TWO DEPENDENCE MEASURES FOR MULTIVARIATE EXTREME VALUE - PDF document

TWO DEPENDENCE MEASURES FOR MULTIVARIATE EXTREME VALUE DISTRIBUTIONS WEISSMAN, ISHAY Technion - Israel Institute of Technology, Israel ieriw01@ie.technion.ac.il 1

Outline: - Introduction - Dependence measures τ 1 , τ 2 - Examples - Relations between τ 1 , τ 2 - Combining two models - Conclusions 2

1. Introduction x ∈ R d , X = ( X 1 , X 2 , · · · , X d ) ∼ G ( x ) , where G is a multivariate extreme value distribution function. WOLOG with Fr´ echet margins: for all j G j ( x ) = exp {− 1 /x } ( x > 0) and exponent function λ ( x ) = − log G ( x ) , G t ( t x ) = G ( x ) tλ ( t x ) = λ ( x ) , ( t > 0) . ⇒ Since for MEVD d � 1 ≤ j ≤ d G j ( x j ) ≥ G ( x ) ≥ min G j ( x j ) , j =1 in our case � 1 max 1 ≤ λ ( x ) ≤ . x j x j (complete dependence) (total independence) 3

The homogeneity of λ implies that λ ( tx ) / Σ( tx j ) − 1 does not depend on t. Define the (generalized) Pickands dependence function A ( v ) = λ ( v − 1 1 , v − 1 2 , · · · , v − 1 ) v ∈ Ω , d where Ω = { v : v j ≥ 0 , Σ v j = 1 } is the unit-simplex. It follows that 1 d ≤ A 0 ( v ) =: max v j ≤ A ( v ) ≤ 1 , λ ( x ) = A ( v )Σ x − 1 , j where v j = x − 1 / Σ x − 1 . j i � 1 d, · · · , 1 � η = A d has an interesting interpretation: � � P 1 ≤ j ≤ d X j ≤ z max = exp {− λ ( z, z, · · · , z ) } = exp {− dη/z } = { exp {− 1 /z }} dη . 4

Hence, θ = dη is the extremal coefficient of ( X 1 , X 2 , · · · , X d ) . θ = 1 ⇔ complete dependence θ = d ⇔ total independence Schlather and Tawn (2002) analyse θ B = | B | η B for all 2 d possible subsets B of { 1 , 2 , · · · , d } . From de Haan and Resnick (1977) and Pickands (1981) � A ( v ) = Ω max v j a j U ( d a ) for a finite positive measure U , � U (Ω) = d and Ω a j U ( d a ) = 1 for all j. The function A is convex because for 0 ≤ α ≤ 1, max { ( αv j + (1 − α ) w j ) a j } ≤ α max v j a j + (1 − α ) max w j a j . 5

1.0 0.9 0.8 A(v) 0.7 0.6 0.5 0.0 0.2 0.4 0.6 0.8 1.0 v Pickands dependence function for the Logistic Model A ( v ) = ( v 1 /α + (1 − v ) 1 /α ) α with α = 0 , . 25 , . 50 , . 75 , 1 . 2. Measures of Dependence : Rescaling η , a natural measure of dependence is � 1 d , · · · , 1 � 1 − A d τ 1 = � � 1 d , · · · , 1 �� max A 1 − A d 6

= d − θ d d − 1 = d − 1(1 − η ) An alternative measure is � Ω (1 − A ( v )) d v τ 2 = max A � Ω (1 − A ( v )) d v � Ω (1 − A 0 ( v )) d v =: S d ( A ) Ω (1 − A ( v )) d v = S d ( A 0 ) . � Which one is preferred? Similar question: mode vs. mean. Expect from dependence measure that for A = αA 0 + (1 − α ) · 1 ⇒ τ = α. Indeed, for this mixture model τ 1 = τ 2 = α. 7

To compute τ 2 we need a formula for S A 0 , the volume above A 0 : d S A 0 2 1 / 4 = . 2500 3 7 / 36 = . 19444 4 . 07986 5 . 02264 Until very recently the challenge was to find a formula for S d ( A 0 ) . My colleague Shmuel Onn derived and proved ( d − 1)! − B d 1 S d ( A 0 ) = d ! where 1 + 1 2 + 1 3 + · · · + 1 � � B d = d is the harmonic sum. 8

Other (bivariate) measures of dependence: In the literature (Beirlant et al, 2004) we encounter τ K = Kendall’s tau = 4 EC ( G 1 ( X 1 ) , G 2 ( X 2 )) − 1 , ρ S = Spearman’s rho = corr ( G 1 ( X 1 ) , G 2 ( X 2 )) , ρ = corr (log G 1 ( X 1 ) , log G 2 ( X 2 )) . Tawn (1988) mentioned τ 1 for d = 2. I have not seen τ 2 . These are all marginal-free and for mixture distributions (not mixture exponents): � � ( U, V ) w.p. 1 − α ( X 1 , X 2 ) = ( U, U ) w.p. α U, V independent, τ K = ρ S = ρ = α. 9

3. Examples. Let V 1 , V 2 , · · · be i.i.d. unit-Fr´ echet. Mixture model : For 0 ≤ α ≤ 1 λ ( x, y ) = α max( x − 1 , y − 1 ) + (1 − α )( x − 1 + y − 1 ) . That is X = max( αV 1 , (1 − α ) V 2 ) Y = max( αV 1 , (1 − α ) V 3 ) . A ( v ) = α max( v, 1 − v ) + (1 − α ) · 1 ( v ∈ [0 , 1]) . τ 1 = τ 2 = α. α 3 α τ K = ρ = 2 − α ≤ ρ S = 4 − α ≤ α. α = τ 1 = τ 2 τ K = ρ ρ S 0 0 0 1 / 4 1 / 5 1 / 7 1 / 2 3 / 7 1 / 3 3 / 4 9 / 13 3 / 5 1 1 1 10

Mixed model : λ ( x, y ) = 1 x + 1 α y − x + y A ( v ) = 1 − α (1 − v ) v τ 1 = α τ 2 = 2 2 , 3 α τ K = 8 tan − 1 ( α/ (4 − α )) 1 / 2 − 2 α 1 / 2 (4 − α ) 1 / 2 ρ = 8 tan − 1 ( α/ (4 − α )) 1 / 2 − 2 − α α 1 / 2 (4 − α ) 3 / 2 4 − α 8 tan − 1 ( α/ (8 − α )) 1 / 2 � � 1 ρ S = 12 + − 3 α 1 / 2 (8 − α ) 3 / 2 8 − α α τ K ρ τ 1 ρ S τ 2 0 0 0 0 0 0 . 25 . 0877 . 0901 . 1250 . 1299 . 1667 . 50 . 1853 . 1958 . 2500 . 2702 . 3333 . 75 . 2947 . 3215 . 3750 . 4222 . 5000 1 . 4184 . 4728 . 5000 . 5874 . 6667 11

de Haan - Resnick model : λ ( x, y, z ) = 1 2 { max( x − 1 , y − 1 ) + max( x − 1 , z − 1 ) + max( y − 1 , z − 1 ) } X 1 = max( V 1 , V 2 ) / 2 X 2 = max( V 1 , V 3 ) / 2 X 3 = max( V 2 , V 3 ) / 2 A ( v ) = 1 2 { max( v 1 , v 2 ) + max( v 1 , v 3 ) + max( v 2 , v 3 ) } η = A (1 / 3 , 1 / 3 , 1 / 3) = 1 / 2 , τ 1 = (3 / 2)(1 − η ) = 3 / 4 τ 2 = 36 7 · 1 8 = 9 14 = . 642857 τ 1 (1 , 2) = τ 2 (1 , 2) = 1 / 2 (Introducing X 3 to the system increases the dependence) 12

Non-symmetric model : X 1 = max( V 1 / 2 , V 2 / 4 , V 3 / 4) X 2 = max(2 V 1 / 3 , V 2 / 3) X 3 = V 3 A ( v ) = max( . 75 v 1 , v 2 ) + max( . 25 v 1 , v 3 ) ( v 1 + v 2 + v 3 = 1) η = 2 / 3 , τ 1 = 1 / 2 , τ 2 = (36 / 7) . 104762 = . 53876 τ 1 (1 , 2) = 3 / 4 = . 75 τ 2 (1 , 2) = 6 / 7 = . 85714 τ 1 (1 , 3) = 1 / 4 = . 25 τ 2 (1 , 3) = 4 / 10 = . 4 13

4. Relations between τ 1 and τ 2 . Theorem. For d = 2, τ 1 ≤ τ 2 . 1.0 0.8 0.6 0.4 0.2 0.0 0.0 0.2 0.4 0.6 0.8 1.0 Proof. Denote 1 − h = A (1 / 2 , 1 / 2)) , ⇒ τ 1 = 2 h. Define the mixture model (green graph) A ∗ ( v ) = τ 1 A 0 ( v ) + 1 − τ 1 , 14

τ ∗ 1 = τ 1 = τ ∗ ⇒ 2 . Since A is convex, A ≤ A ∗ (= at (1 / 2 , 1 / 2)), � � � Ω (1 − A ∗ ) = τ 1 Ω (1 − A ) ≥ Ω (1 − A 0 ) � Ω (1 − A ) τ 2 = Ω (1 − A 0 ) ≥ τ 1 . � This is a perfect proof for d = 2. For d ≥ 3, the picture is misleading, namely, A ≤ A ∗ is not necessarily true. Here is a counter example: de Haan-Resnick model. For v 1 ≥ v 2 ≥ v 3 , v 1 + v 2 + v 3 = 1 , 2 , A ∗ ( v ) = 3 4 v 1 + 1 A ( v ) = v 1 + v 2 4 . Since v 2 ≥ v 3 ⇔ v 2 ≥ (1 − v 1 ) / 2 , 15

A ( v ) − A ∗ ( v ) = 1 4 v 1 + 1 2 v 2 − 1 4 ≥ 0 , with equality when v 1 ≥ 1 / 3 , v 2 = v 3 = (1 − v 1 ) / 2 . For the logistic model A ( v ) = ( v 1 /α + v 1 /α + v 1 /α ) α , 1 2 3 A (1 / 3 , 1 / 3 , 1 / 3) = 3 α − 1 . τ 1 = (3 − 3 α ) / 2 α τ 2 = � Ω (1 − A )36 / 7 0 1 1 1 / 4 . 8420 . 9457 1 / 2 . 6340 . 7670 3 / 4 . 3602 . 4559 1 0 0 16

For d = 2, how big can the difference τ 2 − τ 1 be? 1.0 0.9 0.8 −− − 1−h 0.7 0.6 0.5 0.0 0.2 0.4 0.6 0.8 1.0 Consider all (symmetric, d = 2) models for which A (1 / 2) = 1 − h so that τ 1 = 2 h . 17

All the A functions must be bounded between the green graph and the red one. The green graph corresponds to a mixture model with α = 2 h = τ 1 = τ 2 : X 1 = max(2 hV 1 , (1 − 2 h ) V 2 ) X 2 = max(2 hV 1 , (1 − 2 h ) V 3 ) . The red A corresponds to ”cross over” model: X 1 = max( hV 1 , (1 − h ) V 2 ) X 2 = max((1 − h ) V 1 , hV 2 ) for which τ 2 = 4 h (1 − h ) = 1 − (1 − τ 1 ) 2 . τ 1 = 2 h, ( τ 2 − τ 1 ) = 1 max 4 h occurs at h = 1 / 4 , τ 1 = 1 / 2 , τ 2 = 3 / 4 . 18

To be fair, one could hold the area (volume) constant (i.e. τ 2 ) and let τ 1 vary. For instance, all triangles with height h have τ 2 = 2 h, (0 ≤ h ≤ 1 / 2) but h 1 − h ≤ τ 1 ≤ 2 h = τ 2 . h = 1 / 4 , 1 / 3 ≤ τ 1 ≤ 1 / 2 = τ 2 . 19

Combining two models. X = ( X 1 , · · · , X k ) , Y = ( Y 1 , · · · , Y m ) are combined into Z = ( X 1 , · · · , X k , Y 1 , · · · , Y m ) , k + m = d. To study the dependence measures of Z we must know the dependence between X and Y . If they are independent we can compute τ 1 and τ 2 : A ( v ) = tA 1 ( u ) + (1 − t ) A 2 ( w ) ( v ∈ Ω d ) , t = v 1 + · · · + v k , u ∈ Ω k , w ∈ Ω m , u i = v i v k + i t , 1 ≤ i ≤ k ; w i = (1 − t ) , 1 ≤ i ≤ m. The Jacobian of the transformation ( v 1 , · · · , v d − 1 ) �→ ( t, u 1 , · · · , u k − 1 , w 1 , · · · , w m − 1 ) J = t k − 1 (1 − t ) m − 1 . is 1 − A = t (1 − A 1 ) + (1 − t )(1 − A 2 ) 20

� S ( A ) = (1 − A ( v )) d v = Ω d � 1 � � (1 − A ) d u d w t k − 1 (1 − t ) m − 1 dt = 0 Ω k Ω m � 1 1 � 0 t k (1 − t ) m − 1 dt = (1 − A 1 ( u )) d u ( m − 1)! Ω k � 1 1 � 0 t k − 1 (1 − t ) m dt + (1 − A 2 ( w )) d w ( k − 1)! Ω m = k ! d ! S k ( A 0 ) τ 2 , 1 + m ! d ! S m ( A 0 ) τ 2 , 2 τ 2 = k − B k d − B d τ 2 , 1 + m − B m d − B d τ 2 , 2 where B k = 1 + 1 2 + 1 3 + · · · + 1 k. 21