Definición del proceso de negocio (workflow, objetivos, KPI,

In this paper we have explored the intimate connection between discrete choice models and two-sided matching models, and used results from the literature on matching under imperfectly transferable utility to derive identification and estimation procedures for discrete choice models based on the non-additive random utility specification.

Although it is commonplace to distinguish in the microeconomic literature between

“one-sided” and “two-sided” demand problems, our results show that this distinction does not exist. Although yogurts do not derive utility from being consumed by one consumer or another one, the market adjusts in fashion which is observationally equivalent. Given the matching equivalence, it is indeed just as appropriate to consider a discrete choice problem as one in which consumers choose yogurts, as one in which (fancifully) “yogurts choose consumers”.

The connection between discrete choice and two-sided matching is a rich one, and we are exploring additional implications. For instance, the phenomenon of “multiple discrete choice” (consumers who choose more than one brand, or choose bundles of products on a purchase occasion) is challenging and difficult to model in the discrete choice framework¹⁶ but is quite natural in the matching context, where “one-to-many” markets are commonplace – perhaps the most prominent and well-studied being the National Residents Matching Program for aspiring doctors in the United States (cf. Roth (1984)). We are exploring this

16See Hendel (1999),Dub´e (2004), Fox and Bajari (2013) for some applications.

connection in ongoing work.

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Appendix

1.A

Proofs and additional results

We shall proove stronger versions of theorems 1–4 in a more general setting where we allow for possible ties, that is, when assumption 2 is dropped. In this case, the demand map σ is no longer defined, but should be replaced by a set-valued analog.¹⁷ We first introduce the relevant object, the demand correspondence, before stating our equivalence theorem in this more general setting, and the properties of the inverse demand correspondence.

1.A.1 The demand correspondence

When we drop assumption 2, indifference between two alternatives may occur with positive proba-bility, and the random set of alternatives preferred by the agent may contain several elements. This is for instance the case in an additive random utility model with discrete heterogeneity; indifferences can also arise in vertical differentiation models (see Section 1.5). Hence one cannot define a map σ by (1.2.2). Instead, we can define the demand correspondence Σ (δ) at vector δ as the set of market shares compatible with the optimal choices of consumers when the systematic utilities are δ and some tie-breaking rule is arbitrarily chosen. That is:

Definition 4 (Demand correspondence). The demand correspondence Σ : R^J0 → P (S0) is a function from R^J0 to the power set of S₀ such that Σ (δ), is the set of market shares s such that there is a random variable ˜j valued in J0with probability mass vector s, and such that ˜j maximizes

17See also Pakes and Porter (2015) for a discussion of identification and inference in ARUM models with possible indifferences.

U_εj(δj) over j ∈ J0 almost surely.

We define the inverse demand correspondence by

Σ⁻¹(s) =δ ∈ R^J0: s ∈ Σ (δ)

which is the set of utility vectors δ that rationalize the vector of market shares s ∈ R^J0, which is the identified set of utilities.

Example 1.A.1. Consider the case of an ARUM without heterogeneity: εj = 0 a.s. for j ∈ J0, so that U_εij(δ_j) = δ_j. Then Σ (δ) contains a single element or multiple elements depending on the number of elements contained in arg maxj∈J₀δj. If the argmax has a single element j^∗, then Σ (δ) is the s ∈ S₀ such that s_j^∗= 1 and s_j = 0 for j 6= j^∗. If the argmax is a set B with multiple elements, then Σ (δ) is the set of s ∈ R^J+⁰ such thatP

j∈J₀sj= 1, and j /∈ B implies sj= 0.

The following result is a direct consequence of Strassen’s theorem (Strassen (1965), theorem 5).

It provides a convenient reexpression of Σ (δ).

Proposition 1. _{Let s ∈ R}^J₊⁰ be such that P

j∈J₀sj= 1. Then under assumption 1, the following statements are equivalent:

(i) s ∈ Σ (δ), and (ii) ∀B ⊆ J0, P

j∈Bsj≤ P maxj∈BUεj(δj) ≥ max_j∈J0\BUεj(δj)
.

Proof. Direct implication: Let s be the probability mass vector of a random variable ˜j valued in J0

such that ˜j ∈ J (ε). Then for all B ⊆ J₀, one has

X

j∈B

sj = Pr ˜j ∈ B
≤ Pr (J (ε) ∩ B 6= ∅) . (1.A.1)

Converse implication: Conversely, assume (1.A.1). Then by Strassen’s theorem (Strassen (1965), theorem 5), one can construct ˜j and ε on the same probability space such that ˜j ∈ J (ε) almost surely.

To gain some intuition for (ii), note that the RHS of the inequality is the probability that, for all ε such that the optimizing choices contain some alternative(s) in a set B, those alternatives in B are chosen. This is a upper bound on the actual markets for alternatives in set B.¹⁸

Remark 1.A.1. A necessary condition for the second statement of proposition 1 to hold is

sj≤ P



Uεj(δj) ≥ max

j⁰∈J0

Uεj⁰(δj⁰)



, for all j ∈ J₀ (1.A.2)

which amounts to checking part (ii) in Proposition 1 on the class of singleton subsets. However, this condition is not sufficient as shown in the following example. Consider the case when J0 has three elements and the set J (ε) of optimal alternatives is {j1} wp 1/3, {j1, j2} wp 1/3, and {j3} wp 1/3.

Then s = (2/3, 1/3, 0) satisfies inequalities (1.A.2) for every j ∈ J0. However there is no random variable ˜j valued in J0such that ˜j ∈ J (ε). Indeed, if this were the case, Pr ˜j = j3|J (ε) = {j3}
= 1, thus Pr ˜j = j₃
≥ Pr (J (ε) = {j3}) = 1/3, a contradiction.  Remark 1.A.2. In general Σ (δ) ⊆σ_j(δ) , σj(δ), where we define

σ_j(δ) = P



ε ∈ Ω : Uεj(δj) > max

j⁰∈J0\{j}Uεj(δj⁰)



σ_j(δ) = P



ε ∈ Ω : U_εj(δ_j) ≥ max

j⁰∈J0\{j}U_εj(δ_j⁰)



For instance, if J = {j₁, j₂} and Uεj(δj₁) = Uεj(δj₂) = Uεj(δj₀) for every ε ∈ Ω, then Σ (δ) =

(s1, s2) ∈ R²+: s1+ s2≤ 1 ; indeed, in this case, the agent is indifferent between the three alter-natives in every state of the world, thus any randomized choice is a solution. In this case σ_j(δ) = 0

and σj(δ) = 1. 

Remark 1.A.3. Under assumptions 1 and 2, it follows from proposition 1 that Σ is point-valued, that is Σ (δ) = σ (δ) for all δ. However, it does not mean that σ⁻¹(s) is itself point valued. 

Normalization. Just as in section 1.3, normalization issues will play an important role in our analysis. As a result, we introduce ˜Σ as the correspondence R^J → P (S) induced by Σ under

18A similar inequality is used to generate the upper bound choice probablities in Ciliberto and Tamer’s (2009) study of multiple equilibria in airline entry games.

normalization δ0= 0, hence

Σ˜⁻¹(s) =δ ∈ R^J : s ∈ Σ (δ⁰) for δ⁰∈ R^J0 with δ₋₀⁰ = δ and δ₀⁰ = 0 .

1.A.2 Proof of theorems 1–4

Proof of theorems 1

The proof of theorem 1 follows from the following stronger result, where we have removed assump-tion 2. We state and proof this stronger result.

Theorem 1’. Under assumption 1, consider a vector of market shares s that satisfies sj > 0 andP

j∈J0s_j = 1. Consider a vector δ ∈ R^J0. Then, the two following statements are equivalent:

(i) δ belongs to the identified utility set Σ⁻¹(s) =δ ∈ R^J0 : s ∈ Σ (δ) associated with s in the sense of definition 2 in the discrete choice problem with ε ∼ P , and

(ii) there exists π ∈ M (P, s) and u = maxj∈J₀Uεj(δj) defined by (1.2.1) such that (π, u, −δ) is an equilibrium outcome in the sense of definition 3 in the matching problem, where

fεj(u) = u and gεj(−δ) = −Uεj(δ) . (1.A.3)

Proof (a) From demand inversion to equilibrium matching: Consider δ ∈ ˜Σ⁻¹(s) a solution to the demand inversion problem. Then s_j= P (ε ∈ Ω : U_εj(δ_j) ≥ u (ε)), where

u (ε) = max

j∈J₀Uεj(δj) .

Let us show that we can construct π and set v = −δ such that (π, u, v) is an equilibrium outcome, which is to say it satisfies the three conditions of definition 3.

Let us introduce J (ε) = arg max_j∈J0{U_εj(δ_j)} the set of yogurts that maximize consumer ε’s utility. Then σj(δ) = Pr (j ∈ J (ε)). Let us show that J (ε) has one element with probability one.

Indeed, otherwise, {j, j⁰} ⊆ J (ε) would arise with positive probability for some pair j 6= j⁰, which would imply that there is a positive probability of indifference between j and j⁰, in contradiction

with (1.2.2). Hence for each open set B,

Pr (J (ε) ∩ B 6= ∅) =X

j∈B

Pr (J (ε) = {j}) =X

j∈B

Pr (j ∈ J (ε)) = σj(δ) .

In particular, for each open set B, one has

s (B) ≤ Pr (J (ε) ∩ B 6= ∅) .

By Strassen’s theorem (Strassen (1965), theorem 5)¹⁹, this implies that there is a probability distribution π ∈ M (P, s) such that j ∈ J (ε) on the support of π. Hence π satisfy condition (i) in definition 3. But j ∈ J (ε) implies Uεj(δj) = u (ε). Introducing v (j) = −δj, gεj(v (j)) = −Uε_ij(δj), and fεj(x) = x, one has

f_εj(u (ε)) + gεj(v (j)) ≥ 0

for all (ε, j), with equality on the support of π. Hence, conditions (ii) and (iii) in definition 3 are met, and (π, u, v) is an equilibrium outcome.

(b) From equilibrium matching to demand inversion: Let (π, u, v) be an equilibrium matching in the sense of definition 3, where fεj(x) = x and gεj(y) = −Uε_ij(−y). Then letting δ = −v, one has by condition (ii) that for any ε ∈ Ω and j ∈ J0,

u (ε) − U_εj(δ_j) ≥ 0

thus u (ε) ≥ maxj∈J₀Uεj(δj). But by condition (iii), for j ∈ Supp (π (.|ε)), one has u (ε) = Uεj(δj), thus

u (ε) = max

j∈J₀Uεj(δj) .

Condition (iii) implies that if ˜ε, ˜j
∼ π, then Pr J (˜ε) =˜j
= 1, thus

σj(δ) = P (ε ∈ Ω : J (˜ε) = {j}) = Pr ˜j = j
= sj.

Hence s ∈ ˜Σ (δ), QED.

19Strassen’s theorem is essentially a continuous extension of Hall’s marriage theorem.

Proof of theorem 2

Again, we state and prove a version of theorem 2 where assumption 2 has been removed.

Theorem 2’ (Inverse isotonicity of demand). Under assumption 1, consider s and s⁰in S₀such that s_j ≤ s⁰_j for all j ∈ J . If there are two vectors δ and δ⁰ satisfying (1.2.3) such that s ∈ ˜Σ (δ) and s⁰∈ ˜Σ (δ⁰), then

s ∈ ˜Σ (δ ∧ δ⁰) and s⁰∈ ˜Σ (δ ∨ δ⁰) .

Remark 1.A.4. To the best of our knowledge, this result is novel in the theory of two-sided matchings with imperfectly transferable utility. While in the case of matching with (perfectly) transferable utility, it follows easily from the fact that the value of the optimal assignment problem is a supermodular function in (P, −s), (see e.g. Vohra (2004), theorem 7.20), to the best of our

knowledge it is novel beyond that case²⁰. 

Proof. Assume sj ≤ s⁰_j for all j ∈ J , and let δ and δ⁰ in satisfying (1.2.3) such that s ∈ ˜Σ (δ) and s⁰ ∈ ˜Σ (δ⁰). Let u (ε) = maxj∈J0Uεj(δj) and u⁰(ε) = maxj∈J0Uεj δ_j⁰
. Let δ^∧ = δ ∧ δ⁰ and δ^∨= δ ∨ δ⁰ i.e.

δ_j^∧= min δ_j, δ_j⁰

and δ^∨_j = max δ_j, δ⁰_j
, and let

u^∧(ε) = min (u (ε) , u⁰(ε)) and u^∨(ε) = max (u (ε) , u⁰(ε)) .

(a) Proof of s ∈ ˜Σ (δ ∧ δ⁰): By Strassen’s theorem, the fact that s ∈ ˜Σ (δ) and s⁰ ∈ ˜Σ (δ) is equivalent to the fact that for all A ⊆ J0,

X

j∈A

sj ≤ P {ε ∈ Ω : ∃j ∈ A, u (ε) = Uεj(δj)} , and (1.A.4) X

j∈A

s⁰_j ≤ Pε ∈ Ω : ∃j ∈ A, u⁰(ε) = Uεj δ_j⁰
. (1.A.5)

By the converse implication in Strassen’s theorem, in order to show that s ∈ ˜Σ (δ ∧ δ⁰), it is

20Demange and Gale (1985) show isotonicity in the strong set order with respect to reservation utilities, which is a different result.

sufficient to show that for all A ⊆ J0,

X

j∈A

sj≤ Pε ∈ Ω : ∃j ∈ A, u^∧(ε) = Uεj δ_j^∧
. (1.A.6)

Take A ⊆ J0, and let

A^>=j ∈ A : δj > δ_j⁰

and A^≤=j ∈ A : δj ≤ δ_j⁰

while one defines

Ω^>= {ε ∈ Ω : u (ε) > u⁰(ε)} and Ω^≤ = {ε ∈ Ω : u (ε) ≤ u⁰(ε)} .

By (1.A.5) applied to A = A^>, one has

X

j∈A^>

s_j ≤ X

j∈A^>

s⁰_j ≤ Pε ∈ Ω : ∃j ∈ A^>, u⁰(ε) = Uεj δ_j⁰
;

but if j ∈ A^> and if u⁰(ε) = Uεj δ⁰_j
, then ε ∈ Ω^>. Indeed, otherwise one would have Uεj(δj) ≤ u (ε) ≤ u⁰(ε) = Uεj δ_j⁰
, which would contradict δj > δ_j⁰. Hence, the latter display implies

X

j∈A^>

sj ≤ Pε ∈ Ω^>: ∃j ∈ A^>, u⁰(ε) = Uεj δ_j⁰

thus

X

j∈A^>

sj ≤ Pε ∈ Ω^> : ∃j ∈ A, u^∧(ε) = Uεj δ^∧_j
. (1.A.7)

By (1.A.4) applied to A = A^≤, one has

X

j∈A^≤

s_j≤ Pε ∈ Ω : ∃j ∈ A^≤, u (ε) = U_εj(δ_j) ;

but if j ∈ A^≤ and if u (ε) = Uεj(δj), then ε ∈ Ω^≤. Indeed, otherwise one would have Uεj(δj) =

u (ε) > u⁰(ε) ≥ Uεj δ_j⁰
, which would contradict δj ≤ δ_j⁰. Thus the latter display implies

X

j∈A^≤

sj≤ Pε ∈ Ω^≤: ∃j ∈ A^≤, u (ε) = Uεj(δj) ;

hence

X

j∈A^≤

s_j ≤ Pε ∈ Ω^≤ : ∃j ∈ A, u^∧(ε) = Uεj δ^∧_j
. (1.A.8)

By summation of (1.A.7) and (1.A.8), one obtains (1.A.6), and hence

s ∈ ˜Σ (δ ∧ δ⁰) , QED.

(b) Proof of s⁰∈ ˜Σ (δ ∨ δ⁰): By Strassen’s theorem, the fact that s ∈ ˜Σ (δ) and s⁰ ∈ ˜Σ (δ) is equivalent to the fact that for any Borel subset B ⊆ Ω,

P (B) ≤ X

j∈J₀

s_j1 {∃ε ∈ B : u (ε) = Uεj(δj)} , and (1.A.9)

P (B) ≤ X

j∈J₀

s⁰_j1∃ε ∈ B : u⁰(ε) = Uεj δ⁰_j
. (1.A.10)

By the converse of Strassen’s theorem, in order to show that s⁰ ∈ ˜Σ (δ ∨ δ⁰), it is sufficient to show that for any B ⊆ Ω,

P (B) ≤ X

j∈J₀

s⁰_j1∃ε ∈ B : u^∨(ε) = Uεj δ_j^∨
. (1.A.11)

Take a Borel subset B ⊆ Ω, and let

B^>= {ε ∈ B : u (ε) > u⁰(ε)} and B^≤= {ε ∈ B : u (ε) ≤ u⁰(ε)}

while one defines

J₀^>=j ∈ J0: δj > δ_j⁰

and J₀^≤ =j ∈ J0: δj ≤ δ_j⁰ .

By (1.A.9) applied to B = B^>, one has

P B^>
≤ X

j∈J₀

sj1∃ε ∈ B^>: u (ε) = Uεj(δj) ; (1.A.12)

but if ε ∈ B^>and u (ε) = Uεj(δj), then j ∈ J₀^>; otherwise δj≤ δ⁰_j, and thus u⁰(ε) < u (ε) ≤ Uεj δ_j⁰
, a contradiction. Hence, the sum on the right hand-side of (1.A.12) can be restricted to the elements of J₀^>, which implies

P B^>

≤ X

j∈J₀^>

s_j1 {∃ε ∈ B : u (ε) = U_εj(δ_j)} (1.A.13)

= X

j∈J₀^>

s_j1∃ε ∈ B : u^∨(ε) = U_εj δ_j^∨
,

thus, using the fact that sj ≤ s⁰_j for all j ∈ J₀^>, we deduce that

P B^>
≤ X

j∈J₀^>

s⁰_j1∃ε ∈ B : u^∨(ε) = Uεj δ^∨_j
. (1.A.14)

Next, taking B = B^≤ in (1.A.10) implies that

P B^≤
≤ X

j∈J0

s⁰_j1∃ε ∈ B^≤: u⁰(ε) = Uεj δ_j⁰
; (1.A.15)

but if ε ∈ B^≤ and u⁰(ε) = U_εj δ_j⁰
, then j ∈ J₀^≤; otherwise δ_j > δ_j⁰, and thus U_εj(δ_j) > U_εj δ⁰_j
= u⁰(ε) ≥ u (ε), another contradiction. Hence, (1.A.15) implies

P B^≤

≤ X

j∈J₀^≤

s⁰_j1∃ε ∈ B : u⁰(ε) = Uεj δ⁰_j

(1.A.16)

= X

j∈J₀^≤

s⁰_j1∃ε ∈ B : u^∨(ε) = U_εj δ_j^∨
.

By summation of (1.A.14) and (1.A.16), one obtains (1.A.11), and thus

s⁰ ∈ ˜Σ (δ ∨ δ⁰) , QED.

Let us explore what theorem 2 means in the familiar case of ARUMs, which is equivalent to a model of matching with transferable utility.

Example 1.A.2. In the ARUM case (but in that case only), theorem 2 follows from Topkis (1998) theorem. In this case, Chiong, Galichon, and Shum (2016) have shown that for s ∈ S0

Σ˜⁻¹(s) = arg max

δ:δ0=0





 X

j∈J

δjsj− EP

 max

k∈J0

{δk+ εk}







 .

Note that δ → E^P[maxk∈J₀{δk+ εk}] is submodular. Hence

(δ, s) →X

j∈J

δ_js_j− EP

 max

k∈J₀{δk+ ε_k}



is supermodular in δ and has increasing differences in (δ, s). As a result of Topkis’ theorem, the set-valued map s → ˜Σ⁻¹(s) is increasing in the strong set order, which means that if s ≤ s⁰, δ ∈ ˜Σ⁻¹(s) and δ⁰∈ ˜Σ⁻¹(s⁰), then

δ ∧ δ⁰∈ ˜Σ⁻¹(s) and δ ∨ δ⁰ ∈ ˜Σ⁻¹(s⁰) ,

which exactly recovers the conclusion of theorem 2. However, as soon as the model is no longer an additive random utility model, ˜Σ⁻¹(s) is not obtained by the solution of a maximization problem,

so that Topkis’ theorem cannot be invoked. 

It follows from theorem 2’ that the identified utility set is a lattice whenever nonempty, which implies corollary 1. Indeed, if δ ∈ ˜Σ⁻¹(s) and δ⁰ ∈ ˜Σ⁻¹(s), then δ ∧ δ⁰ ∈ ˜Σ⁻¹(s) and δ ∨ δ⁰ ∈ Σ˜⁻¹(s), QED.

Proof of theorem 3

As before, theorem 3 can be proven without assumption 2, which leads us to formulate a result which will imply theorem 3. Define the domain of ˜Σ⁻¹ as

S₀^dom=s ∈ S0: ˜Σ⁻¹(s) 6= ∅ .

We have:

Theorem 3’. Under assumptions 1, 3, and 4, ˜Σ⁻¹(s) is non-empty for all s ∈ S₀^int.

Remark 1.A.5. While in this paper we make use of theorem 3⁰ to guarantee the existence of an identifying vector of demand shifters δ under very weak restrictions, this result is a contribution to matching theory per se. Indeed, it implies the existence of a solution to the equilibrium transport problem, as introduced in Galichon (2016), definition 10.1.

The proof is based on several lemmas.

Assumption 4 implies that for all η > 0 and ν > 0 there is δ^∗s.t. δ > δ^∗implies Pr (|Xδ− X_δ^∗| > ν) <

η.

Lemma 1. There is a T^∗ such that for T < T^∗ there exists δ^T_j such that

Z exp

U_εj(^δ_j)

T



1 + exp

Uεj(^δ_j)

T

 P (dε) = sj (1.A.17)

and for all T < T^∗, δ^T_j ≥ δ_j where δ_j does not depend on T .

Proof. For T > 0, let

F_j^T(δ) =

Z exp_U

εj(δ) T

 1 + exp_U

εj(δ) T

 P (dε) =

Z 1

1 + exp

−^Uεj_T^(δ) P (dε) .

Assumption 3, part (b) implies:

Fact (a): F_j^T(.) is continuous and strictly increasing.

Next, by assumption 4, there exists δ_j such that δ < δ_j implies Pr (U_εj(δ) > −a) ≤ s_j/2.

Hence, for δ < δ_j

F_j^T(δ) = Z

{Uεj(δ)<−a}

exp_U

εj(δ) T

 1 + exp_U

εj(δ) T

 P (dε) + Z

{Uεj(δ)≥−a}

exp_U

εj(δ) T

 1 + exp_U

εj(δ) T

 P (dε)

≤ 1

1 + exp _T^a
+ sj/2

and taking T^∗= a/ log (1/s_j− 1) if log (1/sj− 1) > 0, and T^∗= +∞ else, it follows that T ≤ T^∗ implies ¹

1+exp(_T^a) ≤ sj/2, hence we get to:

Fact (b): for δ < δ_j and T ≤ T^∗, one has F_j^T(δ) < sj.

Next, by assumption 4, there exists δ_j⁰ such that δ > δ⁰_j implies Pr (Uεj(δ) > 0) ≥ 2sj. Then for δ > δ_j⁰,

F_j^T(δ) ≥ Z

{Uεj(δ)>b}

exp_U

εj(δ) T

 1 + exp_U

εj(δ) T

 P (dε) ≥

Pr (Uεj(δ) > b) 1 + exp (0) ≥ 2sj

2 = sj.

As a result, we get:

Fact (c): for all T > 0 and for δ > δ_j⁰, F_j^T(δ) > sj.

By combination of facts (a), (b) and (c), it follows that for T ≤ T^∗, there exists a unique δ^T_j such that F_j^T

δ^T_j

= s_j and δ^T_j ≤ δ_j⁰, where δ_j⁰ does not depend on T ≤ T^∗.

Let

G^T_j (δj; δ_−j) :=

Z P (dε)

exp

−^Uεj_T^(δj) +P

j⁰∈Jexp

U_εj0(^δj0)^−Uεj(δj) T

 .

Lemma 2. For T < T^∗, if G^T_j 

δ^{T ,k}_j ; δ^{T ,k}_−j

≤ sj, then:

(i) there is a real δ^{T ,k+1}_j ≥ δ^{T ,k}_j such that

G^T_j 

δ_j^{T ,k+1}; δ_−j^{T ,k}

= sj,

(ii) one has G^T_j 

which shows claim (i). To show the second claim, let us note that G^T_j (δ) is decreasing with respect to δj⁰ for any j⁰6= j. Indeed,

is expressed as the expectation of a term which is decreasing in δj⁰. Hence, as δ_−j^{T ,k}≤ δ^{T ,k+1}_−j in the componentwise order, it follows that

G^T_j 

δ_j^{T ,k+1}; δ_−j^{T ,k+1}

≤ G^T_j 

δ^{T ,k+1}_j ; δ^{T ,k}_−j

= sj,

which shows claim (ii).

Because of lemma 2, one can construct recursively a sequence  δ_j^{T ,k}

such that δ^{T ,k+1}_j ≥ δ_j^{T ,k} and

G^T_j  δ_j^{T ,k}

≤ sj, (1.A.18)

From assumption 4, setting η = s0/4 and b = T^∗log (4/s0− 1), one has the existence of δ ∈ R such that δ ≥ δj implies Pr (Uεj(δ) < b) < η.

Lemma 3. For all k ∈ N and T < T^∗, one has

δ_j^k≤ δj (1.A.19)

where δj is a constant independent from T < T^∗.

Proof. By summation of inequality (1.A.18) over j ∈ J , one has

s0 ≤

Z P (dε)

1 +P

j⁰∈Jexp Uεj⁰

 δ^{T ,k}_j0

 /T ≤

Z P (dε)

1 + exp Uεj

 δ_j^{T ,k}

/T

≤ Pr Uεj

 δ_j^{T ,k}

< b +

Z

{^Uεj(^δ_j^{T ,k})≥b}

P (dε) 1 + exp

U_εj δ^{T ,k}_j 

/T

≤ Pr Uεj

δ_j^{T ,k}

< b

+ 1

1 + exp (b/T^∗).

Now assume by contradiction that δ^{T ,k}_j > δj. Then Pr (Uεj(δ) < b) < η = s0/4 and (1 + exp (b/T^∗))⁻¹ = s0/4, and thus one would have

s0≤ s0/4 + s0/4 = s0/2,

a contradiction. Thus inequality (1.A.19) holds.

Lemma 4. Let δ_j^T = limk→+∞δ_j^{T ,k}. One has

G^T_j δ^T_j; δ^T_−j
= sj. (1.A.20)

Proof. One has G^T_j 

δ_j^{T ,k+1}; δ_−j^{T ,k}

= sj; by the fact that δ^{T ,k+1}_j → δ^T_j and δ_−j^{T ,k} → δ_−j^T and by the continuity of G^T_j, it follows (1.A.20).

We can now deduce the proof of theorem .

Proof of theorem 3’. Point (a): lemma 4 implies that one can define

u^T(ε) = T log



1 +X

j∈J

exp Uεj δ^T_j
/T



 and π_εj^T = exp −u^T(ε) + Uεj δ_j^T
T

! ,

and by the same result, one has

Eπ^Tu^T(ε) = Eπ^T Uεj δ_j^T
 .

It follows from lemma 3 that the sequence δ^T_j is bounded independently of T , so by compactness, it converge up to subsequence toward δ⁰_j. Note that δ⁰_j ≤ δ⁰_j ≤ δ⁰_j. We can extract a converging subsequence π^Tn where Tn → 0 and π^Tn → π⁰ in the weak convergence. Mimicking the argument in Villani (2003) page 32, it follows that π⁰∈ M (P, s).

Point (b): Let u⁰(ε) = maxj∈J₀Uεj δ_j⁰
. We have u⁰(ε) ≥ Uεj δ_j⁰
. Let us show that

Eπ⁰u⁰(ε) = Eπ⁰UεJ δ_J⁰
 ,

which will proof the final result. We have Eπ^Tnu^Tn(ε) = Eπ^Tn

hUεj

 δ^T_jⁿi

; let us show that

(i) Eπ^Tnu^Tn(ε) → Eπ⁰u⁰(ε) , and (ii) Eπ^Tn

hUεJ

 δ^T_Jⁿi

→ Eπ⁰UεJ δ_J⁰


Start by showing point (i). We have 0 ≤ u⁰(ε) − u^T(ε) ≤ T_nlog J . As a result, EπTnu^Tn(ε) = EPu^Tn(ε) → E^Pu⁰(ε) = Eπ⁰u⁰(ε).

Next, we show point (ii). One has,

< ν/4. By uniform continu-ity of ε → Uεj(δ) on K, and because δ^T_jⁿ → δ_j⁰, there exists n⁰ ∈ N such that n ≥ n implies

Combining (1.A.21) and (1.A.22), it follows that for n ≥ n⁰⁰, 

Proof of theorem 4

Once again we shall prove this theorem without assumption 2.

Theorem 4’. Under assumptions 1 and 5, ˜Σ⁻¹(s) has a single element for all s ∈ S₀^int.

Proof. Let δ ∈ ˜Σ⁻¹(s). Define Z to be the random vector such that Zj = U_εj⁻¹(Uε0(δ0)), and PZ

to be the probability distribution of Z. By proposition 1, this implies

P



Uε0(δ0) > max

j∈J Uεj(δj)



≤ s0≤ P



Uε0(δ0) ≥ max

j∈J Uεj(δj)



which is equivalent to

P (Zj> δj) ≤ s0≤ P (Zj≥ δj) but because Z has a density, the latter condition is equivalent to

s0= P (Zj≥ δj) .

Now consider δ^min and δ^max the lattice bounds of ˜Σ⁻¹(s). Because of the previous remark, P Zj≥ δ^min_j

= P Zj ≥ δ_j^max
. However, as distribution of Z has full support, the map δ → P (Z_j≥ δj) is strictly increasing in each δj, and as a result of δ_j^min≤ δ^max_j , it follows that δ_j^min= δ_j^max, QED.

1.A.3 Proof of theorem 5

We now provide the proof of theorem 5. In order to prove this result, we shall need a series of

We now provide the proof of theorem 5. In order to prove this result, we shall need a series of

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