Bootstrap independence test for functional linear models

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2.2 Linear independence testGiven a generic H–valued random element H such that E(‖H‖ 2 ) < ∞, its associated covarianceoperator Γ H is defined as the operator Γ H : H → HΓ H (h) = E (〈H − µ H , h〉(H − µ H )) = E (〈H, h〉H) − 〈µ H , h〉µ H ,for all h ∈ H, where µ H ∈ H denotes the expected value of H. From now on, it will be assumedthat E(‖X‖ 2 ) < ∞, and thus, as a consequence of Hölder’s inequality, E(Y 2 ) < ∞. Wheneverthere is no possible confusion, Γ X will be abbreviated as Γ. It is well–known that Γ is a nuclearand self–adjoint operator. In particular, it is a compact operator of trace class and thus, in virtueof the Spectral Theorem Decomposition, there is an orthonormal basis of H, {v j } j∈N , consisting oneigenvectors of Γ with corresponding eigenvalues {λ j } j∈N , that is, Γ(v j ) = λ j v j for all j ∈ N. Asusual, the eigenvalues are assumed to be arranged in decreasing order (λ 1 ≥ λ 2 ≥ . . .). Since theoperator Γ is symmetric and non–negative definite, then the eigenvalues are non–negative.In a similar way, let us consider the cross–covariance operator ∆ : H → R between X and Y givenby∆(h) = E (〈X − µ X , h〉(Y − µ Y )) = E (〈X, h〉Y ) − 〈µ X , h〉µ Y ,for all h ∈ H, where µ Y ∈ R denotes the expected value of Y . Of course, ∆ ∈ H ′ and the followingrelation between the considered operators and the regression parameter Θ is satisfied∆(·) = 〈Γ(·), Θ〉. (4)The Hilbert space H can be expressed as the direct sum of the two orthogonal subspaces inducedby the self–adjoint operator Γ: the kernel or null space of Γ, N (Γ), and the closure of the image orrange of Γ, R(Γ). Thus, Θ is determined uniquely by Θ = Θ 1 +Θ 2 with Θ 1 ∈ N (Γ) and Θ 2 ∈ R(Γ).As Θ 1 ∈ N (Γ), it is easy to check that V ar(〈X, Θ 1 〉) = 0 and, consequently, the model introducedin (1) can be expressed asY = 〈Θ 2 , X〉 + 〈Θ 1 , µ X 〉 + b + ε.Therefore, it is not possible to distinguish between the term 〈Θ 1 , µ X 〉 and the intercept term b,and consequently it is not possible to check whether Θ 1 = 0 or not. Taking this into account, thehypothesis test will be restricted to check{H0 : Θ 2 = 0(5)H 1 : Θ 2 ≠ 0on the basis of the available sample information.Note that in this case, according to the relation between the operators and the regression parametershown in (4), Θ 2 = 0 if, and only if, ∆(h) = 0 for all h ∈ H. Consequently, the hypothesis test in(5) is equivalent to{H0 : ‖∆‖ ′ = 0H 1 : ‖∆‖ ′ (6)≠ 0Remark 1. It should be recalled that, in previous works µ X is assumed to be equal 0. Thus,the preceding reasoning leads to the fact that Θ 1 cannot be estimated based on the informationprovided by X (see, for instance, Cardot, Ferraty, Mas, and Sarda (2003)). Consequently thehypothesis testing is also restricted to the one in the preceding equations. In addition in Cardot,Ferraty, Mas, and Sarda (2003), it is also assumed for technical reasons that R(Γ) is an infinite–dimensional space. On the contrary, this restriction is not imposed in the study here developed.4
Remark 2. Note that another usual assumption is that the intercept term vanishes. Although thisis not common in most of situations, it should be noted that if b = 0 and X is not assumed to becentered (as in this work), then an interesting possibility appears: to check whether Θ 1 = 0 or notby checking the nullity of the intercept term of the model, and thus to check the original hypothesistesting in (2). This open problem cannot be solved with the methodology employed in the currentpaper (or in the previous ones) because the idea is based on checking (6), which is equivalent to therestricted test (5) but not to the unrestricted one in (2).2.3 Testing procedure and asymptotic theoryAccording to the relation between ‖ · ‖ ′ and ‖ · ‖, the dual norm of ∆ ∈ H ′ can be expressedequivalently in terms of the H–valued random element (X − µ X )(Y − µ Y ) as follows‖∆‖ ′ = ‖〈E ((X − µ X )(Y − µ Y )) , ·〉‖ ′ = ‖E ((X − µ X )(Y − µ Y )) ‖.Thus, based on an i.i.d. sample {(X i , Y i )} n i=1drawn from (X, Y ),D = ‖E ((X − µ X )(Y − µ Y )) ‖ = ‖T ‖can be estimated in a natural way by means of its empirical counterpart D n = ‖T n ‖, where T n isthe H–valued random element given byT n = 1 nn∑(X i − X)(Y i − Y ),i=1where X and Y denote as usual the corresponding sample means. The next theorem establishessome basic properties of T n .Theorem 1. Assuming that (1) holds with E(ε) = 0, E(ε 2 ) = σ 2 < ∞ and E(‖X‖ 4 ) < ∞, then1. E(T n ) = E ((X − µ X )(Y − µ Y )) (n − 1)/n2. T n converges a.s.–P to E ((X − µ X )(Y − µ Y )) as n → ∞3. √ n (T n − E ((X − µ X )(Y − µ Y ))) converges in law, as n → ∞, to a centered Gaussian elementZ in H with covariance operatorΓ Z (·) = σ 2 Γ(·) + E ( (X − µ X )〈X − µ X , ·〉〈X − µ X , Θ〉 2) .Proof. Since T n can be equivalently expressed asT n = 1 nn∑(X i − µ X )(Y i − µ Y ) + (X − µ X )(Y − µ Y ),i=1it is straightforward to check item 1. The a.s.–P convergence is a direct application of the SLLNfor separable Hilbert–valued random elements.On the other hand, given that E(‖(X − µ X )(Y − µ Y )‖ 2 ) < ∞, the convergence in law can bededuced by applying the CLT for separable Hilbert–valued random elements (see, for instance, Lahaand Rohatgi (1979)) together with Slutsky’s Theorem. The concrete expression of the operatorΓ Z , that is, Γ Z = Γ (X−µX )(Y −µ Y ) = Γ (X−µX )ε + Γ (X−µX )〈X−µ X ,Θ〉, can be obtained by simplecomputations.In order to simplify the notation, from now on, given any H–valued random element H withE(‖H‖ 2 ) < ∞, Z H will denote a centered Gaussian element in H with covariance operator Γ H .5
Page 1 and 2: Bootstrap independence test for fun
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Bootstrap independence test for functional linear models

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