Pierre Cardaliaguet
email: cardaliaguet@ceremade.dauphine.fr

Rida Laraki
email: rida.laraki@polytechnique.edu

Sylvain Sorin
email: sorin@math.jussieu.fr

A Continuous Time Approach for the Asymptotic Value in Two-Person Zero-Sum Repeated Games 1

Keywords: Jeux stochastiques, jeux répétés, information incomplète, valeur asymptotique, principe de comparaison, inégalités variationnelles, solutions de viscosité, temps continu Stochastic games, repeated games, incomplete information, asymptotic value, comparison principle, variational inequalities, viscosity solutions, continuous time. Classification AMS: Classification JEL 91A15, 91A20, 93C41, 49J40, 58E35, 45B40, 35B51. C73, D82 AMS classification: 91A15, 91A20, 93C41, 49J40, 58E35, 45B40, 35B51. JEL classification: C73, D82 stochastic games, repeated games, incomplete information, asymptotic value, comparison principle, variational inequalities, viscosity solutions, continuous time

Nous nous intéressons à la valeur asymptotique dans les jeux répétés à somme nulle avec une évaluation générale de la suite des paiements d'étapes. Nous montrons l'existence de la valeur asymptotique dans un sens robuste dans les jeux répétés à information incomplète, les jeux de splitting et les jeux absorbants. La technique de preuve consiste (1) à plonger le jeu répété en temps discret dans un jeu en temps continu et (2) à utiliser les solutions de viscosités.

Introduction

We study the asymptotic value of two person zero-sum repeated games. Our aim is to show that techniques which are typical of continuous time games ("viscosity solution") can be used to prove the convergence of the discounted value of such games as the discount factor tends to 0, as well as the convergence of the value of the n-stage games as n → +∞ and to the same limit. The originality of our approach is that it provides the same proof for both classes of problems. It also allows to handle general decreasing evaluations of the stream of stage payoffs, as well as situations in which the payoff varies "slowly" in time. We illlustrate our purpose through three typical problems: repeated games with incomplete information on both sides, first analyzed by [START_REF] Mertens | The Value of Two-Person Zero-Sum Repeated Games with Lack of Information on Both Sides[END_REF] [START_REF] Mertens | The Value of Two-Person Zero-Sum Repeated Games with Lack of Information on Both Sides[END_REF], splitting games, considered by [START_REF] Laraki | The Splitting Game and Applications[END_REF] [START_REF] Laraki | The Splitting Game and Applications[END_REF] and absorbing games, studied in particular by [START_REF] Kohlberg | Repeated Games with Absorbing States[END_REF] [START_REF] Kohlberg | Repeated Games with Absorbing States[END_REF]. For the splitting games, we show in particular that the value of the n-stage game has a limit, which was not known yet.

In order to better explain our approach, we first recall the definition of Shapley operator for stochastic games, and its adaptation to games with incomplete information. Then we briefly describe the operator approach and its link with the viscosity solution techniques used in this paper.

Discounted stochastic games and Shapley operator

A stochastic game is a repeated game where the state changes from stage to stage according to a transition depending on the current state and the moves of the players. We consider the two person zero-sum case. The game is specified by a state space Ω, move sets I and J, a transition probability ρ from I × J × Ω → ∆(Ω) and a payoff function g from I × J × Ω → IR. All sets A under consideration are finite and ∆(A) denotes the set of probabilities on A. Inductively, at stage n = 1, ..., knowing the past history h n = (ω 1 , i 1 , j 1 , ...., i n-1 , j n-1 , ω n ), player 1 chooses i n ∈ I, player 2 chooses j n ∈ J. The new state ω n+1 ∈ Ω is drawn according to the probability distribution ρ(i n , j n , ω n ). The triplet (i n , j n , ω n+1 ) is publicly announced and the situation is repeated. The payoff at stage n is g n = g(i n , j n , ω n ) and the total payoff is the discounted sum n λ(1λ) n-1 g n . This discounted game has a value v λ (Shapley, 1953 [17]). The Shapley operator T(λ, •) associates to a function f in IR Ω the function T(λ, f ) with :

T(λ, f )(ω) = val ∆(I)×∆(J) [λg(x, y, ω) + (1 -λ) ω ρ(x, y, ω)(ω)f (ω)] (1)
where g(x, y, ω) = E x,y g(i, j, ω) = i,j x i y j g(i, j, ω) is the multilinear extension of g(., ., ω) and similarly for ρ(., ., ω), and val is the value operator val ∆(I)×∆(J) = max .

The Shapley operator T(λ, •) is well defined from IR Ω to itself. Its unique fixed point is v λ (Shapley, 1953 [17]). We will briefly write (1) as T(λ, f )(ω) = val{λg + (1λ)Ef }.

Extension: repeated games

A recursive structure leading to an equation similar to the previous one [START_REF] Aumann | Repeated Games with Incomplete Information[END_REF] holds in general for repeated games described as follows:

M is a parameter space and g a function from I × J × M to IR. For each m ∈ M this defines a two person zero-sum game with action spaces I and J for Player 1 and 2 respectively and payoff function g(., ., m). The initial parameter m 1 is chosen at random and the players receive some initial information about it, say a 1 (resp. b 1 ) for player 1 (resp. player 2). This choice is performed according to some initial probability π on A × B × M , where A and B are the signal sets of both players. At each stage n, player 1 (resp. 2) chooses an action i n ∈ I (resp. j n ∈ J). This determines a stage payoff g n = g(i n , j n , m n ), where m n is the current value of the parameter. Then a new value of the parameter is selected and the players get some information. This is generated by a map ρ from I × J × M to probabilities on A × B × M . Hence at stage n a triple (a n+1 , b n+1 , m n+1 ) is chosen according to the distribution ρ(i n , j n , m n ). The new parameter is m n+1 , and the signal a n+1 (resp. b n+1 ) is transmitted to player 1 (resp. player 2). Note that each signal may reveal some information about the previous choice of actions (i n , j n ) and both the previous (m n ) and the new (m n+1 ) values of the parameter.

Stochastic games correspond to public signals including the parameter. Incomplete information games correspond to an absorbing transition on the parameter (which thus remains fixed) and no further information (after the initial one) on the parameter. Mertens, Sorin and Zamir (1994) [START_REF] Mertens | Repeated Games[END_REF] Section IV.3, associate to each such repeated game G an auxiliary stochastic game Γ having the same values that satisfy a recursive equation of the type [START_REF] Aumann | Repeated Games with Incomplete Information[END_REF]. However the play, hence the strategies in both games differ. More precisely, in games with incomplete information on both sides, M is a product space K × L, π is a product probability p ⊗ q with p ∈ P = ∆(K), q ∈ Q = ∆(L) and in addition a 1 = k and b 1 = ℓ. Given the parameter m = (k, ℓ), each player knows his own component and holds a prior on the other player's component. From stage 1 on, the parameter is fixed and the information of the players after stage n is a n+1 = b n+1 = {i n , j n }.

The auxiliary stochastic game Γ corresponding to the recursive structure can be taken as follows: the "state space" Ω is P × Q and is interpreted as the space of beliefs on the true parameter. X = ∆(I) K and Y = ∆(J) L are the type-dependent mixed action sets of the players; g is extended on X × Y × P × Q by g(x, y, p, q) = k,ℓ p k q ℓ g(x k , y ℓ , k, ℓ). Given (x, y, p, q) ∈ X × Y × P × Q, let x(i) = k x k i p k be the probability of action i and p(i) be the conditional probability on K given the action i, explicitly p k (i) = p k x k i x(i) (and similarly for y and q). In this framework the Shapley operator is defined on the set F of continuous concave-convex functions on P × Q by:

T(λ, f )(p, q) = val X×Y {λg(x, y, p, q) + (1 -λ) i,j x(i)y(j)f (p(i), q(j))} (2)
which is the new formulation of T(λ, f )(ω) = val{λg+(1-λ)Ef } and the discounted value v λ (p, q) is the unique fixed point of T(λ, .) on F. These relations are due to Aumann and Maschler (1966) [START_REF] Aumann | Repeated Games with Incomplete Information[END_REF] and [START_REF] Mertens | The Value of Two-Person Zero-Sum Repeated Games with Lack of Information on Both Sides[END_REF] [START_REF] Mertens | The Value of Two-Person Zero-Sum Repeated Games with Lack of Information on Both Sides[END_REF].

Extension: general evaluation

The basic formula expressing the discounted value as a fixed point of the Shapley operator

v λ = T(λ, v λ ) (3)
can be extended for values of games with the same plays but alternative evaluations of the stream of payoffs {g n }.

For example the n-stage game, with payoff defined by the Cesaro mean 1 n n m=1 g m of the stage payoffs, has a value v n and the recursive formula for the corresponding family of values is obtained similarly as

v n = T( 1 n , v n-1 )
with obviously v 0 = 0. Consider now an arbitrary evaluation probability µ on IN ⋆ . The associated payoff in the game is n µ n g n . Note that µ induces a partition Π = {t n } of [0, 1] with t 0 = 0, t n = n m=1 µ m , ... and thus the repeated game is naturally represented as a game played between times 0 and 1, where the actions are constant on each subinterval (t n-1 , t n ) which length µ n is the weight of stage n in the original game. Let v Π be its value. The corresponding recursive equation is now

v Π = val{t 1 g 1 + (1 -t 1 )Ev Πt 1 }
where Π t 1 is the normalization on [0, 1] of the trace of the partition Π on the interval [t 1 , 1]. If one defines V Π (t n ) as the value of the game starting at time t n , i.e. with evaluation µ n+m for the payoff g m at stage m, one obtains the alternative recursive formula

V Π (t n ) = val{µ n+1 g 1 + EV Π (t n+1 )}. ( 4
)
The stationarity properties of the game form in terms of payoffs and dynamics induce time homogeneity

V Π (t n ) = (1 -t n )V Πt n (0) (5)
where, as above, Π tn stands for the normalization of Π restricted to the interval [t n , 1]. By taking the linear extension of V Π (t n ) we define for every partition Π, a function V Π (t) on [0, 1].

Lemma 1 Assume that the sequence µ n is decreasing. Then V Π is C-Lipschitz in t, where C is a uniform bound on the payoffs in the game.

Proof. Given a pair of strategies (σ, τ ) in the game G with evaluation Π starting at time t n in state ω, the total payoff can be written in the form

E ω σ,τ [µ n+1 g 1 + ... + µ n+k g k + .
..] where g k is the payoff at stage k. Assume now that σ is optimal in the game G with evaluation Π starting at time t n+1 in state ω, then the alternative evaluation of the stream of payoffs satisfies, for all τ E ω σ,τ [µ n+2 g 1 + ...

+ µ n+k+1 g k + ...] ≥ V Π (t n+1 , ω). It follows that V Π (t n , ω) ≥ V Π (t n+1 , ω) -|E ω σ,τ [(µ n+1 -µ n+2 )g 1 + ... + (µ n+k -µ n+k+1 )g k + ...]| hence µ n being decreasing V Π (t n , ω) ≥ V Π (t n+1 , ω) -µ n+1 C.
This and the dual inequality imply that the linear interpolation V Π (., ω) is a C Lipschitz function in t.

Asymptotic analysis: previous results

We consider now the asymptotic behavior of v n as n goes to ∞, or of v λ as λ goes to 0.

For games with incomplete information on one side, the first proofs of the existence of lim n→∞ v n and lim λ→0 v λ are due to Aumann and Maschler (1966) [START_REF] Aumann | Repeated Games with Incomplete Information[END_REF], including in addition an identification of the limit as Cav ∆(K) u. Here u(p) = val ∆(I)×∆(J) k p k g(x, y, k) is the value of the one shot non revealing game, where the informed player does not use his information and Cav C is the concavification operator: given φ, a real bounded function defined on a convex set C, Cav C (φ) is the smallest function greater than φ and concave, on C. Extensions of these results to games with lack of information on both sides were achieved by [START_REF] Mertens | The Value of Two-Person Zero-Sum Repeated Games with Lack of Information on Both Sides[END_REF] [START_REF] Mertens | The Value of Two-Person Zero-Sum Repeated Games with Lack of Information on Both Sides[END_REF]. In addition they identified the limit as the only solution of the system of implicit functional equations with unknown φ:

φ(p, q) = Cav p∈∆(K) min{φ, u}(p, q), (6)
φ(p, q) = Vex q∈∆(L) max{φ, u}(p, q)

Here again u stands for the value of the non revealing game:

u(p, q) = val ∆(I)×∆(J) k,ℓ p k q ℓ g(x, y, k, ℓ)
and MZ will denote the corresponding operator

φ = MZ(u). (8)
As for stochastic games, the existence of lim λ→0 v λ is due to Bewley and Kohlberg (1976) [START_REF] Bewley | The Asymptotic Theory of Stochastic Games[END_REF] using algebraic arguments: the Shapley fixed point equation can be written as a finite set of polynomial inequalities involving the variables {λ, x λ (ω), y λ (ω), v λ (ω); ω ∈ Ω} thus it defines a semi-algebraic set in some Euclidean space IR N , hence by projection v λ has an expansion in Puiseux series of λ. The existence of lim n→∞ v n is obtained by an algebraic comparison argument, Bewley and Kohlberg (1976) [START_REF] Bewley | The Asymptotic Solution of a Recursion Equation Occurring in Stochastic Games[END_REF]. The asymptotic values for specific classes of absorbing games with incomplete information are studied in [START_REF] Sorin | Big Match" with Lack of Information on One Side, Part I[END_REF], [START_REF] Sorin | Big Match" with Lack of Information on One Side, Part I[END_REF], (1985) [START_REF] Sorin | Big Match" with Lack of Information on One Side, Part II[END_REF], see also [START_REF] Mertens | Repeated Games[END_REF] [START_REF] Mertens | Repeated Games[END_REF].

Asymptotic analysis: operator approach and comparison criteria

Starting with [START_REF] Rosenberg | An Operator Approach to Zero-Sum Repeated Games[END_REF] [START_REF] Rosenberg | An Operator Approach to Zero-Sum Repeated Games[END_REF] several existence results for the asymptotic value have been obtained, based on the Shapley operator: continuous moves absorbing and recursive games, games with incomplete information on both sides, and for absorbing games with incomplete information on one side, [START_REF] Rosenberg | Zero-Sum Absorbing Games with Incomplete Information on One Side: Asymptotic Analysis[END_REF] [START_REF] Rosenberg | Zero-Sum Absorbing Games with Incomplete Information on One Side: Asymptotic Analysis[END_REF]. We describe here an approach that was initially introduced by Laraki (2001) [START_REF] Laraki | The Splitting Game and Applications[END_REF] for the discounted case. The analysis of the asymptotic behavior for the discounted games is simpler because of its stationarity: v λ is a fixed point of (3). Various discounted game models have been solved using a variational approach (see Laraki [START_REF] Laraki | The Splitting Game and Applications[END_REF], [START_REF] Laraki | Variational Inequalities, System of Functional Equations, and Incomplete Information Repeated Games[END_REF] and [START_REF] Laraki | Explicit Formulas for Repeated Games with Absorbing States[END_REF]). Our work is the natural extension of this analysis to more general evaluations of the stream of stage payoffs including the Cesaro mean and its limit. Recall that each such evaluation can be interpreted as a discretization of an underlying continuous time game. We prove for several classes of games (incomplete information, splitting, absorbing) the existence of a (uniform) limit of the values of the discretized continuous time game as the mesh of the discretization goes to zero. The basic recursive structure is used to formulate variational inequalities that have to be satisfied by any accumulation point of the sequences of values. Then an ad-hoc comparison principle allows to prove uniqueness, hence convergence. Note that this technique is a transposition to discrete games of the numerical schemes used to approximate the value function of differential games via viscosity solution arguments, as developed in Barles-Souganidis [START_REF] Barles | Convergence of approximation schemes for fully nonlinear second order equations[END_REF]. The difference is that: in differential games the dynamics is given in continuous time, hence the limit game is well defined and the question is the existence of its value while here we consider accumulation points of sequences of functions satisfying an adapted recursive equation which is not available in continuous time. Another main difference is that, in our case, the limit equation is singular and does not satisfy the conditions usually required to apply the comparison principles. To sum up, the paper unifies tools used in discrete and continuous time approaches by dealing with functions defined on the product state × time space, in the spirit of [START_REF] Vieille | Weak Approachability[END_REF] [START_REF] Vieille | Weak Approachability[END_REF] for weak approachability or [START_REF] Laraki | Repeated Games with Lack of Information on One Side : the Dual Differential Approach[END_REF] [START_REF] Laraki | Repeated Games with Lack of Information on One Side : the Dual Differential Approach[END_REF] for the dual game of a repeated game with lack of information on one side, see also [START_REF] Sorin | New Approaches and Recent Advances in Two-Person Zero-Sum Repeated Games[END_REF] [START_REF] Sorin | New Approaches and Recent Advances in Two-Person Zero-Sum Repeated Games[END_REF].

Repeated Games with Incomplete Information

Let us briefly recall the structure of repeated games with incomplete information: at the beginning of the game, the pair (k, ℓ) is chosen at random according to some product probability p ⊗ q where p ∈ P = ∆(K) and q ∈ Q = ∆(L). Player 1 knows k while player 2 knows ℓ. At each stage n of the game, player 1 (resp. player 2) chooses a mixed strategy x n ∈ X = (∆(I)) K (resp. y n ∈ Y = (∆(J)) K ). This determines an expected payoff g(x n , y n , p, q).

The discounted game

We now describe the analysis in the discounted case. The total payoff is given by the expectation of n λ(1λ) n g(x n , y n , p, q) and the corresponding value v λ (p, q) is the unique fixed point of T(λ, .) (2) on F ([1], [START_REF] Mertens | The Value of Two-Person Zero-Sum Repeated Games with Lack of Information on Both Sides[END_REF]). In particular, v λ is concave in p and convex in q. We follow here [START_REF] Laraki | The Splitting Game and Applications[END_REF] [START_REF] Laraki | The Splitting Game and Applications[END_REF]. Note that the family of functions {v λ (p, q)} is C-Lipschitz continuous, where C is an uniform bound on the payoffs, hence relatively compact. To prove convergence it is enough to show that there is only one accumulation point (for the uniform convergence on P × Q). Remark that by (3) any accumulation point w of the family {v λ } will satisfy w = T(0, w)

i.e. is a fixed point of the projective operator, see Sorin [START_REF] Sorin | A First Course on Zero-Sum Repeated Games[END_REF], appendix C. Explicitly here: T(0, w) = val X×Y { i,j x(i)y(j)w(p(i), q(j))} = val X×Y E x,y,p,q w(p, q), where p = (p k (i)) and q = (q l (j)). Let S be the set of fixed points of T(0, •) and S 0 ⊂ S the set of accumulation points of the family {v λ } . Given w ∈ S 0 , we denote by X(p, q, w) ⊆ X the set of optimal strategies for player 1 (resp. Y(p, q, w) ⊆ Y for player 2) in the projective game with value T(0, w) at (p, q). A strategy x ∈ X of player 1 is called non-revealing at p, x ∈ N R X (p) if p = p a.s. (i.e. p(i) = p for all i ∈ I with x(i) > 0) and similarly for y ∈ Y. The value of the non revealing game satisfies

u(p, q) = val N R X (p)×N R Y (q) g(x, y, p, q) . (9)
A subset of strategies is non-revealing if all its elements are non-revealing.

Lemma 2 Let w ∈ S 0 and X(p, q, w) ⊂ N R X (p) then w(p, q) ≤ u(p, q).
Proof. Consider a family {v λn } converging to w and x n ∈ X optimal for T(λ n , v λn )(p, q), see

. Jensen's inequality applied to (2) leads to

v λn (p, q) ≤ λ n g(x n , j, p, q) + (1 -λ n )v λn (p, q), ∀j ∈ J . Thus v λn (p, q) ≤ g(x n , j, p, q).
If x ∈ X is an accumulation point of the family {x n }, then x is still optimal in T(0, w)(p, q). Since, by assumption X(p, q, w) ⊂ N R X (p), x is non revealing and therefore one obtains as λ n goes to 0: w(p, q) ≤ g(x, j, p, q), ∀j ∈ J .

So, by [START_REF] Laraki | On the Regularity of the Convexification Operator on a Compact Set[END_REF],

w(p, q) ≤ max x∈N R X (p)
min j∈J g(x, j, p, q) = u(p, q) .

Consider now w 1 and w 2 in S and let (p 0 , q 0 ) be an extreme point of the (convex hull of) the compact set in P × Q where the difference (w 1w 2 )(p, q) is maximal (this argument goes back to [START_REF] Mertens | The Value of Two-Person Zero-Sum Repeated Games with Lack of Information on Both Sides[END_REF] [START_REF] Mertens | The Value of Two-Person Zero-Sum Repeated Games with Lack of Information on Both Sides[END_REF]).

Lemma 3 X(p 0 , q 0 , w 1 ) ⊂ N R X (p 0 ), Y(p 0 , q 0 , w 2 ) ⊂ N R Y (q 0 ).
Proof. By definition, if x ∈ X(p 0 , q 0 , w 1 ) and y ∈ Y(p 0 , q 0 , w 2 ),

w 1 (p 0 , q 0 ) ≤ E x,y,p 0 ,q 0 w 1 (p, q)
and w 2 (p 0 , q 0 ) ≥ E x,y,p 0 ,q 0 w 2 (p, q).

Hence (p, q) belongs a.s. to the argmax of w 1w 2 and the result follows from the extremality of (p 0 , q 0 ).

Proposition 4 lim λ→0 v λ exists.

Proof. Let w 1 and w 2 be two different elements in S 0 and suppose that max w 1w 2 > 0. Let (p 0 , q 0 ) be an extreme point of the (convex hull of) the compact set in P × Q where the difference (w 1w 2 )(p, q) is maximal. Then Lemmas 2 and 3 imply w 1 (p 0 , q 0 ) ≤ u(p 0 , q 0 ) ≤ w 2 (p 0 , q 0 ), hence a contradiction. The convergence of the family {v λ } follows.

Given w ∈ S let Ew(., q) be the set of p ∈ P such that (p, w(p, q)) is an extreme point of the epigraph of w(., q). Lemma 5 Let w ∈ S. Then p ∈ Ew(., q) implies X(p, q, w) ⊂ N R X (p).

Proof. Use the fact that if x ∈ X(p, q, w) and y ∈ N R Y (q) w(p, q) ≤ E x,y,p,q w(p, q) = E x,p w(p, q). Hence one recovers the characterization through the variational inequalities of [START_REF] Mertens | The Value of Two-Person Zero-Sum Repeated Games with Lack of Information on Both Sides[END_REF] [START_REF] Mertens | The Value of Two-Person Zero-Sum Repeated Games with Lack of Information on Both Sides[END_REF] and one identifies the limit as MZ (u).

Proposition 6 lim λ→0 v λ = MZ(u)
Proof. Use Lemma 5 and the characterization of [START_REF] Laraki | The Splitting Game and Applications[END_REF] [START_REF] Laraki | Variational Inequalities, System of Functional Equations, and Incomplete Information Repeated Games[END_REF] or Rosenberg and Sorin (2001) [START_REF] Rosenberg | An Operator Approach to Zero-Sum Repeated Games[END_REF].

The finitely repeated game

We now turn to the study of the finitely repeated game: recall that the payoff of the n-stage game is given by 1 n n k=1 g(x k , y k , p, q) and that v n denotes its value. The recursive formula in this framework is:

v n (p, q) = max x∈X min y∈Y   1 n g(x, y, p, q) + (1 - 1 n ) i,j
x(i)y(j)v n-1 (p(i), q(j))

  = T( 1 n , v n-1 ). ( 10
)
Given an integer n ≥ 1, let Π be the uniform partition of [0, 1] with mesh 1 n and write simply W n for the associate function V Π . Hence W n (1, p, q) := 0 and for m = 0, ..., n -1, W n ( m n , p, q) satisfies:

W n m n , p, q = max x∈∆(I) K min y∈∆(J) L   1 n g(x, y, p, q) + i,j x(i)y(j)W n ( m + 1 n , p(i), q(j))   (11)
Note that W n ( m n , p, q, ω) = 1 -m n v n-m (p, q, ω) and if W n converges uniformly to W, v n converges uniformly to some function v, with W (t, p, q) = (1t) v(p, q). Let T be the set of real continuous functions W on [0, 1]×P ×Q such that for all t ∈ [0, 1], W (t, ., .) ∈ S. X(t, p, q, W ) is the set of optimal strategies for Player 1 in T(0, W (t, ., .)) and Y(t, p, q, W ) is defined accordingly. Let T 0 be the set of accumulation points of the family {W n } for the uniform convergence.

Lemma 7 T 0 = ∅ and T 0 ⊂ T .
Proof. W n (t, ., .) is C-Lipschitz continuous in (p, q) for the L 1 norm since the payoff, given the strategies (σ, τ ) of the players, is of the form k,ℓ p k q ℓ A kℓ (σ, τ ). Using Lemma 1 it follows that the family {W n } is uniformly Lipschitz on [0, 1] × P × Q hence is relatively compact for the uniform norm. Note finally using (10) that T 0 ⊂ T .

We now define two properties for a function W ∈ T and a C 1 test function φ : [0, 1] → IR. [START_REF] Aumann | Repeated Games with Incomplete Information[END_REF] is such that X(t, p, q, W ) is non-revealing and W (•, p, q)φ(•) has a global maximum at t, then u(p, q) + φ ′ (t) ≥ 0.

• P1: If t ∈ [0,
• P2: If t ∈ [0, 1
) is such that Y(t, p, q, W ) is non-revealing and W (•, p, q)φ(•) has a global minimum at t then u(p, q) + φ ′ (t) ≤ 0.

Lemma 8 Any W ∈ T 0 satisfies P1 and P2.

Note that this result is the variational counterpart of Lemma 2.

Proof. Let t ∈ [0, 1), p and q be such that X(t, p, q, W ) is non-revealing and W (•, p, q)φ(•) admits a global maximum at t. Adding the function s → (st) 2 to φ if necessary, we can assume that this global maximum is strict. Let W n k be a subsequence converging uniformly to W . Put m = n k and define θ(m) ∈ {0, . . . , m-1} such that θ(m) m is a global maximum of W m (•, p, q)φ(•) on the set {0, . . . , m -1}. Since t is a strict maximum, one has θ(m) m → t, as m → ∞. From [START_REF] Mertens | The Value of Two-Person Zero-Sum Repeated Games with Lack of Information on Both Sides[END_REF]:

W m θ(m) m , p, q = max x∈X min y∈Y   1 m g(x, y, p, q) + i,j x(i)y(j)W m ( θ(m) + 1 m , p(i), q(j))   .
Let x m ∈ X be optimal for player 1 in the above formula and let j ∈ J be any (non-revealing) pure action of player 2. Then:

W m θ(m) m , p, q ≤ 1 m g(x m , j, p, q) + i x m (i)W m θ(m) + 1 m , p m (i), q .
By concavity of W m with respect to p, we have

i∈I x m (i)W m θ(m) + 1 m , p m (i), q ≤ W m θ(m) + 1 m , p, q hence: 0 ≤ g(x m , j, p, q) + m W m θ(m) + 1 m , p, q -W m θ(m) m , p, q . Since θ(m) m is a global maximum of W (m) (•, p, q) -φ(•) on {0, . . . , m -1} one has: W m θ(m) + 1 m , p, q -W m θ(m) m , p, q ≤ φ θ(m) + 1 m -φ θ(m) m so that: 0 ≤ g(x m , j, p, q) + m φ θ(m) + 1 m -φ θ(m) m .
Since X is compact, one can assume without loss of generality that {x m } converges to some x. Note that x belongs to X(t, p, q, W ) by upper semicontinuity using the uniform convergence of W m to W . Hence x is non-revealing. Thus, passing to the limit one obtains: 0 ≤ g(x, j, p, q) + φ ′ (t).

Since this inequality holds true for every j ∈ J, we also have:

min j∈J g(x, j, p, q) + φ ′ (t) ≥ 0 .
Taking the maximum with respect to x ∈ N R X (p) gives the desired result:

u(p, q) + φ ′ (t) ≥ 0 .
The comparison principle in this case is given by the next result.

Lemma 9 Let W 1 and W 2 in T satisfying P1, P2 and

• P3: W 1 (1, p, q) ≤ W 2 (1, p, q) for any (p, q) ∈ ∆(K) × ∆(L).
Then

W 1 ≤ W 2 on [0, 1] × ∆(K) × ∆(L).
Proof. We argue by contradiction, assuming that max

t∈[0,1],p∈P,q∈Q [W 1 (t, p, q) -W 2 (t, p, q)] = δ > 0 .
Then, for ε > 0 sufficiently small,

δ(ε) := max t∈[0,1],s∈[0,1],p∈P,q∈Q [W 1 (t, p, q) -W 2 (s, p, q) - (t -s) 2 2ε + εs] > 0 . ( 12
)
Moreover δ(ε) → δ as ε → 0. We claim that there is (t ε , s ε , p ε , q ε ), point of maximum in [START_REF] Mertens | Repeated Games[END_REF], such that X(t ε , p ε , q ε , W 1 ) is non-revealing for player 1 and Y(s ε , p ε , q ε , W 2 ) is non-revealing for player 2. The proof of this claim is like Lemma 3 and follows again Mertens Zamir (1971) [START_REF] Mertens | The Value of Two-Person Zero-Sum Repeated Games with Lack of Information on Both Sides[END_REF]. Let (t ε , s ε , p ′ ε , q ′ ε ) be a maximum point of [START_REF] Mertens | Repeated Games[END_REF] and C(ε) be the set of maximum points in P × Q of the function: (p, q) → W 1 (t ε , p, q) -W 2 (s ε , p, q). This is a compact set. Let (p ε , q ε ) be an extreme point of the convex hull of C(ε). By Caratheodory's theorem, this is also an element of C(ε). Let x ε ∈ X(t ε , p ε , q ε , W 1 ) and y ε ∈ Y(s ε , p ε , q ε , W 2 ). Since W 1 and W 2 are in T , we have:

W 1 (t ε , p ε , q ε ) -W 2 (s ε , p ε , q ε ) ≤ i,j x ε (i)y ε (j) [W 1 (t ε , p ε (i), q ε (j)) -W 2 (s ε , p ε (i), q ε (j))] .
By optimality of (p ε , q ε ), one deduces that, for every i and j with x ε (i) > 0 and y ε (j) > 0, (p ε (i), q ε (j)) ∈ C(ε). Since (p ε , q ε ) = i,j x ε (i)y ε (j)(p ε (i), q ε (j)) and (p ε , q ε ) is an extreme point of the convex hull of C(ε) one concludes that (p ε (i), q ε (j)) = (p ε , q ε ) for all i and j: x ε and y ε are non-revealing. Therefore we have constructed (t ε , s ε , p ε , q ε ) as claimed. Finally we note that t ε < 1 and s ε < 1 for ε sufficiently small, because δ(ε) > 0 and W 1 (1, p, q) ≤ W 2 (1, p, q) for any (p, q) ∈ P × Q by P3.

Since the map t → W 1 (t, p ε , q ε ) -(t-sε) 2 2ε
has a global maximum at t ε and since X(t ε , p ε , q ε , W 1 ) is non-revealing for player 1, condition P1 implies that

u(p ε , q ε ) + t ε -s ε ε ≥ 0 . ( 13
)
In the same way, since the map s → W 2 (s, p ε , q ε ) + (tε-s) 2

2ε

εs has a global minimum at s ε and since Y(s ε , p ε , q ε , W 2 ) is non-revealing for player 2, we have by condition P2 that

u(p ε , q ε ) + t ε -s ε ε + ε ≤ 0 .
This latter inequality contradicts [START_REF] Neyman | Repeated games with public uncertain duration process[END_REF].

We are now ready to prove the convergence result for lim n→∞ v n .

Proposition 10 W n converges uniformly to the unique point W ∈ T that satisfies the variational inequalities P1 and P2 and the terminal condition W (0, p, q) = 0. Consequently, v n (p, q) converges uniformly to v(p, q) = W (0, p, q) and W (t, p, q) = (1t)v(p, q), where v = MZ(u).

Proof. Let W ∈ T 0 . From Lemma 8, W satisfies the variational inequalities P1 and P2. Moreover, W (1, p, q) = 0. Since, from Lemma 9, there is at most one function fulfilling these conditions, we obtain convergence of the family {W n }.

Consequently, v n (p, q) converges uniformly to v(p, q) = W (0, p, q) and W (t, p, q) = (1t)v(p, q). In particular if one considers φ(t) = W (t, p, q) as test function, then φ ′ (t) = -v(p, q). Now P1 and P2 reduce to Lemma 2 hence via Lemma 5 to the variational characterization of MZ(u).

General evaluation

Consider now an arbitrarily evaluation probability µ on IN * , with µ n ≥ µ n+1 , inducing a partition Π. Let V Π (t k , p, q) be the value of the game starting at time t k . One has V Π (1, p, q) := 0 and

V Π (t n , p, q) = max x∈X min y∈Y   µ n+1 g(x, y, p, q) + i,j
x(i)y(j)V Π (t n+1 , p(i), q(j))

  . ( 14
)
Moreover V Π belongs to F and is C Lipschitz in (p, q). Lemma 1 then implies that any family of values V Π(m) associated to partitions Π(m) with µ 1 (m) → 0 as m → ∞ has an accumulation point. Denote by T 1 the set of those functions. Then T 1 ⊂ T by ( 14) and lemma 8 extends in a natural way: let V ∈ T 1 and V Π(m) → V uniformly. Let t m n be a global maximum of V Π(m) (., p, q)φ(.) on Π(m). Then t m n → t and one has

0 ≤ g(x n , j, p, q) + 1 µ n (m) V Π(m) t m n+1 , p, q -V Π(m) (t m n , p, q) hence 0 ≤ g(x n , j, p, q) + 1 µ n (m) φ(t m n+1 ) -φ (t m n )
and letting n → ∞ the result follows.

Using Lemma 9 this implies the convergence. Thus:

Proposition 11 V Π(m) converges uniformly to the unique point V ∈ T that satisfies the variational inequalities P1 and P2 and the terminal condition V (0, p, q) = 0. Consequently, v Π(m) (p, q) converges uniformly to v(p, q) = V (0, p, q) and V (t, p, q) = (1-t)v(p, q). Moreover v = MZ(u).

In particular the convergence of {V Π(m) } to the same limit for any family of decreasing partitions allows to use lim λ→0 v λ to characterize the limit.

Splitting games

We consider now the framework of splitting games, [START_REF] Sorin | A First Course on Zero-Sum Repeated Games[END_REF] [START_REF] Sorin | A First Course on Zero-Sum Repeated Games[END_REF], p. 78. Let P and Q be two simplexes (or product of simplexes) of some finite dimensional spaces, and H a C-Lipschitz function from P × Q to IR. The corresponding Shapley operator is defined on continuous saddle (concave-convex) real functions f on P × Q by

T(λ, f )(p, q) = val µ∈M P p ×ν∈M Q q P ×Q [(λH(p ′ , q ′ ) + (1 -λ)f (p ′ , q ′ )]µ(dp ′ )ν(dq ′ )
where M P p stands for the set of Borel probabilities on P with expectation p (and similarly for M Q q ). The associated repeated game is played as follows: at stage n+1 knowing the state (p n , q n ) player 1 (resp. player 2) chooses µ n+1 ∈ M P pn (resp. ν ∈ M Q qn ). A new state (p n+1 , q n+1 ) is selected according to these distributions and the stage payoff is H(p n+1 , q n+1 ). We denote by V λ the value of the discounted game and by v n the value of the n-stage game. A procedure analogous to the previous study of discounted games with incomplete information has been developed by Laraki [START_REF] Laraki | The Splitting Game and Applications[END_REF], [START_REF] Laraki | Variational Inequalities, System of Functional Equations, and Incomplete Information Repeated Games[END_REF], [START_REF] Laraki | On the Regularity of the Convexification Operator on a Compact Set[END_REF].

The discounted game

The next properties are established in [START_REF] Laraki | The Splitting Game and Applications[END_REF] [START_REF] Laraki | Variational Inequalities, System of Functional Equations, and Incomplete Information Repeated Games[END_REF]. Let G be the set of C-Lipschitz saddle functions on P × Q.

Lemma 12

The Shapley operator T(λ, •) maps G to itself and V λ (p, q) is the only fixed point of T (λ, .) in G.

The corresponding projective operator is the splitting operator Ψ:

Ψ(f )(p, q) = val M P p ×νM Q q P ×Q f (p ′ , q ′ )µ(dp ′ )ν(dq ′ ) ( 15
)
and we denote again by S its set of fixed points. Given W ∈ S, P(p, q, W ) ⊂ M P p denotes the set of optimal strategies of player 1 in [START_REF] Rosenberg | An Operator Approach to Zero-Sum Repeated Games[END_REF] for Ψ(W )(p, q). We say that P(p, q, W ) is nonrevealing if it is reduced to δ p , the Dirac mass at p. We use the symmetric notation Q(p, q, W ) and terminology for player 2. We define two properties for functions in S.

• A1: If P(p, q, W ) is non-revealing, then W (p, q) ≤ H(p, q).

• A2: If Q(p, q, W ) is non-revealing, then W (p, q) ≥ H(p, q). Proposition 13 V λ converges uniformly to the unique point V ∈ S that satisfies the variational inequalities A1 and A2.

The link with the MZ operator is as follows: as in Lemma 5 one defines:

• B1: If p ∈ EW (., q), then W (p, q) ≤ H(p, q).

• B2: If q ∈ EW (p, .), then W (p, q) ≥ H(p, q) (where, as before, EV denotes the set of extreme points of a convex or concave map V ). Then one has Ai implies Bi, i = 1, 2 and Proposition 14 Let G ∈ G. Then G satisfies B1 and B2 iff G = MZ(H).

The finitely repeated game

Recall the recursive formula defining by induction the value of the n stage game v n ∈ G, using Lemma 12:

v n (p, q) = val M P p ×M Q q P ×Q [ 1 n H(p ′ , q ′ ) + (1 - 1 n )v n-1 (p ′ , q ′ )]µ(dp ′ )ν(dq ′ ) = T( 1 n , V n-1
). ( 16)

For each integer n ≥ 1, let W n (1, p, q) := 0 and for m = 0, ..., n -1 define W n ( m n , p, q) inductively as follows:

W n m n , p, q = val M P p ×M Q q P ×Q [ 1 n H(p ′ , q ′ ) + W n ( m + 1 n , p ′ , q ′ )]µ(dp ′ )ν(dq ′ ) . (17)
By induction we have W n ( m n , p, q) = 1 -m n v n-m (p, q). Note that W n is the function on [0, 1] × P × Q associated to the uniform partition of mesh 1 n .

Lemma 15 W n is Lipschitz continuous uniformly in n on { m n , m ∈ {0, . . . , n}} × P × Q.

Proof. By Lemma 12 W n (t, ., .) belongs to G for any t. As for Lipschitz continuity with respect to t, we have, if µ is optimal in [START_REF] Shapley | Stochastic Games[END_REF] and by Jensen's inequality:

W n ( m n , p, q) ≤ P ×Q 1 n H(p ′ , q) + W n ( m + 1 n , p ′ , q)dµ(p ′ ) ≤ H ∞ n + W n ( m + 1 n , p, q) .
One gets the reverse inequality W n ( m n , p, q) ≥ -H ∞ n + W n ( m+1 n , p, q) with the symmetric arguments. Therefore W n (•, p, q) is H ∞ -Lipschitz continuous.

Let T be the set of real continuous functions W on [0, 1] × P × Q such that for all t ∈ [0, 1], W (t, ., .) ∈ S. P(t, p, q, W ) is defined as P(p, q, W (t, ., .)) and Q(t, p, q, W ) as Q(p, q, W (t, ., .)). Let T 0 be the set of accumulation points of the family W n . Using [START_REF] Shapley | Stochastic Games[END_REF], we have that T 0 ⊂ T . We introduce two properties for a function W ∈ T and any C 1 test function φ : [0, 1] → IR.

• PS1: If, for some t ∈ [0, 1), P(t, p, q, W ) is non-revealing and W (•, p, q)φ(•) has a global maximum at t, then H(p, q) + φ ′ (t) ≥ 0.

• PS2: If, for some t ∈ [0, 1), Q(t, p, q, W ) is non-revealing and W (•, p, q)φ(•) has a global minimum at t then H(p, q) + φ ′ (t) ≤ 0.

Lemma 16 Any W ∈ T 0 satisfies PS1 and PS2.

Proof. The proof is very similar to the proof of Lemma 8. Let t ∈ [0, 1), p and q be such that P(t, p, q, W ) is non-revealing and W (•, p, q)φ(•) admits a global maximum at t. Adding (•t) 2 to φ if necessary, we can assume that this global maximum is strict. Let W n k be a sequence converging uniformly to W . Write m = n k and define θ(m) ∈ {0, . . . , m-1} such that θ(m) m is a global maximum of W m (•, p, q)φ(•) on {0, . . . , m -1}. Since t is a strict maximum, we have θ(m) m → t. By [START_REF] Shapley | Stochastic Games[END_REF] we have that:

W m θ(m) m , p, q = val M P p ×M Q q P ×Q [ 1 m H(p ′ , q ′ ) + W m ( θ(m) + 1 m , p ′ , q ′ )]µ(dp ′ )ν(dq ′ ).
Let µ m be optimal for player 1 in the above formula and let ν = δ q be the Dirac mass at q. Then:

W m θ(m) m , p, q ≤ P 1 m H(p ′ , q)µ m (dp ′ ) + P W m ( θ(m) + 1 m , p ′ , q)µ m (dp ′ ).
By concavity of W m with respect to p, we have

P W m ( θ(m) + 1 m , p ′ , q)µ m (dp ′ ) ≤ W m ( θ(m) + 1 m , p, q)
Hence:

0 ≤ P H(p ′ , q)µ m (dp ′ ) + m W m θ(m) + 1 m , p, q -W m θ(m) m , p, q .
Since θ(m) m is a global maximum of W m (•, p, q)φ(•) on {0, . . . , m -1} one has:

W m θ(m) + 1 m , p, q -W m θ(m) m , p, q ≤ φ θ(m) + 1 m -φ θ(m) m So that 0 ≤ P H(p ′ , q)µ m (dp ′ ) + m φ θ(m) + 1 m -φ θ(m) m (18)
Since M P p is compact, one can assume without loss of generality that {µ m } converges to some µ. Note that µ belongs to P(t, p, q, W ) by upper semicontinuity and uniform convergence of W m to W . Hence µ is non-revealing: µ = δ p . Thus, passing to the limit in [START_REF] Sorin | Big Match" with Lack of Information on One Side, Part I[END_REF] one obtains:

0 ≤ H(p, q) + φ ′ (t).
The comparison principle in this case is given by the next result.

Lemma 17 Let W 1 and W 2 in T satisfying PS1, PS2 and

• PS3: W 1 (1, p, q) ≤ W 2 (1, p, q) for any (p, q) ∈ ∆(K) × ∆(L). Then W 1 ≤ W 2 on [0, 1] × ∆(K) × ∆(L).
The proof is exactly similar to the proof of Lemma 9.

We are now ready to prove the convergence result for lim n→∞ v n : Proposition 18 W n converges uniformly to the unique point W ∈ T that satisfies the variational inequalities PS1 and PS2 and the terminal condition W (1, p, q) = 0. Consequently, v n (p, q) converges uniformly to v(p, q) = W (0, p, q) and W (t, p, q) = (1t)v(p, q). Moreover v = MZ(H).

Proof. Let W be any limit point of the relatively compact family W n . Then, from Lemma 16, W ∈ T 0 satisfies the variational inequalities PS1 and PS2. Moreover, W (1, p, q) = 0. Since, from Lemma 17, there is at most one map fulfilling these conditions, we obtain convergence. Consequently, v n (p, q) converges uniformly to V (p, q) = W (0, p, q) and W (t, p, q) = (1t)V (p, q). In particular if one choose as test function φ(t) = W (t, p, q), then φ ′ (t) = -V (p, q), so that PS1 and PS2 reduce to A1 and A2. On concludes by using the variational characterization of MZ(u) in Proposition 14.

General evaluation

The same results extend to the general evaluation case defined by a partition Π with µ n decreasing. The existence of V Π is obtained in two steps. We first let V n Π to be 0 on [t n , 1] and define inductively V n Π (t m , ., .) for m < n by

V n Π (t m , p, q) = val M P p ×M Q q P ×Q [µ m+1 H(p ′ , q ′ ) + V n Π (t m+1 , p ′ , q ′ )]µ(dp ′ )ν(dq ′ ). ( 19
)
It follows that V n Π ∈ G by Lemma 12 and converges uniformly to V Π . Then the proof follows exactly the same steps than in Part 2.

Time dependent case

We consider here the case where the function H may depend on the stage. To be able to study the asymptotic behavior one has to define H directly in the limit game: the map H is a continuous real function on [0, 1] × P × Q.

For each integer n, let Z n (1, p, q) := 0 and for m = 0, ..., n -1 define Z n ( m n , p, q) inductively as follows:

Z n m n , p, q = val M P p ×M Q q P ×Q [ 1 n H( m n , p ′ , q ′ ) + Z n ( m + 1 n , p ′ , q ′ )]µ(dp ′ )ν(dq ′ ). ( 20
)
By induction each function Z n ( m n , ., .) is in G and one can show as in Lemma 15 that Z n is uniformly Lipschitz continuous on { m n , m ∈ {0, . . . , n}} × P × Q. Remark : An alternative choice is to replace 1 n H( m n , p ′ , q ′ ) by

m+1 n m n H(t, p ′ , q ′ )dt.
Note that the projective operator is the same than in the autonomous case. Let T be the set of real functions Z on [0, 1] × P × Q such that for all t ∈ [0, 1], Z(t, ., .) ∈ S. We define P(t, p, q, Z) and Q(t, p, q, Z) as before and denote by Z 0 the set of accumulation points of the family Z n . We note that Z 0 ⊂ T . We define two properties for a function Z ∈ T and all C 1 test function φ : [0, 1] → IR.

• PST1: If, for some t ∈ [0, 1), P(t, p, q, Z) is non-revealing and Z(•, p, q)φ(•) has a global maximum at t, then H(t, p, q) + φ ′ (t) ≥ 0.

• PST2: If, for some t ∈ [0, 1), Q(t, p, q, Z) is non-revealing and Z(•, p, q)φ(•) has a global minimum at t then H(t, p, q) + φ ′ (t) ≤ 0.

Lemma 19 Any Z ∈ Z 0 satisfies PST1 and PST2.

Proof. Let t ∈ [0, 1), p and q be such that P(t, p, q, Z) is non-revealing and Z(•, p, q)φ(•) admits a global maximum at t. Adding (•t) 2 to φ if necessary, we can assume that this global maximum is strict. Let Z n k be a sequence converging uniformly to Z. Write m = n k and define θ(m) ∈ {0, . . . , m-1} such that θ(m) m is a global maximum of Z m (•, p, q)-φ(•) on {0, . . . , m-1}. t being a strict maximum θ(m) m → t. By [START_REF] Sorin | A First Course on Zero-Sum Repeated Games[END_REF] we have that:

Z m θ(m) m , p, q = sup µ∈M P p inf ν∈M Q q P ×Q [ 1 m H( θ(m) m , p ′ , q ′ ) + Z m ( θ(m) + 1 m , p ′ , q ′ )]µ(dp ′ )µ(dq ′ ).
Let µ m be optimal for player I in the above formula and let ν = δ q be the Dirac mass at q. Then:

Z m θ(m) m , p, q ≤ P 1 m H( θ(m) m , p ′ , q ′ )µ m (dp ′ ) + P Z n ( θ(m) + 1 m , p ′ , q)µ m (dp ′ ).
By concavity of Z m with respect to p, we have

P Z m ( θ(m) + 1 m , p ′ , q)µ m (dp ′ ) ≤ Z m ( θ(m) + 1 m , p, q).
Hence:

0 ≤ P H( θ(m) m , p ′ , q ′ )µ m (dp ′ ) + m Z m θ(m) + 1 m , p, q -Z m θ(m) m , p, q .
Since θ(m) m is a global maximum of Z ϕ(m) (•, p, q)φ(•) on {0, . . . , m -1} one has:

Z m θ(m) + 1 m , p, q -Z m θ(m) m , p, q ≤ φ θ(m) + 1 m -φ θ(m) m .
M P p being compact, one can assume without loss of generality that {µ m } converges to some µ. Note that µ belongs to P(t, p, q, Z) by upper semicontinuity and uniform convergence of Z n to Z. Hence µ = δ p is non-revealing. Thus, passing to the limit one obtains: 0 ≤ H(t, p, q) + φ ′ (t).

The comparison principle in this case is given by the next result.

Lemma 20 Let Z 1 and Z 2 in T satisfying PS1, PS2 and

• PS3: Z 1 (1, p, q) ≤ Z 2 (1, p, q) for any (p, q) ∈ ∆(K) × ∆(L). Then Z 1 ≤ Z 2 on [0, 1] × ∆(K) × ∆(L).
Proof. We argue by contradiction, assuming that, for some γ > 0 small, max

t∈[0,1],p∈P,q∈Q [Z 1 (t, p, q) -Z 2 (t, p, q) -γ(1 -t)] = δ > 0 .
Then, for ε > 0 sufficiently small,

δ(ε) := max t∈[0,1],s∈[0,1],p∈P,q∈Q [Z 1 (t, p, q) -Z 2 (s, p, q) - (t -s) 2 2ε + -γ(1 -s)] > 0 . (21)
Moreover δ(ε) → δ as ε → 0. Hence as before there is (t ε , s ε , p ε , q ε ), point of maximum in [START_REF] Mertens | Repeated Games[END_REF], such that P(t ε , p ε , q ε , W 1 ) is non-revealing for player I and Q(s ε , p ε , q ε , W 2 ) is non-revealing for player J. Finally we note that t ε < 1 and s ε < 1 for ε sufficiently small, because δ(ε) > 0 and Z 1 (1, p, q) ≤ Z 2 (1, p, q) for any p, q by P3.

Since the map t → Z 1 (t, p ε , q ε ) -(t-sε) 2 2ε has a global maximum at t ε and since P(t ε , p ε , q ε , W 1 ) is non-revealing for player I, condition PST1 implies that

H(t ε , p ε , q ε ) + t ε -s ε ε ≥ 0 . (22)
In the same way, since the map s → W 2 (s, p ε , q ε ) + (tε-s) 2 2ε + γ(1s) has a global minimum at s ε and since Q(s ε , p ε , q ε , W 2 ) is non-revealing for player J, we have by condition PST2 that

H(s ε , p ε , q ε ) + t ε -s ε ε + γ ≤ 0 .
Combining [START_REF] Vieille | Weak Approachability[END_REF] with the previous inequality implies that

H(s ε , p ε , q ε ) -H(t ε , p ε , q ε ) + γ ≤ 0 .
Letting ε → 0, we get a contradiction because s ε and t ε converge (up to some subsequence) to the same limit t.

We are now ready to prove the convergence result for Z n .

Proposition 21 Z n converges uniformly to the unique point Z ∈ T that satisfies the variational inequalities PST1 and PST2 and the terminal condition Z(1, p, q) = 0.

Remark: the same result obviously holds for any sequence of decreasing evaluation.

Proof. Let Z be any limit point of the relatively compact family Z n . Then, from Lemma 19, W ∈ T 0 satisfies the variational inequalities PST1 and PST2. Moreover, Z(1, p, q) = 0. Since, from Lemma 20, there is at most one map fulfilling these conditions, we obtain convergence.

Absorbing games

An absorbing game is a stochastic game where only one state is non absorbing. In the other states one can assume that the payoff is constant (equal to the value) thus the game is defined by the following elements: two finite sets I and J, two (payoff) functions f , g from I × J to [-1, 1] and a function π from I × J to [0, 1] . The repeated game with absorbing states is played in discrete time as usual. At stage m = 1, 2, ... (if absorbtion has not yet occurred) player 1 chooses i m ∈ I and, simultaneously, player 2 chooses j m ∈ J: (i) the payoff at stage m is f (i m , j m ) (ii) with probability 1π (i m , j m ) absorbtion is reached and the payoff in all future stages n > m is g (i m , j m ) and (iii) with probability π (i m , j m ) the situation is repeated at stage m + 1.

Recall that the asymptotic analysis for these games is due to [START_REF] Kohlberg | Repeated Games with Absorbing States[END_REF] [START_REF] Kohlberg | Repeated Games with Absorbing States[END_REF] who also proved the existence of a uniform value in case of standard signaling.

The discounted game

While the spirit of the proof is the same in the general case we first present the discounted case where the argument is more transparent. Define π * (i, j) = 1π(i, j), f * (i, j) = π * (i, j) × g(i, j) and extend bilinearly any ϕ :

I × J → R to R I × R J as follows: ϕ(α, β) = i∈I,j∈J α i β j ϕ(i, j). Theorem 22 As λ → 0, v λ converges to v given by v = val ((x,α),(y,β))∈(∆(I)×R I + )×(∆(J)×R J + ) f (x, y) + f * (α, y) + f * (x, β) 1 + π * (α, y) + π * (x, β) . (23)
Remark : The existence of a value is a part of the Theorem. This formula is simpler than the one established in Laraki [START_REF] Laraki | Explicit Formulas for Repeated Games with Absorbing States[END_REF].

Proof. Consider v 1 an accumulation point of the family {v λ } and let v λn converges to v 1 . We will show that

v 1 ≤ sup (x,α)∈∆(I)×R I + inf (y,β)∈∆(J)×R J + f (x, y) + f * (α, y) + f * (x, β) 1 + π * (α, y) + π * (x, β) . (24)
A dual argument proves at the same time that the family {v λ } converges and that the auxiliary game has a value. Let r λ (x, y) be the discounted payoff induced by a pair of stationary strategies (x, y) ∈ ∆(I) × ∆(J). Then

r λ (x, y) = λf (x, y) + (1 -λ)f * (x, y) λ + (1 -λ)π * (x, y) .
In particular for any x λ optimal for Player 1 one obtains:

v λ ≤ λf (x λ , j) + (1 -λ)f * (x λ , j) λ + (1 -λ)π * (x λ , j) , ∀j ∈ J. ( 25
)
that one can write

v λ ≤ f (x λ , j) + f * ( (1-λ)x λ λ , j) 1 + π * ( (1-λ)x λ λ , j) = c j (λ), ∀j ∈ J. (26)
Note that the ratio f * ( (1-λ)x λ λ , j)/π * ( (1-λ)x λ λ , j) is bounded, hence c j (λ) too. Thus any accumulation point of c j (λ n ) is greater than v 1 . Hence by taking an appropriate subsequence in (26) for each j ∈ J, we obtain :

∃ x ∈ ∆(I) accumulation point of {x λn } s.t. ∀ε > 0, ∃ α = (1-λ)x λ λ ∈ R I + such that v 1 ≤ f (x, j) + f * (α, j) 1 + π * (α, j) + ε, ∀j ∈ J. (27)
Note that by linearity the same inequality holds for any y ∈ ∆(J).

On the other hand, v 1 is a fixed point of the projective operator and x is optimal there, hence

v 1 ≤ π(x, y) v + f * (x, y), ∀y ∈ ∆(J). (28)
Inequality ( 28) is linear thus extends to

π * (x, β) v 1 ≤ f * (x, β), ∀β ∈ R J + . (29)
We multiply (27) by the denominator 1 + π * (α, y) and we add to (29) to obtain the property: ∀ε > 0, ∃ x ∈ ∆(I) and α ∈ R I + such that

v 1 ≤ f (x, y) + f * (α, y) + f * (x, β) 1 + π * (α, y) + π * (x, β) + ε, ∀y ∈ ∆(J), β ∈ R J + ( 30
)
which implies (24), hence the result.

General evaluation

In this section we consider general evaluation probabilities µ = (µ m ) on IN ⋆ such that (µ m ) is non increasing: this later assumption is implicit in all the result below. The payoff corresponding to an evaluation µ is m µ m h m , where h m is the payoff at stage m described above. We denote by v µ the value of this game. Our aim is to show that the v µ have a limit as the "size" of the evaluation probability, i.e., π(µ) := µ 1 = sup m µ m , tends to 0.

Theorem 23 As π(µ) → 0, v µ converges to v given by v = val ((x,α),(y,β))∈(∆(I)×R I

+ )×(∆(J)×R J + ) f (x, y) + f * (α, y) + f * (x, β) 1 + π * (α, y) + π * (x, β) (31)
The proof requires several steps. The main idea is, as before, to embed the original problem into a game on [0, 1]. Recall that µ induces a partition Π = {t m } of [0, 1] with t 0 = 0 and t m = m k=1 µ k for m ≥ 1. Let us denote by W µ (t m ) the value of the game starting at time t m , i.e. with evaluation µ m+k for the payoff h k at stage k. Note that W µ is actually given by W µ (1) = 0 and the recursive formula:

W µ (t m ) = val (x,y)∈∆(I)×∆(J) [µ m+1 f (x, y) + π(x, y)W µ (t m+1 ) + (1 -t m+1 )f * (x, y)] . (32)
Recall that, under our assumption on the monotonicity of the (µ m ), the (linear interpolation of)

W µ is C-Lipschitz continuous in [0, 1]
, where C only depends on the bounds on the payoff (see Lemma 1). Let us set, for any (t, a, b, x, α, y, β)

∈ [0, 1] × R × R × ∆(I) × R I + × ∆(J) × R J + , h(t, a, b, x, α, y, β) = f (x, y) + (1 -t)[f * (α, y) + f * (x, β)] -[π * (α, y) + π * (x, β)] a + b 1 + π * (α, y) + π * (x, β)
We define the lower and upper Hamiltonian of the game as Hence for any ε > 0, there exists x ∈ ∆(I), α ∈ R I + such that ∀y ∈ ∆(J),

H -(t,
∀β ∈ R J + h(t, U (t), φ ′ (t), x, α, y, β) + ε ≥ 0 which implies H -(t, U (t), φ ′ (t)) ≥ 0.
Next we show a comparison principle:

Lemma 25 Let U 1 and U 2 be two continuous functions satisfying R1-R2 and U 1 (1) ≤ U 2 [START_REF] Aumann | Repeated Games with Incomplete Information[END_REF].

Then U 1 ≤ U 2 on [0, 1].
Proof. By contradiction, suppose that there is some t ∈ [0, 1] such that U 1 (t) > U 2 (t). Then, for γ > 0 sufficiently small, max

t∈[0,1] [U 1 (t) -U 2 (t) + γ(t -1)] = δ > 0 .
Let ε > 0 and set

δ(ε) = max (t,s)∈[0,1]×[0,1] [U 1 (t) -U 2 (s) - (t -s) 2 2ε + γ(s -1)] .
Let (t ε , s ε ) be a maximum point in the above expression. Then, δ(ε) → δ as ε → 0 and, for ε sufficiently small, t ε < 1 and

s ε < 1 because U 1 (1) ≤ U 2 (1). From standard arguments, t ε -s ε → 0 as ε → 0. Since the map U 1 (t) -(t-sε) 2 2ε
has a global maximum at t ε ∈ [0, 1), we have by condition R1 that

H -t ε , U 1 (t ε ), t ε -s ε ε ≥ 0 . ( 37
)
In the same way, since the map s → U 2 (s) + (tε-s) 2 2ε γ(s -1) has a global minimum at s ε , we have by condition R2 that

H + s ε , U 2 (s ε ), t ε -s ε ε + γ ≤ 0 . (38)
To simplify the expressions, let us set

U ε 1 = U 1 (t ε ), U ε 2 = U 2 (s ε
) and b ε = tε-sε ε . From (37) and (38) there exists (x ε , α ε ) ∈ ∆(I) × R I + such that:

0 ≤ ε 2 + inf (y,β) h (t ε , U ε 1 , b ε , x ε , α ε , y, β) and (y ε , β ε ) ∈ ∆(J) × R J + such that 0 ≥ -ε 2 + sup (x,α) h (s ε , U ε 2 , b ε + γ, x, α, y ε , β ε ) .
Then, in view of the definition of h, we have

2ε 2 ≥ h (s ε , U ε 2 , b ε + γ, x ε , α ε , y ε , β ε ) -h (t ε , U ε 1 , b ε , x ε , α ε , y ε , β ε ) ≥ (t ε -s ε )[f * (α ε , y ε ) + f * (x ε , β ε )] -[π * (α ε , y ε ) + π * (x ε , β ε )] (U 2 ε -U 1 ε ) + γ 1 + π * (α ε , y ε ) + π * (x ε , β ε ) Now we use U 1 ε -U 2 ε ≥ δ(ε) to obtain 2ε 2 ≥ (t ε -s ε )[f * (α ε , y ε ) + f * (x ε , β ε )] + [π * (α ε , y ε ) + π * (x ε , β ε )] δ(ε) + γ 1 + π * (α ε , y ε ) + π * (x ε , β ε ) ≥ (t ε -s ε )[f * (α ε , y ε ) + f * (x ε , β ε )] 1 + π * (α ε , y ε ) + π * (x ε , β ε ) + min{δ(ε), γ}
Since t εs ε → 0 and the quotient f * (αε,yε)+f * (xε,βε) 1+π * (αε,yε)+π * (xε,βε) remains bounded as ε → 0, we get 0 ≥ min{δ, γ}, which is impossible.

To summarize, we now know that the family (W µ ) has a unique accumulation point U and that this accumulation point is the unique continuous map satisfying R1-R2 and U (1) = 0. The next Lemma, which characterizes the limit function U , completes the proof of Theorem 23:

Lemma 26 Let U (•) be the unique continuous solution to R1-R2 with U (1) = 0. Then U (t) = (1t)v where v is given by (31).

Proof. Let us first show that U is homogeneous in time. This could be obtained by the fact that U is the limit of the W π , but we give here a direct argument. For this we prove that U λ (t) := 1 λ U (λt + (1λ)) equals U (t) for any t ∈ [0, 1] and any λ ∈ (0, 1) by showing that U λ satisfies R1-R2 and U λ (1) = 0. The last point being obvious, let us check for instance that R1 holds for U λ . Since U satisfies R1 for H -, U λ satisfies R1 for H - λ given by

H - λ (t, a, b) = H -(λt + (1 -λ), λa, b) So we just have to show that H - λ (t, a, b) ≥ 0 implies H -(t, a, b) ≥ 0. Assume that H - λ (t, a, b) ≥ 0. Then, for any ε > 0, there exists (x, α) ∈ ∆(I) × R I + such that, for all (y, β) ∈ ∆(J) × R J + , -ε ≤ f (x, y) + (1 -(λt + (1 -λ)))[f * (α, y) + f * (x, β)] -[π * (α, y) + π * (x, β)] λa + b 1 + π * (α, y) + π * (x, β)
Setting α ′ = λα and β ′ = λβ we get

- ε λ ≤ f (x, y) + (1 -t)[f * (α ′ , y) + f * (x, β ′ )] -[π * (α ′ , y) + π * (x, β ′ )] λa + b 1 + π * (α ′ , y) + π * (x, β ′ ) because - ε(1 + π * (α, y) + π * (x, β) 1 + π * (α ′ , y) + π * (x, β ′ ) ≥ - ε λ Therefore there exists (x, α ′ ) ∈ ∆(I) × R I + such that, for all (y, β ′ ) ∈ ∆(J) × R J + , one has h(t, a, b, x, α, y, β) ≥ -ε/λ, i.e., H -(t, a, b) ≥ 0.
Next we identify v := U (0). From the equation satisfied by U (t) = (1t)v we have, using φ(t) = U (t):

H -(t, (1 -t)v, -v) ≥ 0 and H + (t, (1 -t)v, -v) ≤ 0 ∀t ∈ [0, 1] .
Let us choose t = 0. Let ε > 0 and (x, α) be such that for any (y, β)

-ε ≤ f (x, y) + [f * (α, y) + f * (x, β)] -[π * (α, y) + π * (x, β)] v -v 1 + π * (α, y) + π * (x, β) Then v -ε ≤ f (x, y) + f * (α, y) + f * (x, β) 1 + π * (α, y) + π * (x, β) so that v -ε ≤ sup (x,α) inf (y,β) f (x, y) + f * (α, y) + f * (x, β) 1 + π * (α, y) + π * (x, β)
The opposite inequality

v + ε ≥ inf (y,β) sup (x,α) f (x, y) + f * (α, y) + f * (x, β) 1 + π * (α, y) + π * (x, β)
can be established in a symmetric way, which completes the proof of the Lemma.

5 Extensions and comments

Non decreasing evaluations

In stochastic games whith general evaluation, to obtain the same asymptotic limit as the mesh of the partition tends to zero, it is necessary to assume the sequence of evaluation probabilities µ n on IN * to be decreasing: µ n m ≥ µ n m+1 . For example, if the stochastic game oscillates deterministically between state 1 and state 2, the asymptotic occupation measure depends strongly on µ n . For example, if µ n is decreasing, then asymptotically, both states have a total weight of 1/2. However, if {µ n 2m+1 } is decreasing in m and if µ n 2m = (µ n 2m+1 ) 2 , then the asymptotic occupation measure puts a total weight of 1 on the state at stage 1.

However, in all games analyzed in this paper, the monotonicity assumption on µ m is not necessary: the asymptotic value exists and is the same for all evaluation measures. This is due to the irreversibility of these games. In incomplete information repeated games, the results hold because of two reasons: (1) a player is always better off having a private information (which implies concavity of the value function in p and convexity in q), and (2) a player has always the possibility to play a non-revealing strategy. Then V Π is C-Lipschitz continuous: this is the content of Lemma 15. Consequently, the same proof as for decreasing evaluations applies and so the asymptotic value exists in a strong sense and is characterized as the unique solution of the variational inequalities P 1 and P 2. A similar argument shows that the same conclusion holds for splitting games.

In absorbing games, this conclusion holds because once the state changes, it is absorbing. The proof is however more tricky. Let W µ n (t k ) be the value of the game starting at time t k . Then: W µ n (t k ) = val (x,y)∈∆(I)×∆(J) µ n k+1 f (x, y) + π(x, y)W µ n (t k+1 ) + (1t k+1 )f * (x, y) .

As shown in Lemma 1, monotonicity of (µ n m ) in m guarantees that W µ n is C-Lipschitz continuous. Without this assumption, it is not clear how to show uniform Lipschitz continuity. We prove uniform convergence but using different techniques, standard in differential game theory. Namely, consider the Barles-Perthame lower and upper half-relaxed limits. Explicitly, for every t, define W + (t) = lim sup tn→t W µ n (t n ), and similarly W -(t) = lim inf tn→t W µ n (t n ). Then, W + (t) is upper-semi-continuous and W -(t) is lower-semi-continuous. A proof similar to the one given for the decreasing case (with only small modifications) shows that: (1) W + satisfies R1, (2) W -satisfies R2, and (3) any upper-semi-continuous function satisfying R1 is smaller than any lower-semi-continuous function satisfying R2 (whenever they agree on the terminal condition). This implies uniform convergence and uniqueness of the limit.

Observe also that in the three classes of games analyzed in this paper, the existence of the asymptotic value in a strong sense (for all evaluations not necessarily decreasing) is new. Actually, the existence of the uniform value (as in absorbing games, [START_REF] Kohlberg | Repeated Games with Absorbing States[END_REF] [START_REF] Kohlberg | Repeated Games with Absorbing States[END_REF] ) only implies the same asymptotic value for all decreasing evaluations.

A natural question arises: what are the other classes of repeated games for which the asymptotic value is the same for all evaluations? Clearly, this is quite different from the existence of a uniform value. In the example above (stochastic game alternating between states 1 and 2), a uniform value exists but the asymptotic value depends on the sequence of evaluations. In incomplete information repeated games and in splitting games, the uniform value does not exist while there is a "strong" asymptotic value.

Other extensions

More general splitting games. Upper-and lower half-relaxed limits have been used in Laraki [START_REF] Laraki | The Splitting Game and Applications[END_REF] to show the existence of the asymptotic value in discounted splitting games when P and Q are not product of simplexes. Without this assumption, the equi-continuity of the family of discounted values with respect to p and q is not guaranteed. Combining the technique in Laraki [START_REF] Laraki | The Splitting Game and Applications[END_REF] and the continuous time approach allow to show the existence of the asymptotic value for all evaluations under the same general assumptions as the one in Laraki [START_REF] Laraki | The Splitting Game and Applications[END_REF].

Repeated games with public random duration. Neyman and Sorin [START_REF] Neyman | Repeated games with public uncertain duration process[END_REF] studied repeated games with random duration. Those are games in which the weight µ m of period m follows a stochastic process. In our model, this weight is deterministic. Neyman and Sorin [START_REF] Neyman | Repeated games with public uncertain duration process[END_REF] show that when the uniform value exists, then the asymptotic value exists for all random duration. It is plausible to prove existence of an asymptotic value in repeated games with random duration using similar tools. The difference would be in the recursive equation: an additional expectation should be added since the time t k+1 at which the continuation game will start is random and not deterministic.

Repeated games with incomplete information: the dependent case. The result of Mertens and Zamir [START_REF] Mertens | The Value of Two-Person Zero-Sum Repeated Games with Lack of Information on Both Sides[END_REF] holds in a more general framework in which the private information of the players on k ∈ K may be correlated. However one can write a recursive equation on the state space ∆(K). Consequently, the same proof as in the independent case allows to prove existence, uniqueness and characterization of the asymptotic value, for all evaluation coefficients µ.

Conclusion

The main contribution of this approach is to provide a unified treatment of the asymptotic analysis of the value of repeated games: -it applies to all evaluations and shows the interest of the limiting game played on [0, 1]. Further research will be devoted to a formal construction and to the analysis of optimal strategies, -it allows to treat incomplete information games as well as absorbing games. We strongly believe that similar tools will allow to analyze more general classes, -it shows that technics introduced in differential games where the dynamics on the state are smooth can be used in a repeated game framework. On the other hand, the stationary aspect of the payoff functions in repeated games is no longer necessary to obtain asymptotic properties.

  a, b) = sup (x,α)∈∆(I)×R I + inf (y,β)∈∆(J)×R J + h(t, a, b, x, α, y, β)

 [START_REF] Aumann | Repeated Games with Incomplete Information[END_REF]This research was supported by grant ANR-10-BLAN 0112 (Cardaliaguet and Laraki) and ANR-08-BLAN-0294-01 (Sorin). § This research was supported by grant ANR-10-BLAN 0112 (Cardaliaguet and Laraki) and ANR-08-BLAN-0294-01 (Sorin).

The variational characterization of any cluster point U of the family W µ as π(µ) → 0 uses the following properties: for all t ∈ [0, 1) and any C 1 function φ : [0, 1] → R :

Lemma 24 Any accumulation point U (•) of W µ (•) satisfies R1 and R2.

Proof. Let us prove the first variational inequality, the second being obtained by symmetry.

Let t be such that U (•)φ(•) admits a global maximum at t ∈ [0, 1). Adding (•t) 2 to φ if necessary, we can assume that this global maximum is strict. Let µ n = {µ n m } be a sequence of evaluation probabilities on IN ⋆ such that π(µ n ) → 0 and

for n large enough θ(n) + 1 is well defined and from (32) we have

Let x n be optimal for player 1 in the above formula. By compactness one can assume that x n converges to some x (up to a subsequence).

To simplify the notations, we set:

Given j ∈ J we have:

Using the fact that W n (•)-φ(•) has a global maximum at s n the above inequality can be rephrased as

We divide this inequality by 1 + π * (α n , j) so that the quotient is uniformly bounded. Hence, going to the limit and taking subsequences for each j one after the other, we obtain that: for any ε > 0 there exists α such that:

The same inequality holds for any y ∈ ∆(J) instead of j by linearity. Now x is optimal for U (t) leading to 0

and by linearity the same inequality holds for any β ∈ R J + . We multiply (34) by (1 + π * (α, y)) and we add (35) to obtain, ∀y ∈ ∆(J), ∀β ∈ R J


Aresomelanguages'morepragmatic'thanothers?

Introduction

The division of labour between Syntax ('sentence grammar') and Pragmatics (sometimes called 'discourse grammar'), informally defined in (1), is a highly controversialissue,whichcruciallyopposesgenerativeandfunctionallinguists.

(1) a.

Syntax:constraintsbearingonthecombinationofmorphemesand phrasesuptosentencelevel,andtheresultinginterpretiveeffects.

(cf.Johnwants*(something)/Johndrinks{(alcohol)/*(water)). b.

Pragmatics:constraintsbearingontheflowoflinguistic communicationbetweenlanguageusers. (cf."Ihaveasthma"→ Pleaseputoutyourcigarette.) Chomsky'stheoryofAutonomousSyntax(e.g.Chomsky1986)postulatesthatwhatever properties are specific to the Language faculty apply up to Sentence level and are not directly motivated by our world knowledge or the requirements of communication. Underthistheory,trulylinguisticpropertiesareofapurelyformalnatureandappearto us'unmotivated '-justasthemechanismsofhumanvisionappeartous'unmotivated', exceptfromabiologicalviewpoint. FunctionallinguistssuchasBolinger(1979), Garcia(1979), Givón(1979), Kuno(1987), [START_REF] Levinson | Pragmatic reduction of the binding conditions revisited[END_REF], among many others, uphold a different view, according to which the properties of sentence grammar are basically motivated by our world knowledge and theneedsofcommunication.

Onthefunctionalistside,wefurtherfindtheinterestingview-whichIplanto discuss here -according to which natural languages vary as to the relevance of pragmatic factors to their linguistic description. A variant of this idea is that some languages are sentenceoriented while others are discourseoriented (cf. Bamgroongraks 1987 on Thai).) TypicalillustrationsofthistheoreticalstandareLi&Thompson(1975[START_REF] Li | Third-person pronouns and zero-anaphora in Mandarin discours[END_REF]andHuang (1994,2000). Huang(2000)proposesa'newtypology'distinguishing syntacticlanguages(SLs)'frompragmaticlanguages(PLs):

(2) HUANG'S(2000)ASSUMPTIONS '(...)Somelanguagesaremorepragmaticthanothers.Inthesepragmatic languages,syntacticstructureismorecloselyrelatedtosemanticrepresentation and/orpragmaticinformation.'(Huang2000:276) 'TherearesomegroundsforbelievingthatinapragmaticlanguagelikeChinese, JapaneseandKorean,whensyntaxandworldknowledgeclash,worldknowledge frequentlywins.Bywayofcontrast,inasyntacticlanguagelikeEnglish,French andGerman,whenthereisaconflictbetweensyntaxandworldknowledge, syntaxusuallytakestheupperhand(...).'(ibid.:265) InwhatfollowsIwillreviewsomeoftheempiricalevidencegivenbyHuang(2000)(in thewakeofLi&Thompson1975,1979)insupportofthetheoryphrasedin(2).AndI shallcounter-arguethattheassumptionsin(3),ratherthanthosein(2),areontheright track:

(3) MYOWNASSUMPTIONS a. SyntaxandPragmaticsareequallyrelevantforthelanguages whichHuanglabelsSLsandforthoseheclassifiesasPLs b.

Thecontrastsbetweenthesetwogroupsoflanguages alwayspertaintosentence-grammar,cruciallyNOTtothespeakers''world knowledge'ortothewaytheyconstrueinferences. I must emphasize here that should (3) be correct, it would not necessarily follow that Chomsky's'minimalist'theoryofautonomoussyntax(e.g.Chomsky1995)iscorrector optimal.This,inmyopinion,isacompletelyindependentdebate. In section 2, I will survey and discuss the major diagnostic tests proposed by Huang (2000)insupportofhisdistinctionbetweenSLsandPLs.Iwillarguethattheempirical evidence he provides fails to establish that some languages are 'more pragmatic' than others. In the concluding section, I will emphasize the nonexistence of a class of 'pragmatic languages', and briefly suggest why Korean, in particular, may SEEM, at first glance,'morepragmatic'thanFrench.

2.

TheSL/PLdistinction:discussingthediagnostictests Thethreemainpropertieslistedin(4)arethoseonwhichHuangbaseshisdistinction between SLs (represented by English and other Indo-European languages) and PLs (representedbyChinese,JapaneseandKorean). ( 4 5a) and (5b) the implicit subject and object of 'teach' must be bound by the 'teacher'and'pupil'phrases.Oneofthesephrasesisinanoperatorposition(theheadof the relative clause) while the other is in an argument position (the matrix subject). Under Chomsky's syntactic approach, this contrast leads us to identify the zero arguments of 'teach' as differenttypes of zerocategoriesdepending on the position of theirbinders-thezerosaretracesiftheyareboundbytheheadoftherelativeclause, andpronominalsiftheyareboundbythematrixsubject.ButHuangarguesthatthereis no reason to make such a syntactic distinction in Chinese. According to him the two sentences in (5) are each completely ambiguous from a syntactic viewpoint, and only our knowledge of the world (what we know about teachers and pupils) leads us to choose one interpretation over the other: in both (5a) and (5b) we construe the interpretation so that the subject of 'teach' is bound by the 'teacher' phrase, and its object by the 'pupil' phrase. So Huang's assumption is that pragmatics (our world knowledge) is, in Chinese, the crucial factor in ambiguity resolutions involving zero anaphors, whereas in English the crucial factor is syntax (the contrast between whtracesandpronominals).

Icounter-arguethatthemaincontrastbetweentheChineseexamplesin(5)and their English translations lies in morphology, not in the relative prominence of syntax and pragmatics. The English personal pronouns he/him and relative pronouns who/whomareovertandspecifiedforcaseandforthetypeofdependenciestheymay enter, while the Chinese implicit arguments are invisible and hence completely underspecified. Beyond this morphological contrast, there is no difference in the way EnglishandChinesearticulatesyntaxandpragmatics.

Chinese formally distinguishes two sentences, (5a) and ( 5b), each open to two interpretations. Out of the four resulting readings, two are selected as pragmatically felicitous.Englishdistinguishes4sentences(6a,b,c,d),eachopentooneinterpretation. Outofthefourresultingreadings,twoareselectedaspragmaticallyfelicitous(thesame asinChinese): (6) a.

Theteachertstillcannotfindapupilpwhomphetcanteachtp. b.

?#Theteachertstillcannotfindapupilpwhoptpcanteachhimt. c.

?#Thepupilpstillcannotfindateachertwhomthepcanteachtt. d.

Thepupilpstillcannotfindateachertwhotttcanteachhimp. ThesedatathusdonotshowthattheinterpretationofChinesezeropronounsis 'morepragmatic'thanthatofEnglishovertpronouns.Theyonlybringoutthefactthat Chinese uses implicit anaphors in (5) where English uses overt functional nominals in (6), and that zero morphology may be a source of greater ambiguity than overt morphology.

Pronounambiguityresolution

The fact that ambiguities involving anaphora are often resolved on the basis of world knowledgeisinnowayspecifictoZEROanaphora,aswitnessedbytheFrenchexamples in( 7 Leprofdétestel'étudiantqu' il a collé.[kilakole=(7b)] 'Theteacherhatesthestudentwhomheflunked.' In(9)asin(5),twosyntacticanalyses(andinterpretations)(respectivelycorresponding to (7a) and (7b)) are available for each distribution of the 'teacher' and 'student' phrases. But for each distribution of these phrases, one interpretation (hence one structural analysis) is selected on pragmatic grounds -due to what we know about teacher/studentrelations:teachersflunkstudentsandnotconversely.Theseexamples onlyshowthatwhenalinguisticformisequallyopentotwoormoresyntacticanalyses (and semantic interpretations), world-knowledge may contribute to ambiguity resolution. This is a rather well-known fact but it is no truer in Chinese than it is in FrenchorEnglish. 2 OnovertandzeropronounsandpronounspecificationinKorean,cf.Jeon(1989), [START_REF] Shim | Les pronoms coréens ø , kîø et caki dans la phrase et dans le discours:essaidegrammairecomparée[END_REF], [START_REF] Kang | Functional Syntax: Anaphora, Discourse and Empathy[END_REF] Inthiscasethesecondreferentintroducedinthediscourse(laconcierge,thelandlady)is straightforwardly understood as female-denoting (in French because of feminine inflection on the definite article, in English because of the lexical content of the noun landlady). Correlatively, those overt pronouns which are morphologically specified for gender:Frenchil/elle,Englishhe,him,his,her,she)arecontextually unambiguous.But this does not prove ambiguity resolution to be more syntaxbased in French/English thaninChinese.ItboilsdowntothefactthatChinese,unlikeEnglishandFrench,failsto haveinflectionalmorphology.Andthispropertycruciallypertainstowordandsentence grammar,NOTtoworldknowledgeorpragmatics. Syntacticandpragmaticconstraintsplaysimilarrolesinbothtypesoflanguages.World knowledge is only activated for pronoun ambiguity resolution when morphosyntax makestwocompetingreadingsequallyavailable. Huang(2000)emphasizesthatthetheoryofanaphorawhichChomskyandsomeofhis disciples initially built on the basis of English evidence, does not extend to such languages as Chinese. What characterizes the Chomskyan approach to anaphora in general,andtoreflexivesinparticular,isthatitentirelyreliesonstructuralfactors.For instance, it is assumed that there exists a class of natural-language expressions which includeEnglishhimself (herself,etc.) (Korean/adaptedfromNa&Huck1993) koch-eun gughwa-ga olaega-n-da. flower-TOP chrysanthemum-NOM last.long-PRS-DEC Lit.'Flowers,chrysanthemumslastlong.' However, the assumption that 'Chinese-style' topics go unattested in so-called SLs cannotbeupheld.Aswitnessedbytheexamplesin( 22)-( 23),the'Chinese-style'topics in( 20 whichdistinguishChinese-stylefromEnglish-styletopicsonceagaincruciallypertainto sentencegrammar.

Reflexives

3.

Concludingremarks:thesyntax/pragmaticsinterface I am thus led to the conclusion that the distinction proposed by Huang (2000) between SLs and PLs must be discarded as empirically incorrect. The properties on which Huang bases his assumption are actually grounded in morphosyntax, not in pragmatics. As regards anaphora, if Chomsky's Binding Theory appears unable to correctly predict the distribution and interpretation of pronouns and reflexives in various languages (e.g. Chinese), we must conclude that Chomsky's Binding Theory shouldberevised.Butitdoesnotfollowthatsomelanguages,suchasChinese-Japanese-Korean (CJK), are less than others governed by syntactic constraints. In CJK, as in English or French, pronouns and reflexives cannot be freely inserted and interpreted within sentences, they all abide by sentence-internal constraints: for instance, Chinese zeropronouns,justasEnglish/Frenchweakovertpronouns,seemtoresistbeingbound by a co-argument. 5 On the other hand, in English and French, as in CJK, speakers recourse to world knowledge and pragmatic inference to resolve certain types of ambiguities -whenever morphosyntax delivers two or more equally available structuralanalyses.Worldknowledgethenguidesthehearerinselectingonereadingas contextuallyoptimal.Asregardstopic-commentconstructions,theyareequallyattested in English/French and in CJK, the only contrasts lie in morphosyntax, and the morphosyntax of topic marking is quite different in Japanese/Korean, and in Chinese. ThisfurthershedsadoubtontheassumptionthatCJKshouldbegroupedtogetherina singlecategoryof'pragmaticlanguages'

Huang fails to emphasize an important contrast between, e.g., French/English and Korean (and Japanese) regarding the way sentence grammar accommodates Information Structure. A striking fact about Korean and Japanese is that in order to translateanEnglishorFrenchsentenceintotheselanguages,oneMUSTtakeintoaccount InformationStructure.ThisisnottruetothesameextentinsuchlanguagesasEnglish andFrench. Forinstance,aFrenchspeakerassessingthesentencein(27) Minna-ga telebi bo-goiss-eo. '(Heylook)MinnaiswatchingTV.' c.

Minna-neun telebi-leul bo-goiss-eo. 'Minna,sheiswatchingTV.' InFrench,asentencelike(27),utteredunderasingleprosodiccontour,mayfelicitously translate each of the four sentences in (28). This contrast may suggest that Korean is 'morediscourse-oriented'thanFrench.However,thismayonceagainbeshowntoresult from a morphological contrast between French and Korean: in Korean, morphosyntax systematically signals the information status of every nominal (cf. [START_REF] Kwon | Differential function marking, case, and informationstructure:evidencefromKorean[END_REF], while such is not the case in French. This of course does not mean that Information Structure is irrelevant for the syntactic description of such languages as French or English, for there is, as we know, plenty of evidence proving otherwise (cf., e.g.,Lambrecht1994,Erteschik-Shir1997).

The contrast between Korean and French in ( 27)-( 28) is that French morphosyntax may stay neutral with respect to information structure, while Korean morphosyntax must not. Like the other properties discussed in the previous sections, this contrast is groundedin sentence grammar, and in no way supports the claim that Koreanis'morepragmatic'thanFrench.

1

  Abbreviationsusedintheglosses[someofthemareborrowedfromthesources]:ACC=accusative,AUX= (perfect)auxiliary,CL=classifier,DAT=dative,DEC=declarative,DF=definite,DM=demonstrative,F= feminine(gender),HON=honorific,INF=informal(style),LOC=locative,M=masculine(gender),NOM= nominative,PF=perfectivemarker,PL=plural,PASS=passive,POSS=possessive,PRS=present,PST=past, REL=relativemarker,RES=resulativemarker,SG=singular,TOP=topic;1,2,3=first,second,thirdperson. IntheChineseexamplesin(5),zhao+daoaretwoverbs(literally'lookfor'and'arrive')whose combinationinthisnegativecontexttranslatesas'can'tfind'.[ThankstoMarie-ClaudeParisforher feedbackonzhaobudao].

  Li&Thompson(1979)(aswellasHuang2000)claimthattheinterpretationofChinesetype zero pronouns is characteristically calculated on pragmatic grounds, NOT on syntacticgrounds. (10) [...]Therearenostructuralpropertiespredictingtheinterpretationofthe referentforzero-pronouns[...]Theinterpretationofthereferentforthe unrealizedpronounisinferredonthebasisofpragmaticknowledge.' (Li&Thompson1979:312) In(11) 3 IpresentaparadigmofChinesedata(involvingzeropronouns)presentedbyLi &Thompson(1979),withitsFrenchandEnglishtranslationsin(12a-h)and(12a'-h').

  Bamgroongraks, Cholticha, 1987, Sukhothai thai as a discourseoriented language: evidence from zero noun phrases, Ph. D. dissertation, University of Wisconsin-Madison. Bolinger,Dwight,1979,'Pronounsindiscourse',inT.Givon(ed.),DiscourseandSyntax. Huang,Yan,2000,Anaphora:acrosslinguisticstudy,Oxford:OxfordUniversityPress. Jeon, Kyung-Jun . 1989, Relations anaphoriques et ellipses structurales en coréen: essai sur la grammaire de la phrase et la grammaire du discours, Seoul: Hanshin PublishingCo.

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In(13a,b)adjacentreferentialindices(e.g.'yz')indicatethatthetworeadingsareequallyavailable, whileindicesseparatedby'>'(e.g.'y>z')indicatethatone(orseveral)reading(s)is/arepreferred.

ThisgeneralconstraintisfurtherdiscussedinZribi-Hertz(2008).

Thus,themaincontrastsbetween'Chinese-style'topicsandtheirFrench/English counterparts ultimately lie in morphosyntax -the way each type of topic is morphologicallyorlexicallyspecified,whetherthetopicbindsaresumptivemorpheme, and when it does, whether the resumptive is overt or null. Therefore, the properties