Constrained optimal discrimination designs for Fourier regression - PowerPoint PPT Presentation

Constrained optimal discrimination designs for Fourier regression models S. Biedermann, School of Mathematics, University of Southampton Title Page joint work with ◭◭ ◮◮ H. Dette and P. Hoffmann, Department of Mathematics, Bochum ◭ ◮ June, 7th, 2007 Page 1 of 37 Go Back Full Screen Close Quit mODa 8, June 4-8, 2007, Almagro, Spain

0. Contents • The Fourier model Title Page • The design problem: constrained optimal discrimination ◭◭ ◮◮ designs ◭ ◮ • Canonical moments Page 2 of 37 Go Back • Results Full Screen Close Quit

1. The Fourier Regression Model n independent observations Y 1 , . . . , Y n at x 1 , . . . , x n where Y i ∼ N ( g 2 d ( x i ) , τ 2 ) and Title Page ◭◭ ◮◮ d d ◭ ◮ � � g 2 d ( x ) = a 0 + a j sin( jx ) + b j cos( jx ) , x ∈ [ − π, π ] j =1 j =1 Page 3 of 37 Go Back Full Screen a 0 , . . . , a d , b 1 , . . . , b d ∈ I R unknown model parameters, d ∈ I N Close Quit

Used to model periodic phenomena, e.g., in the engineering, physical, biological and medical sciences, and in two-dimensional shape analysis, . . . Title Page ◭◭ ◮◮ ◭ ◮ Page 4 of 37 Go Back Full Screen Close Quit

2. The Design problem Do we really need the full model? Title Page Goal: Model identification ◭◭ ◮◮ Successive F -tests with hypotheses ◭ ◮ Page 5 of 37 H (2 d ) : b d = 0 , H (2 d − 1) : a d = 0 , 0 0 Go Back H (2 d − 2) : b d − 1 = 0 , H (2 d − 3) : a d − 1 = 0 , . . . , 0 0 H (0) Full Screen : a 0 = 0 0 Close in the models g 2 d , g 2 d − 1 , . . . , g 0 until H ( k ) is rejected 0 Quit

Is there a way to influence (maximise) the power of these tests? Title Page The noncentrality parameter of the test for H ( k ) is: 0 ◭◭ ◮◮ k M − 1 k ( σ ) e k ) − 1 δ k ( σ ) = ( e T ◭ ◮ k = 1 , . . . , 2 d Page 6 of 37 where e k = (0 , 0 , . . . , 0 , 1) T ∈ I R k +1 and M − 1 k ( σ ) is the covari- Go Back ance matrix for estimating the full parameter vector in model Full Screen g k and σ is the design of the experiment. Close Quit

   x 1 x 2 . . . x m Example: σ =  ω 1 ω 2 . . . ω m   Title Page 1 sin( x i ) cos( x i )   m � ◭◭ ◮◮   M 2 ( σ ) = ω i sin 2 ( x i ) sin( x i ) sin( x i ) cos( x i )     i =1   ◭ ◮ cos 2 ( x i ) cos( x i ) sin( x i ) cos( x i ) Page 7 of 37   Go Back 1 sin( x ) cos( x ) � π     = sin 2 ( x ) dσ ( x ) Full Screen  sin( x ) sin( x ) cos( x )    − π   cos 2 ( x ) cos( x ) sin( x ) cos( x ) Close Quit

Recall: The non-centrality parameter δ k ( σ ) (and therefore the power of the F -test for H ( k ) 0 ) depends on the design σ Title Page For one test only: Maximise δ k ( σ ) with respect to σ ◭◭ ◮◮ → D ( k ) ֒ 1 -optimality ◭ ◮ (Equivalent to optimal design for estimating the highest order Page 8 of 37 parameter in model g k ) Go Back Full Screen Problem: It is impossible to maximise all δ k ( σ )’s simultan- Close eously Quit

The Constrained Optimality Criterion Define the efficiency of a design σ for discriminating between models g k and g k − 1 as Title Page eff k ( σ ) := δ k ( σ ) k ) , k = 1 , . . . , 2 d ◭◭ ◮◮ δ k ( σ ∗ k is the D ( k ) ◭ ◮ where σ ∗ 1 -optimal design. Page 9 of 37 Assume that testing H (2 d ) is most important, and assign lower 0 Go Back boundaries γ k to each efficiency eff k ( σ ) according to the relative Full Screen importance of the corresponding discrimination problem. Close Quit

A constrained optimal discriminating design σ ∗ maximises eff 2 d ( σ ) Title Page subject to ◭◭ ◮◮ eff k ( σ ) ≥ γ k , k = 2 d − 1 , 2 d − 2 , . . . , 2 d − 2 j − 1 ◭ ◮ for some j ∈ { 0 , . . . , d − 1 } . Page 10 of 37 Go Back Full Screen Close Quit

Constrained optimisation problem : → only numerical solutions possible? ֒ ֒ → idea: rewrite criterion in terms of canonical moments to find Title Page analytical results ◭◭ ◮◮ 1. Show that a symmetric design is optimal ◭ ◮ Page 11 of 37 2. Transform σ into a design ξ σ on [ − 1 , 1] by  Go Back 2 σ ( x ) = 2 σ ( − x ) if 0 < x ≤ π  ξ σ (cos x ) =  σ (0) if x = 0 Full Screen 3. Express matrices M k ( σ ) in terms of ξ σ Close Quit

Example:   1 sin( x ) cos( x ) � π     M 2 ( σ ) = sin 2 ( x ) dσ ( x ) sin( x ) sin( x ) cos( x )     − π   Title Page cos 2 ( x ) cos( x ) sin( x ) cos( x ) ◭◭ ◮◮   √ ◭ ◮ 1 − z 2 1 z � 1   √ √   Page 12 of 37 = 1 − z 2 dξ σ ( z ) 1 − z 2 1 − z 2 z     − 1 √   z 2 1 − z 2 z z Go Back M k is now “almost” a moment matrix, and the efficiencies eff k Full Screen of matrices of such a form can be expressed in terms of canonical Close moments. Quit

3. Canonical Moments (I) Canonical Moments p 1 , p 2 , . . . are transformations of the ordin- ary moments c 1 , c 2 , . . . of a probability measure. Title Page Definition: M class of probability measures with moments ◭◭ ◮◮ c 1 , . . . , c k − 1 , ◭ ◮ c + c − k = max µ ∈M c k ( µ ) , k = min µ ∈M c k ( µ ) Page 13 of 37 ξ probability measure on [ − 1 , 1] with moments c 1 , . . . , c k , . . . . Go Back Then the k th canonical moment p k of ξ is defined as Full Screen p k = c k − c − Close k k − c − c + k Quit

p k = c k − c − k k − c − c + k Title Page Properties: ◭◭ ◮◮ • p k ∈ [0 , 1] for k = 1 , 2 , . . . ◭ ◮ • p k gives the relative position of c k in its moment space Page 14 of 37 given c k − 1 , . . . , c 1 Go Back • moment space for p 1 , . . . , p k is [0 , 1] k Full Screen Close Quit

Example: The moment space for the second moment c 2 on [ − 1 , 1]. 2 = 1 , c − c + 2 = c 2 1 Title Page ◭◭ ◮◮ ◭ ◮ Page 15 of 37 Go Back Full Screen Close Quit

Example: The moment space for the second canonical moment p 2 on [ − 1 , 1]. 2 = 1 , p − p + 2 = 0 (do not depend on the value of p 1 ) Title Page ◭◭ ◮◮ ◭ ◮ Page 16 of 37 Go Back Full Screen Close Quit

Title Page ◭◭ ◮◮ Advantage of canonical moments: Maximisation of a function ◭ ◮ with respect to m canonical moments is maximisation on the Page 17 of 37 m -dimensional cube [0 , 1] m . Go Back Full Screen Close Quit

Properties: • If m is the first index for which p m ∈ { 0 , 1 } then the sequence of canonical moments terminates at p m and the Title Page measure is supported at a finite number of points. ◭◭ ◮◮ • If a sequence p 1 , . . . , p 2 d is given, and there is no such ◭ ◮ m then the measure is not unique. One can always find a Page 18 of 37 measure with these canonical moments by adding arbitrary p i ’s. Go Back Full Screen • Given the canonical moments, the measure can be found by evaluating certain orthogonal polynomials (see section Close 5). Quit

4. Results Theorem 1 : If there exists a constrained optimal discrimin- ating design for ( γ 2 d − 2 j − 1 , . . . , γ 2 d − 1 ), the canonical moments Title Page up to the order 2 d of ξ σ ∗ are given by ( q k = 1 − p k ) ◭◭ ◮◮ ◭ ◮ Page 19 of 37 Go Back Full Screen Close Quit

4. Results Theorem 1 : If there exists a constrained optimal discrimin- ating design for ( γ 2 d − 2 j − 1 , . . . , γ 2 d − 1 ), the canonical moments Title Page up to the order 2 d of ξ σ ∗ are given by ( q k = 1 − p k ) ◭◭ ◮◮ p 2 n − 1 = 1 ◭ ◮ 2 , n = 1 , . . . , d, Page 20 of 37 Go Back Full Screen Close Quit

4. Results Theorem 1 : If there exists a constrained optimal discrimin- ating design for ( γ 2 d − 2 j − 1 , . . . , γ 2 d − 1 ), the canonical moments Title Page up to the order 2 d of ξ σ ∗ are given by ( q k = 1 − p k ) ◭◭ ◮◮ p 2 n − 1 = 1 ◭ ◮ 2 , n = 1 , . . . , d, Page 21 of 37 p 2 n = 1 2 , n = 1 , . . . , d − j − 1 Go Back Full Screen Close Quit

 1 − max { 1 γ 2 d − 2 j +2 n − 1  } , 2 ,   2 2 n � d − j + n − 1  p 2 l q 2 l   l = d − j     if γ 2 d − 2 j +2 n − 1 > γ 2 d − 2 j +2 n    Title Page        max { 1 γ 2 d − 2 j +2 n ◭◭ ◮◮ p 2 d − 2 j +2 n = 2 , } , 2 2 n � d − j + n − 1 p 2 l q 2 l   l = d − j  ◭ ◮    if γ 2 d − 2 j +2 n ≥ γ 2 d − 2 j +2 n − 1 ,      Page 22 of 37        Go Back  n = 0 , . . . , j − 1  Full Screen Close Quit

 1 − max { 1 γ 2 d − 2 j +2 n − 1  } , 2 ,   2 2 n � d − j + n − 1  p 2 l q 2 l   l = d − j     if γ 2 d − 2 j +2 n − 1 > γ 2 d − 2 j +2 n    Title Page        max { 1 γ 2 d − 2 j +2 n ◭◭ ◮◮ p 2 d − 2 j +2 n = 2 , } , 2 2 n � d − j + n − 1 p 2 l q 2 l   l = d − j  ◭ ◮    if γ 2 d − 2 j +2 n ≥ γ 2 d − 2 j +2 n − 1 ,      Page 23 of 37        Go Back  n = 0 , . . . , j − 1  Full Screen Close γ 2 d − 1 p 2 d = 1 − 2 2 j � d − 1 l = d − j p 2 l q 2 l Quit