jmlr jmlr2005 jmlr2005-35 knowledge-graph by maker-knowledge-mining

35 jmlr-2005-Frames, Reproducing Kernels, Regularization and Learning


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Author: Alain Rakotomamonjy, Stéphane Canu

Abstract: This work deals with a method for building a reproducing kernel Hilbert space (RKHS) from a Hilbert space with frame elements having special properties. Conditions on existence and a method of construction are given. Then, these RKHS are used within the framework of regularization theory for function approximation. Implications on semiparametric estimation are discussed and a multiscale scheme of regularization is also proposed. Results on toy and real-world approximation problems illustrate the effectiveness of such methods. Keywords: regularization, kernel, frames, wavelets

Reference: text


Summary: the most important sentenses genereted by tfidf model

sentIndex sentText sentNum sentScore

1 FR Perception, Syst` mes et Information CNRS FRE2645 e INSA de Rouen 76801 Saint Etienne du Rouvray, France Editor: Alex Smola Abstract This work deals with a method for building a reproducing kernel Hilbert space (RKHS) from a Hilbert space with frame elements having special properties. [sent-5, score-1.012]

2 Implications on semiparametric estimation are discussed and a multiscale scheme of regularization is also proposed. [sent-8, score-0.555]

3 The purpose of this paper is to present a method for constructing an RKHS and its associated kernel by means of frame theory (Duffin and Schaeffer, 1952; Daubechies, 1992). [sent-31, score-0.854]

4 A frame of a Hilbert space spans any vector of the space by linear combination of the frame elements. [sent-32, score-1.587]

5 But unlike a basis, a frame is not necessarily linear independent although it achieves stable representation. [sent-33, score-0.759]

6 Since a frame is a more general way to represent elements of Hilbert space, it allows flexibility in the representation of any vector of the space. [sent-34, score-0.774]

7 By giving conditions for constructing arbitrary RKHS from frame elements, our goal is to widen the choice of kernel so that in future applications, one can adapt its RKHS to prior information available concerning a problem at hand. [sent-35, score-0.886]

8 After a short introduction about frame theory, we give conditions for a Hilbert space described by a frame to be an RKHS and then derive the corresponding kernel. [sent-38, score-1.554]

9 Section 5 discusses implication of these results on regularization technique and proposes an algorithm for multiscale approximation. [sent-40, score-0.415]

10 Frames and Reproducing Kernel Hilbert Spaces In this section, we give an introduction to frame theory that will be useful for the remainder of the paper. [sent-64, score-0.759]

11 Definition 1 A set of vectors {φn }n∈Γ is a frame of a Hilbert space H if there exists two constants A > 0 and ∞ > B ≥ A > 0 so that A|| f ||2 ≤ H ∀f ∈ H , ∑| n∈Γ f , φn H |2 ≤ B|| f ||2 . [sent-67, score-0.78]

12 H (4) The frame is said to be tight if A and B are equal. [sent-68, score-0.782]

13 If the set {φn }n∈Γ satisfies the frame condition then the frame operator U can be defined as −→ 2 (5) f −→ { f , φn H }n∈Γ . [sent-69, score-1.557]

14 The reconstruction of f from its frame coefficients needs the definition of a dual frame. [sent-70, score-0.805]

15 1487 (6) R AKOTOMAMONJY AND C ANU Theorem 1 (Daubechies, 1992) Let {φn }n∈Γ be a frame of H with frame bounds A and B. [sent-72, score-1.518]

16 Let us ¯ ¯ define the dual frame {φn }n∈Γ as φn = (U U)−1 φn . [sent-73, score-0.805]

17 (7) (8) n∈Γ 1 ¯ If the frame is tight then φn = A φn . [sent-75, score-0.782]

18 ¯ This theorem also shows that the dual frame {φn }n∈Γ is a family of vectors which allows to recover any f ∈ H , and consequently one can write each vector of the frame and the dual frame as ∀m ∈ Γ, ¯ φm = ∀m ∈ Γ, φm = ∑ ¯ ¯ φm , φn H φn (9) ∑ ¯ φm , φn H φn . [sent-76, score-2.369]

19 (10) n∈Γ and n∈Γ According to this theorem and the above equations, one can note that an orthonormal basis ¯ of H is a special case of frame where A = B = 1, φn = φn and φn = 1. [sent-77, score-0.796]

20 However, as stated by Daubechies (1992), frame redundancy can be statistically useful. [sent-78, score-0.759]

21 For the sake of simplicity, in the following we will call frameable Hilbert space, a Hilbert space H for which there exists a set of vector of H that forms a frame of H . [sent-82, score-0.885]

22 2 A Reproducing Kernel Hilbert Space and Its Frame After this short introduction on frame theory, let us look at the conditions under which a frameable Hilbert space is also a reproducing kernel Hilbert space. [sent-85, score-1.075]

23 (12) Theorem 5 Let H be a Hilbert space and {φn }n∈Γ be a frame of this space. [sent-94, score-0.78]

24 Hence the equation ∀f ∈ H , f= ∑ ¯ f , φn H φn n∈Γ holds in H according to the frame property given in Equation (8) (Mallat, 1998; Daubechies, 1992). [sent-98, score-0.759]

25 H ¯ Furthermore, according to Theorem (1) we know that {φn }n∈Γ is another frame of H . [sent-108, score-0.759]

26 Furthermore, any f ∈ H can be expanded by means of the frame of H , thus according to Equation (14): f (t) = ∑ ¯ f , φn H φn (t) n∈Γ = ¯ f (·), ∑ φn (·)φn (t) n∈Γ (20) H and since H is an RKHS, we have ∀t ∈ Ω, ∀f ∈ H , f (t) = f (·), K(·,t) H . [sent-111, score-0.782]

27 n∈Γ These propositions show that a Hilbert space which can be described by its frame is under general conditions, a reproducing kernel Hilbert space and its reproducing kernel is given by a linear combination of its frame and dual frame product. [sent-113, score-2.715]

28 However, these properties can also be easily shown owing to the frame representation. [sent-117, score-0.789]

29 Here, we are interested in constructing a reproducing kernel Hilbert space together with its frame and discuss about the implications of such result in a functional estimation framework. [sent-125, score-1.022]

30 For that it suffices to prove that the frame {φn } satisfies condition given in Theorem 5. [sent-165, score-0.759]

31 ,N is a frame of H and owing to Theorem 1, the dual ¯ ¯ frame {φn }n=1,. [sent-169, score-1.594]

32 5 −12 −2 0 1 2 3 4 5 x 6 7 8 9 10 −14 0 1 2 3 4 5 x 6 7 8 9 10 Figure 1: Examples of wavelet frame elements (left) anf their dual elements (right). [sent-179, score-1.022]

33 Example 3 Consider a finite set of wavelet on R t − ku0 a j 1 ψ j,k (t) = √ ψ aj aj , j ∈ Z : jmin ≤ j ≤ jmax , k ∈ Z : kmin ≤ k ≤ kmax where (a, u0 ) ∈ R∗ × R, and ( jmin , jmax , kmin , kmax ) ∈ Z4 . [sent-185, score-0.419]

34 Figure (1) plots an example of wavelet frame and dual frame elements for a dilation j = −7. [sent-187, score-1.841]

35 ,N , if endowed with an adequate inner product, is always a frame of the space it spans (see step 2 of the proof of Theorem (8)) . [sent-194, score-0.864]

36 This is not always true for an infinite set of functions and in this case, the frame condition given in Equation (4) and the boundedness of the evaluation functional in Equation (13) have to be verified. [sent-195, score-0.792]

37 Next examples are examples of infinite dimension RKHS which kernels are given explicitly by their frame elements. [sent-196, score-0.791]

38 Hence, {φn (x)}n∈N∗ is a tight frame of H with the frame constant A equals to 1. [sent-199, score-1.541]

39 Let us show that this frame verify the condition given in Theorem (5) in order to prove that H is an RKHS. [sent-200, score-0.759]

40 ≤ Hence, according to the adjoint frame operator U ∗ given in equation (6), for any t ∈ Ω, the function ∑n∈N∗ φn (·)φn (t) is a well-defined function of H . [sent-203, score-0.815]

41 Since according to the frame property given in 1 ¯ ¯ Equation (8), each frame element can be expanded as φk = ∑n∈Γ φk , φn φn , we have φk = αk φk . [sent-220, score-1.541]

42 Hence since the frame and dual frame elements are so that for any t ∈ Ω, we have ¯ (φn (t))2 < ∞. [sent-221, score-1.579]

43 Supposing that U is the frame operator of a either finite or infinite dimensional RKHS, their kernel is based on the statement that the operator Q = U ∗U is a symmetric positive definite operator and the Green function associated to this operator is a Mercer kernel. [sent-234, score-1.012]

44 Thus, the kernel they proposed, named the frame operator kernel, can be expanded with respect to the dual frame elements as ¯ ¯ K(s,t) = ∑ φn (s)φn (t). [sent-235, score-1.72]

45 More recently, Opfer (2004b) has shown that the kernel associated to an RKHS H can be expanded as K(s,t) = ∑ φn (s)φn (t) n∈Γ if and only if the set of functions {φn }n∈Γ is a super tight frame (which is a tight frame with frame bounds equal to 1) of H . [sent-243, score-2.425]

46 This results is a particular case of Theorem (7) since for a super tight ¯ frame each dual frame element is φn = φn . [sent-244, score-1.587]

47 Notice that methods for choosing the appropriate frame elements of the RKHS are not given here. [sent-263, score-0.774]

48 The space spanned by these frame elements associated to L2 (R) inner product is an RKHS. [sent-268, score-0.842]

49 • Since conditions for obtaining a frameable RKHS hold mainly for finite dimensional space (although, it may exists infinite dimensional Hilbert space which frame elements satisfy hypotheses of Theorem (5)), it is fairest to compare the frameable kernel to a finite dimensional kernel. [sent-272, score-1.174]

50 Conditions for constructing frameable kernel are less restricting since the orthogonality of the frame elements are not needed. [sent-276, score-0.974]

51 One can note that for tight frame or orthonormal basis, frameable kernel leads to the following expansion: N K(s,t) = 1 ∑ A ψn (s)ψn (t) n=1 since dual frame elements is equal to frame elements up to a multiplicative constant depending on the frame bound A . [sent-277, score-3.336]

52 Tightness of a frame is a very interesting property since in this case processing the dual frame is no more needed. [sent-278, score-1.564]

53 However, unless we explicitly build the RKHS H so that it is spanned by a tight frame (as in example (5) or in Opfer (2004b)), tightness of a frame needs more constraints on the frame elements than other frames. [sent-279, score-2.357]

54 Thus a tight frame of a space is harder to build than other frame of the same space. [sent-280, score-1.579]

55 n Again in this case, the frame kernel expansion is similar to the Mercer’s kernel one. [sent-282, score-0.917]

56 The main difference between the finite and infinite dimensional case relies on the fact that a finite set of functions {φn } is always a frame of the space it spans (provided that this latter is endowed with an adequate inner product). [sent-283, score-0.882]

57 The mapping is consequently strictly related to the frame elements {φn } and is implicitly defined by them. [sent-291, score-0.774]

58 • Besides, since the kernel has an expansion with regards to the frame elements, the solution of Equation (1) is of easier interpretation. [sent-292, score-0.869]

59 Indeed, although the solution depends on the kernel expression, it can be rewritten as a linear combination of the frame elements. [sent-293, score-0.838]

60 Thus, with frame-based kernel, one has to compute the dual frame elements, (for instance, by means of an iterative algorithm, as the one described in (Grochenig, 1993)). [sent-297, score-0.805]

61 Hence, if the number N of frame elements describing the kernel and the number of data are large, building K becomes rapidly very time-consuming (of an order of N 2 · 2 ). [sent-300, score-0.874]

62 Intuitively, one should use any vector that comes from “good” prior knowledge, in the parametric part of the approximation while leaving in the kernel expansion the other frame elements. [sent-322, score-0.88]

63 Another perspective which follows directly from this finding is a technique of regularization that we call multiscale regularization which is inspired from the multiresolution analysis of Mallat (1998). [sent-333, score-0.523]

64 According to the representer Thej=1 orem (9), fm−k (·) can be written: n fm−k (·) = ∑ ci,m−k Km−k (xi , ·) + i=1 ∑ d j,m−k g j (·) (28) j∈∪m−k Γl l=0 and thus the overall solution of the so-called multiscale regularization is m fˆ(·) = n ∑ ∑ ci,m−k Km−k (xi , ·) + ∑ d j,0 g j (·). [sent-369, score-0.415]

65 k=1 i=1 (29) j∈Γ0 The solution fˆ of the multiscale regularization is the sum of different approximations on nested spaces. [sent-370, score-0.43]

66 Illustrations of the multiscale regularization algorithm on both toy and real-world problems are given in the next section. [sent-387, score-0.415]

67 These values are those proposed by Daubechies (Daubechies, 1992) so that a wavelet set is a frame of L2 (R). [sent-415, score-0.946]

68 For frame-based kernel, if necessary the dual frame is processed using Grochenig’s algorithm. [sent-421, score-0.805]

69 Using prior knowledge on the problem in this context does not improve performance (Sin/Sinc kernel or wavelet kernel compared to gaussian kernel). [sent-428, score-0.397]

70 A justification can be that such kernels use strong prior knowledge (the sin frame element) that is included in the kernel expansion and thus this prior knowledge gets regularized as much as other frame elements. [sent-429, score-1.761]

71 • Gaussian kernel and wavelet basis functions ψ j,k (x) = 1505 1 √ ψ aj x−ku0 a j aj , j ∈ {0, 5} R AKOTOMAMONJY AND C ANU • Wavelet kernel and wavelet basis functions: these functions are the same as in the previous case but the kernel is built only with low dilation wavelet ( j = −10). [sent-435, score-0.997]

72 However, this is still of interest as using wavelet kernel and basis functions does corresponds to prior knowledge that can be reformulated as: “the function to be approximated contains smooth structure (the sin part), irregular structures (the sinc part) and noise”. [sent-453, score-0.55]

73 Thus, prior knowledge on structures which may be easiest to get than prior knowledge on basis function can be easily exploited by means of wavelet span and wavelet kernel. [sent-455, score-0.512]

74 The learning machines are: regularization networks, SVM, semiparametric regularization and multiscale regularization. [sent-459, score-0.645]

75 For semiparametric regularization, the kernel and basis setting was built with a wavelet set given by x − ku0 a j 1 . [sent-486, score-0.408]

76 0069 Table 3: True mean-square-error generalization for regularization networks, SVM, semiparametric regularization networks, and multiscale regularization for f1 and f2 . [sent-511, score-0.753]

77 The kernel is constructed from a set of wavelet frame of dilation jSPH and the basis functions are given by another wavelet set described by jSPL . [sent-512, score-1.307]

78 For multiscale regularization, the setting of the nested spaces are the following: t − ku0 a j ,j=5 , aj a 1 t − ku0 a j F0 = span √ j ψ ,j=0 , aj a 1 t − ku0 a j F1 = span √ j ψ , j = −10 . [sent-513, score-0.527]

79 Comments and analysis of this experiment validating the concept of multiscale approximation are: - semiparametric 2 and multiscale approximation give the best mean-square error. [sent-518, score-0.736]

80 - multiscale approximation balances loss of approximation due to error at each level (see Figure) and flexibility of regularization, thus its performance is better than semiparametric one’s when the multiscale structure of the signal is more pronounced. [sent-525, score-0.736]

81 In this experiment, it seems that leaving the space spanned by wavelet of dilation j = 0 on the parametric span (the space which is not regularized) leads to overfitting. [sent-536, score-0.406]

82 3) shows that multiscale and semiparametric algorithms achieve better approximation of the “wiggles” than nonparametric methods without compromising smoothness in region of the functions where it is needed. [sent-540, score-0.429]

83 For this purpose, two models have been compared, the first one being a regularization networks with a gaussian kernel whereas the other one is a multiscale regularization algorithm with an orthogonal wavelet kernel. [sent-546, score-0.826]

84 The wavelet that has been used is a Symmlet wavelet with 4 vanishing moments (Mallat, 1998). [sent-547, score-0.374]

85 30 (22) Table 4: Averaged normalized mean-square error of estimation of real-world time-series with a gaussian and a wavelet multiscale regularization networks. [sent-563, score-0.639]

86 It shows that for both time-series, the multiscale algorithm performs better than the gaussian regularization networks. [sent-566, score-0.434]

87 This figure shows that the best model for the gaussian regularization networks is rather a smooth model whereas the wavelet multiscale model is far less smooth. [sent-571, score-0.639]

88 Conclusions In this paper, we showed that an RKHS can be defined by its frame elements and conversely, one can construct an RKHS from a frame. [sent-576, score-0.774]

89 These conditions depend on the frame and the dual frame elements of the Hilbert space and under some weak hypothesis, these conditions are easy to check (see example 5) . [sent-579, score-1.63]

90 But the difficulty stands in one question: How to transform prior information on the learning problem to frame elements? [sent-623, score-0.776]

91 • reconstruction from frame elements has been shown to be more robust in presence of noise (Daubechies, 1992). [sent-625, score-0.774]

92 In fact, redundancy attenuates noise effects on the frame coefficients. [sent-626, score-0.759]

93 Thus, this is a good statistical argument for using frame with high redundancy. [sent-627, score-0.759]

94 However, this implies the computing of the dual frame and consequently a higher time complexity of the algorithm. [sent-628, score-0.805]

95 • a multiscale regularization algorithm has been sketched in this paper in order to take advantage of frame kernels. [sent-630, score-1.174]

96 We recall in this appendix a numerical method to process the dual frame of a frameable Hilbert space H with frame elements {φn }n∈Γ . [sent-637, score-1.705]

97 (35) One can also write the operator S as S U ∗U where U is the frame operator defined in equation (5) and (6). [sent-639, score-0.837]

98 Then, in order to process numerically the dual frame of H , one has to apply this algorithm on each element of the frame. [sent-648, score-0.805]

99 One can note that the speed of convergence is highly dependent on frame bounds A and B. [sent-649, score-0.759]

100 Tight frame expansions of multiscale reproducing kernels in Sobolev spaces. [sent-803, score-1.194]


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