Volume 29, Issue 4. Incomplete third-degree price discrimination, and market partition problem. Yann Braouezec ESILV

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1 Volume 29, Issue 4 Incomplete thrd-degree prce dscrmnaton, and market partton problem Yann Braouezec ESILV Abstract We ntroduce n ths paper the "ncomplete" thrd-degree prce dscrmnaton, whch s the stuaton where a monopolst must charge at most k dfferent prces whle the total market s composed of n markets, wth n>k. We thus study the optmal partton problem of the n markets n k groups. As a byproduct, when k=2, we are able to reconsder the so-called (Robnsonan) weak-strong partton. Ctaton: Yann Braouezec, (2009) ''Incomplete thrd-degree prce dscrmnaton, and market partton problem'', Economcs Bulletn, Vol. 29 no.4 pp Submtted: Sep Publshed: November 20, 2009.

2 1 Introducton The common feature of the models n the thrd-degree prce dscrmnaton lterature n monopoly s that there are n local markets, ndependent or not, each characterzed by a local demand functon, lnear or not. The typcal problem (e.g., Schmalensee 1981, Malueg 1993, Varan 1988, Cowan 2008) s to compare, from a welfare pont of vew, the case n whch the monopolst s not allowed to prce dscrmnate,.e., she must charge a unform prce, wth the case n whch she may (thrd-degree) prce dscrmnates,.e., she may charge n dfferent prces. The monopolst charge thus ether one prce, or n (dfferent) prces. Suppose now that for some reason (e.g., regulatory, marketng...), the monopolst must charge at most k dfferent prces, wth 2 k < n so that some markets must be grouped together. The problem of the monopolst s now to fnd the best partton of the n markets n k groups, and the prce assocated to each group that maxmzes the aggregate proft. To solve ths optmzaton problem, the monopolst must thus consder each possble partton A = (A 1,...A k ) of the markets, and then fnd the optmal vector of prces P A = (PA 1,...PA 1 ). The am of ths paper s to provde a smple necessary condton that must be satsfed by the soluton of ths combnatoral-lke optmzaton problem. Do to so, we consder a frm composed of a headquarters, whose job s to propose a "menu" P contanng k prces, and a set of n "experts", whose job s to choose a prce n P. Each expert, mplanted on a local market, perfectly knows her (own) demand functon (hence the name of expert) and chooses the prce n P that maxmzes her (own) proft functon. The menu P nduces thus a partton denoted A(P) of the n experts n at most k dstnct groups. Let P A the vector of prces that maxmzes the aggregate proft of the fxed partton A. When the headquarters proposes P A to the experts, for the partton A to be a possble canddate of the optmzaton problem, P A must nduce the partton A and not another one. When ths s the case, we say that the partton A s admssble. Usng a decentralzed decson process elmnates thus parttons whch are not admssble, but also parttons for whch no prce nduces them. In the partcular case n whch k = 2, we are able to re-examne the so-called Robnsonan weak-strong markets partton (e.g., Robnson 1933, Schmalensee 1981, Cowan 2008), whch s mplctly supposed to be the optmal partton of the markets n two groups. Ths paper s organzed as follows. We frst present the assumptons of the model and the requred termnology; we then defne what an admssble partton s, and we show that such a partton s n general not unque. We prove that an optmal partton exsts and eventually provde a smple way to get a unque admssble partton. 2 The model 2.1 Assumptons We consder a frm whch s n a monopoly stuaton on n separated markets but s allowed to charge at most k dfferent prces, wth 2 k < n. We call ths stuaton an "ncomplete" thrd-degree prce dscrmnaton. One may assume that the frm has n subsdares, and 1

3 that each subsdary j J {1, 2...n} s mplanted n dfferent regon of the world. Let q j (P j ) = (a j P j ) α α > 0 (1) wth a j > 0 j J, be the demand functon of market j, whch s only functon of the prce P j, so that trade among markets s (mplctly) not allowed. Note that when α s lower, equal, or hgher than one, the demand functon s respectvely concave, lnear, or convex. In each subsdary j, there s a local head j J that we call an "expert" because she s assumed to perfectly know her proft functon gven by π j (P j ) = (P j C)(a j P j ) α (2) where C s the constant margnal cost. It s easy to show that the proft functon gven by equaton (2) s sngle-peaked and that P j = a j + αc 1 + α s the monopoly prce of the (local) market j,.e., the prce that maxmzes the proft functon π j (P j ). We shall assume that a j+1 > a j j J so that P1 < P2 <...Pn. For economc nterest, we shall also assume that C < a 1 so that π j (P j ) s a concave functon on ]0, a j [ j J. In our model, the prce decson process of the frm s decentralzed n the followng sense: the headquarters proposes a menu P = (P A1, P A2...P Ak ), wth 0 P A1 P A2... P Ak and each expert j s free to choose one prce n P. Snce each expert must choose one prce, the vector P nduces thus a partton A(P) A = (A 1, A 2,...A k ) of the experts n k groups, where A J s the subset of experts of J who choose the prce P A. By defnton k of a partton, A = J wth A A = for all. Note that for some but not all, =1 A may be empty. 2.2 Set of prces and admssble (consecutve) partton of experts Let V (R k+ ) be the set of prces defned as follows. V (R k+ ) = {P R k+ : 0 P A1 P A2... P Ak P } (4) where P n P, wth P fnte. Note that V (R k+ ) s compact and convex 1, and let Π(A(P)) be the aggregate proft generated by the nduced partton A(P). Π(A(P)) := Π( k A ) = =1 (3) k Π A (P A ) (5) =1 Π A (P A ) = j A π j (P A ) (6) j A π j (P A ) π j (P A ), (7) 1 Compacty follows snce V (R k+ ) s a closed and bounded subset of R k+. Let P, P V (R k+ ) wth P P. Convexty follows snce for all λ [0, 1], λp + (1 λ)p V (R k+ ). 2

4 Equatons (5) and (6) uses the fact that the aggregate proft s an addtve functon. Equaton (7) s an ndvdual ratonalty condton snce each agent j chooses the component P A of P that maxmzes her (own) proft π j (P ). Let us now examne more closely the knd of partton A one may have. Defnton 1. The partton A = (A 1,...A k ) s sad to be consecutve whenever j and j, wth j < j, belong to the set A (.e., they choose the prce P A ), experts j J such that j < j < j also belong to A. Lemma 1. Assume that π j (P j ) = (P j C)(a j P j ) α, wth a j+1 > a j > 0 j J and α > 0. Then, the partton A = (A 1,...A k ) s consecutve. Proof. See the appendx. Let V (A(P)) = {P V (R k+ ) : P nduces the (consecutve) partton A} (8) When k = 2, V (A(P)) s never empty snce for any (consecutve) partton A = (A 1 = {1, 2...h}; A 2 = {h + 1,...n}), one can always choose P = (Ph, P h+1 ). However, when k 3, V (A(P)) may be empty. To see ths, consder the case n whch J = {1, 2...6}, C = 0, P = Pn, and α = 1 so that Pj = a j for all j J. Assume moreover that three (dfferent) 2 prces are allowed,.e., P = (P A1, P A2, P A3 ) V (R 3+ ). Consder now the parameters a 1,...a 6 such that P1 = 1, P2 = 1.9, P3 = 2, P4 = 20, P5 = 20.1, P6 = 21, and the partton A = (A 1 = {1, 2}; A 2 = {3, 4}; A 3 = {5, 6}). If P A2 ]4, 21[, then, at least expert 1,2,3 choose the prce P A1. If P A2 ]0, 4], then, at least experts 4,5 and 6 choose the prce P A3. As a consequence, there s no way to choose P A2 ]0, 21[ to nduce the partton A so that V (A(P)) s empty. When V (A(P)), t s the subset of prces n V (R k+ ) that leaves nvarant the partton A. Let P(A) be the set of (consecutve) parttons for whch V (A(P)). Defnton 2. We say that A P(A) s a feasble V (A(P)) partton for the prce P f P Let Π A (P) be the aggregate proft functon assumng that the partton A s fxed. Defnton 3. We say that A P(A) s an admssble partton f there exsts a vector of prces P A = (P A 1,...P A k ) such that Π A (P A) = 0 Π A (P A ) = j A π j(p A ) = 0 = 1, 2...k (9) P A V (A(P)) (10) We show n lemma A1 n appendx that f PA maxmzes (perhaps locally) the proft functon Π A (P A ) then, PA a j j A, = 1, 2...k. As a consequence, P A has to satsfy the optmalty condton gven by equaton (9). We show n lemma A2 that f P A satsfy 3

5 equaton (9), then, the second order condton for a (local) maxmum s also satsfed. Snce the prce decson process s decentralzed, for the partton A to be admssble, P A must nduce the partton A and not another one,.e., A must be feasble for P A. Let A be an admssble partton and let P(A ) be the set of all admssble parttons. Clearly, P(A ) P(A). We shall denote P A the optmal vector of prces assocated to the admssble partton A. There may ndeed exst (fntely) many admssble parttons 2. To see ths, consder a "smple" partton A = (A = {}, = 1, 2...k 1; A k = {k, k + 1,...n}) and let P A V (Rk+ ) be such that P A = (P A 1 = P1 ;...PA k 1 = P k 1 ; P A k [Pk, P n]. Snce PA = P, each expert chooses the prce P for = 1, 2...k 1. If moreover expert k s such that π k (PA k ) > π k (PA k 1 ), then the partton A s admssble. Of course, there are fntely many such smple parttons and some of them are generally admssble. Defnton 4. We say that A P(A ) s the optmal partton of J f Π(P A ) > Π(P A ) A P(A ) A A (11) Lemma 2. If A P(A ) contans at least one empty group (.e., there exsts at least one {1, 2...k} such that A = ), then, A can not be the optmal partton. Proof. Assume that P = (P A 1,...P A k ) nduces the optmal partton A P(A ) and such that at least A s empty,.e., nobody choose the prce P A. Snce n > k, one can at least fnd one expert j J such that the chosen prce n P s dfferent than her optmal prce P j. Fx now P A = P j. Clearly, ths expert j wll choose ths prce P A, and her proft wll ncrease. But then, the partton change snce now A s not empty. Hence, A can not be the optmal partton. Proposton 1. There exsts a prce P V (R k+ ) that nduces the optmal partton A P(A ) Proof. The aggregate proft functon Π(A(P)) s a contnuous functon snce t s a fnte sum of contnuous functon. Snce V (R k+ ) s a compact (and convex) subset of R k+, Π(A(P)) reaches ts mnmum, and ts maxmum at P, and A s the optmal partton nduced by P so that P V (A (P)). From lemma A1 and A2, we know that P satsfes equaton (9), whch s also a suffcent condton for a maxmum. Let A = (A 1,...A wth k =1 A = J. From lemma 2, A contans k non-empty groups,.e., A k ), = 1...k 2 An analogy may be done wth the coalton structures lterature that assumes free moblty (.e., each agent s free to choose one coalton) n whch there are multple neffcent free moblty equlbra. See Demange (2005). 4

6 2.3 Applcaton when k = 2: a reexamnaton of the weak-strong markets partton The weak-strong markets partton Let Π(P ) = π j (P ) be the proft functon when dscrmnaton s not allowed, and let j J P be the unform monopoly prce. The weak-strong markets partton (see e.g., Cowan 2008, Robnson 1933, Schmalensee 1981, Malueg 1993, Shh et al 1988), denoted W-S s the partton n whch the two sets W and S are defned as follows: W = {j J : P j < P } ; S = {j J : P j > P } (12) Snce π j(p ) < 0 for all j W and π j(p ) > 0 for all j S, when the monopolst s allowed to move from k = 1 to k = 2, t s clear that the proft of the group W (S) ncreases when we decrease (ncrease) the prce. Probably for ths reason, n the thrddegree prce dscrmnaton lterature, t s mplctly beleved that the weak-strong partton s the optmal partton of the n markets n two groups Lnear demand functons Suppose that each expert s endowed wth a lnear demand functon q j (P j ) = a j P j (.e., α = 1) and assume that C = 0 for smplcty, so that the proft functon s smply π j (P j ) = a j P j P 2 j, for P j a j. Let A h be the followng partton A h = (A 1 = {1,...h}; A 2 = {h + 1,...n}) (13) where V (A h (P)) s the set of prces that nduces A h. Snce the partton s consecutve, to nduce the partton A h, t suffces that P A1 and P A2 are such that π h (P A1 ) > π h (P A2 ) and π h+1 (P A1 ) < π h+1 (P A2 ). It s easy to show that the two nequaltes mples that: V (A h (P)) = {(P A1, P A2 ) V (R 2+ ) : (P A1 + P A2 ) ]a h ; a h+1 [} (14) so that V (A h (P) s a convex set. The partton A h wll thus be admssble f there exsts a vector of prces P A h = (PA 1, PA 2 ) such that (PA 1 + PA 2 ) ]a h ; a h+1 [. Our model allows us to examne whether or not the weak-strong partton of the markets n two groups s the optmal partton. In fact, as we shall now show, the weak-strong partton may not be admssble whle, as n Schmalensee (1981), all the markets are served under the unform monopoly prce. To consder such a case, let Ψ(n, h), wth 1 h n 1, and n s a fnte nteger, be the set of parameters a = (a 1,...a n ) R n+, wth a 1 < a 2 <...a n such that: 1. under the unform monopoly prce P, all markets are served,.e., q j (P ) > 0 j J 2. the weak-strong partton s A h = (W = {1,...h}; S = {h + 1,...n} 3. P A h = (P W, P S ) / V (A h(p)) 5

7 If one can fnd a couple (n, h) such that Ψ(n, h) s not empty, ths mples that the weakstrong partton s not always the best way to dvde the set J n two groups. But then, the meanng of the weak-strong partton s not clear. The label weak-strong partton should ndeed be attrbuted only to the optmal partton. We shall now provde an element of Ψ(5, 2). The "dffcult" part s to fnd a such that the weak-strong markets partton s not admssble. To obtan ths property, we thus look for a vector a such that the set V (A 2 (P)) s thn. To ths end, consder the case n whch a 1 = 10, a 2 = 15, a 3 = 15.5, a 4 = 16, a 5 = 20, and recall that Pj = (1/2)a j. It s easy to show that the unform monopoly prce s P = 7.65 so that all the markets are served. As a consequence, the weak-strong markets partton s A 2 = (W = {1, 2}; S = {3, 4, 5}), and P A 2 = (6.25; 8.58). Snce (PW + P S ) = and / ]15; 15.5[, t thus follows that P A 2 / V (A 2 (P)). In fact, snce expert 3 chooses the prce PW, P A 2 V (A 3 (P)). In ths example, P(A ) = {A 1, A 3, A 4 }, and the optmal partton s A 1, whch turns out to be a "smple" one. For the weak-strong markets partton to be the unque admssble partton, the experts of W and S must be "homogeneous" enough n terms of ther optmal prce Pj. To see ths, consder frst the case n whch there are only two experts, a and b, wth demand functon q(x, P ) = x P, for x {a, b}, a < b. It s easy to show that when b < 2a, the proft functon s sngle-peaked, and the unform monopoly prce s P = a+b. Moreover, the two 4 markets are served. Let be a "heterogenety" parameter. To construct the set W and S, at each tme we add an expert n the set W, we also add an expert n S so that the monopoly prce P = a+b 4 remans nvarant. There are thus N experts n each group W and S, and the total number of experts s n = 2N. Ths constructon allows us to control the heterogenety of the optmal prces P j of each group W and S wth the parameter, but also P P j for some j J. When s small enough, then, the weak-strong markets partton s the unque admssble one. The followng proposton gves an example when N = 3. Proposton 2. Let a > 0, b ]a, 2a[, > 0, n = 2N, N = 3. Assume that a j = a + ɛ j and b j = b ɛ j for j = 1, 2, 3, wth ɛ 1 =, ɛ 2 = 0 and ɛ 3 =. When < b a, the weakstrong markets partton gven by W = {a 1, a 2, a 3 }; S = {b 3, b 2, b 1 } s the unque admssble 6 partton. Proof. See the appendx The above proposton s obvously not ntended to gve the most general condtons under whch the weak-strong partton s the unque admssble partton. It s rather to show how one get ths property by restrctng the "heterogenety" of the optmal prce of each group W and S. 3 Concluson In ths paper, we have examned, from the proft-maxmzng monopolst pont of vew, the optmal partton problem of the n markets n k groups. In Braouezec (2009), we consder the welfare problem and we show that, for lnear demands, ncomplete thrd-degree prce dscrmnaton s (generally) optmal from a total welfare pont of vew. 6

8 Appendx Proof of lemma 1. Let P = (P A1,...P Ak ) be the menu of prces. For smplcty, assume that C = 0, and consder experts 1,2 and 3, wth a 1 = a, a 2 = b, and a 3 = c such that 0 < a < b < c. Assume now that experts a and c choose the prce P A. Consder the case P A > P A 1 >... > P A1 > 0. By assumpton, π a (P A ) > π a (P A 1 ), and π c (P A ) > π c (P A 1 ), whch s equvalent to f(x) := P ( ) α A x PA > 1 for x {a, c}. P A 1 x P A 1 Note that we necessarly have a > P A. Snce f (x) > 0 for x > P A, ths mples that 1 < f(a) < f(b) < f(c) for any b between a and c. Consder now the case n whch 0 < P A < P A+1. Now, π a (P A ) > π a (P A+1 ), and π c (P A ) > π c (P A+1 ) s equvalent to f(x) := P ( ) A +1 x α PA+1 < 1 for x {a, c}. Note that a > P A, otherwse expert a P A x P A would have chosen a prce lower than P A. However, t may be the case that c < P A+1, so that c P A+1 = 0. If a > P A+1, snce f (x) > 0, then, 0 < f(a) < f(b) < f(c) < 1. If c < P A+1, then, f(a) = f(b) = f(c) = 0. If c > P A+1 and a < P A+1, then, f(a) = 0 < f(b) < f(c) < 1 for b > P A+1 and f(a) = f(b) = 0 < f(c) < 1 for b < P A+1. In all cases, f a and c choose P A+1, so wll do b Lemma A1. If P A = a j, for some j A, = 1, 2...k, then, ths prce a j can not maxmze the proft functon. Proof. Fx a gven h A and let A h = {j A : j h} and A h+1 = {j A : j h+1}. If j A π j(p h A ) > 0 for P A ]a h 1, a h [, then, j A π j(a h+1 h + ɛ) > 0, ɛ > 0. Note frst that j A π j(p h A ) = π h (P A ) + j A π j(p h+1 A ) > 0 for P A ]a h 1, a h [. Fx ɛ > 0. Snce π h (a h ɛ) < 0 whle π h (P ) = 0 for P > a h and snce j A π j(p h+1 A ) s contnuous for P A [a h ɛ; a h + ɛ], one must have j A π j(p h+1 A ) > 0. As a consequence, the dervatve of the aggregate proft functon can not change of sgn around a pont a j Lemma A2. If P A = (P A 1,...P A k ) solves equaton (9), then, the second order condton s satsfed. Proof. Let A be a set of consecutve ndces and recall that Π A (P A ) = j A π j (P A ). Snce each π j (P A ) s a concave (and twce contnuously dfferentable) functon of P A on ]0, a j [, Π A (P A ) s a concave (and twce contnuously dfferentable) functon of the prce on ]a j, a j+1 [, j A, for = 1,...k. As a consequence, f there exsts PA ]a j, a j+1 [, such that Π A (PA ) = 0, then, by concavty, PA s unque and such that Π A (PA ) < 0. Proof of proposton 2. It s easy (but cumbersome) to construct the proft functon and show that the aggregate proft functon s sngle-peaked f < (2/11)a, so that ths sngle-peakedness property s satsfed for < b a snce b ]a, 2a[. It s also easy to show 6 that Pa 3 < P and Pb 3 > P are satsfed f < b a. By symmetry of each proft functon 2 π j (P ), f γ > γ, then π j (Pj γ ) < π j (Pj + γ) or π j (Pj γ) > π j (Pj + γ ). Let P and P two prces. It thus follows that π j (P ) > π j (P ) f Pj P < Pj P. Consder (PW, P S ), wth PW < P < PS, whch s the optmal vector of prces assocated to the weak-strong partton. Snce the partton s consecutve, f expert a 3 chooses the prce PW, then a 1 and a 2 make the same choce. In the same way, f expert b 3 chooses the prce PS, then, b 1 and b 2 make the same choce. When < b a, P 6 b 3 P > and Pa 3 P >. Snce 7

9 PW [P a 1, Pa 3 ], and Pa 3 Pa 1 = but also PS [P b 3, Pb 1 ], and Pb 3 Pb 1 =, t thus follows that the weak-strong partton s admssble. To show that t s unque, let B S (.e., B s ether {b 3 } or {b 3 ; b 2 }) and consder the consecutve partton A = (W B), (S \B). Note that PW < P W B < P and PS\B > P S > P. The partton A s not admssble snce Pb 3 PS\B < whle P b j PW B > j S,.e., all experts of B choose the prce P S\B and not the prce PW B. Apply the same reasonng for A = (W \ B), (S B), for B W 8

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