Incentivizing High-quality Content from Heterogeneous Users

Transcription

1 Inentvzng Hgh-qualty Content from Heterogeneous Users Inentvzng Hgh-qualty Content from Heterogeneous Users Ynge Xa Unversty of Sene and Tehnology of Chna, Hefe, Chna, Tao Qn Mrosoft esearh, Bejng, Chna, Nengha Yu Unversty of Sene and Tehnology of Chna, Hefe, Chna, Te-Yan Lu Mrosoft esearh, Bejng, Chna, Abstrat Inentve mehansms are very mportant for many Internet serves (e.g., onlne revews and queston-answerng webstes to motvate users to generate hgh-qualty ontent. In ths work, we study the exstene of pure Nash equlbrum (PNE for these mehansms. Most exstng work assumes that users are homogeneous and have the same ablty. However, real-world users are heterogeneous and ther abltes an be very dfferent from eah other due to ther dversty n bakground, ulture, and professon. In ths work, we onsder the followng settng: (1 the users are heterogeneous and eah of them has a prvate type ndatng the best qualty of the ontent he/she an generate; (2 all the users share a fxed total reward. Wth ths settng, we study the exstene of pure Nash equlbrum of several mehansms omposed by dfferent alloaton rules, aton spaes, and nformaton avalablty. We prove the exstene of PNE for some mehansms and the non-exstene for some other mehansms. We also dsuss how to fnd a PNE (f exsts through ether a onstrutve way or a searh algorthm. 1. Introduton More and more Internet webstes rely on users ontrbutons to ollet hgh-qualty ontent, nludng knowledge-sharng serves (e.g., Yahoo! Answers and Quora, onlne produt ommentng and ratng serves (e.g., Yelp, moble app stores, and e-ommere webstes (e.g., Amazon.om. For smplty, we all webstes that rely on User-Generated Content UGC webstes. To attrat more users and nentvze them to ontrbute hgh-qualty ontent, those stes usually gve hgh-qualty ontrbutors some rewards, n terms of ether vrtual value or monetary return. To ollet more rewards, some users mght strategally nterat wth those webstes. Therefore, to maxmze the qualty of the ontent generated from users, a UGC webste needs to arefully desgn ts mehansm and analyze users behavors. We all the mehansms used by those UGC stes UGC mehansms. eently, a lot of effort has been plaed to the desgn and analyss of UGC mehansms [9, 11, 5]. For example, [11] desgns a smple votng rule under a sequental and smultaneous model, n whh both the qualty of ontrbutons and the number of ontrbutors are endogenously determned. [9] studes the rank-based alloaton mehansm and shows that the mehansm always nentvzes hgher qualty equlbra than the proportonal alloaton rule. [10] models the onlne eduaton forums wth two parameters whh represent the frequeny of hekng forums by teahers and students separately. A bref survey about UGC mehansms an be found n [8]. Most of those works assume that users are homogeneous,.e., they are of the same ablty whle ontrbutng to the stes. However, n the real world, users abltes an be very dfferent from eah other due to ther dversty n bakground, ulture, and professon. For example, an experened photographer an wrte a hgh-qualty omment to a photo, whh s very dffult for 1

2 Y. Xa, T. Qn, N. Yu, T.-Y. Lu a non-experened user. Thus, n ths work, we study the game-theoret problem rased n Internet serves wth heterogeneous users. We ntrodue the onept of type for the problem, whh denotes the ablty of a user: the larger the type of a user s, the better ontent he/she an ontrbute to the ste. We further assume that eah user needs to afford a ost to partpate n the game and ontrbute ontent. The ost reflets the effort of ontent generaton, e.g., tme spent on wrtng a revew and the payment for moble network usage. In our work, we assume the osts are bounded. We beleve ths assumpton s more reasonable and pratal than the unbounded-ost assumpton used n [5, 11]. We study two alloaton rules: the top K alloaton rule [12, 5], n whh users wth the hghest K qualtes wll share the reward equally, and the proportonal alloaton rule [11, 12, 16, 4], n whh all the partpants (who make non-zero ontrbuton wll share the reward proportonally to ther ontrbuted qualtes. 1 We onsder two aton spaes: the bnary aton spae, n whh eah user an only hoose to partpate n or not, and the ontnuous aton spae, n whh eah user an hoose the qualty of hs/her ontrbuted ontent. Besdes, we study the problem n both the full-nformaton settng, n whh eah user s type s publly known, and the partal-nformaton settng, n whh eah user only knows hs/her own type and the other player s type follow some dstrbuton F. 1.1 Model In ths subseton, we desrbe the model for analyzng the nentves reated by varous UGC mehansms formally, when ontrbutors are strateg agents wth heterogeneous abltes. There s a set of N strateg users n a UGC webste, and user has a prvate type q [0, 1], whh ndates the best qualty of the ontent he/she an ontrbute to the ste. Wthout loss of generalty, we number the users aordng to the desendng order of ther types,.e., q 1 q 2... q N. Let x (0 x q denote user s aton, whh ndates the qualty of the ontent he/she atually ontrbutes to the ste, and as the orrespondng ost he/she needs to afford. x denotes the aton of all the users,.e., x = (x 1,, x N. x denotes the aton of all the users expeted user ( [N]. In ths work, we onsder the lnear ost for smplty: = x q, where s a onstant denotng the upper bound of the ost. 2 We study two aton spaes. The frst s a bnary aton spae: eah user an only hoose ether not to ontrbute or to ontrbute some ontent wth qualty q (.e., x {0, q }. The seond s a ontnuous aton spae: the qualty x that user ontrbutes to the ste s a real value between 0 and q (.e., x [0, q ]. We say that a user does not partpate n the game f x = 0, and a user partpates n the game f x > 0. The ste has a fxed total reward to be alloated to the ontrbutors, dependng on ther ontrbutons. We study two alloaton rules: the top-k alloaton [12] and the proportonal alloaton [11]. The former alloates K to eah of the users who ontrbute the K largest qualtes (tes break n ommon. Note that f N < K, eah user an stll get K reward only. The latter alloates the x reward to all users proportonally to ther ontrbutons: the reward r alloated to user s j xj f x > 0 and 0 otherwse. Whle analyzng the model, we onsder two settngs: the full-nformaton settng and the partalnformaton settng. In the full-nformaton settng, the types {q } [N] are determnst and known to all the users. In the partal-nformaton settng, the type of eah user s assumed to be drawn from a publly known ontnuous dstrbuton F wth a frst order dervatve. Eah user only knows 1. See [13, 14, 3, 19, 21, 1] for more detals about the proportonal alloaton rule. 2. If eah user has a personalzed ost upper bound C, we an absorb C nto the prvate type q by means of salng: q = q C. 2

3 Inentvzng Hgh-qualty Content from Heterogeneous Users hs/her own type q. 3 In the partal-nformaton settng, we fous on the symmetr strategy and denote t as β(, whose nput s the user s type. We wll study the exstene of pure Nash equlbrum (PNE n the full-nformaton settng and symmetr pure Bayesan Nash equlbrum 4 n the partal-nformaton settng. Wth the above notatons, 1. under the full-nformaton settng, the utlty of user an be wrtten as u (x, x = r (x, x (x, where r (x, x s the reward of user gven hs/her own strategy x and the strateges x of other players. We assume that all the users are ratonal and target at the maxmzaton of ther utltes. We use {x }N =1 to denote an the equlbrum profle. 2. under the partal-nformaton settng, f user wth type q hooses to ontrbute ontent wth qualty θ and the others follow the symmetr strategy β(, the expeted utlty of user s: E{r (θ, β } (θ, where r (θ, β s the reward of user gven the other players strateges β and the expetaton s taken w.r.t. to the dstrbuton F. Eah user wll target at the maxmzaton of hs/her expeted utlty. We use β ( to represent a symmetr pure Bayesan equlbrum strategy. 1.2 Our Man esults We study the exstene of pure Nash equlbrum for several UGC mehansms (wth dfferent alloaton rules, aton spaes and nformaton avalablty. Our man results an be summarzed as follows. For the full-nformaton settng, 1. We prove that the mehansm wth the proportonal alloaton rule and the ontnuous aton spae has a unque PNE. The key of the proof s to onstrut a perturbed game, prove the exstene of PNE for the perturbed game, and then prove that the PNE of the perturbed game wll onverge to the equlbrum of the orgnal game. The unqueness s proven by ontradton. We then desgn an effent algorthm to fnd the PNE for the mehansm. 2. We prove that the PNE exsts for the mehansm wth the top K alloaton rule and the bnary aton spae, as well as the mehansm wth the proportonal alloaton rule and the bnary aton spae. Note there mght be multple equlbra wth dfferent parameters. 3. However, the PNE does not exst for the mehansm wth the top K alloaton rule and the ontnuous aton spae. For the partal-nformaton settng, 1. We onstrutvely prove the exstene of a symmetr PNE for the mehansm wth the top K alloaton rule and the ontnuous aton spae. The key of the proof s to onstrut a smple but (maybe nfeasble symmetr strategy and then onvert t to a feasble symmetr equlbrum strategy by repeated albraton. Our proof also provdes a method to onstrut a symmetr PNE. 2. We prove the exstene of a symmetr PNE for the mehansm wth the proportonal alloaton and ontnuous aton spae when the type spae s dsrete. 3. It s a knd of Bayesan game (see [7, 17, 20] for more about Bayesan games. What s more, our settng s the nterm stage n a Bayesan game (see the Appendx F of [15],.e., eah partpant knows hs/her own type, but does not know the others types. 4. For the purpose of notaton onssteny, n the partal-nfomaton settng, we stll use PNE to represent the pure Bayesan Nash equlbrum. 3

4 Y. Xa, T. Qn, N. Yu, T.-Y. Lu 3. For the bnary aton spae, we prove the exstene of the ut-off equlbrum strateges for the mehansms wth both the top K alloaton rule and the proportonal alloaton rule. We also prove the unqueness of the symmetr ut-off equlbrum strategy for the top K alloaton rule under ertan rumstanes. 2. Full-nformaton Settng In ths seton, we study the mehansms n the full-nformaton settng. In ths settng, the type q of any user s known to all the users. Ths settng orresponds to the real-world senaros where the users are famlar wth eah other. For example, onsderng a professonal mathematal queston posted n Yahoo! Answer, there wll be only a few users n the ommunty who an answer the queston and they know eah other qute well. Combnng the dfferent hoes of the alloaton rule and the aton spae, there are four mehansms of our nterest. 2.1 M 1 : Top K Alloaton, Bnary Aton Spae It s not dffult to verfy the PNE for ths sheme. parameters (, K, : There are three ases dependng on the 1. < K: no user wll ontrbute ontent to the ste and the equlbrum s x = 0,. 2. = K: there are many equlbra: any group of k K users ontrbutng to the ste s an equlbrum. 3. > K: the frst K users wll ontrbute to the ste. If there are some users wth the same type, we need to number them aordng to any pre-spefed order. That s, x = q, 1 K and x = 0, > K. 2.2 M 2 : Top K Alloaton, Contnuous Aton Spae There does not exst PNE for ths mehansm under the full-nformaton settng aordng to the followng dsussons. 1. Zero-partpatng and more than K people partpatng n are obvously not an equlbrum. 2. If the equlbrum s onstruted by fewer than K people partpatng n, there s at least one person would ould get postve utlty by makng a postve ontrbuton. Ths ontradts wth the onept of equlbrum. 3. If there are K people partpatng n, all the ontrbutors wll generate ontent wth qualty ε 0 but ε 0, thus the equlbrum strategy does not exst. 2.3 M 3 : Proportonal Alloaton, Bnary Aton Spae It turns out that PNE exsts and there an be multple equlbra for ths mehansm. 1. If <, there exsts a unque equlbrum n whh nobody wll ontrbute: x = 0,. 2. If =, multple equlbra exst: (1 nobody ontrbutng s an equlbrum, and (2 any sngle user ontrbutng s also an equlbrum. 3. If >, Nash equlbrum exsts. Denote j as the ndex satsfyng q j j k=1 q k > and qj+1 j+1 k=1 q k smultaneously. Then x = q, 1 j and x = 0, > j ompose an equlbrum. There an exst multple equlbra. Consder an example wth parameters N = 3, = 4, = 4

5 Inentvzng Hgh-qualty Content from Heterogeneous Users 1, q = (0.9247, , One an verfy that both x = (0.9247, , 0 and x = (0.9247, 0, are equlbra. 2.4 M 4 : Proportonal Alloaton, Contnuous Aton Spae In ths subseton, we frst prove the exstene of the PNE for mehansm M 4, then show that the PNE of M 4 s unque, and fnally present an algorthm to fnd the unque PNE. Throughout ths subseton, we wll frequently use the funton s(, whose output s the sum of eah element of the nput vetor The Exstene of the PNE for the Mehansm M 4 Theorem 1 For the full-nformaton settng, there exsts a PNE for the mehansm M 4. Proof. Consder an aton profle {x } [N]. Denote s(x = j x j. The utlty of user s x u (x, x = x + s(x x, f x > 0; q 0, f x = 0. (1 We use u to represent u (x, x when the ontext s lear. Sne the utlty funton s not ontnuous at x = 0, t s not the n-person onave game defned n [18] and we annot dretly use Theorem 1 n t. To overome ths dffulty, we onsder an ɛ-perturbed game [6], n whh the aton of eah user s lower bounded by a small quantty ɛ > 0. The aton spae of the ɛ-perturbed game (denoted as X ɛ s N =1 [ɛ, q ], n whh the utlty funton, defned n Eqn. (1, s ontnuous. It s obvous that X ɛ s a onvex, losed and bounded set n the Euldan spae. By the Theorem 1 n [18], we know at least one equlbrum strategy, denoted as x ɛ, exsts for the ɛ-perturbed game. Sne X ɛ s ompat, we an always fnd a seres of {ɛ n } n=1 0 whose orrespondng x ɛn onverges. Denote the lmt pont as x 0. Next we use three steps to prove that x 0 s a PNE of the orgnal game. Before gettng nto the three steps, we frst derve the best-response strategy of user gven x. Gven x > 0, the frst order dervatve of u w.r.t x s u x u (x, x x s(x = (x + s(x. (2 2 q = 0 has a soluton n the regon ( s(x, +. Denote the soluton as x (q, x, whh s: qs(x x (q, x = s(x. (3 x (q, x alulated from Eqn. (3 an be smaller than zero (for the orgnal game, or smaller than ɛ for the ɛ-perturbed game or larger than q, whh s not a feasble aton. Thus, x (q, x annot be dretly seen as the best-response strategy of user nether for the orgnal game nor for the ɛ-perturbed game. In the orgnal game, we make the followng analyss. 1. If x (q, x 0, by the monotonty of u x, we know that user annot be better off by makng postve ontrbuton. Then the best-response strategy of user s not to ontrbute ontent. Aordngly, f the equlbrum profle {x }N =1 exsts, gven x = 0, we an get that 0. u (x,x x x=x 2. If x (q, x (0, q, t means that user annot proftably devate from ontrbutng quantty x (q, x, whh s just the best-response strategy of user. Aordngly, f the equlbrum profle {x }N =1 exsts, gven x (0, q, we an get that u(x,x x = 0. x=x 5

6 Y. Xa, T. Qn, N. Yu, T.-Y. Lu 3. If x (q, x q, t means that user annot get more reward by generatng ontent wth qualty less than q,.e. 0, and the best-response strategy of user s q. Aordngly, u x x q f the equlbrum profle {x }N =1 exsts, gven x = q, we an get that u(x,x x x q 0. Smlarly, n an ɛ-perturbed game, [N], gven x ɛ, f the best-response strategy of user s to ontrbute ontent wth qualty ɛ, (.e., x ɛ = ɛ, t means that s(x ɛ (ɛ + s(x ɛ 2 q 0. (4 (Step 1 : In ths step, we show that x 0 s not a zero vetor. We prove ths by ontradton. Otherwse, N =1 xɛn ould be nfntely lose to zero as n, from whh we an dretly get that x ɛn 1 and x ɛn 2 are strtly less than q 1 and q 2 respetvely. As x ɛn 0, we know at least one of the followng two ases would happen: 1. There exst an j [N] and a subsequene 5 {ɛ nk } of {ɛ n } s.t. x ɛn k j = ɛ nk : For user j, we have But as N =1 xɛn k 0, we an also obtan that s(x ɛn k j (s(x ɛn k j + 0. (5 ɛn k 2 q j s(x ɛn k j (s(x ɛn k j + ɛn k 2 q j = s(x ɛn k j s(x ɛn k j + ɛn k N =1 xɛn k q j 2 N =1 xɛn k q j +, (6 whh s a ontradton wth Eqn. (5. 2. If the above ase does not hold, there s a subsequene {ɛ nk } of {ɛ n } s.t. x ɛn k 1 > ɛ nk, x ɛn k ɛ nk : 6 Set N =3 xɛn k = δ nk and Q = q 1 + q 2. By Eqn. (2 we obtan: x ɛn k 2 + δ nk 2 + δ nk = 0 2 q 1 x ɛn k 1 + δ nk 2 + δ nk = 0. 2 q 2 (x ɛn k 1 + x ɛn k (x ɛn k 1 + x ɛn k 2 > From the above two equatons, we an get x ɛn k 1 + x ɛn k 2 = 1 ( 2 Q + ( Q 2 + 4δ nk δnk, (8 Q whh wll not onverge to zero as n k. Ths ontradts wth the assumpton that N =1 xɛn ould be nfntely lose to zero as n. (Step 2 : In ths step, we show that the vetor x 0 has at least two non-zero elements,.e., x If x 0 0 = 1, we an assume only the th element of x 0 s strtly larger than zero ( [N]. That s, there exsts an nteger N 1, n > N 1, we have s(x ɛn 5. We use { } to represent the nfnte sequene {} ndex=1. 6. (x ɛn k 1, x ɛn k 2 are nteror ponts at ths tme. 7. We use x 0 to represent the number of non-zero elements n the vetor x. ( x0 2 + s(x ɛn 2 q 0. (9 (7 6

7 Inentvzng Hgh-qualty Content from Heterogeneous Users Due to the assumpton that only x 0 s larger than zero, we have lm n s(x ɛn = 0, we know lm n s(x ɛn ( x0 2 + s(x ɛn 2 q = q < 0, whh ontradts wth Eqn. (9. Therefore, we an onlude that x 0 0 2, from whh we know that [N], s(x 0 > 0. (Step 3 : We wll prove x 0 s a PNE of the orgnal game by ontradton. We just need to prove that none of the users ould proftably devate from x 0 unlaterally. If not a PNE, there exsts a user would lke to devate (1 from x 0 = 0 to x > 0; (2 from x0 (0, q to x [0, q ]\{x 0 }; (3 from x 0 = q to x < q. 1. As x ɛn 0 when n, ɛ > 0, there exsts a postve nteger N 2 (ɛ, n > N 2 (ɛ, x ɛn < ɛ. By settng ɛ = x 2, we have: s(x ɛn 0. (s(x ɛn + x 2 2 q Thus, we obtan s(x 0 (x + ( s(x 0 = lm s(x0 2 q n (x + s(x0 2 s(x ɛn ( x 2 + s(x ɛn 2 ( + lm n s(x ɛn ( x 2 + s(x ɛn 2 q. (10 It s obvous that lm n s(x ɛn = s(x0. Aordng to Step 1, we an get that s(x0 > 0 [N]. Thus, n Eqn. (10, the frst term s strtly less than zero, and the seond term s no more than zero. Therefore, user should not ontrbute ontent wth qualty x. 2. If x < x0, set δ = x0 x 2. Then for suffently large n, we have Smlarly, we have s(x 0 (x + ( s(x 0 = lm s(x0 2 q n (x + s(x ɛn s(x0 2 (x + δ + s(x ɛn s(x ɛn (x + δ + s(x ɛn 2 q 0. (11 2 ( + lm n s(x ɛn (x + δ + s(x ɛn 2 q n whh the frst tem s strtly larger than zero, and the seond tem s no less than zero. Hene, t s better for user to ontrbute more than x. Smlar ontradton ould be fnd at the ase when x > x0. 3. Set δ = q x 2. For suffently large n, we have s(x ɛn (s(x ɛn + δ + x 2 q 0. Thus, we obtan s(x 0 (x + ( s(x 0 = lm s(x0 2 q n (x + s(x ɛn s(x0 2 (x + δ + s(x ɛn 2 ( + lm n s(x ɛn (x + δ + s(x ɛn 2 q Also, the frst tem s strtly larger than zero, and the seond tem s no less than zero. Thus, user should not ontrbute ontent wth qualty less than q. Therefore, we an onlude that x 0 must be the equlbrum strategy of the orgnal game. Besdes, we know n the equlbrum, there are at least two users who ontrbute postve qualty ontent.,, 7

8 Y. Xa, T. Qn, N. Yu, T.-Y. Lu The Unqueness of the PNE for the Mehansm M 4 Theorem 2 The mehansm M 4 has a unque PNE. To prove the unqueness, we need the followng lemmas that reveal the strutures of the PNE: Lemma 3 In an equlbrum profle, a user wth a strtly larger type wll ontrbute no less than a user wth a smaller type. That s, f q < q j, we have x x j n an equlbrum. Proof. The proof s by ontradton. Suppose there exst two users and j wth q < q j but x > x j n an equlbrum. (Note n ths assumpton, x > 0. Defne A = k,j x k. One an easly verfy that q (A + x j < q j(a + x. (12 The value of x and x j have four possble ases: Case 1 : 0 < x < q, 0 < x j < q j: Sne {x }N =1 s an equlbrum profle, we have A + x j (x + x j + A + x = 0 and A2 q (x + x j + = 0. A2 q j After some dervatons, we get q (A + x j = q j(a + x, whh ontradts wth (12. Case 2 : x = q, 0 < x j < q j: Smlar to Case 1, we an get A + x j (x + x j + A + x 0 and A2 q (x + x j + = 0. A2 q j After some dervatons, we get q (A + x j q j(a + x, whh ontradts wth (12. Case 3 : 0 < x < q, x j = 0: We an obtan A + x j (x + x j + A + x = 0 and A2 q (x + x j + 0. A2 q j After some dervatons, we get q (A + x j q j(a + x, whh ontradts wth (12. Case 4 : x = q, x j = 0: We know A + x j (x + x j + A + x 0 and A2 q (x + x j + 0. A2 q j After some dervatons, we get q (A + x j q j(a + x, whh ontradts wth (12. The proofs of the followng three lemmas are ether smlar to the proof of Lemma 3 or very easy. We leave them to Appendx A.1, A.2 and A.3. Lemma 4 In an equlbrum profle, users wth the same type wll ontrbute ontent wth the same qualty. That s, f q = q j, then we have x = x j n an equlbrum. Lemma 5 Consder two users wth the types q q j. If x have x j = q j n the same equlbrum. = q n an equlbrum, then we also Lemma 6 If < 4, then n an equlbrum, [N], x < q. Let s(x = N =1 x denote the total qualty of the ontent ontrbuted by all the users. The followng three lemmas llustrate the propertes of s(x n an equlbrum. Lemma 7 In an equlbrum profle x, f x (0, q, then we have x = s(x q s 2 (x. (13 8

9 Inentvzng Hgh-qualty Content from Heterogeneous Users Lemma 8 In an equlbrum profle x, f x = q [m] and x j < q j j {m + 1,..., N} n whh 1 m < N, we have q m+1 ( < s(x q m 2 ( (14 Lemma 9 In an equlbrum profle, f x n > 0 and x n+1 = 0 (n < N, we have q n+1 s(x < q n. (15 Lemma 7 an be easly obtaned from Eqn. (2. The proofs of Lemma 8 and Lemma 9 an be found n Appendx A.4 and A.5. If there are multple equlbra of the mehansm M 4, by Lemma 7, Lemma 8 and Lemma 9, we have the followng three orollares: (the proof of Corollary 12 s n Appendx A.6 Corollary 10 If x and y are both equlbra for the mehansm M 4 wth more than one users ontrbutng ther types and there are more users who ontrbute ther types n y, we have s(x > s(y. Corollary 11 If x and y are both equlbra for the mehansm M 4 and there are more users who make postve ontrbuton n y,.e., x 0 < y 0, we have s(x > s(y. Corollary 12 If x and y are both equlbra for the mehansm M 4, then > 1 s.t. x, y (0, q, we have: (1 f s(x > s(y, then x < y ; (2 f s(x = s(y, then x = y. Now we ould start to prove Theorem 2 by ontradton. If M 4 has two dfferent equlbra x and y, by the above lemmas, x and y must be of one the followng sx ases: (S1 0 x, y q and x 0 = y 0 ; 8 (S2 0 x, y q and x 0 y 0 ; (S3 0 x, y q, x 0 = y 0, and the top m (1 m < x 0 users n both x and y ontrbute ther types; (S4 n both x and y, the top m (1 m x 0 users ontrbute ther types, and x 0 y 0 ; (S5 n both x and y there are more than one users who ontrbute ther types and there are more users who ontrbute ther types n y than n x ; (S6 there are some users ontrbutng ther types n x and all the partpants ontrbute strtly less than ther types n y. Next we prove the mpossbltes of the sx ases: (S1 Suppose x 0 = y 0 = n. Summng u x = 0 (defned n Eqn. (2 over from 1 to n and makng smple alulatons, we obtan s(x = (n 1 Smlarly, we an obtan (n 1 n k=1 1 n k=1 1 s(y = for y, thus s(x = s(y. By Lemma 7 we know x = y [n]. q k Aordngly, x = y. (S2 W.l.o.g., suppose x 0 = n 1, y 0 = n 2 and n 1 < n 2. By Corollary 11 we know s(x > s(y. Then by Corollary 12 we an obtan x < y {2,, n 1}. Meanwhle, y > 0 = x {n 1 + 1,, n 2 }, and we an get s(x 1 < s(y 1. It s easy to see that s(y 1 s 2 (y q 1 > s(x 1 s 2 (x q 1 = 0. Thus n y, user 1 an proftably devate by ontrbutng ontent wth qualty more than y 1. Ths ontradts wth the assumpton that y s a PNE. q k x means x 0. So are,,. 9

10 Y. Xa, T. Qn, N. Yu, T.-Y. Lu (S3 Suppose x = y = q [m] and x 0 = y 0 = n. After some smple dervaton we an get that both s(x and s(y are the s nm defned n Eqn. (20. By Lemma 7 we an get x = y {m + 1,, n}, and thus x = y. (S4 Suppose x = y = q [m], x 0 = n 1, and y 0 = n 2. There are two possble sub ases. 1. n 1 = m, n 2 > m: It s obvous that s(x < s(y, but aordng to Corollary 11, we have s(x > s(y, whh s a ontradton. Smlar ontradton ould also be found when n 2 = m, n 1 > m. 2. n 1 > n 2 > m: By Corollary 11, we an obtan s(x < s(y. But gven s(x < s(y, by Corollary 12, we get x > y {m + 1,, n 2 }, whh leads to s(x > s(y. Smlar dsusson ould be appled to the ase that n 2 > n 1 > m. (S5 Assume that there exst two ndexes m 1 and m 2 (1 m 1 < m 2 < N suh that x = q [m 1 ], x < q {m 1 + 1,, N}, yj = q j j [m 2] and yj < q j j {m 2 +1,, N}. 1. Consder the ase that there exst an n 1 {m 1 + 1,, N} and an n 2 {m 2 + 1,, N} suh that x (0, q {m 1 + 1,, n 1 } and y j (0, q j j {m 2 + 1,, n 2 }. By Corollary 10 we know s(x > s(y. When s(x > s(y, t s also obvous that n 1 > m 2. Then, (a by Corollary 11 we know n 1 n 2 ; (b by Corollary 12 we an obtan x < y {m 2 + 1,, n 1 }. Note that y k = q k > x k k {m 1 + 1,, m 2 } and y has more non-zero elements, we wll obtan s(x < s(y, whh ontradts wth s(x > s(y. 2. Consder the ase that yj = 0 j {m 2+1,, N} and x (0, q {m 1 +1,, n 1 }. By Corollary 10 we know s(x > s(y. Then by Corollary 11 we an obtan n 1 m 2. Whle x < q {m 1 + 1,, n 1 }, but y = q [m 2 ], we an obtan s(x < s(y, whh s a ontradton. Smlar ontradton ould also be found for the ase that x 0 = m 1 and y 0 > m If x = 0 {m 1 + 1,, N} and y j = 0 j {m 2 + 1,, N}, t s obvous that s(y > s(x. But aordng to Corollary 10, s(x > s(y. 4. If m 2 = N, we add a vrtual ontrbutor wth type q N+1 suh that mn{s(x, s(y } > q N+1 and q N > q N+1. Then (x, 0 and (y, 0 are both equlbra of the new game wth N + 1 players. And we ould use the above dsussons to handle ths ase. (S6 Suppose x = q, [m], x 0 = n 1, and y 0 = n If s(x s(y, by Corollary 11 we have n 1 n 2. If m n 2, t s obvous that s(x > s(y, whh ontradts wth the assumpton; that s, m has to be smaller than n 2. By Corollary 12 we have x y {m + 1,, n 2}, thus s(x > s(y, whh s also a ontradton. 2. Now we onsder the ase of s(x > s(y. By Corollary 11 we have n 1 n 2. If [m], we have x > y and s(x s(y. 9 Thus, f m 2, {2,, m}, s(y s(x (x + s(x 2 < s(x (y + s(x 2 s(y (y + s(y 2, 10 (16 s(x 9. Else, we wll have (y > +s(y 2 q (x 0, whh mples that user wants to ontrbute +s(x 2 q more quantty and t ontradts wth the assumpton that y s a PNE. 10. The seond holds beause s(x s(y y 2. 10

11 Inentvzng Hgh-qualty Content from Heterogeneous Users we ould nfer that s(y (y + s(x > s(y 2 q (x + 0. (17 s(x 2 q Therefore, n y, user s nentvezed to ontrbute ontent wth qualty more than y {2,, m}, whh ontradts wth the assumpton y s a PNE. If m = 1, as well as s(x > s(y, we know x < y {2,, n 1 }, then s(x 1 < s(y 1. But aordng to we know whh s a ontradton. s(x 1 s(x 1 (x 1 + s(x 1 2 q 1 0 s(y 1 (y 1 + s(y 1 2 q 1 = 0, s(x > s(y = s(y 1, q 1 q 1 Therefore, we have dsussed the mpossbltes of all the sx ases. Now we an onlude that the mehansm M 4 has a unque pure Nash equlbrum. ( Fndng the PNE of M 4 Gven Mehansm M 4, n whh N users ompete for the reward, we an ndue an n-loal game,.e., only the frst n users ompete for the reward. Denote the PNE of the ndued n-loal game as x (n. In the equlbrum profle of an ndued n-loal game, f every user ontrbutes postve qualty ontent, then we an use Algorthm 1 to fnd the PNE leveragng the followng three equatons. Algorthm 1: Algorthm to fnd the PNE of an ndued n-loal game Input: Users types: q = (q 1, q 2,, q n where q 1 q 2 q n. Output: The PNE of the ndued n-loal game: x (n = (x (n 1, x(n 1 Calulate the y n [n] wth Eqn. (19 ; 2 f 0 y n q [n] then 3 x (n y n and return x (n ; 4 else 5 for m 1 : n 1 do 6 f q m q m+1 then 7 Calulate x nm {m + 1,, n} wth Eqn. (21; 8 f {m + 1,, n}, x nm [0, q ] then 9 [m], x (n 10 x (n q [n] and return x (n ; q and {m + 1,, n}, x (n 2,, x(n n. x nm. eturn x (n (n 1 y n = n 1 n 1 [1 k=1 q k q n 1 ] (19 k=1 q k m n (n m 1 + (n m Q ma nm Q m = q, A nm =, s nm = (20 q 2A nm =1 x nm = s nm =m+1 q s 2 nm (21 11

12 Y. Xa, T. Qn, N. Yu, T.-Y. Lu Note that f the n-loal game has a PNE x (n s.t. 0 < x (n < q, then t ould be alulated by Eqn. (19; f the PNE of the n-loal game s the top m (1 m n users ontrbutng ther types whle the left n m users ontrbutng below ther types, x (n ould be dervated by Eqn. (21. Dervatons from Eqn. (19 to Eqn. (21 ould be found n Appendx A.8 and Appendx A.9. Theorem 13 If the PNE x (n of an ndued n-loal game satsfes x (n 0, the output of Algorthm 1 s the PNE of the ndued n-loal game. Proof. We prove ths theorem by ontradton. Denote the output of Algorthm 1 as x. 11 Lemma 4 shows that users wth the same type wll ontrbute ontent wth the same qualty. If 0 x q or x = q, we an easly verfy that x s the PNE. Hene we only need to onsder the ase that the top m ( m 1 users ontrbute ther types n x but x s not the equlbrum. In the real equlbrum profle (denoted as x, we set the top m users ontrbute ther types. It s ertan that the m m + 1. Sne the ( m + 1-th user ontrbutes below hs types n x, we an easly verfy that there are two possble ases: q m+1 1. s( x > 2 ( : In the real equlbrum, the m + 1th user wll ontrbute hs type, by Lemma 8 we know s( x > s(x q 2 {m + 1,, n}. Aordng to Lemma 7, we know x dereases wth s(x n [ q 2,. As a result, x x { m + 1,, n}. Thus, s(x should be no less than s( x, whh s a ontradton. 2. s( x < q m+1 2 (1 1 4 : Aordng to Eqn. (20, we have (n m 1 + (n m ( m =1 q( n = m+1 2 n = m+1 q 1 q < q m+1 ( (22 After some dervatons we an get that we only need to dsuss the followng ase: (detals ould be found n Appendx A.10 m( (n m 1 < (1 1 4 n = m+1 q m q. (23 Note n the real equlbrum, the m-th user wll ontrbute hs/her type. We an verfy that q m 2 (1 1 4 q 1 2 (1 1 4 s(x q m 2 ( (24 and q m 2 (1 1 4 s(x < q n. (25 Therefore we know the rght hand sde of Eqn. (23 s smaller than ( m q m n q n + 2 (m m( (n m. (26 q q = m+1 =m+1 We have ( m + 2(n m 1 < (m m( (n m. ( To smplfy notatons, we omt the supersrpts (n n ths proof. That s, x should be x (n, and x should be x (n. 12

13 Inentvzng Hgh-qualty Content from Heterogeneous Users Denote a = It s easy to get 1 a < 2. earrangng Eqn. (27, we an get 2a m 2 m 2 < m(a 2 ( m + 1(a 2. (28 After some dervatons, we an get m < 1, whh s a ontradton wth the settng that m 1. Aordng to the next lemma, an equlbrum of the ndued loal game an be extended to an equlbrum of the orgnal game (.e., Mehansm M 4 wth N users under ertan ondtons. Lemma 14 If {x (n } [n] s an equlbrum of the ndued n-loal game (n < N and n =1 x(n q n+1, then {x } [N], where x = x(n, 1 n and x = 0, n + 1 N, s an equlbrum of the orgnal game. The proof of Lemma 14 s n Appendx A.7. Therefore, the equlbrum of the orgnal game an be found by Algorthm 2: Algorthm 2: Algorthm to fnd the PNE of the orgnal game Input: Users types: q = (q 1, q 2,, q N where q 1 q 2 q N. Output: The PNE of the orgnal game: x = (x 1, x 2,, x N. 1 for n 2 : N do 2 f q n q n+1 (when n = N, dretly go to the next lne then 3 Calulate the PNE x (n of the ndued n-loal game by Algorthm 1; 4 Verfy whether t ould be extended to the PNE of the orgnal by Lemma 14; 5 If yes, x x(n [n] and x 0 {n + 1,, N}, return x. 3. Partal-nformaton Settng In ths seton, we nvestgate the exstene of the symmetr pure Bayesan Nash equlbrum for the UGC mehansms under the partal-nformaton settng. We dsuss all of the four mehansms here. 3.1 M 5 : Top K Alloaton, Bnary Aton Spae We only onsder the ase that > K and omt the margnal ase K here. We set F s a ontnuous and strtly nreasng funton. We defne the ut-off strategy for the partal-nformaton settng and bnary aton spae as follows: If a user s type τ s no smaller than a threshold τ, he/she wll hoose to ontrbute ontent to the ste; otherwse, he/she wll not partpate n the game. Mathematally, a ut-off strategy β( s spefed by a threshold parameter τ,.e., β(τ = { τ, f τ τ ; 0, f τ < τ. (29 Let funton T (τ denote the probablty that a user wth qualty τ s n one of the top K ontrbutors. Clearly, f N K 0, we have T (τ = 1; when N K 1, we have K 1 ( N 1 T (τ = F (τ N 1 j (1 F (τ j. (30 j Intutvely, a user wth hgher qualty s more lkely to get the reward: j=0 13

14 Y. Xa, T. Qn, N. Yu, T.-Y. Lu Lemma 15 T (τ s a non-dereasng funton of τ. Proof. We only need to dsuss the non-trval ase,.e. N K 1. ( T (τ N 2 = (N 1f(τ F (τ N K 1 (1 F (τ K 1 0 (31 τ K 1 Thus T (τ s a non-dereasng funton of τ. Then we an onstrut a symmetr ut-off equlbrum strategy for M 5 : Theorem 16 Let τ denote the soluton of Eqn. (32. T (τ = 0 (32 K [N], we have that M 5 has a unque symmetr ut-off equlbrum strategy, 12 whh s shown as below. β (q = { q, f q τ ; 0, f q < τ. (33 Proof. The proof onssts of two steps. (Step 1 The exstene of a symmetr ut-off equlbrum strategy: Followng the strategy, f a user wth type τ hooses to partpate n the game, the probablty that he/she ould get the reward s P (τ, τ = K 1 j=0 K 1 j=0 ( N 1 F (τ N 1 j (1 F (τ j, f τ τ ; j ( N 1 F (τ N 1 j (1 F (τ j, f τ < τ. j If τ τ, we have P (τ, τ P (τ, τ = K, and the user s expeted utlty f he/she partpates n the game s K P (τ, τ K = 0. (35 K If τ < τ, the expeted utlty for the user s 0 f he/she partpates n the game. Thus nobody an proftably devate the strategy unlaterally. (Step 2: The unqueness of the symmetr ut-off equlbrum strategy: Sne we set that F s a strtly nreasng and ontnuous funton n M 5, Eqn. (32 has a unque soluton. Let us onsder another symmetr ut-off strategy wth threshold τ a : 1. If K T (τ a > 0, a user wth type τ < τ a ould proftably devate by partpatng, beause hs/her expeted utlty s K P (τ, τ a > If K T (τ a < 0, we have K P (τ a, τ a < 0. By the ontnuty of P (τ, τ a w.r.t τ, t s ertan that there exsts a τ u suh that τ (τ u, τ a, K P (τ, τ a < 0. Thus, a user wth type fallng n (τ u, τ a an proftably devate by not ontrbutng ontent. Therefore, suh a ut-off strategy s not an equlbrum strategy, and we onlude that M 5 has a unque ut-off PNE. By the monotonty of T (τ, one an easly use the bnary searh algorthm to fnd the τ numerally. 12. We mean that among all the symmetr ut-off strateges, there s a unque equlbrum strategy. (34 14

15 Inentvzng Hgh-qualty Content from Heterogeneous Users 3.2 M 6 : Top-K Alloaton, Contnuous Aton Spae We frst gve a general desrpton to a symmetr equlbrum strategy, then prove the exstene of PNE when the dstrbuton F s unform, fnally generalze the result to any dstrbuton. Let us onsder a symmetr nreasng strategy β( : eah user wth type q wll ontrbute to the ste wth qualty β(q. T (τ (defned n Eqn. (30 s the probablty that a user wth type τ wns the game gven all the users adopt the same nreasng strategy β(. Suppose that users j follow the nreasng symmetr equlbrum strategy β(. If user pretends that hs/her type s τ and ontrbutes β(τ, then hs/her expeted utlty 13 s The frst order dervatve of u s u (τ; q = β(τ T (τ. (36 K q u (τ; q τ = K T (τ τ β (τ q. (37 If β(q s an equlbrum strategy for user, hs/her expeted utlty should be maxmzed at τ = q. That s, we should have u (τ; q = 0. (38 τ τ=q Note that β(0 = 0. Solvng the above equaton, we get ( (N 1 N 2 τ β(τ = zf (z N K 1 (1 F (z K 1 df (z (39 K K 1 Then we have the followng results. 0 Lemma 17 If β(τ τ, τ [0, 1], then the funton β( n the above equaton s a symmetr equlbrum strategy. Proof. Frst, t s obvous that β(τ n an nreasng funton w.r.t τ. Then we prove that nobody wants to devate the strategy β( unlaterally. If user j wants to generate some ontent wth qualty β(τ [0, β(q j (.e., τ < q j, by Eqn. (37 and Eqn. (38, we obtan T (τ K β (τ q > T (τ K β (τ τ = 0, (40 hene the utlty of user nreases w.r.t β(τ when β(τ [0, β(q. Smlarly, the utlty of user dereases w.r.t β(τ when β(τ (β(q, β(1]. Also, user should not ontrbute ontent wth qualty n (β(1, q ] (f β(1 < q, beause f others follow the strategy β(, eah of them wll ontrbute ontent wth qualty at most β(1. Therefore, we an onlude that β( s the nreasng equlbrum strategy. However, t s possble that β(τ expressed by Eqn. (39 s larger than τ. For example, f F s the unform dstrbuton over [0, 1], β(τ an be wrtten as below. ( β(τ = K 1 ( N 2 (N 1 ( 1 K k 1 K 1 τ N k (41 K K 1 k N k k=0 13. Here we use u (τ, q to denote the expeted utlty of user gven that all the users follow the symmetr strategy β(, and he/she pretends hs/her type s τ whle hs/her true type s q. 15

16 Y. Xa, T. Qn, N. Yu, T.-Y. Lu Then we have β(1 = N K K N, (42 whh 14 mght be larger than 1. If β(τ > τ for some τ [0, 1], β(τ wll not be an equlbrum strategy anymore. We need to albrate β(τ. For ease of desrpton, we frst llustrate how to make albraton when F s the unform dstrbuton, and then extend the result to any dstrbuton. Wth some dervatons, one an get that the equaton β(τ = τ has at most two solutons n the regon (0, 1] when the dstrbuton F s unform. If there exst two solutons (denoted as τ 1 and τ 2 (τ 1 < τ 2, there wll be an τ p (> τ 1 that satsfes β (τ p = 1. Then we have: Theorem 18 If F s the unform dstrbuton over [0, 1] and β(τ = τ has two solutons n (0, 1], the followng β ( funton s an equlbrum, where τ 1 and τ p are defned above. β(τ, τ [0, τ 1 ]; β (τ = τ, τ (τ 1, τ p ]; (43 β(τ β(τ p + τ p, τ (τ p, 1]. Proof. Frst, suppose τ [0, τ 1 ]. Sne β(τ τ, we have that β (τ = β(τ s the best response of type τ. Seond, t s lear that the frst order dervatve β (τ s larger than 1 for any τ (τ 1, τ p. Suppose that all other users follow strategy β ( exept user wth q (τ 1, τ p, and suppose he/she pretends to have a type τ. 1. If τ (τ 1, τ p and τ q, we have and u (τ; q τ > K u (τ; q = K T (τ τ q, T (τ τ β (τ T (τ q K τ β (τ = 0. τ Therefore, the larger τ s, the larger utlty he/she wll get. However, sne the ontrbuted qualty s upper bounded by hs/her type q, the best hoe for the user s to take the aton τ = q. 2. If τ [0, τ 1 ], we have u (τ; q τ = K T (τ τ β (τ > T (τ q K τ β (τ = 0. τ Thus the user should pretend to have a type τ 1, whh s, however, stll worse than revealng the true type q. Thus, for any x n (τ 1, τ p ], the best response s β (τ = τ. Thrd, for any τ n (τ p, 1], we have β (τ 1. Integratng β (τ from τ p to x and usng β(τ p = τ p, we get β (τ τ p = β(τ β(τ p. It s easy to verfy that β (τ τ for any τ n (τ p, 1]. Therefore, we have that β (τ = β(τ+τ p β(τ p s the best response for any τ n (τ p, 1]. Thus, the theorem s proved. 16

17 Inentvzng Hgh-qualty Content from Heterogeneous Users β(x β(τ β (τ = β(τ τ [0,τ1] β (τ = τ τ (τ1,τp] β(τ β (τ = β(τ τ [0,τ1] β (τ = τ τ (τ1,τp] β (τ = β(τ β(τp+τp τ (τp,1] Strategy Strategy Strategy τ1 τp τ Type / τ (a β(τ τ [0, 1] τ1 τp τ Type / τ (b Calbraton from τ 1 to τ p τ1 τp τ Type / τ ( β (τ τ [0, 1] Fgure 1: Illustraton of albraton Fg. 1 shows an equlbrum strategy of a UCG mehansm wth N = 11, K = 5, = 1, = 8. Fg. 1(a shows the orgnal β(τ τ [0, 1]. We an see that β(τ s not feasble n (τ 1, 1]. Fg. 1(b shows the albraton from τ 1 to τ p. The fnal equlbrum strategy β (τ s shown n Fg. 1(. Next we generalze the above results. For a general dstrbuton F over [0, 1], we frst ntalze β (τ = β(τ, τ [0, 1] and then albrate β (τ as follows. 1. Chek whether β (τ > τ startng from τ = 0 to τ = Suppose [τ 1, τ 2 ] s the frst nterval that β (τ > τ, and τ p s the pont n ths nterval satsfyng β (τ p = 1. Let o denote the value of β (τ at τ p (.e., o = β (τ p, and then albrate β (τ = τ, τ [τ 1, τ p ] and β (τ = β (τ o + τ p, τ (τ p, 1]. 3. Contnue to hek whether β (τ > τ startng from τ = τ p to τ = 1. If there s stll some nterval wth β (τ > τ, we albrate β (τ as shown n the prevous step. 4. We repeat the hekng and albratng proedure untl β (τ τ, τ [0, 1]. After the albraton proess, we obtan an equlbrum strategy β (τ from β(τ, whh s shown n Eqn. (39, for any dstrbuton F. Therefore we have the followng theorem. Theorem 19 M 6 has at least one symmetr PNE. 3.3 M 7 : Proportonal Alloaton, Bnary Aton Spae Now we study the exstene of PNE for the mehansm wth the proportonal alloaton rule and the bnary aton spae n the partal-nformaton settng. Suppose that users j follow the ut-off strategy defned n Eqn. (29. Then the expeted utlty 15 of user an be wrtten as follows f he/she partpates n the game (x = q. u (q ; τ = N 1 k=0 ( N 1 k F (τ N 1 k (1 F (τ k r (q, k; τ (44 In the above equaton, r (q, k; τ s the expeted reward of user gven another k users partpatng n the game whose qualtes are larger than τ,.e., 0, f q = 0; r (q, k, τ, f q > 0 and k = 0; = 1 (45 1 q... df (τ 1 τ...df (τ k τ, f q > 0 and k > 0, τ τ q + τ 1 + τ τ k 14. The alulaton of β(1 ould be found n Appendx A Here u (q ; τ denotes the expeted utlty of user wth type q gven the threshold of the ut-off strategy s τ. 17

18 Y. Xa, T. Qn, N. Yu, T.-Y. Lu where F (τ τ represents the ondtonal probablty dstrbuton of the type τ gven τ τ. If Eqn. (29 s a symmetr ut-off equlbrum strategy, the best response of user s also to follow the strategy gven that all the other users follow the strategy. That s, > 0, f q > τ ; u (q, τ s = 0, f q = τ ; (46 < 0, f q < τ. It s not dffult to get that 1. u (q, τ nreases w.r.t. q, 2. u (0, 0 = < 0 and u (1, 1 = > 0. We an verfy u (τ, τ s ontnuous w.r.t τ [0, 1] (see Appendx A.12 for proof. Therefore, there exsts a τ satsfyng the three ondtons n Eqn. (46; n turn, ths τ makes Eqn. (29 a symmetr ut-off equlbrum strategy. Thus, we have the followng theorem. Theorem 20 M 7 has at least one PNE f >. 3.4 M 8 Proportonal Alloaton, Contnuous Aton Spae We only onsder the ase that users types are dsrete and take values from a fnte set T = {t 1, t 2,, t M }, n whh t > 0 [M]. The type s the same as what we defne n Subseton 1.1, whh denotes the best possble qualty of the ontent that a user an ontrbute to the ste. Eah ontrbutor has the same belef that the dstrbuton of the others types are: ( t1 t 2 t M, (47 p 1 p 2 p M where p > 0 [M] and M =1 p = 1. Eah ontrbutor knows hs/her own type. We wll prove the exstene of the symmetr PNE under the partal nformaton settng. Lke M 4, we frst formulate a perturbed game, prove that t has a PNE, and then show that the PNE of the perturbed games wll onverge to the PNE of the orgnal game. Consder the symmetr strategy for the frst N 1 users: anyone wth type t wll ontrbute ontent wth qualty β [M]. Suppose the N-th user wth type t ( T wll ontrbute ontent wth qualty θ. When θ = 0, hs/her expeted utlty s 0; when θ > 0, hs/her expeted utlty (denoted as u(θ, t, β s u(θ, t, β = M 1=1 M N 1 =1 p 1 p N 1 θ β β N 1 + θ θ t, (48 where β s the symmetr strategy of the other N 1 users,.e., β = (β 1, β 2,, β M. The frst order dervatve of u(θ, t, β s u(θ, t, β θ = p 1 p (β N β N 1 (β 1,, β N 1 + θ 2 t. (49 N 1 The user s goal s to maxmze hs/her expeted utlty: If u (0 0 (denote u(θ, t, β as u(θ when the ontext s lear, hs/her best-response strategy s to make no ontrbuton; f u (t 0, the user should generate ontent wth qualty t; f u (0 > 0 and u (t < 0, then u (θ = 0 has a unque soluton n (0, t and denote t as θ, whh s exatly the best-response strategy. Thus, we ould get 18

19 Inentvzng Hgh-qualty Content from Heterogeneous Users that the best-response strategy of the user wth type t gven others symmetr strategy β (denoted B(t, β s 0, f u (0 0; B(t, β = θ, f u (0 > 0 and u (t < 0; (50 t, f u (t 0. In order to avod the ase that all the users generatng zero-qualty ontent smultaneously, whh wll make Eqn. (48 meanngless, we onsder an ɛ-perturbed game, n whh any user has to ontrbute ontent wth qualty at least ɛ(> 0. In that ase, the best-response strategy beomes { ɛ, f B(t, β ɛ; B ɛ (t, β = B(t, β, others. (51 Lemma 21 B ɛ (t, β s a ontnuous funton w.r.t β n [ɛ, 1] M for any ɛ (0, Now defne a perturbed mappng B ɛ (t, β from T ɛ = [ɛ, t 1 ] [ɛ, t 2 ] [ɛ, t M ] to tself, where β ɛ = B(t, β [M]. T ɛ s a onvex and ompat set. Aordng to Brouwer fxed pont theorem [2], the mappng B ɛ (t, β has at least one fxed pont n T ɛ. We ould fnd a sequenes of {ɛ n } 0 and get the orrespondng seres of the fxed ponts. Defne the fxed ponts as {β ɛ1, β ɛ2,, β ɛn, }. Sne these ponts are n the ompat spae, w.l.o.g, we ould set the seres as a onvergent sequene and denote the lmt pont as β 0. We wll prove that β 0 s the symmetr Bayesan Nash equlbrum of the orgnal game. That s, a user wth type t wll ontrbute ontent wth qualty β 0. Lemma 22 All the elements n β 0 are non-zero,.e., β 0 Proof. We prove ths lemma by ontradton. If β ɛn nteger N 3 (ɛ, n > N 3 (ɛ, β ɛn u(β ɛn 0, for any [M]. 0, then ɛ (0, 1, there exsts a postve < ɛ. Then the utlty of some user wth type t and strategy β ɛn s, t, β ɛn = pn 1 N + p 1 p N 1 βɛn (β ɛn + + β ɛn N 1 + β ɛn 1 t β ɛn, (52 n whh the s over ( 1,, N 1 {1, 2,, M} N 1 \{,,, }. But f he/she ontrbutes ontent wth qualty 2β ɛn, hs/her utlty s and u(2β ɛn u(2β ɛn, t, β ɛn u(β ɛn p N 1 (N 1, t, β ɛn = 2pN 1 N p 1 p N 1 βɛn (β ɛn + + β ɛn N 1 + 2β ɛn, q, β ɛn pn 1 (N 1 β ɛn N(N + 1 t 1 2 t β ɛn. (53 > pn 1 (N 1 ɛ. (54 N(N + 1 t If ɛ < t N(N+1, we know n the ɛ n -perturbed game where n > N 3 (ɛ, some user wth type t an be better off by ontrbutng ontent wth qualty 2β ɛn, whh ontradts wth that assumpton that β ɛn s a PNE of the ɛ n -perturbed game. Lemma 23 In the orgnal game, nobody ould get more utlty by unlaterally devatng from β 0. Proof. We prove ths lemma by ontradton. In the proof, we wll frequently use the sum over the ndexes 1,, N 1. For ease of referene, denote M M 1=1 N 1 =1 as p 1 p N 1 θ β s 1 + +β s N 1 +θ θ z s +θ, n whh s atually over M N 1 terms, and s s the supersrpt. If β 0 s not the symmetr PNE, there must exst some user wth t who an proftably devate β The proof ould be found n Appendx A

20 Y. Xa, T. Qn, N. Yu, T.-Y. Lu 1. If x ɛn t, but some user wants to devate to t δ 0 n the orgnal game, n whh δ > 0: When n s suffently large, we have z ɛn (z ɛn + t δ 2 2 t > 0, (55 and z 0 (z 0 + t δ2 t ( z 0 = lm n (z 0 + t δ2 ( z ɛn + lm (z ɛn + t δ 2 2 n (z ɛn z ɛn + t δ 2 2 t. (56 Sne z 0 > 0, n Eqn. (56, the frst term s strtly larger than zero, and the seond term s no less than zero. Thus we an nfer that z 0 (z 0+t δ2 t s strtly larger than zero, whh means that user should ontrbute ontent wth qualty more than t δ. Thus, t δ s not the best-response strategy for user n the orgnal game. We fnd a ontradton. 2. If β ɛn β 0 (0, t, but he/she wants to devate to β (< β0 n the orgnal game: set δ = β0 β 2. We an easly get that when n s suffently large, and z 0 (z 0 + β 2 t ( z 0 = lm n (z 0 + β 2 z ɛn (z ɛn + β + δ2 t > 0 (57 ( z ɛn (z ɛn + β + + lm δ2 n (z ɛn z ɛn + β + δ2 t n whh the frst term s strtly postve and the seond one s non-negatve. Therefore we know that user should ontrbute more than quantty β. Smlar dsusson ould be appled to the ase β > β0. That s, nobody ould proftably devate from β 0 unlaterally. Therefore, we an onlude that Theorem 24 M 8 has at least one symmetr PNE., (58 4. Conlusons and Future work In ths paper, we have studed the UGC mehansms under a new framework: users are heterogeneous and the best qualtes users an ontrbute are dfferent from eah other. Under ths framework, we have onsdered several mehansms nvolvng two alloaton rules, two aton spaes, and two nformaton settngs. We proved the exstene of multple PNE for some mehansms, the exstene and unqueness of PNE for some mehansms, and the non-exstene of PNE for some other mehansms. As for the future work, there are qute a few nterestng ponts worth nvestgatng. Frst, we plan to ondut the effeny analyss for those mehansms whose equlbra are proven to exst. Seond, we plan to study the mxed Nash equlbrum for the UGC mehansms. Thrd, we wll nvestgate more general ost funtons (e.g., onvex funtons. Fourth, we wll make omparsons between dfferent UCG mehansms and dentfy the best one for pratal use. 20

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