s P ( Reorganizing the aboe two equations yields (W5 Q(s P ( + d V ( + d V ( d Q(s P (
2 Taking the limit of d to zero, we obtain dv ( d = n Q(s P (. Moing d to the right hand side, integrating both sides from to, and assuming V ( = 0 (the lowest type gets zero payoff, we get (W6 V ( = Q(s P (xdx, for [, ]. Notice the expected payment from an adertiser of type is n Q(s P ( V ( by (W. So the expected payment from one bidder is E [ (W7 ] Q(s P ( V ( = = = [ [ Q(s P ( Q(s Q(s P (f( ( F ( ] Q(s P (tdt f(d ] Q(s P ( ( P ( F ( f(d f( The total expected reenue from all adertisers is n times the aboe. With strictly increasing bidding functions, the equilibrium probability for an adertiser of type to win the th share is the likelihood that of his competitors hae higher aluation than his and the rest of the competitors hae lower aluation, and thus P ( can be specified as in (7. d Proof of Lemma. (a For =, 2,..., n, (W8 α = n = n = n P ( [f ( ( F (] d = n P ( d [ ( F (] ( F ( dp ( ( n n F (n ( F ( [(n (n F (] f ( d where the third step is due to integration by parts. We can easily erify α > 0. 2
3 For = 2, 3,..., n, (W9 α α = n {(n F ( n 2 ( F ( ( n n F (n ( F ( [(n (n F (]}f ( d Denoting A ( (n F ( ( n n ( F ( 2 [(n (n F (], we can rewrite (W9 as α α = n [ F (] F (n A ( f ( d. We argue that A( singlecrosses zero from below on [, ]. To see, let 0 be the solution to (n (n F ( = 0. We can erify that A ( < 0, A ( increases in for 0, and A ( is positie for all > 0. Thus A ( crosses zero only once from below, implying [ F (] F ( n A ( f ( also singlecrosses zero from below on (,. Denoting the crossing point of the latter as c, we hae (W0 α α = n ( F ( F ( n A ( f ( d > n c ( F ( F ( n A ( f ( d = n c [P 2 ( P + ( + ( P (] f ( d where the last equality results from substituting the definition of A ( and rearranging terms. The right side of (W0 is zero because for =,..., n, (W P ( f ( d = ( n n F ( n [ F (] df ( = ( n n x n ( x dx 0 = ( n n (n n n = n where the second step is due to integration by substitution and the third step is due to repeated integration by parts. Therefore, α α > 0 for = 2, 3,..., n. 3
4 We next show that α α n > 0. { α α n = n F ( n [ F (] n } d [ ( F (] = n + n > n + n = n + n [ F (] (n [ F ( n 2 + ( F ( n 2] f(d [ F (] (n [ F ( n 2 + ( F ( n 2] f(d = 0 ( n + n n where the second step is due to integration by parts and the fourth step is due to (W. (b Denote h (x np (x f (x. By (W, h (x dx =. Thus we can regard h (x as a probability density function. We next show that for =, 2,..., n, h (x firstorder stochastically dominates h + (x. h (x h + (x { = nf (x ( n n F (xn [ F (x] ( n n F (xn [ F (x] } = ( n f (x F (x n [ F (x] [nf (x (n ] Denote c as the solution to nf (x (n = 0. Because h (x < h + (x for any x (, c, h (x dx < h + (x dx for (, c. Because h (x > h + (x for any x ( c,, h (x dx > h + (x dx for ( c,, which implies h (x dx < h + (x dx for ( c, (note that h (x dx = h (x dx. In all, we hae h (x dx < h + (x dx for any (,, implying that h (x firstorder stochastically dominates h + (x. According to the property of firstorder stochastic dominance (e.g., Proposition 6.D. at page 95 of MasColell, Whinston, and Green (995, if J (x is an increasing function of x, h (x J (x dx > h + (x J (x dx. Therefore α > α +. 4
5 Proof of Lemma 2. Assume the optimal share structure is (s, s 2,..., s n. Denote k+ s σ and notice that s k k+ k σ s k+ + 0 because of the sizeorder constraint. (s k +, s k +2,..., s k+ must be the solution to the following maximization problem: (W2 (W3 max k+ α Q (s, subect to: s k +... s k+ and k+ s k s k + and s k+ s k+ + s σ We will work on the related maximization problem without constraint (W3 and check (W3 later. The Lagrangian function then can be written as L = k+ α Q (s + µ ( σ k+ s + k+ γ (s s + where µ and γ are Lagrange multipliers. Hence, the KuhnTucher conditions are (let γ k 0, γ k+ 0 (W4 α Q (s µ + γ γ = 0, for = k +,..., k+ Aeraging (W4 for the first l shares and the remaining shares, respectiely, we hae (W5 ( k +l l α Q (s + γ k +l = ( k+ k+ k l α Q (s γ k +l. = k +l+ By definition, k+ is the maximizer for the aerage return factor starting from k +, so (W6 l k +l α k+ k+ k l α = k +l+ Also note that Q (s is nondecreasing in. Therefore, we hae l k +l α Q (s k+ k+ k l = k +l+ α Q (s. If γ k +l = 0, (W5 can hold only if (W6 holds in equality and s k + =... = s k+. In other words, if any γ = 0 ( k < < k+, we must hae s k + =... = s k+. Otherwise, we hae γ > 0 for all k < < k+, which implies s k + =... = s k+ 5
6 by the KuhnTucker condition. So, regardless, we hae s k + =... = s k+ = k+ k σ, which naturally satisfies constraint (W3. Proof of Lemma 3. It is sufficient to show that if ŝ s then ŝ + s +, for any. If and + are located in the same plateau, ŝ = ŝ + and s = s +, and ŝ + s + holds triially. So we assume and + are located in plateau k and k +, respectiely. If s + = 0, the result holds triially. Suppose s + > 0 (so that ŝ s > s + > 0. By Proposition 2, we hae ˆᾱ k Q (ŝ ˆᾱ k+ Q (ŝ + and ᾱ k Q (s = ᾱ k+ Q (s +. Together with condition (7, we hae (W7 Q (ŝ Q (ŝ + ˆᾱ k+ ˆᾱ k ᾱk+ ᾱ k = Q (s Q (s + which implies Q (ŝ Q (s Q (ŝ + Q (s +. Note that Q (ŝ Q (s by concaity of Q( and ŝ s. So we hae Q (ŝ + Q (s +, which implies ŝ + s +. Lemma W (Ranking of α ( 0 Under the MHR condition, for any marginal type 0 (,, if α ( 0 > 0, α ( 0 > α + ( 0 ; if α ( 0 0, α + ( 0 < 0. Proof. For the case 0 (,, define h (x x 0 h (x. Following steps in the h (tdt 0 proof of Lemma (b, we can similarly show that h i (x x 0 firstorder stochastically dominates h i+ (x x 0. Thus, (W8 h (x x 0 J (x dx > h + (x x 0 J (x dx, for =, 2,..., n. 0 0 (W9 Substituting h (x x 0 with α ( 0 = 0 h (x J (x dx > h (x 0 h (tdt and rearranging, we hae 0 h (t dt h + (x J (x dx = 0 h + (t dt 0 0 h (t dt 0 h + (t dt α +( 0. 6
7 Suppose α ( 0 > 0. If α + ( 0 0, from 0 h (t dt > 0 h + (t dt > 0 (because h (t firstorder stochastically dominates h + (t, we hae α ( 0 > α + ( 0. If α + ( 0 < 0, it is easy to see α ( 0 > α + ( 0. If α ( 0 0, (W9 implies α + ( 0 < 0. References [] MasColell, Andreu, Michael D. Whinston, and Jerry R. Green (995, Microeconomic Theory. Oxford Unersity Press, New York. 7
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