A Bidding Game with Heterogeneous Players

Transcription

1 A Bidding Game with Heterogeneous Players Alberto Bressan and Deling Wei Department of Mathematics, Penn State University University Park, Pa 6802, USA s: December 2, 203 Abstract A one-sided limit order book is modeled as a noncooperative game for n players Agents offer various quantities of an asset at different prices p [0,P], competing to fulfill an incoming order, whose size X is not known a priori Players can have different payoff functions, reflecting different beliefs about the fundamental value of the asset and probability distribution of the random variable X In [4] the existence of a Nash equilibrium was established by means of a fixed point argument The main issue discussed in the present paper is whether this equilibrium can be obtained from the unique solution to a two-point boundary value problem, for a suitable system of discontinuous ODEs Some additional assumptions are introduced, which yield a positive answer In particular, this is the case when there are exactly 2 players, or when all n players assign the same exponential probability distribution to the random variable X A counterexample shows that these assumptions cannot be removed Keywords: optimality conditions, discontinuous ODE, optimal pricing strategy, bidding game, Nash equilibrium, limit order book Introduction This paper is concerned with a continuum model of the limit order book in a stock market, viewed as anoncooperativegame fornplayers As in[3]ourmain goal isto studytheexistence of a Nash equilibrium, determining the optimal bidding strategies of the various agents who submit limit orders We assume that an external buyer asks for a random amount of X > 0 of shares of a certain asset This external agent will buy the amount X at the lowest available price, as long as this price does not exceed a given upper bound P One or more sellers offer various quantities of this asset at different prices, competing to fulfill the incoming order, whose size is not known a priori Having observed the prices asked by his competitors, each seller must determine an optimal

2 strategy, maximizing his expected payoff Of course, when other sellers are present, asking a higher price for a stock reduces the probability of selling it The model introduced in [3] was extended in [4], assuming that agents differ from each other in various respects Each agent assigns a different probability distribution to the random variable X, based on his own beliefs An optimistic seller expects a large incoming order, which will fill most of the outstanding bids A pessimistic seller will expect a small order, filling only the lowest priced bids In the following, we denote by ψ i (s = Prob{X > s} ( the probability distribution assigned by the i-th player to the random variable X Each agent assigns a different fundamental value p i to the assets he is putting on sale In other words, to the i-th agent it would be indifferent to sell his assets at unit price p i or to keep them Existence of a Nash equilibrium, in this more general setting, was recently proved in [4] by means of a topological technique However, this technique did not provide information about the uniqueness of the solution, or how to construct it Aim of the present paper is to show that, in several cases, this Nash equilibrium solution can be found by solving a a Boundary Value Problem for a system of ODEs Let κ i be the total amount of assets offered for sale by the i-th agent We use the Lagrangian variable β [0,κ i ] to label one particular asset By a pricing strategy for the i-th seller we mean a nondecreasing map φ i : [0,κ i ] [0,P] To compute the expected payoff achieved by this strategy, let φ j, j =,φ n, j i be the pricing strategies adopted by the other agents, and define ( F j (p = meas {β [0,κ j ]; φ j (β < p} (2 Moreover, let Φ i (p = j i F j (p (3 be the total amount of assets put on sale at price < p by all the other agents The expected payoff for the i-th player is then measured by J i (φ i,φ i = κi 0 (φ i (β p i ψ i (β +Φ i (φ i (β dβ (4 The integrand in (4 contains two factors The term φ i (β p i is the difference between the price at which the asset β is put on sale and its actual value to the i-th player The term ψ i (β+φ i (φ i (β is the subjective probability (according to the i-th player that the asset β will be actually sold For future reference, we record some basic assumptions on the probability distributions ( 2

3 (A All maps s ψ i (s, i =,,n are continuously differentiable and satisfy ψ i (0 =, ψ i (+ = 0, ψ i(s < 0 for all s > 0, (5 (lnψ i (s 0 for all s > 0 (6 Example The assumptions (A are satisfied if ψ i (s = e αs or if ψ i (s = (+s α, for some α > 0 On the other hand, (6 fails if ψ i (s = e s2 Definition Let (φ,,φ n be an n-tuple of pricing strategies such that no two players put a positive amount of assets for sale at exactly the same price We say that these strategies provide a Nash equilibrium if, calling Φ,,Φ n the corresponding functions in (2-(3, one has J i (φ i,φ i J i (φ i,φ i (7 for every i =,,n and every other pricing strategy φ i : [0,κ i ] [0,P] for the i-th player Remark As a basic modeling assumption, the asset will always be bought from whoever seller offers the lowest price However, if two or more sellers put a positive amount of asset for sale exactly at the same price, one needs to specify which of the agents has selling priority This would require an additional discussion of the model However, in a Nash equilibrium this situation never happens, because the player that does not have priority can always improve his expected payoff by slightly reducing his price In the general case where agents have different payoff functions, the existence of at least one Nash equilibrium was recently established in [4] The proof relied on a sequence of discrete approximations, combined with a topological fixed point argument Apart from the case of players with the same payoff functions, studied in[3], the uniqueness of Nash equilibria remains an open problem In the present paper we seek a more explicit way to construct the functions (F,,F n in (2, and hence determine the equilibrium solution Toward this goal, we introduce a boundary value problem for a a system on n ODEs, determining the functions F j These equations are obtained by adding some auxiliary inequalities to the set of necessary conditions for optimality derived in [3] Compared with classical literature, this problem is far from standard It consists of a system of ODEs F i (p = Q i(p,f (p,,f n (p, i =,,n, (8 where the right hand sides are discontinuous along the hyperplanes where F j = κ j The boundary data take the form together with where p A [0,P] is a point to be determined F i (P = κ i for all i except at most one, (9 F i (p A = 0 i =,,n, (0 3

4 In Section 2 we give a precise description of the right hand side of (8, and study its properties In Section 3 we prove the existence and uniqueness of the solution to the boundary value problem (8 (0 As shown by a counterexample, this solution may not yield a Nash equilibrium, in general This is because the ODEs in (8 are obtained by imposing some additional inequalities which are not implied by the optimality conditions Our main result, proved in Section 4, provides various sufficient conditions in order that the solution to (8 (0 yield a Nash equilibrium In particular, this is always the case if either (i there are exactly two players, or (ii in (4 the probability distribution functions have the form ψ (s = = ψ n (s = e λs, while p,,p n can be arbitrary In recent literature, models of the limit order book have been studied in [, 5, 8, 9, ] For a general introduction to game theory and Nash equilibria we refer to [2, 6, 2] 2 An algebraic problem The main goal of this section is to study a class of functions Q i appearing on the right hand side of the ODEs (8 As a first step, we consider a set of linear equations with constraints Lemma Given n 2 numbers 0 < a a 2 a n, (2 there exists a unique n-tuple (x,x 2,,x n of non-negative numbers with the following properties: x j a i i =,,n, (22 j i x i > 0 = x j = a i, (23 j i n x i > 0 (24 i= Proof Consider the integer m = min k {2,,n}; k k a j a k+, (25 where a n+ = +, so that the inequality in (25 is always satisfied when k = n It is straightforward to check that all conditions (22 (24 are satisfied by setting x i = m a j (m 2a i i =,,m, (26 j m, j i j= x m+ = x m+2 = = x n = 0 (27 4

5 2 It remains to prove that the solution is unique As a first step, we claim that any solution (x,,x n must be of the form (26-(27, for some integer m {2,,n} Indeed, assume that x l = 0 but x k > 0 for some l < k Then (22 implies x j < j k n x j a l a k j= Hence (23 is not satisfied when i = k 3 For a given m, the formulas (26-(27 uniquely determine the n-tuple (x,,x n To complete the proof of uniqueness, it remains to show that there exists at most one integer m such that this n-tuple satisfies the conditions (22 (24 Assume, on the contrary, that (x,,x m,0,,0 and (x,,x m,0,,0 are two distinct solutions, with 2 m < m n, x m > 0 Since (x,,x m,0,,0 is a solution, by (22 it follows m m a j = j= m x j a m+ (28 j= This implies x m = = m m m a j (m 2a m j= m a j +a m+ + j= m j=m+2 a j (m 2a m ( m (m a m+ +a m+ +(m m 2a m (m 2a m ( m (m 2a m (m 2a m 0 This contradicts the assumption x m > 0, proving that the solution is unique Corollary Given any n-tuple a = (a,a 2,,a n of strictly positive numbers (not necessarily ordered as in (2, there exists a unique vector x = (x,,x n satisfying the conditions (22-(24 Indeed, we can always find a permutation π of the set of indices {,,n} such that and apply Lemma 0 < a π( a π(2 a π(n, (29 5

6 Given a = (a,a 2,,a n, we shall denote by G(a = (G (a,,g n (a = (x,,x n (20 the unique solution of (22-(24 Observe that, if k,l are two indices such that a k = a l, then by uniqueness it follows G k (a = G l (a The next lemma collects some properties of the map G Lemma 2 The map G = (G,,G n is Lipschitz continuous and quasi-monotone Namely, given two n-tuples a = (a,,a n and ã = (ã,,ã n, if then G i (a G i (ã a i = ã i, a j ã j for all j i, (2 Proof By Corollary, the map G is well defined As shown by the proof of Lemma, for any given a there exists a unique subset of indices I(a {,,n} with cardinality m = #I(a such that a j (m 2a i > 0 if i I(a, m x i = G i (a = j I(a, j i (22 This implies the a priori bounds 0 if i / I(a 0 x i = G i (a j a j (23 2 From the equations (22-(24 it follows that the map G has closed graph Namely, given sequences a ν = (a ν,,aν n, x ν = (x ν,,xν n, ν such that a ν ā, x ν x, a ν = G(x ν for every ν, it follows that x = G(ā Being a locally bounded function with closed graph, G is continuous 3 The conclusion of the lemma is clearly a consequence of the following claim: (C Fix any i {,,n} and let any (n -tuple of numbers (a,,a i,a i+,,a n be given Then the maps s G j (a,, a i, s, a i+,, a n (24 are Lipschitz continuous and monotone Namely, G i is decreasing, while all other G j for j i are increasing wrt s 6

7 A proof of (C will be worked out in the remaining steps 4 Up to a permutation of indices, it is not restrictive to assume that i = n and 0 < a a 2 a n Call a(s = (a,, a n,s By the continuity of G it follows that the maps s G j (a(s are all continuous If ]s,s + [ is an open interval where the set of indices I(a(s remains constant, from the formula (22 it follows that the maps G j are all uniformly Lipschitz continuous, with s G i (a(s decreasing while s G j (a(s increasing for all j i 5 To complete the proof of (C, it remains to show that, as s increases, there are finitely many values 0 < s < s 2 < < s N such that the set I(a(s is constant on each open interval ]s k, s k [ In turn, for any i {,,n }, it suffices to show that the set I(a(s changes only finitely many times as s ranges in the subinterval ]a i,a i+ [ This clearly follows as a consequence of (25 Remark 2 According to (25, if a k j ka j, then x k = 0 Hence, for i k, the values x i = G i (a,,a n provide the unique solution to the system of (n equations x j a i i {,,n}\{k},, (25 j i,k x i > 0 = x j = a i, (26 j i,k x i > 0 (27 i k obtained from the (22 (24 by removing a k Lemma 3 Consider any n-tuple a = (a,,a n of strictly positive numbers, and let a = (a,a k,a k+,a n be the (n -tuple obtained by removing the entry a k For i =,,n, let x i = G i (a be the solution of (22 (24 Moreover, for i k let x i = G i (a be the solution of the corresponding system of (n equations obtained by removing a k Then x j x j for all j k (28 Moreover, if x k > 0, then there are at least two distinct indices k,k 2 {,,n}\{k} such that x k > x k, x k 2 > x k2 (29 Proof For each i k, as a consequence of Remark 2 we have x i x i = a k s G i(a,,a k,s,a k+,,a n ds (220 7

8 Clearly the integrand is non-negative Recalling (26 we see that s G i(a,,a k,s,a k+,,a n = m(s if x i (s > 0 and x k (s > 0, 0 otherwise (22 Here we use the notation x j (s = G j (a,,a k,s,a k+,,a n, while m(s denotes the number of non-zero components in this solution: m(s = #{j; x j (s 0} From the representation (220 it is clear that x j x j for all j k, so that (28 holds 2 To prove (29, we observe that it is not restrictive to assume that the a i are arranged in increasing order, as in (2 We consider two cases Case : k 3, so that m(a k k 3 In this case we have x (a k > 0 x 2 (a k > 0 By continuity, thereexists ε > 0 such that x (s,x 2 (s > 0 for all s [a k,a k +ε] By (220-(22, this implies x x ak +ε a k s G (a,,a k,s,a k+,,a n ds ε n > 0 The same estimate holds for x 2 The conclusion thus holds with k =, k 2 = 2 Case 2: k {,2} To fix the ideas, assume k = 2, the case k = being entirely similar We claim that, when s = a 3, Indeed, the inequality introduced in (25 G i (a,s,a 3,a 4,,a n > 0 for i {,2,3} l l a j a l+ j= here cannot be satisfied when l = 2 Hence m(s 3 By continuity, we can find ε > 0 such that m(s 3 for s [a 3 ε, a 3 ] For i {,3} we now have x i x i = a3 a 3 ε a 2 s G i(a,s,a 3,,a n ds s G i(a,s,a 3,,a n ds Choosing k =, k 2 = 3, we reach the desired conclusion ε n > 0 (222 Remark 3 The previous analysis shows that the maps G i (a,,a n can be defined also in thecase wheresome of the a i take the value+, provided that there exist at least two distinct indices j k such that a j,a k < Indeed, one can simply define G i (a = 0 if a i = + For future use, we recall here a standard comparison lemma for solutions to systems of ODEs, originally proved in [7] We recall that a map G = (G,,G n : IR R n IR n is called quasimonotone if for every i {,,n} the following holds 8

9 If y i = ỹ i and y j ỹ j for all j i, then G i (t,y,,y n G i (t,ỹ,,ỹ n Lemma 4 Assume that the map G : IR IR n IR n is Lipschitz continuous and quasimonotone Let t y(t = (y,,y n (t and t ỹ(t = (ỹ,,ỹ n (t be two solutions of the same systems of ODEs ẏ = G(t,y (223 Then the following comparison properties holds (i If then y i (t 0 ỹ i (t 0 for all i =,,n, (224 y i (t ỹ i (t for all i =,,n, t t 0 (225 (ii If in addition to (224 one has the strict inequality y h (t 0 < ỹ h (t 0 for some index h {,,n}, then y h (t < ỹ h (t for all t t 0 3 The two-point boundary value problem Given an n-tuple of pricing strategies (φ,,φ n, consider the functions ( F i (p = meas {β [0,κ i ]; φ i (β < p} i {,,n}, (3 F(p = n F j (p (32 j= If (φ,,φ n provide a Nash equilibrium in the sense of Definition, then the analysis in [3] has shown that the following optimality conditions must hold: For every i {,,n}, if F i (p > 0 then F j (p = j i ψ i (F(p (p p i ψ i (33 (F(p For reader s convenience, we briefly recall how (33 is obtained For a given i {,,n}, if F i (p > 0 this means that the i-th player is putting something on sale at price p, hence φ i (β = p for some β [0,κ i ] The optimality of the strategy φ i implies that, by making a small perturbation φ i (β = p+h, the expected payoff does not increase Therefore 0 = d ( (p+h (p+h p i ψ i β + F j dh j i h=0 ( = ψ i β + ( F j (p +(p p i ψ i β + F j (p j i j i = ψ i (F(p+(p p i ψ (F(p i F j (p 9 j i F j(p j i

10 This yields (33 For a rigorous derivation under general assumptions we refer to [3] In the remainder of this section we shall construct functions F,,F n which satisfy these optimality conditions In the next section, under some additional assumptions, we will prove that these functions provide a Nash equilibrium solution to the bidding game Without loss of generality, we assume that 0 < p p 2 p n < P (34 Given F(p as in (32, define a i (p,f(p = ψ i (F(p (p p i ψ i (F(p if p > p i and F i (p < κ i, + if p p i or F i (p = κ i (35 Our main goal is to prove the existence of a unique solution to the following free boundary value problem ( F i (p = G i a (p,f(p,,a n (p,f(p p [p A,P], i =,,n, (36 F (p A = = F n (p A = 0, (37 F i (P = κ i for all i with the exception of at most one index i (38 Here G,,G n are the functions introduced at (20 The lowest asking price p A ]0,P[ is regarded as a free boundary The index i in (38, corresponding to the unique player that puts a positive amount of assets for sale at the top price P, needs also to be determined as part of the solution Remark 4 The necessary conditions for optimality proved in [3] yield the implication F i (p > 0 = F j (p = j i ψ i (F(p (p p i ψ i (F(p These conditions alone do not uniquely determine a system of ODEs for the functions F i For example, at each p one could choose any two indices j,k and set F j(p = ψ k (F(p (p p k ψ k (F(p, F k (p = ψ j (F(p (p p j ψ j, F (F(p i(p = 0 for i / {j,k} Applying Lemma with a i defined at (35, we can uniquely determine the values x i = F i (p, provided that we impose the additional inequalities F j (p j i ψ i (F(p (p p i ψ i, i =,,n, (39 (F(p corresponding to (22 However, one should keep in mind that these additional inequalities do not follow from the optimality conditions There may be Nash equilibria that do not satisfy 0

11 (39, while the unique solution of the boundary value problem (36 (38 may not yield a Nash equilibrium This issue will be discussed in detail in Section 4 Theorem For i =,,n, let the quantities κ i > 0 and the prices p i as in (34 be given, together with functions ψ i satisfying in (5-(6 Then the boundary value problem (36 (38 has a unique solution Proof Because of (35, the right hand sides of the ODEs in (36 are piecewise smooth, with discontinuities occurring when F i = κ i Our construction will thus be achieved by an inductive algorithm, which restarts every time where the solution reaches a discontinuity INITIAL STEP Consider an initial point q 0 ]p 2,P[, whose precise value will be determined later We begin by solving the system of ODEs (36 with initial data F (q 0 = F 2 (q 0 = = F n (q 0 (30 Since q 0 > p 2, we have a (q 0,0 <, a 2 (q 0,0 < Hence the right hand sides of the ODEs in (36 are well defined and locally Lipschitz continuous By the analysis in Section 2, this Cauchy problem has a unique local solution defined for p q 0 This solution can be continued up to the point } q = P min {p > q 0 ; F i (p = κ i for some index i (3 Here and in the sequel we use the notation a b = min{a,b} INDUCTIVE STEP Now assume that q 0 < q < < q ν have been determined, and the solution has been constructed on the interval [q 0,q ν ] If either (i q ν = P or (ii the set of indices {i; F i (q ν < κ i, q ν > p i } contains less than two elements, then the construction stops In the opposite case, we consider the set of indices I ν = {i; Fi (q ν < κ i } (32 The equations (35-(36 now yield a system of N ν = #Iν (ie, the cardinality of the set I ν differential equations for the components F i, i I ν In addition, F i (p = 0, F i(p = κ i for all i / I ν (33 This Cauchy problem, with initial data provided by the inductive step at p = q ν, has a unique local solution, defined for p q ν This solution can be continued up to the point { } q ν+ = P min p > q 0 ; F i (p = κ i for some index i I ν (34 This achieves the inductive step Clearly the algorithm must terminate after at most n 2 steps, yielding a unique solution to (36-(30, defined on some maximal interval [q 0,q ] In the remainder of the proof we will

12 show that there exists a unique value for the minimum ask price p A such that, setting q 0 = p A, one has q = P and (38 holds More precisely, writing q = q (q 0 to stress the dependence of q on the initial point q 0, one has p A = inf {q0 [0,P]; q (q 0 = P} (35 2 In this step we prove that, for any solution of the boundary value problem (36 (38 which is defined on the entire interval [p A,P], all functions F i are Lipschitz continuous with a uniform Lipschitz constant (independent of p A Toward this goal, choose constants m 0,m such that 0 < m 0 ψ i(s ψ i (s m for all i =,,n, 0 s K = n κ j (36 j= Introducing the constant we claim that δ { = (P p n exp 2Km } m 2, (37 0 F i(p = 0 for all i =,,n, p < p i +δ (38 Notice that, if (38 holds, then (33 implies F j(p max i max p p i +δ ψ i (F(p (p p i ψ (F(p m δ (39 This provides the uniform upper bound on the Lipschitz constant of all functions F j, j =,n In the remainder of this step we thus work toward a proof of (38 To fix the ideas, fix an index i and assume p j < p j = p j+ = = p i = = p k < p k+ for some indices j i k Assume that F i (p i +ε = G i (a (p i +ε,,a n (p i +ε > 0 (320 In this case, as shown by the proof of Lemma, the sum of the two smallest elements in the n-tuple (a,,a n must be greater than a i Hence, min 2 (a (p i +ε,,a n (p i +ε > a i(p i +ε 2 = ψ i(f(p 2εψ i (F(p m 0 2ε (32 Here and in the sequel we use the notation min 2 (a,,a n to denote the second smallest element of the n-tuple (a,,a n Next, assume that a j (p i +ε > m 0 2ε (322 2

13 for some index j For p > p i +ε, if a j (p < then m p p j a j (p = To estimate the right hand side of (323 we consider two cases If p j p i then ψ j (F(p (p p j ψ j (F(p m 0 p p j (323 a j (p m 0 p p j m 0 p p i (324 On the other hand, if p j < p i then the assumption (322 implies Therefore, for p p i +ε we have m p i +ε p j a j (p i +ε > m 0 2ε p p i p p j ε p i +ε p j m 0 2m, a j (p m 0 p p i p p i p p j From (32, 324, and (325 we conclude Observing that min 2 (a (p,,a n (p n i= m 2 0 2m (p p i and setting δ 0 = P pn, from (326 we deduce K Hence P n p i +ε i= proving our claim (38 F i (pdp pi +δ 0 p i +ε F i(p min 2 (a (p,,a n (p n i= F i (pdp pi +δ 0 m 2 0 2m (p p i (325 p i +ε for all p > p i +ε (326 m 2 0 2m (p p i dp = m2 0 2m ln ( δ0 ε { ε δ 0 exp 2Km } m 2, ( By choosing q 0 = P ε with ε > 0 sufficiently small, it is clear that the solution of (36- (30 is well defined on [q 0,P] and satisfies F i (P < κ i for every i Hence the set on the right hand side of (35 is nonempty and the value p A is well defined We claim that, if the minimum asking price p A is defined as in (35, then solution of the Cauchy problem (36-(37 satisfies the terminal condition (38 as well Indeed, consider a decreasing sequence of initial points q ν p A, with q (q ν = P for each ν Let (F,ν,,F n,ν be the corresponding solution to (36 with initial data F,ν (q ν = = F n,ν (q ν = 0, (328 3

14 defined on the interval [p ν,p] By the uniform Lipschitz continuity of the functions F i,ν, proved in step 2, we can extract a subsequence converging to an n-tuple of Lipschitz functions (F,,F n It is straightforward to check that these functions provide a solution to the Cauchy problem (36 (37 on the interval [p A,P] Hence q (p A = P and the infimum in (35 is actually attained as a minimum To prove that this solution (F,,F n also satisfies the terminal condition (38, we assume that, on the contrary, F j (P < κ j and F l (P < κ l for two distinct indices j l We can then find a constant ε > 0 such that F j (P < κ j ε, F l (P < κ l ε (329 Consider a strictly increasing sequence of initial points q 0,ν p A For each ν let F ν = (F,ν,,F n,ν be the corresponding solution to the Cauchy problem (36 with initial data (328 Byassumption, thissolution isdefinedonsomemaximalinterval [q 0,ν, q ν]withq ν < P By quasi-monotonicity, the sequence of solutions is monotone decreasing More precisely, for any two indices µ < ν and any i {,,n} we have F i,ν (p F i,µ (p for all p [q µ, q (q µ ] [q ν, q (q ν ] We now observe that the pointwise limit F = ( F,, F n, defined as F i (p = inf ν F i,ν(p p A p < supq (q ν (330 ν provides a solution to the Cauchy problem (36-(37 By uniqueness, F = F and hence q (q ν P as ν A contradiction is now obtained as follows By the analysis in step 2, the derivatives F i,ν remain uniformly bounded In particular, we can assume F j,ν(p M, F l,ν (p M (33 for some constant M and all ν Choose δ > 0 such that Mδ < ε/2 By (330 and (329, for ν sufficiently large we have F j,ν (P δ < F j (P δ+ ε 2 κ j ε 2, F l,ν(p δ F l (P δ+ ε 2 κ l ε 2 By (33, this implies F j,ν (p < κ j, F l,ν (p < κ l for all p [P δ, P] Therefore, q (q ν = P, the minimality of p A This contradiction proves our claim, ie the terminal condition (38 is satisfied 4 It now remains to prove that the solution to the boundary value problem (36 (38 is unique Clearly, as soon as the initial point p A is chosen, the solution to the Cauchy problem (36-(37 is unique Consider two starting points p A < pã and let F, F be the solutions to the corresponding problems F i (p = G i ( a (p,f(p,,a n (p,f(p F (p = = F n (p = 0 p [0,p A ], 4 p [p A,P], i =,,n, (332

15 ( F i (p = G i a (p, F(p,,a n (p, F(p F (p = = F n (p = 0 p [pã,p], i =,,n, p [0,pÃ], To prove uniqueness for the boundary value problem it suffices to show: (333 (C For every p ]p A,P] one has F i (p F i (p for all i {,,n} (334 Moreover, there exist at least two indices j,k (possibly varying with p, such that F j (p < F j (p, Fk (p < F k (p (335 Indeed, if F = (F,,F n is a solution to the boundary value problem (36 (38, then F cannot be a second solution, because for some indices j,k F j (P < F j (P κ j, Fk (P < F k (P κ k, hence the terminal condition (38 fails The uniqueness of the solution thus follows from our claim (C To prove (C, we proceed by induction For i =,,n, define { } P i = min p pã; F i (p = κ i, P i = min { p pã; Fi (p = κ i } (336 Rearranging these values in increasing order, we can write {pã, P,,P n, P,, P n } = {τ 0,τ,,τ N }, with pã = τ 0 < τ < < τ N = P We compare the solutions of the two Cauchy problems (332, (333 The inequalities in (334 can be proved by induction on l =,,N Indeed, they trivially hold when p = pã Assuming that (334 holds for p = τ l, by (i in Lemma 4 and the quasi-monotonicity of the right hand sides of (332-(333 we conclude that the same inequalities are true for p [τ l, τ l+ ] The strict inequalities in (335 will also be proved by induction on the intervals [τ l,τ l+ [ INITIAL STEP For every τ [p A, pã], by Lemma there are two indices j k (possibly depending on τ such that F j (τ > 0, F k (τ > 0 For every p ]p A, pã], integrating over the interval [p A, p] we conclude that there are at least two indices j k such that F j (p > 0 = F j (p, F k (p > 0 = F k (p (337 5

16 INDUCTIVE STEP Assume that the inequality in (335 has been proved for all p ]p A, τ l [, for some 0 l < N We show that it remains valid on the interval [τ l, τ l+ [ as well For any p, define the sets of indices L(p = {i; F i (p < κ i } = {i; p < P i } To achieve the inductive step, consider any p ]τ l,τ l [ By the inductive hypothesis, there exists two indices j k such that Two cases will be considered CASE : τ l / { P j, P k } Observe that this implies F j ( p < F j ( p, Fk ( p < F k ( p (338 F j (p < κ j, Fk (p < κ k for all p < τ l+ In this case, using part (ii of Lemma 4 we first conclude that the inequalities (338 hold for all p [ p,τ l ] A second application of Lemma 4 shows that the same strict inequalities hold also for p [τ l, τ l+ [ CASE 2: τ l { P j, P k } To fix the ideas, assume τ l = P j = min{p; Fj (p = κ j } Observe that in this case we must have F j (p = κ j for all p τ l Otherwise the relations F j ( p < F j ( p, Fj (τ l = F j (τ l would provide a contradiction with part (ii of Lemma 4 We are thus in the situation shown in Fig κ i F j F ~ j Fh ~ F h p A p~ A _ ~ τ p τ =P j p Figure : At p = τ l = P j the functions F j and F j become equal However, during the previous interval [τ l,τ l ] there are at least two other indices h,h such that F h > F h and F h > F h on a set with positive measure We claim that there exists at least two distinct indices h,h {,,n} such that F h (τ l < F h (τ l, Fh (τ l < F h (τ l (339 6

17 Suppose on the contrary that (339 fails Then there exists an index k such that F i (τ l = F i (τ l for all i k (340 To achieve a contradiction, observe that (340 implies F i (p = F i (p for all p [ p, τ l [, i L(p, i k (34 Since F j ( p < F j (τ l = κ j, there is a subset S [ p,τ l ] of positive measure such that F j (p > 0 for every p S By (29 in Lemma3, for every p S we can findat least one index h L(p, h k (possibly depending on p, such that F h (p > F h (p (342 This is clearly in contradiction with (34 We thus conclude that (339 holds If now τ l = P, we are done Otherwise, using again part (ii of Lemma 4 we conclude that F h (p > F h (p, F h (p > F h (p for all p [τ l, τ l+ [ This completes the inductive step in the proof of our claim (C 4 Computing the Nash equilibrium Let (F,,F n be a solution of the boundary value problem (36 (38 These functions determine a unique n-tuple of bidding strategies Namely, for every p P, the value F i (p determines the total amount of assets put on sale by the i-th player at price < p By (38 there can be at most one player, say the agent i, who puts a positive amount of assets for sale exactly at the price P However, as explained in Remark 4, this solution to the boundary value problem does not necessarily yield a Nash equilibrium We further illustrate this point by an example Example 2 Consider a bidding game for three sellers We assume p =, p 2 = p 3 = 4, ψ (s = e s, ψ 2 (s = ψ 3 (s = e 4s, (4 The values 0 < κ < κ 2 = κ 3 and P will be chosen later, so that the minimum ask price will turn out to be p A = 5 (42 Let (F,F 2,F 3 be the solution to the BVP (36 (38 constructed in Theorem Since Players 2 and 3 have the same payoff function, by uniqueness we have F 2 (p = F 3 (p for all p In a small interval of the form [p A,p A +δ], by (33 it follows 2F 2 (p = p p = G (p, F (p+f 2 (p = 4(p p 2 = G 2(p = G 3 (p (43 7

18 Therefore F (p = 4(p p 2 2(p p, F 2 (p = F 3 (p = 2(p p (44 We choose κ = ε > 0 sufficiently small Observe that, when p p A = 5, by (4 the right hand sides of (44 take the values F i (p /8 Therefore, we can uniquely determine the value q(ε = min {p > p A ; F (p = ε} = 5+8ε+o(ε (45 For p > q(ε one has F (p ε, F 2(p = F 3(p = Given ε > 0 sufficiently small and P >> 5 we set 4(p 4 (46 κ = ε, κ 2 = κ 3 = Notice that, (45 and (47 together imply q(ε 5 dp P 2(p + dp q(ε 4(p 4 (47 κ 2 = κ 3 = ln(p 4+O(ε (48 4 Withtheabovechoiceofκ i, thetriple(f,f 2,F 3 providestheuniquesolutiontotheboundary value problem We claim that, if P is sufficiently large, the above solution is not a Nash equilibrium, because the strategy of Player is not optimal Indeed, when Player puts on sale a total amount κ = ε of assets at price p p A = 5, for ε > 0 small his expected payoff is J (ε = 4ε+o(ε (49 On the other hand, if he puts all his assets for sale at the top price P, his expected payoff is ε J (ε = (P exp{ κ κ 2 s}ds 0 = (P e 2κ ε+o(ε = P P 4 ε+o(ε (40 When P sufficiently large we have J (ε > J (ε, hence the first strategy is not optimal In the remainder of this section we seek additional conditions, which guarantee that the n- tuple of strategies (F,,F n obtained by solving the boundary value problem (36 (38 provides a Nash equilibrium to the bidding game Toward this goal, we shall use Lemma 5 A sufficient condition in order that the strategy φ i for the i-th player be optimal is ( } (φ i (β p i ψ i (β+φ i (φ i (β = max {(p p i ψ i β+φ i (p for ae β [0,κ i ] p [0,P] (4 8

19 Proof Intuitively, the above statement should be clear If φ i yields the maximum expected payoff from the sale of each single asset β [0,κ i ], then φ( is optimal To prove the lemma, for any admissible strategy ϕ : [0,κ i ] [0,P] we simply observe that κi J(ϕ,Φ i = (ϕ(β p i ψ i (β +Φ i (ϕ(β dβ 0 = = κi 0 κi 0 max p [0,P] ( } {(p p i ψ i β +Φ i (p dβ (φ i (β p i ψ i (β +Φ i (φ i (β dβ = J(φ i,φ i To proceed further, we need to introduce an additional condition: (H For every i {,,n} and every p < P, the following implication holds If G i (a ( p,,a i ( p,a n ( p > 0 for some p ]p i,p], then for every p ] p,p[ one has ψ i (F(p G i (a (p,,a i (p, (p p i ψ i (F(p,a i+(p,,a n (p > 0 (42 Roughly speaking, the assumption (H means that, if i-th player puts some asset for sale at price p, then he continues to put assets for sale at every price p [ p,q i ], for some q i such that F i (q i = κ i The only reason for which he does not put assets for sale at prices p > q i is that he simply does not have anything more to sell In this case, recalling (35, we have a i (p = ψ i (F(p (p p i ψ i (F(p if p < q i, + if p [q i, P] (43 Lemma 6 Under the same assumptions as in Theorem, let p (F,F n (p be a solution to the boundary value problem (36 (38 If the condition (H holds, then the corresponding pricing strategies yield a Nash equilibrium to the bidding game with payoffs (3-(4 Proof For every i {,,n} we need to show that the strategy φ i of the i-th player is a best reply to the strategies adopted by all other players By Lemma 5, this is the case if (4 holds Calling E(β = d [ ] (p p i ψ i (β+φ i (p dp = ψ i (β+φ i (p+(p p i ψ i (β+φ i(p Φ i (p, (44 our conclusion will be reached by proving that { E(p 0 if p < φ(β, E(p 0 if p > φ(β (45 9

20 2 When p < φ i (β, since F i (p β < κ i and the equations (36 are satisfied, recalling (22 we have Φ i (p = F j (p a ψ i (F(p i(p = (p p i ψ (46 j i i (F(p Inserting (46 in (44 and recalling that ψ i < 0, for p < φ i(β we thus obtain d [ ] (p p i ψ i (β +Φ i (p dp ψ i (β +Φ i (p+(p p i ψ i(β +Φ i (p ψ i (F(p (p p i ψ i (F(p [ = ψ i(β ψi (β +Φ i (p +Φ i (p ψ i (β +Φ i(p ψ ] i(f(p ψ i (F(p 0 (47 Indeed, β + Φ i (p F(p Moreover, by the assumption (6 the map s ψ i (s/ψ i (s is non-increasing 3 Next, assume p > φ i (β Using the assumption (H with p = φ i (β, we obtain Φ i(p ψ i (F(p (p p i ψ i, (48 (F(p with equality holding if F i (p < κ i A similar computation as in (47 now yields E(p ψ i (β +Φ i (p+(p p i ψ i (β +Φ i(p [ = ψ i (β +Φ ψi (β +Φ i (p i(p ψ i (β +Φ i(p ψ ] i(f(p ψ i (F(p ψ i (F(p (p p i ψ i (F(p 0 (49 Indeed, in this second case we have β +Φ i (p F(p This establishes the second inequality in (45, completing the proof Using Lemma 6, we can give a number of sufficient conditions in order that the pricing strategies constructed in Theorem provide a Nash equilibrium Theorem 2 Under the same assumptions as in Theorem, let p (F,F n (p be a solution to the boundary value problem (36 (38 These strategies provide a Nash equilibrium to the bidding game if any one of the following assumptions holds (i n = 2 (ii For every i,j {,,n} and max{p i,p j } < p < P, one has d dp ln ψ i (F(p ψ j (F(p (p p i ψ i ln (F(p (p p j ψ j (F(p 0 (420 (iii ψ (s = = ψ n (s = e λs for some λ > 0 20

21 Proof In the case of two players, let (F,F 2 be the solution to the boundary value problem (36-(38 Then F (p = ψ 2 (F(p (p p 2 ψ 2 (F(p = a 2(p > 0, F 2 (p = ψ (F(p (p p ψ (F(p = a (p > 0, (42 for all p ]p A,P[ Hence the assumption (H in Lemma 6 trivially holds Part (i of the theorem is thus a consequence of Lemma 6 2 To prove part (ii, let the inequality (420 hold To understand the basic case, for i =,,n define α i (p = ψ i (F(p (p p i ψ i (F(p if p > p i, + if p p i (422 After a permutation of indices, we can assume that α (p α n (p for all p (423 Indeed, by (420, if α i ( p = α j ( p < then α i (p = α j (p < for all p p Hence the difference α i (p α j (p can never change sign Note that (423 implies Otherwise, if i < j but p i > p j, then lim a j(p = p p i + in contradiction with (423 p p n (424 ψ j (F(p i (p i p j ψ j (F(p i < lim p p i + a i(p = +, Next, assuming that for some p, we claim that G i (α ( p,,α n ( p > 0 (425 G i (α (p,,α n (p > 0 for all p [ p, P[ (426 Indeed, as shown in the proof of Lemma, one has G i (α (p,α n (p > 0 if and only if k k α j (p > α k+ (p j= k = 2,,i Equivalently, this holds if and only if k k j= α j (p α k+ (p > k = 2,,i (427 2

22 By (423 and (420 it follows that, for j k +, ( αj (p ln 0, α k+ (p d dp ( αj (p α k+ (p 0 Hence, if (427 holds for p = p, the same holds for all p p 3 The argument in the previous step shows that, if a i (p = α i (p for all i =,,n, then the assumption (H in Lemma 6 is satisfied By applying Lemma 6 we conclude that the solution (F,,F n of the boundary value problem yields a Nash equilibrium To complete the proof of (ii, we need to consider the case where the a i (p defined at (43 do not necessarily coincide with the α i (p in (422 This happens precisely when p > p i, F i (p = κ i and a i (p = Assume that, at some point p, one has G i (a ( p,,a i ( p,,a n ( p > 0 (428 Clearly this implies a i ( p = α i ( p < Notice that, if j < i and a j ( p =, then a j (p = for all p p For notational convenience, for any p p define b j (p = { if j < i and aj ( p =, α j (p otherwise For p p we then have G i (a (p,,a i (p,α i (p,a i+ (p,,a n (p G i (b (p,,b i (p,α i (p,b i+ (p,,b n (p > 0 (429 Indeed, the first inequality is a consequence of the quasi-monotonicity of the maps G i, proved in Lemma 2 The second inequality is obtained from (428, using the arguments in step 2, after discarding the components j < i for which a j ( p = Having proved that the assumption (H holds, by an application of Lemma 6 we conclude that the solution (F,,F n to the boundary value problem yields a Nash equilibrium 4 To prove part (iii of the theorem we show that, if ψ (s = = ψ n (s = e λs, then (420 holds Indeed, to fix the ideas assume p i p j For p > p i we then have d dp ln λ(p p i ln λ(p p j = d [ ] ln(p p j ln(p p i = 0 dp p p j p p i Therefore (iii follows as a special case of (ii Remark 5 (uniqueness As proved in Theorem, the solution to the two-point boundary value problem(36 (38 is always unique Under the additional assumption(h, this solution yields a Nash equilibrium However, this does not necessarily imply that the Nash equilibrium is unique Indeed, in principle there may be other Nash equilibria which are not obtained by solving our system of ODEs 22

23 In the case n = 2, the uniqueness of the Nash equilibrium follows easily from the necessary conditions (42 We conjecture that uniqueness also holds under the assumption (420 in Theorem 2 It remains an open problem to understand if there can be multiple Nash equilibria, in cases where the assumptions (i (iii in Theorem 2 fail Acknowledgment This research was partially supported by NSF, with grant DMS-08702: Problems of Nonlinear Control References [] K Bach and S Baruch, Strategic liquidity provision in limit order markets Econometrica, 8 (203, [2] A Bressan, Noncooperative differential games Milan J of Mathematics, 79 (20, [3] A Bressan and G Facchi, A bidding game in a continuum limit order book, SIAM J Control Optim, to appear [4] A Bressan and G Facchi, Discrete bidding strategies for a random buying order SIAM J Financial Math, submitted [5] R Cont, S Stoikov, and R Talreja, A stochastic model for order book dynamics, Operations Research 58 (200, [6] E J Dockner, S Jorgensen, N V Long, and G Sorger, Differential games in economics and management science Cambridge University Press, 2000 [7] E Kamke, Zur Theorie der Systeme gewöhnlicher Differentialgleichungen II, Acta Math 58 ( [8] A Obizhaeva, and J Wang, Optimal trading strategy and supply/demand dynamics, Journal of Financial Markets, to appear [9] T Preis, S Golke, W Paul, and J J Schneider, Multi-agent-based order book model of financial markets Europhysics Letters 75 (2006, [0] S Predoiu, G Shaikhet, and S Shreve, Optimal execution in a general one-sided limitorder book SIAM Journal on Financial Mathematics 2 (200, [] I Rosu, A dynamic model of the limit order book The Review of Financial Studies 22 (2009, [2] J Wang, The Theory of Games Oxford University Press,