Indices de pouvoir et d interaction en théorie des jeux coopératifs : une app. une approche par moindres carrés

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1 Indices de pouvoir et d interaction en théorie des jeux coopératifs : une approche par moindres carrés University of Luxembourg Mathematics Research Unit, FSTC Paris, Le 18 Janvier 2011

2 Cooperative games and pseudo-boolean functions Cooperative games Set of players : N = {1,..., n}. Game : f : 2 N R. For a coalition S of players f (S) is the worth of S (usually f ( ) = 0, not additive). Pseudo-Boolean functions These are functions f : {0, 1} n R. Identifying S N and 1 S {0, 1} n, games are pseudo-boolean functions. A pseudo-boolean function can be written f (x) = v(s) x i x i ) = S N i S i N\S(1 a(s) x i. S N i S Multilinear extensions (Owen,1972) f : [0, 1] n R : x a(s) x i. S N i S

3 Power indexes Two problems : to share the benefits of the game or to measure the importance/influence of a player on the outcome of the game. The Shapley power index (L.S. Shapley 1953): φ Sh (f, i) = T i (n t 1)!t! n! (f (T i) f (T )) = T i (n t 1)!t! i f (T ), n! The discrete derivative : i f (T ) = f (T i) f (T \ i) The Banzhaf power index (J. Banzhaf 1965): φ B (f, i) = n 1 T i (f (T i) f (T )) = 2 T n 1 i i f (T )

4 Interaction indexes I To measure the interaction among players i and j: The Banzhaf interaction index (Owen (1972), Murofushi-Soneda (1993)) I B (f, {i, j}) = 1 2 n 2 T N\{i,j} Note that, for T N \ {i, j}, (f (T ij) f (T i) f (T j) + f (T )). ij f (T ) = f (T ij) f (T i) f (T j) + f (T ) = (f (T ij) f (T )) (f (T i) f (T )) (f (T j) f (T )) = (f (T ij) f (T j)) (f (T i) f (T )). For f (x) = T N a(t ) i T x i and S N, S f (x) = T S a(t ) i T \S x i

5 Interaction indexes II To measure the interaction among players in coalition S: The Banzhaf interaction index of S (Roubens (1996)) I B (f, S) = 1 2 n s S f (T ). T N\S The Shapley interaction index of S (Grabisch (1997)) I Sh (f, S) = T N\S (n t s)!t! (n s + 1)! S f (T ). Probabilistic interaction index of S (Grabisch,Roubens, see also Fujimoto, Kojadinovic, Marichal (2006)) I(f, S) = pt S S f (T ), with p S T 0 and T ps T = 1. T N\S

6 Alternative expressions of interactions Expressions in terms of the Möbius transform I B (f, S) = ( ) t s 1 a(t ), 2 T S I Sh (f, S) = T S 1 a(t ). t s + 1 In terms of the derivatives of f I B (f, S) = (D S f )( 1 2 [0,1] ) = D S f (x) dx, n I Sh (f, S) = D S f (x,..., x) dx. [0,1]

7 Main properties I B (f, S) i S Alternative representations The map f (I B (f, S) : S N) is a linear bijection : f (x) = S N ( xi 1 ). 2 Symmetry-anonymity If π S n and π(f )(x 1,..., x n ) = f (x π(1),..., x π(n) ), then Dummy players I(π(f ), π(s)) = I(f, S). A player i is dummy in f if f (T i) = f (T ) + f (i) f ( ) for T N \ i. If i is dummy, then I(f, i) = f (i) and I(f, S) = 0 S i. Some axiomatic characterizations use these properties.

8 Banzhaf power index and linear model Given a pseudo-boolean f, consider a linear model for f : f 1 (x) = a + a 1 x a n x n The power of i in f is given by a i. Randomization : all the coalitions are equally likely to form. Least squares method : find f 1 that minimizes x {0,1} n (f (x) g(x)) 2 among all linear models g. Theorem (Hammer-Holzman (1992)) In the solution of the least squares problem, a i = I B (f, i).

9 Banzhaf interaction index and multi-linear model V k : space of pseudo-boolean functions of degree k at most V k = {g : g(x) = c(s) x i } S N,s k i S For each f, find f k = s k a k(s) i S x i V k that minimizes (f (x) g(x)) 2 x {0,1} n among all g V k. Theorem (Grabisch-Marichal-Roubens (2000)) ( ) t s 1 (1 ) t sa(t a k (S) = a(s) + ( 1) k s ). k s 2 a s (S) = I B (f, S) (!). T S,t>k

10 Weighted least squares w(s) : the probability that coalition S forms : w(s) = Pr(C = S) Under independence p i = Pr(C i) = S i w(s) (0, 1) and w(s) = p i (1 p i ) i S i N\S Associated weighted least squares problem Find the unique f k V k that minimizes the (squared) distance x {0,1} n w(x) among all functions g V k. ( ) 2 f (x) g(x) = w(s) ( f (S) g(s) ) 2 S N Rmk : The distance is associated to the inner product f, g = x {0,1} n w(x)f (x)g(x), and w is defined by p = (p 1,..., p n ).

11 First solution of the least squares problem The use of independence Guoli Ding et al (2010) B k = {v S : S N, s k}, where v S : {0, 1} n R is given by v S (x) = x i p i pi (1 p i S i ) = i S\T ( p i) T S i S pi (1 p i ) i T forms an orthonormal basis for V k. The projection f k = T N t k f, v T v T. x i The index f, v S I B,p (f, S) = i S pi (1 p i )

12 First properties The index characterizes the projection : f k V k is the best kth approximation of f iff I B,p (f, S) = I B,p (f k, S) S : s k. (Hint : this is equivalent to f, v S = f k, v S.) The map f I B,p (f, S) is linear. The map f (I B,p (f, S) : S N) is a bijection f k (x) = I B,p (f, T ) i T(x i p i ), k = n (!) T N,t k The index and the multilinear extension of f : I B,p (f, S) = (D S f )(p)

13 Explicit formulas From f (x) = T N a(t ) i T x i Explicit expression of the index I B,p (f, S) = a(t ) T S Proof : Just compute the derivatives of f. i T \S p i Explicit expression of the approximation f k (x) = S N,s k a k(s) i S x i a k (S) = a(s) + ( 1) k s T S,t>k ( )( t s 1 k t i T \S p i ) a(t ) Proof : Use expression of f k and I B,p (f, S), expand and do some algebra.

14 The index as an expected value An expected value of the discrete derivative I B,p (f, S) = E( S f ) = w(x) S f (x). x {0,1} n Proof : Use S f (x) = T S a(t ) i T \S x i, independence and explicit formula for I B,p (f, S). An average (As a probabilistic interaction index) I B,p (f, S) = pt S ( S f )(T ), T N\S where pt S = Pr(T C S T ) = i T p i i (N\(S T )) (1 p i). Proof : Use ( S f )(T ) = ( S f )(T \ S). Interpretation : pt S = Pr(C = S T C S) = Pr(C = T C N \ S)

15 Further properties Null players A player i is null for f if f (T i) = f (T ) for all T N \ i. If S contains a null player then I B,p (f, S) = 0. Dummy coalitions D N is dummy for f if f (T ) = f (T D) + f (T (N \ D)) f ( ) for every T N. If D is dummy for f, if K D and K \ D : I B,p (f, K) = 0. Symmetry of the index An index I is symmetric if I(π(f ), π(s)) = I(f, S) for all permutations π. I B,p is symmetric if and only w or is symmetric i.e. p 1 = = p n.

16 Back to Banzhaf and Shapley A link with the Banzhaf index I B,p (f, S) = I B,p (f, T ) T S (p i p i ), set p i or p i to 1 2 (!) i T \S Another link with the Banzhaf index I B (f, S) = I B,p (f, S) dp [0,1] n Proof : Just use explicit expressions and integrate. Interpretation : take an average over p if it is not known. A link with the Shapley index I Sh (f, S) = 1 0 I B,(p,...,p) (f, S) dp. Interpretation : average if the players behave in the same (unknown) way

17 Part II : Interactions among variables of functions over [0, 1] n.

18 Problem I: Influence of variables The problem Consider f : [0, 1] n R. How to measure the influence of variable x k over f (x 1,..., x k,..., x n )? Applications x i : partial score of an alternative with respect to criterion i. f (x 1,..., x n ) : global score. A simple example f (x 1,..., x n ) = n i=1 c ix i + c 0. A simple answer I(f, k) = c k.

19 More examples The geometric mean f (x 1,..., x n ) = ( n i=1 x ) 1 n i ; Weighted geometric means f (x 1,..., x n ) = n i=1 x ci i ; Weighted maxima f (x 1,..., x n ) = max i N min(c i, x i ); Lovasz extensions f (x 1,..., x n ) = T N a T min i T x i...

20 A solution to problem I Step I Given a function f, approximate f by a linear model of the form f 1 = c 0 + c 1 x c n x n. Restriction: f L 2 ([0, 1] n ). Step II Define I(f, k) = c k. Explicit expression I(f, k) = 12 f (x) ( x k 1 ) d x. [0,1] 2 n Alternative definition Consider a linear model of the form f {k} = c 0 + c k x k.

21 The worth to improve index The concept of influence of subsets of variables was defined by M. Grabisch and C. Labreuche. Axioms : I(, k) is linear on L 2 ([0, 1] n ); I(, k) is continuous on L 2 ([0, 1] n ); Step evaluation : value of the index on particular threshold functions; Normalization: value of the index for weighted means. The result The influence index (worth to improve) criterion k: W k = 6 f (x)(2x k 1)d x. [0,1] n In general for S {1,..., n}, W S = 3 2 S f (x) ( x i x i ) [0,1] i S(1 ) d x. n i S

22 Problem II : Interactions among variables The problem Consider f : [0, 1] n R. How to measure the interactions among {x i : i S} in f (x 1,..., x k,..., x n )? A simple example Consider a multilinear function f : [0, 1] 2 R: (x 1, x 2 ) c 0 + c 1 x 1 + c 2 x 2 + c 12 x 1 x 2. A simple answer The interaction between x 1 and x 2 within f is defined by c 12.

23 Least squares approximations The approximation problems Denote by M s the space of mutlilinear functions g : [0, 1] n R of degree less than or equal to s, i.e. of the form g(x) = x i. T N, T s c(t ) i T The standard L 2 distance of functions f, h L 2 ([0, 1] n ) is d(f, h) 2 ( ) 2d = f (x) h(x) x. [0,1] n For f L 2 ([0, 1] n ), find f s M s that minimizes d(f, g) for g in M s. Alternative problem : Define M S as the space of multilinear functions g(x) = T S c(t ) i T x i. For every f L 2 ([0, 1] n ), compute f S.

24 Interaction indexes First index For f L 2 ([0, 1] n ) and s n write f s (x) = a s (T ) x i. T N, T s i T Define I 1 (f, S) = a s (S) for S N, s = S. Second index For f L 2 ([0, 1] n ) and S N write f S (x) = a S (T ) x i. T S i T Define I 2 (f, S) = a S (S) for S N.

25 Computations The result For every f L 2 ([0, 1] n ) and S N we have I 1 (f, S) = I 2 (f, S) := I(f, S) = 12 S [0,1] n f (x) i S ( x i 1 ) d x. 2 Proof The space L 2 ([0, 1] n ) is an inner product space w.r.t. f, g = f (x)g(x)d x. [0,1] n Define v T (x) = 12 T /2 i T ( x i 1 ), T N. 2 Then f s = T s f, v T v T and f S = T S f, v T v T.

26 Examples f (x) = 1 n n x i. i=1 I(f, k) = 1 n ; I(f, S) = 0 if S 2. I(f, k) = ( 1 I(f, S) = ( 1 n+1 ) n 6 n+1 2n+1 ; ) n ( 6 2n+1) s. f (x) = ( n i=1 x ) 1 n i I(f, S) = i N f (x) = n i=1 x ci i 1 6c i c i + 1 c i + 2. i S

27 Interpretations as average of derivatives For x [0, 1] n, D k f (x) = local influence of x k to f at x; D k D j f (x) = local interaction of x j and x k within f at x; Theorem If f L 2 ([0, 1] n ) is smooth enough, then I(f, S) = q S (x) D S f (x) dx, [0,1] n where q S (x) = 6 S i S x i(1 x i ) is a p.d.f. on [0, 1] n. Proof : Integration by parts.

28 Interpretations as average of difference quotients Define the discrete derivative S h f (x) of f at x [0, 1]n by {i} h f (x) = f (x + h ie i ) f (x). + induction. The difference quotient is then Qh S f (x) = S h f (x) i S h i Theorem For every S N, we have I(f, S) = p S (x, y S ) Qy xf S (x) dy S dx, x [0,1] n y S [x S,1] where p S (x, y S ) = 6 S i S (y i x i ) is a p.d.f on the integration domain.

29 Properties I The index is linear and continuous. Link with the Banzhaf index If g : {0, 1} n R is pseudo-boolean and f = ḡ is the multilinear extension, then I(f, S) = I B (g, S). Symmetry The index I is symmetric: we have I(π(f ), π(s)) = I(f, S).

30 Properties II Ineffective variables If x i is ineffective for f, then I(f, S) = 0 whenever i S. Dummy subsets of variables D N is dummy for f : [0, 1] n R if f (x) = f (0 D x N\D ) + f (0 N\D x D ) f (0). Then if S D and S \ D, I(f, S) = 0. S-Increasing functions If f is S-increasing for some S N ( S y xf (x) 0 for y x), then I(f, S) 0.

31 More examples Multiplicative functions If f is multiplicative : f (x) = i N f i(x i ) then I(f, S) I(f, ) = I(f i, i) I(f i, ). i S Discrete Choquet integrals If f (x) = T N a(t ) min i T x i, then I(f, S) = 6 S T S a(t ) B( S + 1, T + 1), where B(p, q) = 1 0 t p 1 (1 t) q 1 dt = (p 1)!(q 1)! (p + q 1)!.

32 Normalized index Goal : to compare indexes for different functions. Problem : we need to normalize the index. Observation : the index is defined (for S ) as a covariance : I(f, S) = f E(f ), 12 S /2 (v T E(v T )) We define the normalized index as the Pearson coefficient: r(f, S) = I(f, S) 12 S /2 σ(f ). r(am, i) = 1/ n; r(min N x k, i) = 3 ; n(n+2) r(gm, {i}) = (( 3 (n+1) 2 n ) 1/2. 2n+1 n(n+2)) 1

33 Thanks for your attention arxiv: arxiv: