Chapter 5. Nucleolus. Core(v) := {x R n x is an imputation such that e(x, S) 0 for all coalitions S}.

Transcription

1 Chapter 5 Nucleolus In this chapter we study another way of selecting a unique efficient allocation that forms an alternative to the Shapley value. Fix a TU-game (N, v). Given a coalition S and an allocation x we put e(x, S) := v(s) x(s) and call e(x, S) the excess of S at x. Intuitively, given a coalition S, e(x, S) measures the amount of dissatisfaction (or complaint ) of S from the allocation x. Indeed, the larger e(x, S) is the smaller x(s) is. Using the excess function the core can be alternatively defined by: Core(v) := {x R n x is an imputation such that e(x, S) 0 for all coalitions S}. Given an allocation x, by varying S over all coalitions, we obtain the excess function e(x, ) that we abbreviate to e(x) and view as a sequence of m := 2 N 1 reals. In other words, each allocation x determines a sequence e(x) R m. Before we proceed let us illustrate this definition on the One seller, two buyers TU-game from Example 2. Take the imputation x := (1.5, 0, 0.5) (that lies in the core of the game). We represent e(x) using the following table, in which for readability we also list the function v and where we abbreviate the coalition {1} to 1, etc: 32

2 S v(s) e(x, S) We can now reorder the obtained excesses in the decreasing order: (0, 0, 0, 0.5, 0.5, 0.5, 1.5), so by putting the excesses of the most dissatisfied coalitions first. This way we associated with the imputation x the sequence of diminishing complaints. In turn, for the imputation y := (1, 1, 0) we have S v(s) e(y, S) Here the sequence of diminishing complaints is (1, 0, 0, 1, 1, 1, 1). We notice that x leads to a less problematic sequence of (diminishing) complaints than y. Indeed, the largest complaint about the imputation x is 0, while the largest complaint about the imputation y is 1. In general, we would like to identify an imputation for which the resulting sequence of diminishing complaints is least problematic. To make this idea precise we shall use the lexicographic order lex on the sequences of reals of length m := 2 n 1 defined by where x lex y iff x lex y or x = y, (x 1,..., x m ) lex (y 1,..., y m ) iff k m (x k < y k and i k x i = y i ). 33

3 Obviously, for the above two sequences we have (0, 0, 0, 0.5, 0.5, 0.5, 1.5) lex (1, 0, 0, 1, 1, 1, 1, 1). In turn, the reader may check that for the imputation z := (1, 0, 1) the sequence of diminishing complaints is (0, 0, 0, 0, 1, 1, 1) and we have (0, 0, 0, 0.5, 0.5, 0.5, 1.5) lex (0, 0, 0, 0, 1, 1, 1), that is the sequence of diminishing complaints for the imputation x is less problematic than the one for the imputation z. Given a sequence f R m of reals we denote by f its reordering from the largest to the smallest element. We shall apply this procedure to the excess functions e(x) (that we view as sequences). For example, we just showed above that in the One seller, two buyers TU-game for x = (1.5, 0, 0.5), y = (1, 1, 0) and z = (1, 0, 1) we have both e(x) lex e(y) and e(x) lex e(z). The nucleous of a TU-game is then defined as the following set: {x R n x is an imputation and e(x) lex e(y) for all imputations y}. Informally, an imputation x is in the nucleolus if it yields the least problematic sequence of complaints. The following crucial theorem shows that nucleolus determines a unique imputation. Theorem 14 For each superadditive TU-game the nucleolus consists of a single imputation. We split the proof into two parts. First we show that the nucleolus is non-empty. We shall need the following standard result. Recall that for a function f : A B and C B, f(a) := {f(x) x A} and f 1 (C) := {x A f(x) C}. Theorem 15 Let A be a non-empty compact subset of R n. Then (i) A function f : A R is continuous iff for every closed subset B of R the set f 1 (B) is compact. 34

4 (ii) If f : A R m is continuous, then f(a) is a non-empty compact subset of R m. Using it and the Extreme Value Theorem 9 mentioned in Chapter 3 we now establish the following general result about the lexicographic ordering lex. Theorem 16 Let A be a non-empty compact subset of R n. Then the set is non-empty. {x A x lex y for all y A} In fact, the above set has exactly one element, that is the lexicographic ordering on a non-empty compact subset of R n has a unique minimum. But we shall not this observation. Proof. For i {1,..., m} denote by π i : A R the projection function. That is, π i (x 1,..., x n ) = x i. Clearly, each projection function is continuous. Below we use the following notation, given a function f : A R and a non-empty subset B of A: argmin x B f(x) := {y B f(y) = min x B f(x)}. That is, argmin x B f(x) is the set of the elements of B on which the function f reaches the minimum on B. The set argmin x B f(x) is non-empty only when the minimum on B is actually reached for some x. First, we define inductively the subsets A 0,..., A m of R n as follows: A 0 := A, A i+1 := argmin x Ai π i+1 (x). That is, A i+1 is the set of elements of A i whose (i + 1)st component is minimal. We prove by induction that each set A i is a non-empty compact subset of R n. By assumption the claim holds for i = 0. Suppose it holds for some i {0,..., m 1}. By the Extreme Value Theorem 9 min x Ai π i+1 (x) exists, so the set A i+1 is non-empty. Moreover, alternatively A i+1 = (π i+1 ) 1 ({min x Ai π i+1 (x)}) A i. So by Theorem 15(i) A i+1 is a compact subset of R n. 35

5 We conclude that A m is a non-empty set. But alternatively A m = {x A x lex y for all y A}, which concludes the proof. To clarify why the above result is relevant for us we now establish the following lemma. Lemma 17 Consider a superadditive TU-game. (i) The set {x x is an imputation} is a non-empty compact subset of R n. (ii) The set {e(x) x is an imputation} is a non-empty compact subset of R m. Proof. (i) The set of imputations is non-empty, since by the superadditivity n i=1 v(i) v(n). It is also straightforward to see that it is closed. Finally, it is bounded. Indeed, for each imputation x and for all i {1,..., n} we have v(i) x(i) by the individual rationality. Moreover, by the efficiency x(n) = v(n), so, again by the individual rationality 1, x(i) + j i v(j) v(n), i.e., x(i) v(n) j i v(j). (ii) The non-emptiness follows directly from (i). Further, consider the function f from the set Imp of all imputations to R defined by f(x) := e(x). There are continuous functions f 1,..., f m from Imp to R such that f(x) := (f 1 (x), f 2 (x),..., f m (x)). Indeed, for each i {1,..., m} define f i to be the function that assigns to an imputation x the ith largest excess e(x, S), i.e., the ith element of e(x). One can easily check that f i is continuous. For example, take the One seller, two buyers TU-game and the already considered imputation x = (1.5, 0, 0.5). Then e(x) = (0, 0, 0, 0.5, 0.5, 0.5, 1.5), so we have in particular f 3 (x) = 0 and f 4 (x) = 0.5. So f itself is continuous and by Theorem 15(ii) the set f(imp) = {e(x) x is an imputation} is non-empty and compact. 1 j i is a shorthand for the summation over all j {1,..., n}, j i. 36

6 Proof of Theorem 14, part I. We prove now that the nucleolus is non-empty. This turns out be an immediate consequence of Theorem 16 and Lemma 17(ii). Indeed, fix a superadditive TU-game (N, v). Take m = 2 n 1 and as A R n the set {e(x) x is an imputation}. By Lemma 17(ii) A is a nonempty compact subset of R m, so by Theorem 16 the set {e(x) x is an imputation and e(x) lex e(y) for all imputations y} is non-empty. Hence the nucleolus is non-empty. To prove that the nucleolus has at most one element we shall use the following result about the lexicographic order. Theorem 18 Let A be a non-empty convex subset of R m. Then the set {x A x lex y for all y A} has at most one element. Proof. We start by listing some general properties of the lexicographic order. Claim 3 For all x R m, x lex x. Proof. Left as Exercise 1. Claim 4 For all x R m and permutations σ of {1,..., m}, xσ lex x, where for i {1,..., m}, (xσ) i := x σ(i). Proof. By Claim 3 xσ lex (xσ). Also (xσ) = x. Claim 5 For all x, y, u, z R m and α > 0 x lex y and u lex z imply αx lex αy and αu lex αz, x lex y and u lex z imply x + u lex y + z, x lex y and u lex z imply x + u lex y + z. 37

7 Proof. Left as Exercise 2. We can now prove the theorem. Suppose that the set {x A x lex u for all u A} has at least two elements. So for some x, y A, x y we have x = y and for all u A, x lex u and y lex u. Take some α (0, 1). For some permutation σ (αx + (1 α)y) = (αx + (1 α)y)σ = αxσ + (1 α)yσ. By Claim 4 xσ lex x and yσ lex y. We first show that actually xσ = x and yσ = y. Suppose otherwise. Then either xσ lex x or yσ lex y, so by Claim 5 αxσ + (1 α)yσ lex αx + (1 α)y, and hence by the choice of σ and the fact that x = y we obtain (αx + (1 α)y) lex x. But A is convex, so αx + (1 α)y A. This means that we found u A such that u lex x, which contradicts the choice of x. So indeed xσ = x and yσ = y. Hence xσ = yσ and consequently x = y which yields a contradiction. To clarify why the above result is relevant for us we now establish the following lemma analogous to Lemma 17. Lemma 19 Consider a superadditive TU-game. (i) The set {x x is an imputation} is a non-empty convex subset of R n. (ii) The set {e(x) x is an imputation} is a non-empty convex subset of R m. Proof. (i) Immediate by the definition of an imputation. (ii) Take two imputations x and y and consider the corresponding excess functions e(x) and e(y). Take some α [0, 1] and consider the function αe(x) + (1 α)e(y). For all coalitions S we have (αe(x) + (1 α)e(y))(s) = αe(x, S) + (1 α)e(y, S) = α(v(s) x(s)) + (1 α)(v(s) y(s)) = 38

8 v(s) (αx + (1 α)y)(s) = e(αx + (1 α)y, S), which shows that αe(x) + (1 α)e(y) = e(αx + (1 α)y). But by (i) αx + (1 α)y is an imputation, so we conclude that αe(x) + (1 α)e(y) is an imputation. Proof of Theorem 14, part II. We prove now that the nucleolus has at most one element. By Theorem 18 and Lemma 19(ii) the set {e(x) x is an imputation and e(x) lex e(y) for all imputations y} has at most one element. But for all imputations x and y if x y, then e(x) e(y). Consequently the set {x x is an imputation and e(x) lex e(y) for all imputations y}, i.e., the nucleolus, also has at most one element. The following general result clarifies the relation between the nucleolus and the core. Note that we do not require here that the nucleolus is nonempty. Theorem 20 For each TU-game with a non-empty core the nucleolus is included in the core. Proof. Take an imputation y that lies in the core of a TU-game. Then for all coalitions S we have e(y, S) 0. Let x be an element of the nucleous. Then e(x) lex e(y). The first elements of e(x) and e(y) are respectively max(e(x)) and max(e(y)), so we have max(e(x)) max(e(y)). Consequently for all coalitions S we have e(x, S) max(e(x)) max(e(y)) 0, i.e., x lies in the core. We noted in Chapter 4 that the Shapley value does not need to be an element of the core. So the above result shows that nucleolus and Shapley value are in general different concepts. We can also use it to find nucleolus. As an illustration let us return to the One seller, two buyers TU-game. This game is clearly superadditive so by Theorem 14 we can identify the nucleolus with its unique element. We show that the imputation x = (1.5, 0, 0.5) is its nucleolus. We noticed in Chapter 39

9 3 that the core of this game equals {(t, 0, t 2) t [1, 2]}. So by the above theorem we know that the nucleolus is of the form x t = (t, 0, t 2) for some t [1, 2]. So e(x t ) is defined as follows: S e(x t, S) 1 t t t t Now, t [1, 2] implies that t [ 2, 1], 1 t [ 1, 0] and t 2 [ 1, 0], so the corresponding ordered sequence e(x t ) is either (0, 0, 0, 1 t, t 2, t 2, t) or (0, 0, 0, t 2, t 2, 1 t, t). Suppose that 1 t 0.5. Then, since (1 t) + (t 2) = 1, we have either 1 t > 0.5 > t 2 or t 2 > 0.5 > 1 t. In both cases the sequence (0, 0, 0, 0.5, 0.5, 0.5, 1.5) is lex than the corresponding reordered version of (0, 0, 0, 1 t, t 2, t 2, t). So e(x) lex e(x t ), hence x t is not the nucleolus, contrary to its choice. So 1 t = 0.5, i.e., x t = x, which proves that x is indeed the nucleolus. Exercise 1 Prove Claim 3, i.e. that for all x R m, x lex x. Exercise 2 Prove Claim 5. Exercise 3 Prove that for all x, y R m and α [0, 1] (αx + (1 α)y) lex αx + (1 α)y. 40