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 Wasilkowski

The problem We consider the following integration problem for a class F of ∞ -variate functions f (x) = f ( x 1 , x 2 , . . . ) : � � I ( f ) = D N f (x) ρ ( x j ) dx , j ∈ N 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] .

Each function f ∈ F is assumed to have an expansion of the form � f (x) = f u (x) , | u | < ∞ where, for each x ∈ D N , the decomposition is an absolutely convergent sum over the subsets u of N with finite cardinality | u | , and each f u depends only on the variables x j with j ∈ u . (In particular, f ∅ is a constant function.) Example: ANOVA or anchored ANOVA expansions.

The ∞ -variate integral is defined formally by d � � � I ( f ) := lim f u (x) ρ ( x j ) dx d →∞ D d u ⊆{ 1 ,...,d } j =1 � = lim I u ( f u ) d →∞ u ⊆{ 1 ,...,d } � I u ( f u ) , = | u | < ∞ where � � I u ( f u ) := D | u | f u (x u ) ρ u (x u ) dx u with ρ u (x u ) = ρ ( x j ) . j ∈ u

Assumption on f u Each term f u in the expansion of f is assumed to belong to some normed space F u , and moreover to have bounded norm, � f u � F u ≤ B u , for known positive numbers B u .

Example Suppose D = [ − 1 2 , 1 2 ] and F u is the RKHS with kernel � K u (x u , y u ) = K ( x j , y j ) , j ∈ u where K ( x, y ) = | x | + | y | − | x − y | . 2 The functions f u ∈ F u have square-integrable mixed first derivatives, and vanish at x if x j = 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 B u = µ ( | u | !) b 1 � ( κj ) − b 2 . j ∈ u for some b 2 > max( b 1 , 0) and some µ, κ > 0 .

More assumptions All functions in F u are measurable with respect to ρ (x u ) dx u , and the functionals I u are continuous, i.e., � � � � � C u := � I u � F u = sup D | u | f u (x u ) ρ u (x u ) dx u � < ∞ . � � � � f u � F u ≤ 1 The numbers B u and C u satisfy � C u B u < ∞ . | u | < ∞

Then the integration problem is well defined: The integration functional I on F defined by � � � I ( f ) = D N f (x) ρ ( x j ) dx := I u ( f u ) , j ∈ N | u | < ∞ is well defined, because � sup |I ( f ) | ≤ sup | I u ( f u ) | f ∈F f ∈F | u | < ∞ � ≤ � I u � F u sup � f u � F u f u ∈F u | u | < ∞ � ≤ C u B u < ∞ . | u | < ∞

The idea of the MDM algorithm Knowing the numbers B u , 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 2 ε . 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 f u at arbitrary points x u in the domain D | u | , at a cost that depends only on | u | : specifically, that we can evaluate f u (x u ) for x u ∈ D | u | at cost £ ( | u | ) , where £ : { 0 , 1 , 2 , . . . } → [0 , ∞ ] is known and non-decreasing.

The algorithm The MDM for the integral I ( f ) with f = � | u | < ∞ f u has the form � A ε ( f ) := A u ,n u ( f u ) , u ∈U ( ε ) that is, we are allowed to use a different cubature rule for each subset u ∈ U ( ε ) , n u � A u ,n u ( f u ) = w u ,i f u ( t u ,i ) . i =1 The rule A u ,n u and the values of n u are chosen in such a way that � 1 | I u ( f u ) − A u ,n u ( f u ) | ≤ 2 ε . u ∈U ( ε )

The total error For f ∈ F the total error then satisfies � � |I ( f ) − A ε ( f ) | ≤ | I u ( f u ) | + | I u ( f u ) − A u ,n u ( f u ) | u / ∈U ( ε ) u ∈U ( ε ) 1 1 ≤ 2 ε + 2 ε = ε , and the (information) cost is � cost( A ε ) = n u £ ( | u | ) . (1) u ∈U ( ε )

Finding the active set U ( ε ) Recall that for each u we already assumed � | u | < ∞ C u B u < ∞ . Now we assume more: namely that the sequence ( C u B u ) u has some known rate of decay. Specifically, we assume that � � ( C u B u ) 1 /α < ∞ � α 0 := sup α : > 1 . | u | < ∞ | u | < ∞ ( C u B u ) 1 / 2 < ∞ then α 0 ≥ 2 . Example: If � Then for any α ∈ (1 , α 0 ) we define ε/ 2 � � u : ( C u B u ) 1 − 1 /α > U ( ε ) = U ( ε, α ) := . � | v | < ∞ ( C v B v ) 1 /α Problems remain: this needs a) an infinite sum, and b) an infinite number of decisions.

This works because ... ε/ 2 � � u : ( C u B u ) 1 − 1 /α > U ( ε ) = . � | v | < ∞ ( C v B v ) 1 /α

This works because ... ε/ 2 � � U ( ε ) C = u : ( C u B u ) 1 − 1 /α ≤ . � | v | < ∞ ( C v B v ) 1 /α � � | I u ( f u ) | ≤ C u B u ∈U ( ε ) ∈U ( ε ) u / u / � ( C u B u ) 1 − 1 /α ( C u B u ) 1 /α = u / ∈U ( ε ) ε/ 2 � | v | < ∞ ( C v B v ) 1 /α ( C u B u ) 1 /α ≤ � u / ∈U ( ε ) ≤ ε 2 , as required.

Example For our standard example of the anchored RKHS and anchored ANOVA, we easily compute: C u = � I u � F u = C | u | C 0 := � I � F = 12 − 1 / 2 , = 12 −| u | / 2 . 0 For the bound on the norm of f u assume B u = ( | u | !) b 1 µ � ( κj ) − b 2 j ∈ u for some b 2 > max( b 1 , 0) and some µ, κ > 0 . Then ln(1 /ε ) � � u ∈U ( ε,α ) | u | = O max ε → 0 . as ln(ln(1 /ε ))

Example (continued) For the anchored ANOVA decomposition, � f (x) = f u (x u ) | u | < ∞ it is known (Kuo, IHS, Wasilkowski and Wo´ zniakowski, 2010) that � ( − 1) | u |−| v | f ( y v , 1 , y v , 2 , . . . ) , f u (x u ) = v ⊆ u with  x j if j ∈ v   y v ,j = 0 if j / ∈ v .   Suppose that the cost of a single evaluation of f ( y v , 1 , y v , 2 , . . . ) is $( | v | ) , which is non-decreasing. For example, $( k ) = k .

Then the cost of evaluating f u (x u ) is Then the cost of evaluating f u (x u ) is | u | � | u | � $( k ) ≤ 2 | u | $( | u | ) . � � £ ( | u | ) = $( | v | ) = k k =0 v ⊆ u Thus the cost is small whenever max | u | is small. For example, if $( k ) = e O ( k ) then � O (1 / ln(ln(1 /ε ))) � 1 u ∈U ( ε ) £ ( | u | ) = max . ε

The algorithm: trapezoidal-Smolyak , Recall � f = f u | u | < ∞ and � A ε ( f ) := A u ,n u ( f u ) . u ∈U ( ε ) For u ∈ U ( ε ) we choose A u ,n u to be a trapezoidal-Smolyak algorithm for dimension | u | : d � � A | u | ,n u = ( U i j − U i j − 1 ) , j =1 i ∈ N | u | , | i |≤ κ u where | i | = i 1 + i 2 + · · · + i d , and { U i } i ≥ 1 are (composite) trapezoidal rules with 2 i + 1 equally spaced points, and U 0 = 0 . The choice of κ ≥ | u | determines n u , n u ≥ 2 κ u −| u | +1 .

Theorem Let F u be our standard anchored RKHS, and take the rule A u ,n u to be a trapezoidal Smolyak rule, with κ u given, for u ∈ U ( ε ) , by κ u = . . . . Assume that £ ( d ) = e O ( d ) . Then 3 / 2   � 1+ O (1 / ln(ln(1 /ε ))) � 1  � B 1 / 2 cost( A ε ) ≤ 4  u ε u ∈U ( ε )

Implementation? In progress!

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