In this introductory chapter we review some current methods of numerical inte-gration in order to put subsequent chapters into a wider context. This serves asmotivation for later investigations.
The problem of numerical integration occurs in applications from physics, chem-istry, finance, biology, computer graphics, and others, where one has to computesome integral (for instance, an expectation value) which cannot be done analyti-cally. Hence, one has to resort to numerical methods in this case. We shall, in thefollowing, consider only the standardised problem of approximating an integral ofthe form ∫
[0,1]sf (x) dx.
The books of Fox [81], Tezuka [256], Glasserman [85] and Lemieux [154], andthe surveys of Keller [120] and L’Ecuyer [151] deal more directly with questionsarising from applications.
1.1 The one-dimensional case
Let us consider the case s = 1 first. Let f : [0, 1] → R be a Riemann integrablefunction. We proceed now as follows. Take a sample of N points x0, . . . , xN−1 inthe interval [0, 1) and calculate the average function value at those points, i.e.
1
N
N−1∑n=0
f (xn).
As approximation to the integral, we use the value
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The question arises as to how large the approximation error is using this method,i.e. how large is the value∣∣∣∣∣
∫ 1
0f (x) dx − 1
N
N−1∑n=0
f (xn)
∣∣∣∣∣?Intuitively, we expect the integration error to depend on two quantities, namely,
� on the quadrature points x0, . . . , xN−1 ∈ [0, 1), and� on the function f .
Let us consider these two points in turn. The quadrature points should have nobig gaps in between, otherwise large portions of the function are not considered inthe approximation. Hence, {x0, . . . , xN−1} should be well distributed in [0, 1). Forinstance, assume we want to integrate the function f : [0, 1] → R given by
f (x) =
⎧⎪⎪⎨⎪⎪⎩
0 if x ≤ 1
2,
1 if x >1
2.
If all the points x0, . . . , xN−1 are in the interval [0, 1/2], i.e. the points are not welldistributed in [0, 1), then we obtain
1
N
N−1∑n=0
f (xn) = 0
as an approximation to the integral∫ 1
0f (x) dx = 1
2,
see Figure 1.1.Hence we obtain an integration error∣∣∣∣∣
1
N
N−1∑n=0
f (xn) −∫ 1
0f (x) dx
∣∣∣∣∣ = 1
2.
Of course the error also depends strongly on the integrand f itself, and, inparticular, on the smoothness and a suitable norm of the integrand f , which in asense measures how greatly f varies; for instance, constant functions are always
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Figure 1.1 Example of badly distributed quadrature points.
0.2 0.4 0.6 0.8 1
0.5
1
1.5
2
Figure 1.2 Example of the strongly varying function f (x) = 1 + cos(2πkx) withk = 10.
integrated exactly, using this method. On the other hand, assume that we havean integrand f which varies strongly, like the function f (x) = 1 + cos(2πkx) inFigure 1.2 for some large value of k. If we choose for N = k the points x0, . . . , xN−1
as xn = (2n + 1)/(2N) for 0 ≤ n < N , then they can be said to be ‘well’ distributedin [0, 1), but we still obtain a large integration error. Indeed, we have
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Hence, again we obtain a large integration error∣∣∣∣∣1
N
N−1∑n=0
f (xn) −∫ 1
0f (x) dx
∣∣∣∣∣ = 1.
Remark 1.1 We see later in Chapter 2 that the integration error can indeed bebounded by a product of a quantity which measures the distribution properties ofthe points x0, . . . , xN−1 and a quantity which measures how greatly the integrandf varies.
1.2 The general case
Now let us consider the case where s ∈ N. Let f : [0, 1]s → R be, say, a Riemannintegrable function. We want to approximate the value of the integral
∫ 1
0· · ·
∫ 1
0f (x1, . . . , xs) dx1 · · · dxs =
∫[0,1]s
f (x) dx.
For this purpose we proceed as in the case s = 1; i.e. we choose quadrature pointsx0, . . . , xN−1 ∈ [0, 1)s and approximate the integral via the average function valueof f at those N points; i.e.
∫[0,1]s
f (x) dx ≈ 1
N
N−1∑n=0
f (xn).
Again, we want to estimate the absolute value of the integration error∣∣∣∣∣∫
[0,1]sf (x) dx − 1
N
N−1∑n=0
f (xn)
∣∣∣∣∣ .
Now the question arises as to how we should choose the quadrature pointsx0, . . . , xN−1. Considering the case s = 1, a solution which suggests itself fors > 1 would be to choose the points on a centred regular lattice. For s = 1we would choose, as above, the points xn = (2n + 1)/(2N ) for 0 ≤ n < N .In general, for m ∈ N, m ≥ 2, the centred regular lattice c
m is given by thepoints
xk =(
2k1 + 1
2m, . . . ,
2ks + 1
2m
)(1.1)
for all k = (k1, . . . , ks) ∈ Ns0 with |k|∞ := max1≤i≤s |ki | < m (hence, we have
N = ms points). An example of a centred regular lattice is shown in Figure 1.3.
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Figure 1.3 Centred regular lattice c6 in [0, 1)2, i.e. s = 2 and m = 6.
As mentioned above, we need to make some assumptions on the smoothnessof the integrand f . We therefore assume that the integrand f is continuous. Inthis case we can introduce the following concept as a measure of how much thefunction f varies.
Definition 1.2 For a continuous function f : [0, 1]s → R, the modulus of conti-nuity is given by
Mf (δ) := supx, y∈[0,1]s
|x− y|∞≤δ
|f (x) − f ( y)| for δ ≥ 0,
where | · |∞ is the maximum norm, i.e. for x = (x1, . . . , xs) we set |x|∞ :=max1≤i≤s |xi |.
If we assume that the function f is uniformly continuous on [0, 1]s , then we havelimδ→0+ Mf (δ) = 0. Note that for any function f , its modulus Mf is non-decreasingand subadditive. Recall that a function f is non-decreasing if f (x) ≤ f (y) for allx ≤ y, and that a function f is subadditive if f (x + y) ≤ f (x) + f (y) for all x, y
in the domain of f .Furthermore, for non-constant functions f , the smallest possible order of Mf
is Mf (δ) = O(δ) as δ → 0+. Recall that we say h(x) = O(g(x)) as x → 0 if andonly if there exist positive real numbers δ and C such that |h(x)| ≤ C|g(x)| for|x| < δ.
For k = (k1, . . . , ks) ∈ Ns0 with |k|∞ < m, let Qk = ∏s
i=1[ki/m, (ki + 1)/m).Then each point of the centred regular lattice (1.1) is contained in exactly oneinterval Qk, namely the point xk (see again Figure 1.3).
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Now let f : [0, 1]s → R be a continuous function and let xk for k ∈ Ns0 with
|k|∞ < m be the points of a centred regular lattice. Then we have∣∣∣∣∣∫
[0,1]sf (x) dx − 1
ms
∑k∈N
s0
|k|∞<m
f (xk)
∣∣∣∣∣ =∣∣∣∣∣
∑k∈N
s0
|k|∞<m
∫Qk
f (x) − f (xk) dx
∣∣∣∣∣
≤∑k∈N
s0
|k|∞<m
∫Qk
Mf (|x − xk|∞) dx
≤ ms
∫B( 1
2m )Mf (|x|∞) dx, (1.2)
where B(ε) := {x ∈ Rs : |x|∞ ≤ ε}.Assume that the function f is in addition Lipschitz continuous (for example, it
suffices if f has partial derivatives), i.e. there is a real number Cf > 0 such that
Mf (δ) ≤ Cf δ for all δ > 0.
Then, using (1.2), we have∣∣∣∣∣∫
[0,1]sf (x) dx − 1
ms
∑k∈N
s0
|k|∞<m
f (xk)
∣∣∣∣∣ ≤ ms
∫B( 1
2m )Cf |x|∞ dx
≤ Cf
2m= Cf
2N1/s, (1.3)
where the last inequality can be obtained by estimating |x|∞ ≤ 1/(2m).This result cannot be improved significantly for uniformly continuous functions.
Before we show the corresponding result, let us give some examples. For instance,take s = 1 and consider the function f (x) = c(1 + cos(2πNx))/(2N) for someconstant c > 0 (see Figure 1.4). Notice that f ′(x) = −cπ sin(2πNx), hence theLipschitz constant is Cf = sup0≤x≤1 |f ′(x)| = cπ and the modulus of continuitysatisfies Mf (δ) ≤ cπδ for all δ > 0. Thus, unlike the function itself, the Lipschitzconstant and the modulus of continuity do not depend on N . If we consider theLipschitz constant or the modulus of continuity of f as a measure of how stronglyf varies, then this measure does not depend on N . Hence, we have a family offunctions which all vary equally strongly. Let us now consider the integration errorsof these functions.
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Motivated by the above examples we show the following unpublished result dueto G. Larcher. In the following we call a uniformly continuous function M : R+
0 →R+
0 , where R+0 = {x ∈ R : x ≥ 0}, which is non-decreasing, subadditive, and for
which we have limδ→0+ M(δ) = 0 a modulus.
Theorem 1.3 For any modulus M and any x0, . . . , xN−1 in [0, 1)s , there is auniformly continuous function f : [0, 1]s → R with modulus of continuity Mf ≤M , such that
∣∣∣∣∣∫
[0,1]sf (x) dx − 1
N
N−1∑n=0
f (xn)
∣∣∣∣∣ ≥ N
∫B
(1
2N1/s
) M(|x|∞) dx.
Proof Consider the Voronoi diagram {V0, . . . , VN−1} of x0, . . . , xN−1 withrespect to the maximum norm, i.e.
Vn = {x ∈ [0, 1]s : |x − xn|∞ = min0≤j<N
|x − xj |∞}
for 0 ≤ n < N , and define f : [0, 1]s → R by f (x) := M(|x − xn|∞) for x ∈ Vn.Then f is uniformly continuous since M is continuous, {V0, . . . , VN−1} is a Voronoidiagram, and f is defined on a compact domain.
We show that Mf ≤ M . Let x, y ∈ [0, 1]s and assume that f (x) > f ( y). If x, yare in the same Voronoi cell, say Vn, then we have
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where we used that M is subadditive and non-decreasing. If x, y are not in thesame Voronoi cell, say x ∈ Vn and y ∈ Vk with n �= k, then we have
|f (x) − f ( y)| = M(|x − xn|∞) − M(| y − xk|∞)
≤ M(|x − xk|∞) − M(| y − xk|∞)
≤ M(max(|x − xk|∞ − | y − xk|∞, 0))
≤ M(|x − y|∞),
where we again used that M is subadditive and non-decreasing. Hence we haveMf ≤ M .
It remains to show the lower bound on the integration error. We have
∣∣∣∣∣∫
[0,1]sf (x) dx − 1
N
N−1∑n=0
f (xn)
∣∣∣∣∣ =N−1∑n=0
∫Vn
M(|x − xn|∞) dx. (1.4)
Let Wn := {x ∈ [0, 1]s : |x − xn|∞ ≤ 1/(2N1/s)
}. Then we have
N−1∑n=0
∫Vn
M(|x − xn|∞) dx
=N−1∑n=0
(∫Vn∩Wn
M(|x − xn|∞) dx +∫
Vn\Wn
M(|x − xn|∞) dx)
. (1.5)
Let y ∈ Vn \ Wn for some n and let x ∈ Wk \ Vk for some k. Then we have| y − xn|∞ > 1/(2N1/s) and |x − xk|∞ ≤ 1/(2N1/s). Since M , by definition, isnon-decreasing, it follows that M(| y − xn|∞) ≥ M(|x − xk|∞).
We also have∑N−1
n=0 λs(Vn) = 1 and∑N−1
n=0 λs(Wn) ≤ 1, where λs is the s-dimensional Lebesgue measure. Hence, we have
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Combining Theorem 1.3 with (1.2), we obtain the following result fromG. Larcher which states that the centred regular lattice yields the smallest pos-sible integration error for the class of uniformly continuous functions with a givenmodulus of continuity.
Corollary 1.4 Let N = ms and let M be any modulus. Then we have
infP
supf
∣∣∣∣∣∫
[0,1]sf (x) dx − 1
N
N−1∑n=0
f (xn)
∣∣∣∣∣ = N
∫B
(1
2N1/s
) M(|x|∞) dx,
where the infimum is extended over all point sets P consisting of N points in[0, 1)s and the supremum is extended over all uniformly continuous functionsf : [0, 1]s → R with modulus of continuity Mf = M . Moreover, the infimum isattained by the centred regular lattice.
The problem in the upper bounds (1.2) and (1.3), respectively, is that the inte-gration error depends strongly on the dimension s. For large s the convergence ofN−1/s to 0 is very slow as N → ∞. This phenomenon is often called the curse ofdimensionality. The question arises as to whether one can choose ‘better’ quadra-ture points x0, . . . , xN−1, i.e. for which the integration error depends only weakly(or not at all) on the dimension s. The question can be answered in the affirmative,which can be seen by the following consideration.
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