Correlation matrices and Jacobi polynomials

Transcription

1 June 18, 2014

2 Program Correlation matrices Given margins and given correlation matrices Extreme points of the correlation matrices The Gasper distribution

3 Correlation matrices. Denote by R n the set of semipositive definite matrices of order n with diagonal (1,1,...,1). An element Σ = (r ij ) of R n is called a correlation matrix. The reason is : if (X 1,...,X n ) µ then cor(x i,x j ) = r ij define an element R(µ) of R n. The set R n is closed and bounded in the space of symmetric matrices of order n since r ij 1 when Σ = (r ij ) R n.

4 Example, n = 3. R(x,y,z) = 1 z y z 1 x y x 1 det R(x,y,z) = 1 x 2 y 2 z 2 +2xyz. The surface R(x,y,z) = 0 is the Kummer surface.

5 The Kummer surface 1 x 2 y 2 z 2 +2xyz = 0 Figure: R 3 is limited by this surface restricted to the cube [ 1,1] 3. It contains the simplex with vertices (1,1,1),(1, 1, 1),( 1, 1,1),( 1,1, 1) and is invariant by a finite group of rotations of order 4.

6 Distributions with given margins and given correlation Let ν 1,...,ν n be probabilities on R (with second moments) and R R n. Problem: describe the set S of probabilities µ on R n with prescribed margins ν j and prescribed correlation R(µ) = R. The set S is compact and convex but difficult to grasp. Questions 1. Is S non empty? 2. Can we construct at least one element of S? 3. Extreme points of S? 1) is difficult in general, 2) is not so hard in many cases, 3) is very difficult. This lecture considers the problem when ν 1 =...,ν n = ν is uniform on [0,1] (in this case the elements of S are called copulas) and more generally when ν is a symmetric beta distribution.

7 A side remark about Gaussian copulas If (X 1,...,X n ) N(0,Σ) with Σ R n and if Φ is the distribution function of N(0, 1), consider the distribution µ of (Φ(X 1 ),...,Φ(X n )). Then µ is a copula whose correlation R(Σ) is close to Σ. However the map Σ R(Σ) from R n into itself is injective but not surjective. In other terms there are correlation matrices which cannot be the correlation of such a Gaussian copula. Example (no time for the proof -computational and not difficult)

8 Main theorem Theorem For n = 3,4,5 and for any correlation matrix R there exists a copula of [0,1] n having R as correlation matrix. Furthermore, this is also true while replacing the uniform distribution on [0,1] by the beta distribution β k,k if k 1/2.

9 the four steps of the proof 1. Enough is to prove it when R is an extreme point of the compact convex set R n of the correlation matrices; 2. The set of extreme points of R n is not very well known. It contains elements of rank r if and only if r(r +1)/2 n. (Ycart s theorem). In particular if n = 3,4,5 the extreme points of R n have rank 1 or Correlations matrices of rank 2 have the following characteristic form: There exists numbers α 1,...,α n such that R = (cos(α j α k )) 1 j,k n. 4. If c = (cosα 1,...,cosα n ) T and s = (sinα 1,...,sinα n ) T a probability with correlation R must be concentrated on the linear two dimensional subspace H of R n generated by c and s.

10 A particular case Figure: We cut the cube by the plane P x +y +z = 0, obtaining an hexagon H. The circle C is inscribed in H.

11 Archimedes the Great The vertices of the cube are (± 3,± 3,± 3). S is the sphere with C as one of its large circle. We take a point uniformly on S. We project it on P (so that it falls inside C). The three coordinates of that point are each uniformly distributed on [ 3, 3] by Archimedes. The correlation is E(XY) = E(YZ) = E(ZX) = 1/2, E(X 2 ) = E(Y 2 ) = E(Z 2 ) = 1. We have seen that this correlation matrix was unattainable by a Gaussian copula.

12 The crucial part Theorem. Let k 1/2. If R = (cos(α i α j )) 1 i,j n there exists a probability distribution µ k on R n such that the correlation matrix of µ k is R and such that X j β k,k when X = (X 1,...,X n ) µ k.

13 The Gasper distribution in R 2 Recall that (p,q,r) = det 1 r q r 1 p q p 1 = 1 p 2 q 2 r 2 +2pqr For 1 < r < 1 consider the ellipse E r = {(x,y); (x,y,r) = 0} and its interior U r. The ellipse is tangent to the sides of the square [ 1,1] 2. When k > 1/2 we now introduce the Gasper probability on U r φ k,r (dx,dy) = 2k 1 2π (1 r2 ) 1 4 k 2 (x,y,r) k 3 2 1Ur (x,y)dxdy.

14 The singular case of the Gasper distribution When k = 1/2 we consider a singular probability φ 1/2,r concentrated on the ellipse E r which can be defined by φ 1/2,r = lim k 1/2 φ k,r.

15 Gasper distribution and Jacobi polynomials The Gasper distribution φ k,r appears as a Lancaster distribution for the pair (ν k,ν k ) where ν k (dx) = 21 2k Γ(2k) Γ 2 (1 x 2 ) k 1 1 (k) ( 1,1) (x)dx on [ 1,1]. More specifically consider the sequence (Q n ) n=0 of the orthonormal polynomials for the weight ν k. Thus Q n is the Jacobi polynomial Pn k 1,k 1 normalized such that 1 1 Q 2 n(x)ν k (dx) = 1.

16 The Gasper function K For 1/2 < k denote K(x,y,z) = n=0 Q n (x)q n (y)q n (z). Q n (1) This series converges if x, y, z < 1 and its sum is zero when (x,y) is not in the interior U r of the ellipse E r. With this notation we have φ k,r (dx,dy) = K(x,y,r)ν k (dx)ν k (dy). This result is essentially due to Gasper (1971) (with credits to Sonine, Gegenbauer and Moller). See Koudou (1995) and (1996) and Letac (2009) for details. The consequences of this last formula is that the correlation of φ k,r is exactly r.

17 The distribution µ k appears. Now introduce X = (X 1,...,X n ) satisfying the two conditions: X is concentrated on the plane H (X 1,X 2 ) φ k,cos(α1 α 2 )

18 The correlation matrix of X We claim that it is R = (cos(α i α j )) 1 i,j n This is saying that q(θ) = E( θ,x 2 ) n e iα j θ j 2 = E( θ,x 2 ) θ,c 2 θ,s 2 j=1 is zero. If θ H we have q(θ) = 0 since X is concentrated on H. If θ is in the two dimensional space F defined by θ 3 = θ 4 =... = θ n = 0 we have q(θ) = 0 since the correlation of (X 1,X 2 ) is cos(α 1 α 2 ). Since H and F are supplementary spaces let us write θ = u +v where u H and v F. Therefore θ,x = v,x and θ,c 2 + θ,s 2 = v,c 2 + v,s 2 = E( v,x 2 ).

19 The margins of X We claim that X i ν k for all i. Enough is to prove that for i j then (X i,x j ) φ k,cos(αi α j ), which implies that X i ν k. To see this we observe that there exists a pair (A,B) of real random variables such that X = Ac +Bs. In particular (X i,x j ) = (Acosα i +B sinα i,acosα j +B sinα j ). We now compute the joint distribution from the fact that the distribution φ k,cos(α1 α 2 ) of (X 1,X 2 ) is known, and we obtain that the density of (A,B) is C(1 a 2 b 2 ) k 1 2 when restricted to the unit disc. The important point is that the distribution of (A,B) does not depend on (α 1,α 2 ). As a consequence (X i,x j ) φ k,cos(αi α j ) and the proof is complete.

20 Conclusion 1. We have shown that if R = (r ij ) 1 i,j n R n has rank 2, there exists a probability µ k on R n such that if (X 1,...,X n ) µ k then (X i,x j ) φ(k,r ij ) In particular this implies that the correlation of µ k is R and X j ν k 2. doing k = 1 we get the uniform distribution ν 1 on ( 1,1). 3. From Ycart theorem, all extreme points of R n have rank 1 or 2 for n = 3,4,5. 4. Therefore we can construct a copula with an arbitrary correlation matrix if the dimension is Extension to the non symmetric beta β a,b (with a,b 1/2 are possible from the Gasper paper.