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Reproducing Kernel Tutorial

Reproducing Kernel Tutorial

Fred J. Hickernell

November 15, 2021
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  1. Reproducing Kernels
    Fred J. Hickernell
    Department of Applied Mathematics Center for Interdisciplinary Scientific Computation
    Office of Research
    Illinois Institute of Technology [email protected] mypages.iit.edu/~hickernell
    Thanks to Mac Hyman for the invitation
    Thanks to many students and collaborators
    Slides available at speakerdeck.com/fjhickernell/reproducing-kernel-tutorial
    Please interrupt and ask questions
    November 15, 2021

    View Slide

  2. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    What Can We Do with Reproducing Kernel Hilbert Spaces?
    Use the Reisz Representation Theorem to derive error bounds for algorithms for linear problems,
    such as integration, function approximation, solving linear differential equations
    Derive optimal algorithms
    Determine how fast the error bounds decay to zero as the computational effort increases, and even
    whether convergence depends significantly on the number of variables
    Include trends, BUT I have not prepared that for today.
    Derive a parallel analysis using Gaussian processes where the reproducing kernel is interpreted as a
    covariance kernel
    2/20

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  3. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    What Can We Do with Reproducing Kernel Hilbert Spaces?
    Use the Reisz Representation Theorem to derive error bounds for algorithms for linear problems,
    such as integration, function approximation, solving linear differential equations, BUT
    You must be able to solve the problem for your reproducing kernel
    You must pick a kernel that matches your input function
    You may need to tune the kernel parameters
    Derive optimal algorithms, BUT it takes O(n3) operations to compute the weights
    Determine how fast the error bounds decay to zero as the computational effort increases, and even
    whether convergence depends significantly on the number of variables
    Include trends, BUT I have not prepared that for today.
    Derive a parallel analysis using Gaussian processes where the reproducing kernel is interpreted as a
    covariance kernel , BUT I have not prepared that for today
    2/20

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  4. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Reproducing Kernels for Functions on {1, . . . , d}, aka Vectors
    Let F := all functions on {1, . . . , d} “=” Rd
    Pick a symmetric, positive definite (positive eigenvalues) matrix W ∈ Rd×d to define an inner product
    ⟨f, h⟩ := fTWh, ∀f, h ∈ F, where f = f(t) d
    t=1
    3/20

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  5. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Reproducing Kernels for Functions on {1, . . . , d}, aka Vectors
    Let F := all functions on {1, . . . , d} “=” Rd
    Pick a symmetric, positive definite (positive eigenvalues) matrix W ∈ Rd×d to define an inner product
    ⟨f, h⟩ := fTWh, ∀f, h ∈ F, where f = f(t) d
    t=1
    Reproducing kernel, K, is defined by K(t, x) d
    t,x=1
    = K := W−1, and has the properties
    Symmetry K(t, x) = K(x, t) because W is symmetric and thus so is K
    Positive Definiteness K(xi
    , xj
    ) n
    i,j=1
    is positive definite for any distinct x1
    , . . . , xn ∈ {1, . . . , d}
    Belonging K(·, x) = xth column of K =: Kx ∈ F
    Reproduction ⟨K(·, x), f⟩ = KT
    x
    Wf = ex
    f = f(x) since K := W−1; ex
    := (0, . . . , 0, 1
    xth position
    , 0, . . .)T
    3/20

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  6. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Reproducing Kernels for Functions on {1, . . . , d}, aka Vectors
    Let F := all functions on {1, . . . , d} “=” Rd
    Pick a symmetric, positive definite (positive eigenvalues) matrix W ∈ Rd×d to define an inner product
    ⟨f, h⟩ := fTWh, ∀f, h ∈ F, where f = f(t) d
    t=1
    Reproducing kernel, K, is defined by K(t, x) d
    t,x=1
    = K := W−1, and has the properties
    Symmetry K(t, x) = K(x, t) because W is symmetric and thus so is K
    Positive Definiteness K(xi
    , xj
    ) n
    i,j=1
    is positive definite for any distinct x1
    , . . . , xn ∈ {1, . . . , d}
    Belonging K(·, x) = xth column of K =: Kx ∈ F
    Reproduction ⟨K(·, x), f⟩ = KT
    x
    Wf = ex
    f = f(x) since K := W−1; ex
    := (0, . . . , 0, 1
    xth position
    , 0, . . .)T
    Riesz Representation Theorem says that for any linear function, LINEAR, there is a representer g such
    that LINEAR(f) = ⟨g, f⟩ = gTWf. Note



    g(1)
    .
    .
    .
    g(d)



    = g = KWg =



    KT
    1
    Wg
    .
    .
    .
    KT
    d
    Wg



    =



    ⟨K(·, 1), g⟩
    .
    .
    .
    ⟨K(·, d), g⟩



    =



    LINEAR(K(·, 1))
    .
    .
    .
    LINEAR(K(·, d))



    3/20

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  7. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Reproducing Kernels for Functions on {1, . . . , d}, aka Vectors
    Let F be a vector space of functions
    define an inner product
    W is gone
    Reproducing kernel, K, has the properties
    Symmetry K(t, x) = K(x, t)
    Positive Definiteness K(xi
    , xj
    ) n
    i,j=1
    is positive definite for any distinct x1
    , . . . , xn ∈ {1, . . . , d}
    Belonging K(·, x) ∈ F
    Reproduction ⟨K(·, x), f⟩ = f(x)
    Riesz Representation Theorem says that
    LINEAR(f) = ⟨g, f⟩



    g(1)
    .
    .
    .
    g(d)



    =



    LINEAR(K(·, 1))
    .
    .
    .
    LINEAR(K(·, d))



    3/20

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  8. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Reproducing Kernels for Functions on General Domains [1]
    Suppose that (F, ⟨·, ·⟩) is a Hilbert space of functions on Ω for which function evaluation is bounded.
    Then there exists a unique reproducing kernel K : Ω × Ω → R for which
    K(t, x) = K(x, t)
    symmetry
    , K(·, x) ∈ F
    belonging
    , f(x) = ⟨K(·, x), f⟩
    reproduction
    ∀t, x ∈ Ω, f ∈ F
    K(X, X) = K(xi
    , xj
    ) n
    i,j=1
    is positive definite for any n × d X with distinct rows lying in Ω
    F is the completion of {c1
    K(·, x1
    ) + · · · + cn
    K(·, xn
    ) : n ∈ N, c ∈ Rn}; any K satisfying the above implies F
    4/20

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  9. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Reproducing Kernels for Functions on General Domains [1]
    Suppose that (F, ⟨·, ·⟩) is a Hilbert space of functions on Ω for which function evaluation is bounded.
    Then there exists a unique reproducing kernel K : Ω × Ω → R for which
    K(t, x) = K(x, t)
    symmetry
    , K(·, x) ∈ F
    belonging
    , f(x) = ⟨K(·, x), f⟩
    reproduction
    ∀t, x ∈ Ω, f ∈ F
    K(X, X) = K(xi
    , xj
    ) n
    i,j=1
    is positive definite for any n × d X with distinct rows lying in Ω
    F is the completion of {c1
    K(·, x1
    ) + · · · + cn
    K(·, xn
    ) : n ∈ N, c ∈ Rn}; any K satisfying the above implies F
    Riesz Representation Theorem says that for any bounded LINEAR : F → R there exists a representer
    g ∈ F such that LINEAR(f) = ⟨g, f⟩ for all f ∈ F. What is g?
    4/20

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  10. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Reproducing Kernels for Functions on General Domains [1]
    Suppose that (F, ⟨·, ·⟩) is a Hilbert space of functions on Ω for which function evaluation is bounded.
    Then there exists a unique reproducing kernel K : Ω × Ω → R for which
    K(t, x) = K(x, t)
    symmetry
    , K(·, x) ∈ F
    belonging
    , f(x) = ⟨K(·, x), f⟩
    reproduction
    ∀t, x ∈ Ω, f ∈ F
    K(X, X) = K(xi
    , xj
    ) n
    i,j=1
    is positive definite for any n × d X with distinct rows lying in Ω
    F is the completion of {c1
    K(·, x1
    ) + · · · + cn
    K(·, xn
    ) : n ∈ N, c ∈ Rn}; any K satisfying the above implies F
    Riesz Representation Theorem says that for any bounded LINEAR : F → R there exists a representer
    g ∈ F such that LINEAR(f) = ⟨g, f⟩ for all f ∈ F. What is g?
    g(x) =
    reproduction
    ⟨K(·, x), g⟩ =
    symmetry
    ⟨g, K(·, x)⟩ =
    representer
    LINEAR K(·, x) ∀x ∈ Ω
    4/20

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  11. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Reproducing Kernels for Functions on General Domains [1]
    Suppose that (F, ⟨·, ·⟩) is a Hilbert space of functions on Ω for which function evaluation is bounded.
    Then there exists a unique reproducing kernel K : Ω × Ω → R for which
    K(t, x) = K(x, t)
    symmetry
    , K(·, x) ∈ F
    belonging
    , f(x) = ⟨K(·, x), f⟩
    reproduction
    ∀t, x ∈ Ω, f ∈ F
    K(X, X) = K(xi
    , xj
    ) n
    i,j=1
    is positive definite for any n × d X with distinct rows lying in Ω
    F is the completion of {c1
    K(·, x1
    ) + · · · + cn
    K(·, xn
    ) : n ∈ N, c ∈ Rn}; any K satisfying the above implies F
    Riesz Representation Theorem says that for any bounded LINEAR : F → R there exists a representer
    g ∈ F such that LINEAR(f) = ⟨g, f⟩ for all f ∈ F. What is g?
    g(x) =
    reproduction
    ⟨K(·, x), g⟩ =
    symmetry
    ⟨g, K(·, x)⟩ =
    representer
    LINEAR K(·, x) ∀x ∈ Ω
    ∥g∥2 = ⟨g, g⟩ =
    representer
    LINEAR(g) = LINEAR·· LINEAR· K(·, ··)
    Do not need the definition of ⟨·, ·⟩ to compute g and ∥g∥
    4/20

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  12. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Squared Exponential Kernel on R
    The squared exponential (aka Gaussian) kernel for
    univariate functions takes the form
    K(t, x) = A exp −γ2 |t − x|2 , t, x ∈ R
    corresponds to the Hilbert space of functions with
    norm [2, (6.18)]
    ∥f∥2 = A

    π

    m=0
    R
    f(m)(x) 2
    dx
    m!4mγ2m+1
    which means that functions have all deriviatives
    square integrable.
    5/20

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  13. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Squared Exponential Kernel on Rd
    The squared exponential kernel for d-variate functions takes the
    form
    K(t, x) = A exp −γ2
    1
    |t1
    − x1
    |2 − · · · − γ2
    d
    |td
    − xd
    |2 ,
    t, x ∈ Rd
    corresponds to the Hilbert space of functions with norm
    ∥Dmf∥2
    2
    :=
    Rd
    ∂∥m∥1 f(x)
    ∂xm1
    1
    · · · ∂xmd
    d
    2
    dx
    ∥f∥2 = A

    π
    m∈Nd
    0
    ∥Dmf∥2
    2
    ∥m∥1
    ! 4∥m∥1
    d
    k=1
    γ2mj
    k
    which means that functions have all deriviatives square inte-
    grable. This kernel is stationary. It is isotropic if γ1
    = · · · = γd
    .
    6/20

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  14. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Matérn Kernels
    A popular family of kernels with a range of smooth-
    ness depending on r with an associate norm that is
    not simple to write down:
    Kr
    (t, x) = A ∥t − x∥r
    2
    Mod Bessel Secr
    (γ ∥t − x∥2
    )
    K1/2
    (t, x) = A1/2
    exp(−γ ∥t − x∥2
    ) not very smooth
    K3/2
    (t, x) = A3/2
    (1 + γ ∥t − x∥2
    ) exp(−γ ∥t − x∥2
    ) somewhat smoother
    7/20

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  15. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    The Centered Discrepancy Kernel [3]
    A reproducing kernel used to analyze cubatures gives the
    weighted centered discrepancy takes the form
    K(t, x) :=
    d
    k=1
    1 +
    γk
    2
    |tk
    − 1/2| + |xk
    − 1/2| − |tk
    − xk
    | .
    t, x ∈ [0, 1]d
    which corresponds to the Hilbert space for functions defined on
    [0, 1]d with the following norm:
    ∥Dmf∥2
    2
    :=
    Rd
    ∂∥m∥1 f(x)
    ∂xm1
    1
    · · · ∂xmd
    d
    2
    xj=1/2 for mj=0 j s.t. mj>0
    dxj
    ∥f∥2 := A
    ∥m∥∞⩽1
    ∥Dmf∥2
    2
    γk
    Mixed partial derivatives of up to order one in each coordinate
    must be square integrable.
    8/20

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  16. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    The Delta Kernel
    A reproducing kernel with an uncountable basis is
    K(t, x) :=
    1 + γ, t = x,
    1, otherwise,
    t, x ∈ [0, 1]d
    which corresponds to the Hilbert space for functions
    that are a constant everywhere except possibly at a
    countable number of points.
    I(f) =
    [0,1]d
    f(x) dx
    ∥f∥2 := |I(f)|2 +
    x∈[0,1]d
    |f(x) − I(f)|2
    γ
    This Hilbert space has an uncountable basis.
    9/20

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  17. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Hilbert Spaces of Signed Measures [4]
    Let M be the Hilbert spaced of measures on Ω that is the completion of
    {c1
    δx1
    + · · · + cn
    δxn
    : n ∈ N, c ∈ Rn} under the norm induced by
    ⟨µ, ν⟩M
    :=
    Ω×Ω
    K(t, x) (µ × ν)(dt × dx)
    and δx
    is the Dirac measure, i.e.,

    f(t) δx(dt) = f(x). There exists a one-to-one and onto, isometric (I
    think) mapping T : M → F defined as
    T(µ)(x) :=

    K(t, x) µ(dt) ∀x ∈ Ω, µ ∈ M
    such that ⟨T(ν), f⟩ =

    f(x) ν(dx).
    If T(νx) is the representer for the solution of a differential equation at x, is νx
    the Green’s function?
    10/20

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  18. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Can We Findlike the W for functions on {1, . . . , d}
    Let M be the Hilbert spaced of measures on Ω that is the completion of
    {c1
    δx1
    + · · · + cn
    δxn
    : n ∈ N, c ∈ Rn} under the norm induced by
    ⟨µ, ν⟩M
    :=
    Ω×Ω
    K(t, x) (µ × ν)(dt × dx)
    and δx
    is the Dirac measure, i.e.,

    f(t) δx(dt) = f(x).
    Is there a measure ω on Ω × Ω such that
    Ω×Ω
    K(t, s)K(u, x) ω(ds × du) = K(t, x) ∀t, x ∈ Ω?
    This would be like the W for functions on {1, . . . , d}
    11/20

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  19. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Separable Hilbert Spaces, i.e., Those with Countable Bases
    Hilbert spaces of functions on Ω with countable bases can be written in terms of an L2(Ω) basis
    f(x) =
    k
    f(k)φk(x),

    φk(x)φl(x) dx = δk,l
    and the reproducing kernel is
    K(t, x) =
    k
    λkφk(t)φk(x), note that

    K(x, x) dx =
    k
    λk < ∞, so λk → 0
    ⟨f, g⟩ :=
    k
    f(k)^
    g(k)
    λk
    , since this implies ⟨K(·, x), f⟩ :=
    k
    λkφk(x)f(k)
    λk
    =
    k
    φk(x)f(k) = f(x)
    If we formally define the distribution m(x) =
    k
    f(k)φk(x)
    λk
    , then

    K(t, x)m(t) dt =
    Ω k,l
    λkφk(t)φk(x)
    f(l)φl(t)
    λl
    dt =
    k
    φk(x)f(k) = f(x)
    This m(x)dx seems to be the µ(dx) that gets mapped into f
    It would seem that W(t, x) =
    k
    φk(t)φk(x)
    λk
    , which is not a convergent series
    12/20

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  20. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Error for Approximating Linear Functionals
    Suppose that
    Linear SOL : F → R is the desired solution (integral, derivative at a point, etc.)
    X = (x1
    , . . . , xn
    )T is the array of data sites
    APPX,α(f) = α1
    f(x1
    ) + · · · + αn
    f(xn
    ) = αTf(X) is the approximation
    Then the approximation error has a tight upper bound of
    SOL(f) − APPX,α(f)
    linear, bounded
    = ⟨g, f⟩ ⩽ ∥g∥
    badness of APPX
    ∥f∥
    badness of f
    where g(x) = SOL − APPX,α
    K(·, x)
    ∥g∥2 = SOL − APPX,α
    ··
    SOL − APPX,α
    ·
    K(·, ··)
    = SOL·· SOL· K(·, ··) − 2αT SOL K(X, ·) + αTK(X, X)α
    APPX
    badness does not require ⟨·, ·⟩, but only K K(X, ·) = K(xi
    , ·) n
    i=1
    , K(X, X) = K(xi
    , xj
    ) n
    i,j=1
    Optimal weights are α = K(X, X)−1 SOL K(X, ·) ; optimal data sites, X, are hard nonlinear optimization
    13/20

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  21. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    An Example of the Cubature Error Bound [3]
    For the problem and non-optimal approximation
    SOL(f) =
    [0,1]d
    f(x) dx, APPX
    =
    1
    n
    n
    i=1
    f(xi
    ),
    and the reproducing kernel
    K(t, x) :=
    d
    k=1
    1 +
    γk
    2
    |tk
    − 1/2| + |xk
    − 1/2| − |tk
    − xk
    | t, x ∈ [0, 1]d
    the error bound is |SOL(f) − APPX
    | ⩽ BAD(x1
    , . . . , xn
    ) BAD(f), where
    BAD2(x1
    , . . . , xn
    ) = ∥representer of the error∥2
    =
    13
    12
    d

    2
    n
    n
    i=1
    d
    k=1
    1 +
    γk
    2
    |xik
    − 1/2| − |xik
    − 1/2|2
    +
    1
    n2
    n
    i,j=1
    d
    k=1
    1 +
    1
    2
    |xik
    − 1/2| + xjk
    − 1/2 − xik
    − xjk
    Requires O(dn2) operations to compute
    BAD(f) = ∥f − f(1/2, . . . , 1/2)∥, which is impractical to compute
    14/20

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  22. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Optimal Function Approximation
    Consider function evaluation at x, i.e., SOLx(f) = f(x). In this case,
    SOL··
    x
    SOL·
    x
    K(·, ··) = K(x, x), SOLx
    K(X, ·) = K(X, x)
    and the optimal algorithm is
    APPx,X,opt
    (f) = K(x, X)K(X, X)−1f(X)
    f(x) − K(x, X)K(X, X)−1f(X) 2
    ⩽ K(x, x) − K(x, X)K(X, X)−1K(X, x) ∥f∥2
    ⩽ K(x, x) − K(x, X)K(X, X)−1K(X, x)
    only depends on X
    f − K(·, X)K(X, X)−1f(X)
    best approximation to f
    2
    APP·,X,opt
    (f) is in the Hilbert space, and even in the span of K(·, x1
    ), . . . , K(·, xn
    ), so
    K(·, X)K(X, X)−1f(X)
    best approximation
    ⊥ f − K(·, X)K(X, X)−1f(X)
    error of approximation
    The optimal linear approximation for an arbitrary linear functional is just the linear functional applied to the
    optimal function approximation
    K(X, X) can be ill-conditioned for smooth kernels and lots of data
    15/20

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  23. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Cardinal Functions
    To visualize the effect of an additional data point on the function approximation, we plot
    APP·,X,opt
    (f) = K(·, X)K(X, X)−1f(X) for f(X) = ei
    (all data are zero but one)
    The cardinal functions of the smoother kernel is more oscillatory
    16/20

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  24. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Why Is the Optimal Approximation Linear?
    Fix x ∈ Ω. Let
    BX,f(X),R
    = {g ∈ F : ∥g∥2
    ⩽ R2 + ∥APP·,X,opt
    (f)∥2 , g(X) = f(X)} functions that look like f
    BX,⊥,R
    = {h ∈ F : ∥h∥ ⩽ R, h(X) = 0} functions that vanish at the data sites
    = {h ∈ F : ∥h∥ ⩽ R, ⟨h, K(·, x1
    )⟩ = · · · = ⟨h, K(·, xn
    )⟩ = 0}
    Any g ∈ BX,f(X),R
    may be written as g = APP·,X,opt
    (f) + g⊥
    with g⊥
    ∈ BX,⊥,R
    yopt
    := argmin
    y∈R
    ERR(y)
    ERR(y) := max
    g∈BX,f(X),R
    |g(x) − y| = max
    g⊥∈BX,⊥,R
    |APPx,X,opt
    (f) + g⊥
    (x) − y|
    Since for every g⊥
    ∈ BX,⊥,R
    it also is true that −g⊥
    ∈ BX,⊥,R
    , the optimal choice of y is APPx,X,opt
    (f).
    17/20

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  25. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    Tuning the Kernel Parameters
    Virtually all reproducing kernels have parameters, θ, that govern smoothness and shape. To ensure that
    your function is typical for the reproducing kernel Hilbert space, F, one should likely tune these
    parameters from the function data. Here is a proposal:
    θopt
    = argmin
    θ
    log f(X)TKθ(X, X)−1f(X)
    squared norm of the minimum norm interpolant
    +
    1
    n
    log(det(Kθ(X, X)))
    This corresponds to choosing θ to minimize the volume of the ellipsoidal solid in Rn consisting of all
    possible function data whose minimum-norm interpolants have an Fθ
    -norm no greater than that of the
    observed interpolant.
    It also corresponds to using empirical Bayes when working in the Gaussian process setting with
    covariance kernels, Kθ
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  26. Thank you
    These slides are available at
    speakerdeck.com/fjhickernell/reproducing-kernel-tutorial

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  27. Background Rep Ker & Riesz Rep Thm Kernel Ex Assoc Measures Error Bds References
    References
    1. N. Aronszajn. Theory of Reproducing Kernels. Trans. Amer. Math. Soc. 68, 337–404 (1950).
    2. Rasmussen, C. E. & Williams, C. Gaussian Processes for Machine Learning. (online version at
    http://www.gaussianprocess.org/gpml/) (MIT Press, Cambridge, Massachusetts, 2006).
    3. H., F. J. A Generalized Discrepancy and Quadrature Error Bound. Math. Comp. 67, 299–322 (1998).
    4. H., F. J. Goodness-of-Fit Statistics, Discrepancies and Robust Designs. Statist. Probab. Lett. 44,
    73–78 (1999).
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