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Infinite-dimensional integration by the Multivariate Decomposition - - PowerPoint PPT Presentation

Infinite-dimensional integration by the Multivariate Decomposition Method Ian Sloan i.sloan@unsw.edu.au The University of New South Wales ICERM Workshop on IBC and Stochastic Computation, 2014 Joint with F Kuo, D Nuyens, L Plaskota, G


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Infinite-dimensional integration by the Multivariate Decomposition Method

Ian Sloan

i.sloan@unsw.edu.au The University of New South Wales ICERM Workshop on IBC and Stochastic Computation, 2014 Joint with F Kuo, D Nuyens, L Plaskota, G Wasilkowski

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The problem

We consider the following integration problem for a class F of ∞-variate functions f(x) = f(x1, x2, . . .): I(f) =

  • DN f(x)
  • j∈N

ρ(xj) dx, where D is a bounded or unbounded Borel subset of R and ρ is a probability density on D. E.g. D = [0, 1], ρ(x) = 1 ∀ x ∈ [0, 1].

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Each function f ∈ F is assumed to have an expansion of the form f(x) =

  • |u|<∞

fu(x), where, for each x ∈ DN, the decomposition is an absolutely convergent sum over the subsets u of N with finite cardinality |u|, and each fu depends only on the variables xj with j ∈ u. (In particular, f∅ is a constant function.) Example: ANOVA or anchored ANOVA expansions.

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The ∞-variate integral is defined formally by I(f) := lim

d→∞

  • Dd
  • u⊆{1,...,d}

fu(x)

d

  • j=1

ρ(xj) dx = lim

d→∞

  • u⊆{1,...,d}

Iu(fu) =

  • |u|<∞

Iu(fu), where Iu(fu) :=

  • D|u| fu(xu) ρu(xu) dxu

with ρu(xu) =

  • j∈u

ρ(xj).

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Assumption on fu

Each term fu in the expansion of f is assumed to belong to some normed space Fu, and moreover to have bounded norm, fuFu ≤ Bu, for known positive numbers Bu.

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Example

Suppose D = [− 1

2, 1 2] and Fu is the RKHS with kernel

Ku(xu, yu) =

  • j∈u

K(xj, yj), where K(x, y) = |x| + |y| − |x − y| 2 .

The functions fu ∈ Fu have square-integrable mixed first derivatives, and vanish at x if xj = 0 for some j ∈ u.

For this space a recent paper (Kuo, IHS, Schwab, SINUM 2012) showed, for a class of PDE problems with random coefficients, that Bu = µ (|u|!)b1

j∈u

(κj)−b2. for some b2 > max(b1, 0) and some µ, κ > 0.

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More assumptions

All functions in Fu are measurable with respect to ρ(xu) dxu, and the functionals Iu are continuous, i.e., Cu := IuFu = sup

fuFu≤1

  • D|u| fu(xu) ρu(xu) dxu
  • < ∞.

The numbers Bu and Cu satisfy

  • |u|<∞

Cu Bu < ∞.

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Then the integration problem is well defined:

The integration functional I on F defined by I(f) =

  • DN f(x)
  • j∈N

ρ(xj) dx :=

  • |u|<∞

Iu(fu), is well defined, because sup

f∈F

|I(f)| ≤ sup

f∈F

  • |u|<∞

|Iu(fu)| ≤

  • |u|<∞

IuFu sup

fu∈Fu

fuFu ≤

  • |u|<∞

Cu Bu < ∞ .

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The idea of the MDM algorithm

Knowing the numbers Bu, we determine in advance an active set U(ε). The active set contains all the finite subsets of N that we need to retain to determine I(f) to within an error of no more than 1

2ε.

How to determine the active set? Later!

The subsets not in U(ε) will be ignored, that is, will be approximated by zero. We assume that we can sample fu at arbitrary points xu in the domain D|u|, at a cost that depends only on |u|: specifically, that we can evaluate fu(xu) for xu ∈ D|u| at cost £(|u|), where £ : {0, 1, 2, . . . } → [0, ∞] is known and non-decreasing.

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The algorithm

The MDM for the integral I(f) with f =

|u|<∞ fu has the form

Aε(f) :=

  • u∈U(ε)

Au,nu(fu), that is, we are allowed to use a different cubature rule for each subset u ∈ U(ε), Au,nu(fu) =

nu

  • i=1

wu,ifu(tu,i). The rule Au,nu and the values of nu are chosen in such a way that

  • u∈U(ε)

|Iu(fu) − Au,nu(fu)| ≤

1 2ε.

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The total error

For f ∈ F the total error then satisfies |I(f) − Aε(f)| ≤

  • u/

∈U(ε)

|Iu(fu)| +

  • u∈U(ε)

|Iu(fu) − Au,nu(fu)| ≤

1 2ε + 1 2ε = ε,

and the (information) cost is cost(Aε) =

  • u∈U(ε)

nu£(|u|).

(1)

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Finding the active set U(ε)

Recall that for each u we already assumed

|u|<∞ CuBu < ∞.

Now we assume more: namely that the sequence (CuBu)u has some known rate of decay. Specifically, we assume that α0 := sup

  • α :
  • |u|<∞

(CuBu)1/α < ∞

  • > 1.

Example: If

|u|<∞(CuBu)1/2 < ∞ then α0 ≥ 2.

Then for any α ∈ (1, α0) we define U(ε) = U(ε, α) :=

  • u : (CuBu)1−1/α >

ε/2

  • |v|<∞(CvBv)1/α
  • .

Problems remain: this needs a) an infinite sum, and b) an infinite number of decisions.

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This works because ...

U(ε) =

  • u : (CuBu)1−1/α >

ε/2

  • |v|<∞(CvBv)1/α
  • .
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This works because ...

U(ε)C =

  • u : (CuBu)1−1/α ≤

ε/2

  • |v|<∞(CvBv)1/α
  • .
  • u/

∈U(ε)

|Iu(fu)| ≤

  • u/

∈U(ε)

CuBu =

  • u/

∈U(ε)

(CuBu)1−1/α(CuBu)1/α ≤

  • u/

∈U(ε)

ε/2

  • |v|<∞(CvBv)1/α (CuBu)1/α

≤ ε 2 , as required.

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Example

For our standard example of the anchored RKHS and anchored ANOVA, we easily compute: C0 := IF = 12−1/2, Cu = IuFu = C|u| = 12−|u|/2. For the bound on the norm of fu assume Bu = (|u|!)b1 µ

  • j∈u

(κj)−b2 for some b2 > max(b1, 0) and some µ, κ > 0. Then max

u∈U(ε,α) |u| = O

  • ln(1/ε)

ln(ln(1/ε))

  • as

ε → 0.

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Example (continued)

For the anchored ANOVA decomposition, f(x) =

  • |u|<∞

fu(xu) it is known (Kuo, IHS, Wasilkowski and Wo´

zniakowski, 2010) that

fu(xu) =

  • v⊆u

(−1)|u|−|v|f(yv,1, yv,2, . . .) , with yv,j =      xj if j ∈ v 0 if j / ∈ v. Suppose that the cost of a single evaluation of f(yv,1, yv,2, . . .) is $(|v|), which is non-decreasing. For example, $(k) = k.

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Then the cost of evaluating fu(xu) is

Then the cost of evaluating fu(xu) is £(|u|) =

  • v⊆u

$(|v|) =

|u|

  • k=0

|u| k

  • $(k) ≤ 2|u| $(|u|).

Thus the cost is small whenever max|u| is small. For example, if $(k) = eO(k) then max

u∈U(ε) £(|u|) =

1 ε O(1/ ln(ln(1/ε))) .

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The algorithm: trapezoidal-Smolyak

Recall , f =

  • |u|<∞

fu and Aε(f) :=

  • u∈U(ε)

Au,nu(fu).

For u ∈ U(ε) we choose Au,nu to be a trapezoidal-Smolyak algorithm for dimension |u|: A|u|,nu =

  • i∈N|u|, |i|≤κu

d

  • j=1

(Uij − Uij−1), where |i| = i1 + i2 + · · · + id, and {Ui}i≥1 are (composite) trapezoidal rules with 2i + 1 equally spaced points, and U0 = 0. The choice of κ ≥ |u| determines nu, nu ≥ 2κu−|u|+1.

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Theorem

Let Fu be our standard anchored RKHS, and take the rule Au,nu to be a trapezoidal Smolyak rule, with κu given, for u ∈ U(ε), by κu = . . . . Assume that £(d) = eO(d). Then cost(Aε) ≤ 4 1 ε 1+O(1/ ln(ln(1/ε)))  

u∈U(ε)

B1/2

u

 

3/2

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Implementation?

In progress!

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Reference

FY Kuo. D Nuyens, L Plaskota, IH Sloan, GW Wasilkowski “The multivariate decomposition method for infinite-dimensional integration” (under construction).