Multidimensional interpolation. 1 An abstract description of interpolation in one dimension

Transcription

1 Paul Klein Department of Economics University of Western Ontario FINNISH DOCTORAL PROGRAM IN ECONOMICS Topics in Contemporary Macroeconomics August 2008 Multidimensional interpolation 1 An abstract description of interpolation in one dimension This discussion is based on that in Miranda and Fackler (2002), but is a little bit more explicit Function approximation in one dimension involves defining an approximating function f of a given function f : A R, where A R The function f is a member of a finite-dimensional set of functions with basis {ψ i } n i=0 This means that, for any x A, we have f(x) = In vector notation, we may write (1) as n θ j ψ j (x) (1) j=0 f(x) = ψ(x)θ (2) where ψ(x) = ψ 0 (x) ψ 1 (x) ψ n (x) 1

2 and θ = θ 0 θ 1 θ n The interpolation problem in one dimension is to find {θ i } n i=0 such that f(x i ) = n θ j ψ j (x i ) (3) j=0 for all i = 0, 1,, n In vector notation, we may write (3) as y = Ψθ (4) where and ψ 0 (x 0 ) ψ 1 (x 0 ) ψ 2 (x 0 ) ψ n (x 0 ) ψ 0 (x 1 ) ψ 1 (x 1 ) ψ n (x 1 ) Ψ = ψ 0 (x 2 ) ψ 0 (x n ) ψ n (x n ) f(x 0 ) f(x y = 1 ) f(x n ) In the case of Chebyshev interpolation, we have an explicit expression for Ψ 1 k that case is particularly simple To state this precisely, define M via Ψ = 1 M so Then, in the case of Chebyshev polynomial interpolation, we have Ψ 1 = 1 1 T n + 1 2M T 2

3 2 Interpolation in d dimensions Now suppose f : A R where A R d Then we have a set of n k + 1 basis functions {ψ k i } n k i=0 for each dimension k = 1, 2,, d We also have a grid {xk i } n k i=0 dimension k = 1, 2,, d for each The set of known function values is naturally organized in a multi-dimensional array as follows Y (i 1, i 2,, i d ) = f(x 1 i 1, x 2 i 2,, x d i d ) We would like (2) and (4) to be valid in the multi-dimensional case as well To achieve that, we begin by defining y = Y (:) This is Matlabese for hyper-columnwise vectorization What does this mean exactly? In two dimensions it is just the usual vec operator In three dimensions, you first take the two-dimensional matrices Y (:, :, i) for i = 0, 1,, n 3, apply the vec operator to each of them, then stack the results on top of each other, with vec(y (:, :, 0)) on top of course 1 In four dimensions, Y (:) begins with vec of Y (:, :, 0, 0), followed by vec of Y (:, :, 1, 0), followed by vec of Y (:, :, 0, 1), followed by vec of Y (:, :, 1, 1) In d dimensions, the order is defined by incrementing the first index until the end is reached, then increment the second index by one step, reset the first index and again increment it until the end, then increment the second index one step etc etc until the end of the first index is reached Then increment the third index by one step, reset the second index etc etc Meanwhile, we define, for k = 1, 2,, d, ψ k 0(x k 0) ψ k 1(x k 0) ψ k 2(x k 0) ψ k n k (x k 0) ψ k 0(x k 1) ψ k 1(x k 1) ψ k n k (x k 1) Ψ k = ψ k 0(x k 2) ψ k 0(x k n k ) ψ k n k (x k n k ) 1 0-based indexing is not actually allowed in Matlab, but it is allowed in Fortran 3

4 and Finally, define Ψ = Ψ d Ψ d 1 Ψ 1 ψ(x) = ψ d (x d ) ψ d 1 (x d 1 ) ψ 1 (x 1 ) where x = (x 1, x 2,, x d ) With these definitions, (2) and (4) remain valid in the multidimensional case as well in the following sense If θ solves (4) then for all 0 i 1 n 1, 0 i 2 n 2,, 0 i d n d we have ψ(x 1 i 1, x 2 i 2,, x d i d )θ = Y (i 1, i 2,, i d ) Notice also that solving (4) in the multi-dimensional case is not as hard as it looks, since Ψ 1 = Ψ 1 d Ψ 1 d 1 Ψ Why it works It is surprisingly hard to show that this works in general But let s show it in the simplest possible non-trivial case Let the basis functions in each of two dimensions be 1 and x and 1 and y, respectively Let the grids be {x 0, x 1 } and {y 0, y 1 } Organize the known function values in a 2 2 matrix Z Notice first that the interpolating function is written as f(x, y) = ψ(x, y)θ where ψ(x, y) = 1 y 1 x = 1 x y xy Now form the matrix Ψ via 1 y 0 Ψ y Ψ x = 1 y 1 1 x 0 1 x 1 = 1 x 0 y 0 x 0 y 0 1 x 1 y 0 x 1 y 0 1 x 0 y 1 x 0 y 1 1 x 1 y 1 x 1 y 1 4

5 Now consider the expression Ψθ Evidently it should equal but this is precisely vec(z) Z 00 Z 10 Z 01 Z 11, 3 The curse of dimensionality The number of basis functions in ψ(x) = ψ d (x d ) ψ d 1 (x d 1 ) ψ 1 (x 1 ) is d k=1 which grows very rapidly in d So the question is whether one can economize a bit One approach is the following, where we will focus on the two-dimensional case to keep things clear Suppose the variables are x and y and the basis functions in the x dimension are 1, x, and x 2 Similarly, in the y dimension they are 1, y, and y 2 Then there are nine elements in ψ y (y) ψ x (x), including x 2 y 2 But perhaps we wanted a quadratic approximation, and x 2 y 2 is not quadratic but quartic So we should just use those products where the sum of the exponents is less than or equal to two, ie the following: 1, x, y, xy, x 2 and y 2 That s only six terms This remark can be generalized, but that would be a bit tedious, wouldn t it? Another approach is to use a sparse grid The pioneering paper here was Smolyak (1963) Most economists who know about this approach learned it from Malin et al (2007) n k 5

6 References Malin, B, D Krueger, and F Kubler (2007, October) Computing Stochastic Dynamic Economic Models with a Large Number of State Variables: A Description and Application of a Smolyak-Collocation Method NBER Technical Working Papers 0345, National Bureau of Economic Research, Inc Miranda, M J and P L Fackler (2002) Applied Computational Economics and Finance MIT Press Smolyak, S A (1963) Quadrature and interpolation formulas for tensor products of certain classes of functions Soviet Math Dokl 4,