The Mixed Strategy Nash Equilibrium of the Television News Scheduling Game

Transcription

1 The Mxed Strategy Nash Equlbrum of the Televson News Schedulng Game Jean Gabszewcz Dder Laussel Mchel Le Breton July 007 Abstract We characterze the unque mxed-strategy equlbrum of an extenson of the "televson news shedulng game" of Cancan, Bergström and Blls (995) where vewers want to watch the rst newscast broacast after they return home. A fracton of the vewers record randomly one of the newscast to watch them n case they are too late. At equlbrum, nether of the two statons broadcasts ts evenng news n the rst part of the evenng and the densty functon s strctly decreasng. Introducton Cancan, Bergström and Blls (995) have analyzed the problem of schedulng evenng televson newscasts as an "Hotellng locaton problem wth drectonal constrants". In ths smple but ngenous framework, statons have to decde the tme at whch they are gong to broadcast ther evenng news, gven that vewers, who go back at d erent tmes n the evenng, watch the rst newscast broadcast after they return home. They showed that ths spec c game has no pure strategy Nash equlbrum. Obvously what characterzes an "Hotellng locaton problem wth drectonal constrants" s that consumers care not only about how far they are from the d erent rms but also about on whch sde the rms are located. Cancan, Blls and Bergström pont out that the schedulng of arlne departures s another example of ths type of games: a busness traveller wll take the latest ght whch allows hm/her to arrve at hs/her destnaton at a spec c tme but no ght whch arrves after ths tme. Another possble example ("voters leanng towards the left") s n the eld of poltcal economy : each voter s characterzed by hs/her deal polcy t and votes for the canddate who s the closest on the left of ths deal pont but never for a canddate on the rght of t. Introducng drectonal constrants of ths knd changes n a substantal way the outcome of the locaton game when the players maxmze ther "audence" (or the number of voters). Indeed, n the usual Hotellng model wthout drectonal constrants, t s well-known that the statons would broadcast ther news at the medan tme among the deal tmes of the populaton of vewers.

2 When one looks at the tmes at whch the evenng news are broadcast n four European countres, one nds a rather mportant dsperson, except may be n France where the range s rather narrow: from 9.30 (FR3) to 9H 54 (M6) and 0H (TF and A) It s much larger n RFA: 8.45 (RTL), 0.00 (ARD),.45 (ZDF) to name only the man Channels. In Great Brtan the broadcastng tmes are rather late: 0.00 (BBC 4),.00 (BBC ),.30 (BBC ). Fnally n Italy, they are as follows : 8.30 (Studo Aperto), 8.55 (TG 4), 9.00 (TG 3), 0.00 (TG and TG 5), 0.30 (TG). If the statons are ndeed audence-maxmzers, ths dsperson s an argument whch seems to plead aganst the usual mnmum-d erentaton result of the usual Hotellng model wthout drectonal constrants and to suggest a possble mxed-strategy equlbrum. In ths paper we show that there s unque mxed-strategy Nash equlbrum of the followng extenson of the CBB game. We assume that an exogenous fracton of the vewers record randomly one of the two evenng news : these vewers watch the recorded news n case where they reach ther home too late to watch one of them on lve; the CBB game corresponds to the case where = 0. We characterze ths equlbrum when the vewers are unformly dstrbuted on the nterval [0; ]. At equlbrum nether of the two statons broadcasts ts evenng news n the rst part of the evenng; dependng upon the value of, t vares between 4 and e : The densty s shown to be decreasng. Further, for all tme T the probablty of broadcastng news before T ncreases wth. The Non-Exstence of a Pure Strategy Equlbrum Suppose that there are vewers who are unformly dstrbuted on an nterval [0; T ] It s supposed that a gven type t vewer comes back home at tme t and watches the rst newscast whch s broadcast after hs/her return f he/she can do so whle a fracton [0; ] of those mssng the last news wll watch them as they have been wse enough to record one of them whch they randomly recorded before leavng home There are two statons whch broadcast the evenng news and each staton (= ; ) has to choose at whch tme T [0; T ] t wll broadcast ts evenng news n order to maxmze ts audence. Let T = for the sake of smplcty but wthout any loss of generalty. The payo s of the two statons are smply de ned as In some cases t s d cult to have a clearcut dtncton between and evenng and nght news. If audence-maxmzng s a debatable assumpton for characterzng free-to-ar Channels behavor n general we thnk that t s a rather sensble "ntermedate assumpton" regardng more spec cally the evenng news whch may be preceded and followed but not nterrupted by advertsng slots Havng more people watchng the news generally means more people watchng the advertsng slots before and after the news.

3 A = T + ( T ), A = T T + ( T ) ; f T < T () A = T T + ( T ) ; A = T + ( T ), f T > T A = A = + ( ) T f T = T = T Ths game has no pure strategy Nash equlbrum. If ts compettor j broadcasts ts evenng news before T m the best reply of staton s to broadcast ts own news at T =,.e. as late as possble. If j broadcasts ts news after T m the best strategy for s to broadcast ts evenng news just before j. 3 The Mxed-Strategy Equlbrum We rst characterze the unque symmetrcal equlbrum wth absolutely contnuous dstrbuton functons. Then, we show that ths s n fact the unque mxed-strategy equlbrum of ths game. In what follows, we denote by F the cumulatve and by f the densty. Proposton If both medas choose absolutely contnuous dstrbuton functons wth a connected support, the broadcastng news game has a unque,symmetrcal, p T mxed strategy equlbrum where () both rms select F (T ) = for all h h T ( ) ; and F (T ) = 0 for all T 0; ( ) when 6= 0 and () both rms select F (T ) = ln(t ) + for all T e ; and F (T ) = 0 for all T 0; e when = 0: Proof : At any such mxed strategy Nash equlbrum, gven the strategy f(t j ) of staton j staton must have the same expected pay-o whatever ts broadcastng tme T provded t belongs to the nterval [; ] : Z T (T ( ) T j + Z )f(t j)dt j + T + T T j f(t j )dt j = C where C s the constant expected audence. D erentatng wth respect to T we obtan ( T ) f(t )+( Z T Z ) f(t j )dt j ( T ) + T f(t )+ f(t j )dt j = 0 T.e. after smpl catons 3

4 T f(t ) + F (T )) = 0 so that the equlbrum mxed strategy F satsfy the rst order lnear d erental equaton: whch admts as a soluton df dx = x F (x) x F (x) = p x when 6= 0 () or F (x) = ln x when = 0 Obvously must be strctly postve. On the other hand, we must have = as f ths dd not hold true, by choosng a broadcastng tme equal to a rm would trvally have a larger expected audence. Now = mples = therefore : when 6= 0 and = e when = 0: and F () = ( ) = 0 leadng to =. We note that = s decreasng wth respect to. Ths means that the broadcastng of news s lkely to shft later n the evenng f the proporton of households holdng and usng a recorder decreases. Note that converges to e when tends to 0:We plot below the probablty that a staton broadcasts ts evenng news before tme t when = 0. The medan tme at whch a staton broadcasts ts news s approxmately 0:60653 Fgure We now show that the absolutely contnuous equlbrum descrbed n proposton s unque as there are no equlbra wth atoms and (or) dsconnected supports. Lemma A staton s equlbrum dstrbuton functon cannot have an atom at some x < 4

5 Proof. Suppose that the contrary s true. Then we show that there exsts > 0 such that PrfT j [x; x + ]g = 0: Indeed there always exsts an arbtrarly small but strctly postve " such that the expected pay-o of j when playng x " s larger than when playng x + " : there s ndeed a rst-order gan from dscontnuously reducng the probablty that staton s gong to broadcast ts news before staton j whch straghtforwardly outweghs a second-order loss from reachng a slghtly lower audence by broadcastng a bt earler. Now staton would clearly gan by choosng a slghtly later broadcastng tme than x (the probablty that T > T j would be unchanged whle the audence would be ncreased) and so the equlbrum dstrbuton functon of cannot have an atom at x: Lemma 3 The supports of the equlbrum dstrbuton functons are connected. Proof. Suppose to the contrary that staton 0 s equlbrum dstrbuton functon has a "hole" between x and y where y > x. It s straghtforward to show that then, there exsts some " such that staton j would be better o by broadcastng at y rather than at some t [x "; y) : the dscontnuous ncrease n audence compensates (and outweghts for t > x ") the ncrease n the probablty that T j > T : Thus j s equlbrum dstrbuton must have a hole between x " and y: Applyng the same argument, we can now show that staton 0 s equlbrum dstrbuton functon must have a hole begnnng from some x, where > 0; to y, hence a contradcton. It remans to show that the connected equlbrum dstrbuton functons cannot have an atom at : Lemma 4 The equlbrum dstrbuton functons are atomless. 5

6 Proof. From Lemmas and, the equlbrum dstrbuton functons are connected and cannot have atoms at some t < : Suppose that the dstrbuton functon of player has an atom at,.e. staton broadcasts ts news at T = wth a operablty > p > 0: From Lemma the support of ts dstrbuton functon must be an nterval [ ; ] wth R f(t )dt = p: On the other hand the support of j 0 s dstrbuton functon must be a sem-open nterval [ j ; ) (gven that s dstrbuton has an atom at, staton j wll never broadcast at T j = ).. Snce j has the same equlbrum expected pay-o from broadcastng at any tme n [ ; ) t must be 3 that f (T ) = p T T T so that R f (T )dt = p entals that = p : Applyng the same reasonng to staton j R j f j (T j )dt j = entals j = < : However ths s mpossble n equlbrum snce staton j would earn a greater pay-o by broadcastng ts new at T j = rather than at any j T j < 4 : Our last result descrbes the comparatve statcs of the equlbrum wth respect to. Proposton For all [0; ], let F (x) = over the nterval h ;. Then f < 0, F domnates F 0 accordng to rst degree stochastc domnance. p x Proof Let t. Under ths change of varable, the functon F, now denoted, s de ned as follows : We now prove that : (t) = t t x t wth t [; +[ (t) 0 (t) 0 Straghtforward calculatons lead to : (t) = + Therefore (t) 0 a(t) b(t) where : (t ) ln x t e ln x t Note that : a(t) e ln x (t ) ln x t and b(t) + t a 0 (t) = ln x t e ln x t > 0 snce x 3 See above equaton () and use = : 4 Of course a smlar argument holds when = 0: 6

7 and : b 0 (t) = t t 3 ln x < 0 snce t Ths mples that (s) 0 for all s [; +[ lm t! a(t) lm t! b(t). Snce lm a(t) = lm b(t) = t! t! the concluson follows and the proof s completed. 4 Concluson We have shown that the "televson news schedulng game" has a unque strategy equlbrum and we have characterzed ths equlbrum. The nterest of ths game goes beyond the smple understandng of evenng news schedulng by TV- Channels. The model ntroduced by Cancan, Bergström and Blls s an extreme case of Hotellng locaton problems wth drectonal constrants when movng towards the left or towards the rght s completely mpossble. A more general analyss of Hotellng locaton games wth drectonal constrants where movng towards the rght s more costly than movng towards the left, or the reverse, would certanly be nterestng. wth possble applcatons n poltcal economy. Ths s on our research agenda. References [] Cancan M., Bergström T. and A. Blls, 995, Hotellng Locaton Problems wth Drectonal Constrants: An Applcaton to Televson News Schedulng, Journal of Industral Economcs, 43, -4. 7