Introduction to polynomials chaos with NISP

Michael Baudin (EDF)Jean-Marc Martinez (CEA)

Version 0.4January 2013

Abstract

This document is an introduction to polynomials chaos with the NISP module for Scilab.

Contents

1 Orthogonal polynomials 41.1 Definitions and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Orthogonal polynomials for probabilities . . . . . . . . . . . . . . . . . . . . . . . 71.3 Hermite polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.4 Legendre polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.5 Laguerre polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.6 Chebyshev polynomials of the first kind . . . . . . . . . . . . . . . . . . . . . . . . 191.7 Accuracy of evaluation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.8 Notes and references . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2 Multivariate polynomials 242.1 Occupancy problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.2 Multivariate monomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.3 Multivariate polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.4 Generating multi-indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.5 Multivariate orthogonal functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.6 Tensor product of orthogonal polynomials . . . . . . . . . . . . . . . . . . . . . . 322.7 Multivariate orthogonal polynomials and probabilities . . . . . . . . . . . . . . . . 362.8 Notes and references . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3 Polynomial chaos 393.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.2 Truncated decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.3 Univariate decomposition examples . . . . . . . . . . . . . . . . . . . . . . . . . . 433.4 Generalized polynomial chaos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.5 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.6 Notes and references . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

1

4 Thanks 48

A Gaussian integrals 48A.1 Gaussian integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48A.2 Weighted integral of xn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

Bibliography 51

2

Copyright c© 2013 - Michael BaudinThis file must be used under the terms of the Creative Commons Attribution-ShareAlike 3.0

Unported License:

http://creativecommons.org/licenses/by-sa/3.0

3

1 Orthogonal polynomials

In this section, we define the orthogonal polynomials which are used in the context of polynomialchaos decomposition. We first present the weight function which defines the orthogonality of thesquare integrable functions. Then we present the Hermite, Legendre, Laguerre and Chebyshevorthogonal polynomials.

1.1 Definitions and examples

We will denote by I an interval in R. The interval [a, b] is the set of real numbers such that x ≥ aand x ≤ b, for two real numbers a and b. The open interval (−∞,+∞) is also considered as aninterval, although its boundaries are infinite. The half-open intervals (−∞, b) and (a,+∞) arealso valid intervals.

Definition 1.1. ( Weight function of R) Let I be an interval in R. A weight function w is anonnegative integrable function of x ∈ I.

Example (Weight function for Hermite polynomials) The weight function for Hermite polyno-mials is

w(x) = exp

(−x

2

2

), (1)

for x ∈ R. This function is presented in the figure 1. Clearly, this function is differentiable andnonnegative, since exp(−x2) > 0, for any finite x. The integral∫

Rexp(−x2/2)dx =

√2π (2)

is called the Gaussian integral (see the proposition A.1 for the proof).

Definition 1.2. ( Weighted L2 space in R) Let I be an interval in R and let L2w(I) be the set of

functions g which are square integrable with respect to the weight function w, i.e. such that theintegral

‖g‖2 =

∫I

g(x)2w(x)dx (3)

is finite. In this case, the norm of g is ‖g‖.

Example (Functions in L2w(R)) Consider the Gaussian weight w(x) = exp(−x2/2), for x ∈ R.

Obviously, the function g(x) = 1 is in L2w(R).

Now consider the function

g(x) = x (4)

for x ∈ R. The proposition A.3 (see in appendix) proves that∫Rx2 exp(−x2/2)dx =

√2π. (5)

Therefore, g ∈ L2w(R).

4

Figure 1: The Gaussian weight function f(x) = exp(−x2/2).

Definition 1.3. ( Inner product in L2w(I) space) For any g, h ∈ L2

w(I), the inner product of gand h is

〈g, h〉 =

∫I

g(x)h(x)w(x)dx. (6)

Assume that g ∈ L2w(I). We can combine the equations 3 and 6, which implies that the L2

w(I)norm of g can be expressed as an inner product:

‖g‖2 = 〈g, g〉 . (7)

Example (Inner product in L2w(R)) Consider the Gaussian weight w(x) = exp(−x2/2), for x ∈ R.

Then consider the function g(x) = 1 and h(x) = x. We have∫Rg(x)h(x)w(x)dx =

∫Rx exp(−x2/2)dx (8)

=

(∫x≤0

x exp(−x2/2)dx+

∫x≥0

x exp(−x2/2)dx

). (9)

However, ∫x≤0

x exp(−x2/2)dx = −∫x≥0

x exp(−x2/2)dx, (10)

by symmetry. Therefore, ∫Rg(x)h(x)w(x)dx = 0, (11)

which implies 〈g, h〉 = 0.

5

The L2w(I) space is a Hilbert space.

Definition 1.4. ( Orthogonality in L2w(I) space) Let I be an interval of R. Let g and h be two

functions in L2w(I). The functions g and h are orthogonal if

〈g, h〉 = 0. (12)

Definition 1.5. ( Polynomials) The function Pn is a polynomial if there is a sequence of realnumbers αi ∈ R, for i = 1, 2, ..., n such that :

Pn(x) =n∑i=0

αixi, (13)

where x ∈ I. In this case, the degree of the polynomial is n.

Definition 1.6. ( Orthogonal polynomials) The set of polynomials {Pn}n≥0 are orthogonal poly-nomials if Pn is a polynomial of degree n and:

〈Pi, Pj〉 = 0 (14)

for i 6= j.

We assume that, for all orthogonal polynomials, the degree zero polynomial P0 is equal to one:

P0(x) = 1, (15)

for any x ∈ I.

Proposition 1.7. ( Integral of orthogonal polynomials) Let {Pn}n≥0 be orthogonal polynomials.We have ∫

I

P0(x)w(x)dx =

∫I

w(x)dx (16)

Moreover, for n ≥ 1, we have ∫I

Pn(x)w(x)dx = 0 (17)

Proof. The equation 16 is the straightforward consequence of 15. Moreover, for any n ≥ 1, wehave ∫

I

Pn(x)w(x)dx =

∫I

P0(x)Pn(x)w(x)dx (18)

= 〈P0(x), Pn(x)〉 (19)

= 0, (20)

by the orthogonality property.

The figure 2 presents the function Hen(x)w(w), for n = 0, 1, ..., 5, where w is the Gaussianweight w(x) = exp(−x2/2). The proposition 1.7 states that the only n for which the integral isnonzero is n = 0. Indeed, for n = 2, 4, i.e. for n even, we can see that the area of the segmentsabove zero is equal to the area of the segments below zero. This is more obvious for n = 1, 3, 5,i.e. for n oddd, because the function Hen(x) is antysymmetric, i.e. Hen(−x) = −Hen(x).

6

Figure 2: The function Hen(x)w(x), for n = 0, 1, ..., 5, and the Gaussian weight w(x) =exp(−x2/2).

1.2 Orthogonal polynomials for probabilities

In this section, we present the properties of Pn(X), when X is a random variable associated withthe orthogonal polynomials {Pn}n≥0.

Proposition 1.8. ( Distribution from a weight) Let I be an interval in R and let w be a weightfunction on I. The function:

f(x) =w(x)∫

Iw(x)dx

, (21)

for any x ∈ I, is a distribution function.

Proof. Indeed, its integral is equal to one:∫I

f(x)dx =1∫

Iw(x)dx

∫I

w(x)dx (22)

= 1, (23)

which concludes the proof.

Example (Distribution function for Hermite polynomials) The distribution function for Hermitepolynomials is

f(x) =1√2π

exp

(−x

2

2

), (24)

for x ∈ R. The function f(x) is presented in the figure 3.

7

Figure 3: The Gaussian distribution function f(x) = 1√2π

exp(−x2/2).

Proposition 1.9. ( Expectation of orthogonal polynomials) Let {Pn}n≥0 be orthogonal polyno-mials. Assume that X is a random variable associated with the probability distribution functionf , derived from the weight function w. We have

E(P0(X)) = 1 (25)

Moreover, for n ≥ 1, we have

E(Pn(X)) = 0 (26)

Proof. By definition of the expectation,

E(P0(X)) =

∫I

P0(x)f(x)dx (27)

=

∫I

f(x)dx (28)

= 1, (29)

by the equation 22. Moreover, for any n ≥ 1, we have

E(Pn(X)) =

∫I

Pn(x)f(x)dx (30)

=1∫

Ix(x)dx

∫I

Pn(x)w(x)dx (31)

= 0, (32)

where the first equation derives from the equation 21, and the second equality is implied by17.

8

Proposition 1.10. ( Variance of orthogonal polynomials) Let {Pn}n≥0 be orthogonal polynomials.Assume that x is a random variable associated with the probability distribution function f , derivedfrom the weight function w. We have

V (P0(X)) = 0. (33)

Moreover, for n ≥ 1, we have

V (Pn(X)) =‖Pn‖2∫Iw(x)dx

. (34)

Proof. The equation 33 is implied by the fact that P0 is a constant. Moreover, for n ≥ 1, we have:

V (Pn(X)) = E((Pn(X)− E(Pn(X)))2

)(35)

= E(Pn(X)2

)(36)

=

∫I

Pn(x)2f(x)dx (37)

=1∫

Iw(x)dx

∫I

Pn(x)2w(x)dx, (38)

where the first equality is implied by the equation 26.

Figure 4: Histogram of N = 10000 outcomes of Hen(X), where X is a standard normal randomvariable, for n = 0, 1, ..., 5.

Example In order to experimentally check the propositions 1.9 and 1.10, we consider Hermitepolynomials. We generate 10000 pseudo random outcomes of a standard normal variable X, and

9

compute Hen(X). The figure 4 presents the empirical histograms of Hen(X), for n = 0, 1, ..., 5.The histogram for n = 0 is centered onX = 1, sinceHe0(x) = 1. This confirms that E(He0(X)) =1 and V (He0(X)) = 0. The histograms for n = 1, 3, 5 are symmetric, which is consistent withthe fact that E(Hen(X)) = 0, for n ≥ 1. The following Scilab session presents more detailednumerical results. The first column prints n, the second column prints the empirical mean, thethird column prints the empirical variance and the last column prints n!, which is the exact valueof the variance.

n mean(X) variance(X) n!

0. 1. 0. 1.

1. 0.0041817 0.9978115 1.

2. - 0.0021810 2.0144023 2.

3. 0.0179225 6.1480027 6.

4. 0.0231110 24.483042 24.

5. - 0.0114358 112.13277 120.

Proposition 1.11. Let {Pn}n≥0 be orthogonal polynomials. Assume that x is a random variableassociated with the probability distribution function f , derived from the weight function w. Fortwo integers i, j ≥ 0, we have

E(Pi(X)Pj(X)) = 0 (39)

if i 6= j, and

E(Pi(X)2) = V (Pi(X)). (40)

Proof. We have

E(Pi(X)Pj(X)) =

∫I

Pi(x)Pj(x)f(x)dx (41)

=1∫

Iw(x)dx

∫I

Pi(x)Pj(x)w(x)dx (42)

=〈Pi, Pj〉∫Iw(x)dx

. (43)

If i 6= j, the orthogonality of the polynomials implies 39. If, on the other hand, we have i = j,then

E(Pi(X)2) =〈Pi, Pi〉∫Iw(x)dx

(44)

=‖Pi‖2∫Iw(x)dx

. (45)

We then use the equation 34, which leads to the equation 40.

In the next sections, we review a few orthogonal polynomials which are important in thecontext of polynomial chaos. The table 5 summarizes the results.

10

Distrib. Support Poly. w(x) f(x) ‖Pn‖2 V (Pn)

N (0, 1) R Hermite exp(−x2

2

)1√2π

exp(−x2

2

) √2πn! n!

U(−1, 1) [−1, 1] Legendre 1 12

22n+1

12n+1

E(1) R+ Laguerre exp(−x) exp(−x) 1 1

Figure 5: Some of properties of orthogonal polynomials

1.3 Hermite polynomials

In this section, we present Hermite polynomials and their properties.Hermite polynomials are associated with the Gaussian weight:

w(x) = exp(−x2/2), (46)

for x ∈ R. The integral of this weight is:∫Rw(x)dx =

√2π. (47)

The distribution function for Hermite polynomials is the standard Normal distribution:

f(x) =1√2π

exp

(−x

2

2

), (48)

for x ∈ R.Here, we consider the probabilist polynomials Hen, as opposed to the physicist polynomials

Hn [17].The first Hermite polynomials are

He0(x) = 1, (49)

He1(x) = x (50)

The remaining Hermite polynomials satisfy the recurrence:

Hen+1(x) = xHen(x)− nHen−1(x), (51)

for n = 1, 2, ....

Example (First Hermite polynomials) We have

He2(x) = xHe1(x)−He0(x) (52)

= x · x− 1 (53)

= x2 − 1. (54)

Similarily,

He3(x) = xHe2(x)− 2He1(x) (55)

= x(x2 − 1)− 2x (56)

= x3 − 3x. (57)

11

He0(x) = 1He1(x) = xHe2(x) = x2 − 1He3(x) = x3 − 3xHe4(x) = x4 − 6x2 + 3He5(x) = x5 − 10x3 + 15x

Figure 6: Hermite polynomials

Figure 7: The Hermite polynomials Hen, for n = 0, 1, 2, 3.

12

n c0 c1 c2 c3 c4 c5 c6 c7 c8 c90 11 12 -1 13 -3 14 3 -6 15 15 -10 16 -15 45 -15 17 -105 105 -21 18 105 -420 210 -28 19 945 -1260 378 -36 1

Figure 8: Coefficients of Hermite polynomials Hen, for n = 0, 1, 2, ..., 9

The first Hermite polynomials are presented in the figure 6.The figure 7 plots the Hermite polynomials Hen, for n = 0, 1, 2, 3.The figure 8 presents the coefficients ck, for k ≥ 0 of the Hermite polynomials, so that

Hen(x) =n∑k=0

ckxk. (58)

The Hermite polynomials are orthogonal with respect to the weight w(x). Moreover,

‖Hen‖2 =√

2πn!. (59)

Hence,

V (Hen(X)) = n!, (60)

for n ≥ 1, where X is a standard normal random variable.The following HermitePoly Scilab function creates the Hermite polynomial of order n.

function y=HermitePoly(n)

if (n==0) then

y=poly(1,"x","coeff")

elseif (n==1) then

y=poly ([0 1],"x","coeff")

else

polyx=poly ([0 1],"x","coeff")

// y(n-2)

yn2=poly(1,"x","coeff")

// y(n-1)

yn1=polyx

for k=2:n

y=polyx*yn1 -(k-1)* yn2

yn2=yn1

yn1=y

end

end

endfunction

13

The script:

for n=0:10

y=HermitePoly(x,n);

disp(y)

end

produces the following output:

1

x

2

- 1 + x

3

- 3x + x

2 4

3 - 6x + x

3 5

15x - 10x + x

7

1.4 Legendre polynomials

In this section, we present Legendre polynomials and their properties.Legendre polynomials are associated with the unity weight

w(x) = 1, (61)

for x ∈ [−1, 1]. We have ∫ 1

−1w(x)dx = 2. (62)

The associated probability distribution function is:

f(x) =1

2, (63)

for x ∈ [−1, 1]. This corresponds to the uniform distribution in probability theory.The first Legendre polynomials are

P0(x) = 1, (64)

P1(x) = x (65)

The remaining Legendre polynomials satisfy the recurrence:

(n+ 1)Pn+1(x) = (2n+ 1)xPn(x)− nPn−1(x), (66)

for n = 1, 2, ....

14

P0(x) = 1P1(x) = xP2(x) = 1

2(3x2 − 1)

P3(x) = 12(5x3 − 3x)

P4(x) = 18(35x4 − 30x2 + 3)

P5(x) = 18(63x5 − 70x3 + 15x)

Figure 9: Legendre polynomials

Example (First Legendre polynomials) We have

2P2(x) = 3xP1(x)− P0(x) (67)

= 3x2 − 1, (68)

which implies

P2(x) =1

2(3x2 − 1). (69)

Similarily,

3P3(x) = 5xP2(x)− 2P1(x) (70)

= 5x1

2(3x2 − 1)− 2x (71)

=15

2x3 − 5

2x− 2x (72)

=15

2x3 − 9

2x (73)

=1

2(15x3 − 9x). (74)

Finally, we divide both sides of the equation by 3 and get

P3(x) =1

2(5x3 − 3x). (75)

The first Legendre polynomials are presented in the figure 9.The Legendre polynomials are orthogonal with respect to w(x) = 1. Moreover,

‖Pn‖2 =2

2n+ 1(76)

Furthermore,

V (Pn(X)) =1

2n+ 1, (77)

for n ≥ 1, where X is a random variable uniform in the interval [−1, 1].The figure 10 plots the Legendre polynomials Pn, for n = 0, 1, 2, 3.The following LegendrePoly Scilab function creates the Legendre polynomial of order n.

15

Figure 10: The Legendre polynomials Pn, for n = 0, 1, 2, 3.

function y=LegendrePoly(n)

if (n==0) then

y=poly(1,"x","coeff")

elseif (n==1) then

y=poly ([0 1],"x","coeff")

else

polyx=poly ([0 1],"x","coeff")

// y(n-2)

yn2=poly(1,"x","coeff")

// y(n-1)

yn1=polyx

for k=2:n

y=((2*k-1)* polyx*yn1 -(k-1)* yn2)/k

yn2=yn1

yn1=y

end

end

endfunction

1.5 Laguerre polynomials

In this section, we present Laguerre polynomials and their properties.Laguerre polynomials are associated with the weight

w(x) = exp(−x), (78)

16

for x ≥ 0.The figure 11 plots the exponential weight function.

Figure 11: The Exponential weight.

We have: ∫ +∞

0

w(x)dx = 1. (79)

Hence, the associated probability distribution function is:

f(x) = exp(−x), (80)

for x ≥ 0. This is the standard exponential distribution in probability theory.The first Laguerre polynomials are

L0(x) = 1, (81)

L1(x) = −x+ 1 (82)

The remaining Laguerre polynomials satisfy the recurrence:

(n+ 1)Ln+1(x) = (2n+ 1− x)Ln(x)− nLn−1(x), (83)

for n = 1, 2, ....The first Laguerre polynomials are presented in the figure 12.

Example (First Laguerre polynomials) We have

2L2(x) = (3− x)L1(x)− L0(x) (84)

= (3− x)(1− x)− 1 (85)

= 3− x− 3x+ x2 − 1 (86)

= x2 − 4x+ 2. (87)

17

L0(x) = 1L1(x) = −x+ 1L2(x) = 1

2!(x2 − 4x+ 2)

L3(x) = 13!

(−x3 + 9x2 − 18x+ 6)L4(x) = 1

4!(x4 − 16x3 + 72x2 − 96x+ 24)

L5(x) = 15!

(−x5 + 25x4 − 200x3 + 600x2 − 600x+ 120)

Figure 12: Laguerre polynomials

Hence,

L2(x) =1

2(x2 − 4x+ 2). (88)

Similarily,

3L3(x) = (5− x)L2(x)− 2L1(x) (89)

= (5− x)1

2(x2 − 4x+ 2)− 2(1− x). (90)

Therefore,

6L3(x) = (5− x)(x2 − 4x+ 2)− 4(1− x) (91)

= 5x2 − 20x+ 10− x3 + 4x2 − 2x− 4 + 4x (92)

= −x3 + 9x2 − 18x+ 6, (93)

which implies

L3(x) =1

6(−x3 + 9x2 − 18x+ 6). (94)

The figure 13 plots the Laguerre polynomials Ln, for n = 0, 1, 2, 3.The Laguerre polynomials are orthogonal with respect to w(x) = exp(−x). Moreover,

‖Ln‖2 = 1. (95)

Furthermore,

V (Ln(X)) = 1, (96)

for n ≥ 1, where X is a standard exponential random variable.The following LaguerrePoly Scilab function creates the Laguerre polynomial of order n.

function y=LaguerrePoly(n)

if (n==0) then

y=poly(1,"x","coeff")

elseif (n==1) then

y=poly ([1 -1],"x","coeff")

else

polyx=poly ([0 1],"x","coeff")

18

Figure 13: The Laguerre polynomials Ln, for n = 0, 1, 2, 3.

// y(n-2)

yn2=poly(1,"x","coeff")

// y(n-1)

yn1=1-polyx

for k=2:n

y=((2*k-1-polyx)*yn1 -(k-1)* yn2)/k

yn2=yn1

yn1=y

end

end

endfunction

1.6 Chebyshev polynomials of the first kind

In this section, we present Chebyshev polynomials of the first kind and their properties.It is important to mention that Chebyshev polynomials of the second kind are not presented

in this document.Chebyshev polynomials are associated with the weight

w(x) =1√

1− x2, (97)

for x ∈ [−1, 1]. This weight does not correspond to a particular distribution in probability theory.The first Chebyshev polynomials are

T0(x) = 1, (98)

T1(x) = x (99)

19

T0(x) = 1T1(x) = xT2(x) = 2x2 − 1T3(x) = 4x3 − 3xT4(x) = 8x4 − 8x2 + 1T5(x) = 16x5 − 20x3 + 5x

Figure 14: Chebyshev polynomials of the first kind

The remaining Chebyshev polynomials satisfy the recurrence:

Tn+1(x) = 2xTn(x)− Tn−1(x), (100)

for n = 1, 2, ....The first Chebyshev polynomials are presented in the figure 14.

Example (First Chebyshev polynomials) We have

T2(x) = 2xT1(x)− T0(x) (101)

= 2x2 − 1. (102)

Similarily,

T3(x) = 2xT2(x)− T1(x) (103)

= 2x(2x2 − 1)− x (104)

= 4x3 − 2x− x (105)

= 4x3 − 3x. (106)

The figure 15 plots the Chebyshev polynomials Tn, for n = 0, 1, 2, 3.The following ChebyshevPoly Scilab function creates the Chebyshev polynomial of order n.

function y=ChebyshevPoly(n)

if (n==0) then

y=poly(1,"x","coeff")

elseif (n==1) then

y=poly ([0 1],"x","coeff")

else

polyx=poly ([0 1],"x","coeff")

// y(n-2)

yn2=poly(1,"x","coeff")

// y(n-1)

yn1=polyx

for k=2:n

y=2* polyx*yn1 -yn2

yn2=yn1

yn1=y

end

end

endfunction

20

Figure 15: The Chebyshev polynomials Tn, for n = 0, 1, 2, 3.

1.7 Accuracy of evaluation

In this section, we present the accuracy issues which appear when we evaluate the orthogonalpolynomials with large values of the polynomial degree n. More precisely, we present a funda-mental limitation in the accuracy that can be achieved by the polynomial representation of theorthogonal polynomials, while a sufficient accuracy can be achieved with a different evaluationalgorithm.

The following LaguerreEval function evaluates the Laguerre polynomial of degree n at pointx. One one hand, the LaguerrePoly function creates a polynomial which can be evaluated withthe horner function, while, on the other hand, the function LaguerreEval directly evaluates thepolynomial. As we are going to see soon, although the recurrence is the same, the numericalperformances of these two functions is rather different.

function y=LaguerreEval(x,n)

if (n==0) then

nrows=size(x,"r")

y=ones(nrows ,1)

elseif (n==1) then

y=1-x

else

// y(n-2)

yn2=1

// y(n-1)

yn1=1-x

for k=2:n

y=((2*k-1-x).*yn1 -(k-1)* yn2)/k

21

yn2=yn1

yn1=y

end

end

endfunction

Consider the value of the polynomial L100 at the point x = 10. From Wolfram Alpha, whichuses an arbitrary precision computer system, the exact value of L100(10) is

13.277662844303454137789892644070318985922997400185621...

In the following session, we compare the accuracies of LaguerrePoly and LaguerreEval.

-->n=100;

-->x=10;

-->L=LaguerrePoly(n);

-->y1=horner(L,x)

y1 =

3.695D+08

-->y2=LaguerreEval(x,n)

y2 =

13.277663

We see that the polynomial created by LaguerrePoly does not produce a sufficient accuracy,while LaguerreEval seems to be accurate. In fact, the value produced by LaguerreEval hasmore than 15 correct decimal digits.

In order to see the trend when x and n increases, we compute the number of common digits,in base 2, in the evaluation of Ln(x), computed from LaguerrePoly and LaguerreEval. SinceScilab uses double precision floating point numbers with 64 bits and 53 bits of precision, thisnumber varies from 53 (when the two functions return exactly the same value) to 0 (when thetwo functions return two value which have no digit in common). The figure 16 plots the numberof common digits, when n ranges from 1 to 100 and x ranges from 1 to 21.

We see that the number of correct digits produced by LaguerrePoly decreases when x and nincrease.

The explanation for this fact is that the polynomial representation of the Laguerre polynomialforces to use n = 100 additions of positive and negative powers of x. This computation is ill-conditionned, because it involves large numbers with different signs.

For example, the following session presents some of the coefficients of L100.

-->L

L =

2 3 4 5 6

1 - 100x + 2475x - 26950x + 163384.37x - 627396x + 1655628.3x

7 8 9 10

- 3176103.3x + 4615275.2x - 5242040.9x + 4770257.2x

[...]

95 96 97 98

- 7.28D-141x + 3.95D-144x - 1.68D-147x + 5.25D-151x

99 100

- 1.07D-154x + 1.07D-158x

22

Figure 16: The number of common digits in the evaluation of Laguerre polynomials, for increasingn and x.

We see that the coefficients signs alternate.Consider n real numbers yi, for i = 1, 2, ..., n, representing the monomial values which appear

in the polynomial representation. We are interested in the sum

S = y1 + y2 + ...+ yn.

We know from numerical analysis [8] that the condition number of a sum is

|y1|+ |y2|+ ...|yn||y1 + y2 + ...yn|

.

Hence, when the sum S is small in magnitude, but involves large terms yi, the condition numberis large. In other words, small relative errors in yi are converted into large relative errors in S,and this is what happens for polynomials.

Hence, the polynomial representation should be avoided in practice to evaluate the orthogonalpolynomial, although they are a convenient way of using them in Scilab. On the other hand, astraightforward evaluation of the polynomial value based on the recurrence formula can producea good accuracy.

1.8 Notes and references

The classic [1] gives a brief overview of the main results on orthogonal polynomials in the chap-ter 22 ”Orthogonal polynomials”. More details on orthogonal polynomials are presented in [3],especially the chapter 12 ”Legendre functions” and chapter 13 ”More special functions”.

The book [2] has three chapters on orthogonal polynomials, including quadrature, and otheradvanced topics.

One of the references on the topic is [15], which covers the topic in depth.

23

Cell 1 Cell 2 Cell 31 | *** | - | - |2 | - | *** | - |3 | - | - | *** |4 | ** | * | - |5 | ** | - | * |6 | * | ** | - |7 | - | ** | * |8 | * | - | ** |9 | - | * | ** |10 | * | * | * |

Figure 17: Placing d = 3 balls into p = 3 cells. The balls are represented by stars ∗.

In the appendices of [11], we can find a presentation of the most common orthogonal polyno-mials used in uncertainty analysis.

2 Multivariate polynomials

In this section, we define the multivariate polynomials which are used in the context of polynomialchaos decomposition. We consider the regular polynomials as well as multivariate orthogonalpolynomials.

2.1 Occupancy problems

In the next section, we will count the number of monomials of degree d with p variables. Thisproblems is the same as placing d balls into p cells, which is the topic of the current section.

Consider the problem of placing d = 3 balls into p = 3 cells. The figure 17 presents the 10ways to place the balls.

Let us denote by αi ≥ 0 the number of balls in the i-th cell, for i = 1, 2, ..., p. Let α =(α1, α2, ..., αp) be the associated vector. Each particular configuration is so that the total numberof balls is d, i.e. we have

α1 + α2 + ...+ αp = d. (107)

Example (Case where d = 8 and p = 6) Consider the case where there are d = 8 balls todistribute into p = 6 cells. We represent the balls with stars ”*” and the cells are represented bythe spaces between the p + 1 bars ”|”. For example, the string ”|***|*| | | |****|” represents theconfiguration where α = (3, 1, 0, 0, 0, 4).

Proposition 2.1. ( Occupancy problem) The number of ways to distribute d balls into p cells is:((pd

))=

(p+ d− 1

d

). (108)

The left hand side of the previous equation is the multiset coefficient.

24

Proof. Within the string, there is a total of p+1 bars : the first and last bars, and the intermediatep−1 bars. We are free to move only the p−1 bars and the d stars, for a total of p+d−1 symbols.Therefore, the problem reduces to finding the place of the d stars in a string of p+ d− 1 symbols.From combinatorics, we know that the number of ways to choose k items in a set of n items is(

nk

)=

n!

k!(n− k)!, (109)

for any nonnegative integers n and k, and k ≤ n. Hence, the number of ways to choose the placeof the d stars in a string of p+ d− 1 symbols is defined in the equation 108.

2.2 Multivariate monomials

The four one-dimensional monomials 1, x, x2 can be used to create a multivariate monomial, justby multiplying the appropriate components. Consider, for example, p = 3 and let x ∈ R3 be apoint in the three-dimensional space. The function

x2x23 (110)

is the monomial associated to the vector α = (0, 1, 2), which entries are the exponents of x =(x1, x2, x3).

Definition 2.2. ( Multivariate monomial) Let x ∈ Rp. The function Mα(x) is a multivariatemonomial if there is a vector α = (α1, α2, ..., αp) of integers exponents, where αi ∈ {0, 1, ..., p},for i = 1, 2, ..., p such that

Mα(x) = xα11 x

α22 ...x

αpp , (111)

for any x ∈ Rp.

The vector α is called a multi-index.The previous equation can be rewritten in a more general way:

Mα(x) =

p∏i=1

xαii , (112)

for any x ∈ Rp.

Definition 2.3. ( Degree of a multivariate monomial) Let M(x) be a monomial associated withthe exponents α, for any x ∈ Rp. The degree of M is

d = α1 + α2 + ...+ αp. (113)

The degree d of a multivariate monomial is also denoted by |α|.The figure 18 presents the 10 monomials with degree d = 2 and p = 4 variables.

Proposition 2.4. ( Number of multivariate monomials) The number of degree d multivariatemonomials of p variables is: ((

pd

))=

(p+ d− 1

d

). (114)

Proof. The problem can be reduced to distributing d balls into p cells: the i-th cell representsthe variable xi where we have to put αi balls. It is then straightforward to use the proposition2.1.

The figure 19 presents the number of monomials for various d and p.

25

α1 α2 α3 α4

2 0 0 01 1 0 01 0 1 01 0 0 10 2 0 00 1 1 00 1 0 10 0 2 00 0 1 10 0 0 2

Figure 18: The monomials with degree d = 2 and p = 4 variables.

p\d 0 1 2 3 4 51 1 1 1 1 1 12 1 2 3 4 5 63 1 3 6 10 15 214 1 4 10 20 35 565 1 5 15 35 70 126

Figure 19: Number of degree d monomials with p variables.

2.3 Multivariate polynomials

Definition 2.5. ( Multivariate polynomial) Let x ∈ Rp. The function P (x) is a degree d multi-variate polynomial if there is a set of multivariate monomials Mα(x) and a set of real numbersβα such that

P (x) =∑|α|≤d

βαMα(x), (115)

for any x ∈ Rp.

We can plug the equation 112 into 115 and get

P (x) =∑|α|≤d

βα

p∏i=1

xαii , (116)

for any x ∈ Rp.

Example (Case where d = 2 and p = 3) Consider the degree d = 2 multivariate polynomial withp = 3 variables:

P (x1, x2, x3) = 4 + 2x3 + 5x21 − 3x1x2x3 (117)

for any x ∈ R3. The relevant multivariate monomials exponents α are:

(0, 0, 0), (0, 0, 1), (118)

(1, 0, 0), (1, 1, 1). (119)

26

The only nonzero coefficients β are:

β(0,0,0) = 4, β(0,0,1) = 2, (120)

β(1,0,0) = 5, β(1,1,1) = −3. (121)

Proposition 2.6. ( Number of multivariate polynomials) The number of degree d multivariatepolynomials of p variables is:

P pd =

(p+ dd

). (122)

Proof. The proof is by recurrence on d. The only polynomial of degree d = 0 is:

P0(x) = 1. (123)

Hence, there are:

1 =

(p0

)(124)

multivariate polynomials of degree d = 0. This proves that the equation 122 is true for d = 0.The polynomials of degree d = 1 are:

Pi(x) = xi, (125)

for i = 1, 2, ..., p. Hence, there is a total of 1 + p polynomials of degree d = 1, which writes

1 + p =

(p+ 1

1

). (126)

This proves that the equation 122 is true for d = 1.Now, assume that the equation 122 is true for d, and let us prove that it is true for d+ 1. For

a multivariate polynomial P (x) of degree d+ 1, the equation 115 implies

P (x) =∑|α|≤d+1

βαMα(x), (127)

=∑|α|≤d

βαMα(x) +∑|α|=d+1

βαMα(x), (128)

for any x ∈ Rp. Hence, the number of multivariate polynomials of degree d + 1 is the sum ofthe number of multivariate polynomials of degree d and the number of multivariate monomialsof degree d+ 1. From the equation 114, the number of such monomials is((

pd+ 1

))=

(p+ dd+ 1

). (129)

Hence, the number of multivative polynomials of degree d+ 1 is:(p+ dd+ 1

)+

(p+ dd

). (130)

27

p\d 0 1 2 3 4 5 61 1 2 3 4 5 6 72 1 3 6 10 15 21 283 1 4 10 20 35 56 844 1 5 15 35 70 126 2105 1 6 21 56 126 252 4626 1 7 28 84 210 462 924

Figure 20: Number of degree d polynomials with p variables.

However, we know from combinatorics that(nk

)+

(n

k − 1

)=

(n+ 1k

), (131)

for any nonnegative integers n and k. We apply the previous equality with n = p+d and k = d+1,and get (

p+ dd+ 1

)+

(p+ dd

)=

(p+ d+ 1d+ 1

), (132)

which proves that the equation 122 is also true for d+ 1, and concludes the proof.

The figures 20 and 21 present the number of polynomials for various d and p.One possible issue with the equation 115 is that it does not specify a way of ordering the P p

d

multivariate polynomials of degree d. However, we will present in the next section a constructiveway of ordering the multi-indices is such a way that there is a one-to-one mapping from the singleindex k, in the range from 1 to P p

d , to the corresponding multi-index α(k). With this ordering,the equation 115 becomes

P (x) =

P pd∑

k=1

βkMα(k)(x), (133)

for any x ∈ Rp.

Example (Case where d = 2 and p = 2) Consider the degree d = 2 multivariate polynomial withp = 2 variables:

P (x1, x2) = β(0,0) + β(1,0)x1 + β(0,1)x2 + β(2,0)x21 + β(1,1)x1x2 + β(0,2)x

22 (134)

for any x1, x2 ∈ R2. This corresponds to the equation

P (x1, x2) = β1 + β2x1 + β3x2 + β4x21 + β5x1x2 + β6x

22 (135)

for any x1, x2 ∈ R2.

28

Figure 21: Number of degree d polynomials with p variables.

2.4 Generating multi-indices

In this section, we present an algorithm to generate all the multi-indices α associated with adegree d multivariate polynomial of p variables.

The following polymultiindex function implements the algorithm suggested by Lemaıtre andKnio in the appendix of [11]. The function returns a matrix a with p columns and P p

d rows. Eachrow a(i,:) represents the i-th multivariate polynomial, for i = 1, 2, ..., P p

d . For each row i, theentry a(i,j) is the exponent of the j-th variable xj, for i = 1, 2, ..., P p

d and j = 1, 2, ..., p. Thesum of exponents for each row is lower or equal to d. The first row is the zero-degree polynomial,so that all entries in a(1,1:p) are zero. The rows 2 to p+1 are the first degree polynomials: allcolumns are zero, except one entry which is equal to 1.

function a=polymultiindex(p,d)

// Zero -th order polynomial

a(1,1:p)=zeros(1,p)

// First order polynomials

a(2:p+1,1:p)=eye(p,p)

P=p+1

pmat =[]

pmat (1:p,1)=1

for k=2:d

L=P

for i=1:p

pmat(i,k)=sum(pmat(i:p,k-1))

29

end

for j=1:p

for m=L-pmat(j,k)+1:L

P=P+1

a(P,1:p)=a(m,1:p)

a(P,j)=a(P,j)+1

end

end

end

endfunction

For example, in the following session, we compute the list of multi-indices corresponding tomultivariate polynomials of degree d = 3 with p = 3 variables.

-->p=3;

-->d=3;

-->a=polymultiindex(p,d)

a =

0. 0. 0.

1. 0. 0.

0. 1. 0.

0. 0. 1.

2. 0. 0.

1. 1. 0.

1. 0. 1.

0. 2. 0.

0. 1. 1.

0. 0. 2.

3. 0. 0.

2. 1. 0.

2. 0. 1.

1. 2. 0.

1. 1. 1.

1. 0. 2.

0. 3. 0.

0. 2. 1.

0. 1. 2.

0. 0. 3.

There are 20 rows in the previous matrix a, which is consistent with the table 20. In theprevious session, the first row correspond to the zero-th degree polynomial, the rows 2 to 4corresponds to the degree one monomials, the rows from 5 to 10 corresponds to the degree 2monomials, whereas the rows from 11 to 20 corresponds to the degree 3 monomials.

2.5 Multivariate orthogonal functions

In this section, we introduce the weighted space of square integrable functions in Rp, so that wecan define the orthogonality of multivariate functions.

Definition 2.7. ( Interval of Rp) An interval I of Rp is a subspace of Rp, such that x ∈ I if

x1 ∈ I1, x2 ∈ I2, ..., xp ∈ Ip, (136)

30

where I1, I2,..., Ip are intervals of R. We denote this tensor product by:

I = I1 ⊗ I2 ⊗ ...⊗ Ip. (137)

Now that we have multivariate intervals, we can consider a multivariate weight function onsuch an interval.

Definition 2.8. ( Multivariate weight function of Rp) Let I be an interval in Rp. A weightfunction w on I is a nonnegative integrable function of x ∈ I.

Such a weight can be created by making the tensor product of univariate weights.

Proposition 2.9. ( Tensor product of univariate weights) Let I1, I2, ..., Ip be intervals of R. As-sume that w1, w2, ..., wp are univariate weights on I1, I2, ..., Ip. Therefore, the tensor product func-tion

w(x) = w1(x1)w2(x2)...wp(xp), (138)

for any x ∈ R, is a weight function of x ∈ I.

Proof. Indeed, its integral is∫I

w(x)dx =

∫I

w1(x1)w2(x2)...wp(xp)dx1dx2...dxp (139)

=

(∫I1

w1(x1)dx1

)(∫I2

w2(x2)dx2

)...

(∫Ip

wp(xp)dxp

). (140)

However, all the individual integrals are finite, since each function wi is, by hypothesis, a weighton Ii, for i = 1, 2, ..., p. Hence, the product is finite, so that the function w is, indeed, a weighton I.

Example (Multivariate Gaussian weight) In the particular case where all weights are the same,the expression simplifies further. Consider, for example, the Gaussian weight

exp(−x2/2), (141)

for x ∈ R. We can then consider the tensor product:

w(x) = w(x1)w(x2)...w(xp), (142)

for x1, x2, ..., xp ∈ R.

Example (Multivariate Hermite-Legendre weight) Consider the weight for Hermite polynomialsw1(x1) = exp(−x21/2) for x1 ∈ R, and consider the weight for Legendre polynomials w2(x2) = 1for x2 ∈ [−1, 1]. Therefore,

w(x) = w1(x1)w2(x2) (143)

for x ∈ I is a weight on I = R⊗ [−1, 1].

31

In the remaining of this document, we will not make a notation difference between the uni-variate weight function w(x) and the multivariate weight w(x), assuming that it is based on atensor product if needed.

Definition 2.10. ( Multivariate weighted L2 space in Rp) Let I be an interval in Rp. Let w be amultivariate weight function on I. Let L2

w(I) be the set of functions g which are square integrablewith respect to the weight function w, i.e. such that the integral

‖g‖2 =

∫I

g(x)2w(x)dx (144)

is finite. In this case, the norm of g is ‖g‖.

Definition 2.11. ( Multivariate inner product in L2w(I) space) Let I be an interval of Rp. Let w

be a multivariate weight function on I. For any g, h ∈ L2w(I), the inner product of g and h is

〈g, h〉 =

∫I

g(x)h(x)w(x)dx. (145)

Let I be an interval of Rp and assume that g ∈ L2w(I). We can combine the equations 144

and 145, which implies that the L2w(I) norm of g can be expressed as an inner product:

‖g‖2 = 〈g, g〉 . (146)

Proposition 2.12. ( Multivariate probability distribution function) Let I be an interval of Rp.Let w be a multivariate weight function on I. Assume {Xi}i=1,2,...,p are independent randomvariables associated with the probability distribution functions fi, derived from the weight functionswi, for i = 1, 2, ..., p. Therefore, the function

f(x) = f1(x1)f2(x2)...fp(xp), (147)

for x ∈ I is a probability distribution function.

Proof. We must prove that the integral of f is equal to one. Indeed,∫I

f(x)dx =

∫I

f1(x1)f2(x2)...fp(xp)dx1dx2...dxp (148)

=

(∫I1

f1(x1)dx1

)(∫I2

f2(x2)dx2

)...

(∫Ip

fp(xp)dxp

)(149)

= 1, (150)

where each integral is equal to one since, by hypothesis, fi is a probability distribution functionfor i = 1, 2, ..., p.

2.6 Tensor product of orthogonal polynomials

In this section, we present a method to create multivariate orthogonal polynomials.Consider the cas where p = 2 and let’s try to create a bivariate Hermite orthogonal polyno-

mials. Therefore, we consider the inner product:

〈g, h〉 =

∫R2

g(x1, x2)h(x1, x2)w(x1)w(x2)dx1dx2, (151)

32

associated with the multivariate Gaussian weight w(x).We can make the tensor product of some Hermite polynomials, which leads, for example, to:

Ψ1(x1, x2) = He1(x1) (152)

Ψ2(x1, x2) = He1(x1)He2(x2), (153)

associated with the multivariate Gaussian weight w(x). These polynomials are of degree 2. Wemay wonder if these two polynomials are orthogonal. We have

〈Ψ1,Ψ2〉 (154)

=

∫R2

Ψ1(x1, x2)Ψ2(x1, x2)w(x1)w(x2)dx1dx2 (155)

=

∫R2

He1(x1)2He2(x2)w(x1)w(x2)dx1dx2 (156)

=

(∫RHe1(x1)

2w(x1)dx1

)(∫RHe2(x2)w(x2)dx2

). (157)

By the orthogonality of Hermite polynomials, we have∫RHe2(x2)w(x2)dx2 =

∫RHe0(x2)He2(x2)w(x2)dx2 (158)

= 0. (159)

Hence,

〈Ψ1,Ψ2〉 = 0, (160)

which implies that Ψ1 and Ψ2 are orthogonal. However,

‖Ψ1‖2 =

∫R2

He1(x1)2w(x1)w(x2)dx1dx2 (161)

=

(∫RHe1(x1)

2w(x1)dx1

)(∫Rw(x2)dx2

)(162)

=√

2π‖He1‖2. (163)

Definition 2.13. ( Tensor product of orthogonal polynomials) Let I be a tensor product intervalof Rp and let w the associated multivariate tensor product weight function. Let φ

α(k)i

(xi) be a family

of univariate orthogonal polynomials. The associated multivariate tensor product polynomials are

Ψk(x) =

p∏i=1

φα(k)i

(xi) (164)

for k = 1, 2, ..., P pd , where the degree of Ψk is

d =

p∑i=1

α(k)i . (165)

33

d Multi-Index Polynomial

0 α(1) = [0, 0] Ψ1(x) = He0(x1)He0(x2) = 11 α(2) = [1, 0] Ψ2(x) = He1(x1)He0(x2) = x11 α(3) = [0, 1] Ψ3(x) = He0(x1)He1(x2) = x22 α(4) = [2, 0] Ψ4(x) = He2(x1)He0(x2) = x21 − 12 α(5) = [1, 1] Ψ5(x) = He1(x1)He1(x2) = x1x22 α(6) = [0, 2] Ψ6(x) = He0(x1)He2(x2) = x22 − 13 α(7) = [3, 0] Ψ7(x) = He3(x1)He0(x2) = x31 − 3x13 α(8) = [2, 1] Ψ8(x) = He2(x1)He1(x2) = (x21 − 1)x23 α(9) = [1, 2] Ψ9(x) = He1(x1)He2(x2) = x1(x

22 − 1)

3 α(10) = [0, 3] Ψ10(x) = He0(x1)He3(x2) = x32 − 3x2

Figure 22: Multivariate Hermite polynomials of p = 2 variables and degree d = 3.

Figure 23: The multivariate Hermite polynomials, with degree d = 2 and p = 2 variables.

34

Example (Multivariate Hermite polynomials) The figure 22 presents the multivariate Hermitepolynomials of p = 2 variables and degree d = 3. The figure 23 presents the multivariate Hermitepolynomials, with degree d = 2 and p = 2 variables.

Proposition 2.14. ( Multivariate orthogonal polynomials) The multivariate tensor product poly-nomials defined in 2.13 are orthogonal.

Proof. We must prove that, for two different integers k and `, we have

〈Ψk,Ψ`〉 = 0. (166)

By the definition of the inner product on L2w(I), we have

〈Ψk,Ψ`〉 =

∫I

Ψk(x)Ψ`(x)w(x)dx (167)

By assumption, the multivariate weight function w on I is the tensor product of univariate weightfunctions wi on Ii. Hence,

〈Ψk,Ψ`〉 =

∫I

p∏i=1

φα(k)i

(xi)

p∏i=1

φα(`)i

(xi)

p∏i=1

wi(xi)dx (168)

=

p∏i=1

∫Ii

φα(k)i

(xi)φα(`)i

(xi)wi(xi)dxi. (169)

In other words,

〈Ψk,Ψ`〉 =

p∏i=1

⟨φα(k)i, φ

α(`)i

⟩. (170)

However, the multi-indice ordering implies that, if k 6= `, then there exists an integer i ∈{1, 2, ..., p}, such that

α(k)i 6= α

(`)i . (171)

By assumption, the φα(k)i

(xi) polynomials are orthogonal. This implies⟨φα(k)i, φ

α(`)i

⟩= 0, (172)

which concludes the proof.

We can use the equation 170 to compute the L2w(I) norm of Ψk. Indeed,

‖Ψk‖2 = 〈Ψk,Ψk〉 (173)

=

p∏i=1

⟨φα(k)i, φ

α(k)i

⟩(174)

=

p∏i=1

‖φα(k)i‖2. (175)

This proves the following proposition.

35

Proposition 2.15. ( Norm of multivariate orthogonal polynomials) The L2w(I) norm of the mul-

tivariate orthogonal polynomials defined in 2.13 is:

‖Ψk‖2 =

p∏i=1

‖φα(k)i‖2. (176)

Example (Norm of the multivariate Hermite polynomials) Consider the multivariate Hermitepolynomials in the case where p = 2. The figure 22 indicates that

Ψ8(x) = He2(x1)He1(x2), (177)

for x1, x2 ∈ R. Hence, the equation 176 implies:

‖Ψ8‖2 = ‖He2‖2‖He1‖2. (178)

The norm of the univariate Hermite polynomial is given by the equation 59. Hence,

‖Ψ8‖2 =√

2π · 2! ·√

2π · 1! (179)

= 4π. (180)

For any p ≥ 1, we have

‖Ψk‖2 =

p∏i=1

√2πα

(k)i ! (181)

= (2π)2/pp∏i=1

α(k)i !. (182)

2.7 Multivariate orthogonal polynomials and probabilities

In this section, we present the properties of Ψk(X), where X is a multivariate random variableassociated with multivariate orthogonal polynomials {Ψk}k≥1.

Proposition 2.16. ( Expectation of multivariate orthogonal polynomials) Let I be a tensor prod-uct interval of Rp and let w the associated multivariate tensor product weight function. Let Ψk

be the family of tensor product orthogonal multivariate polynomials, associated with the weight w.Let X be a multivariate random variable associated with the f multivariate probability distributionfunction, where {Xi}i=1,2,...,p are independent random variables. Therefore,

E(Ψ1(X)) = 1 (183)

and

E(Ψk(X)) = 0, (184)

for k > 1.

36

Proof. Indeed,

E(Ψ1(X)) =

∫I

Ψ1(x)f(x)dx (185)

=

∫I

f(x)dx (186)

= 1, (187)

since Ψ1(x) = 1, for any x ∈ I. Moreover, for any k > 1, we have

E(Ψk(X)) =

∫I

Ψk(x)f(x)dx (188)

=

∫I

p∏i=1

φα(k)i

(xi)

p∏i=1

fi(xi)dx1dx2...dxp (189)

=

p∏i=1

∫Ii

φα(k)i

(xi)fi(xi)dxi (190)

=

p∏i=1

E(φα(k)i

(xi)). (191)

However, the proposition 1.9 implies:

E(φα(k)i

(X))

= 1 (192)

if α(k)i = 0 and

E(φα(k)i

(X))

= 0 (193)

if α(k)i ≥ 1.

Since k > 1, there is at least one integer i ∈ {1, 2, ..., p} such that α(k)i ≥ 1. If this was not

true, this would imply that α(k)i = 0, for i = 1, 2, ..., p. This implies that k = 1, which contradicts

the hypothesis.Therefore, there is at least one integer i such that the expectation is zero, so that the product

in the equation 191 is also zero.

Proposition 2.17. ( Variance of multivariate orthogonal polynomials) Under the same hypothesesas in proposition 2.16, we have

V (Ψ1(X)) = 0 (194)

and

V (Ψk(X)) =

p∏i=1

E(φα(k)i

(Xi)2), (195)

for k > 1.

37

Proof. Obviously, the random variable Ψ1(X) = 1 has a zero variance. Furthermore, for k > 1,we have

V (Ψk(X)) = E((Ψk(X)− E(Ψk(X)))2

)(196)

= E(Ψk(X)2

), (197)

since E(Ψk(X)) = 0. We then use the tensor product definition of both Ψk and f , and finally getto the equation 195.

For k > 1, we have

E(φα(k)i

(Xi)2)

= V(φα(k)i

(Xi))

+ E(φα(k)i

(Xi))2. (198)

Therefore, based on the propositions 1.9 and 1.10, we can use the equation 195 to evaluate thevariance of Ψk.

Proposition 2.18. ( Covariance of multivariate orthogonal polynomials) Under the same hy-potheses as in proposition 2.16, for any two integers k and `, we have

Cov(Ψk(X),Ψ`(X)) = 0, (199)

if k 6= `.

Proof. Assume that k > 1. Therefore,

Cov(Ψ1(X),Ψk(X)) = E ((Ψ1(X)− µ1) (Ψk(X)− µk)) , (200)

where µ1 = E(Ψ1(X)) and µk = E(Ψk(X)). However, we have Ψ1(X) = E(Ψ1(X)) = 1. Hence,

Cov(Ψ1(X),Ψk(X)) = 0. (201)

The same is true if we consider

Cov(Ψk(X),Ψ1(X)) = Cov(Ψ1(X),Ψk(X)) (202)

= 0, (203)

by the symmetry property of the covariance.Assume now that k and ` are two integers such that k, ` > 1. We have,

Cov(Ψk(X),Ψ`(X)) = E ((Ψk(X)− µk) (Ψ`(X)− µ`)) (204)

= E (Ψk(X)Ψ`(X)) , (205)

since, by the proposition 2.16, we have µk = µ` = 0. Hence,

Cov(Ψk(X),Ψ`(X)) =

∫I

Ψk(x)Ψ`(x)f(x)dx (206)

=1∫

Iw(x)dx

∫I

Ψk(x)Ψ`(x)w(x)dx (207)

=1∫

Iw(x)dx

〈Ψk,Ψ`〉 (208)

by the definition of the weight function w. By the proposition 2.14, the polynomials Ψk and Ψ`

are orthogonal, which concludes the proof.

38

2.8 Notes and references

The stars and bars proof used in section 2.1 is presented in Feller’s book [7], in the section”Application to occupancy problems” of the chapter 2 ”Elements of combinatorial analysis”. Thebook [11] and the thesis [10] present spectral methods, including the multivariate orthogonalpolynomials involved in polynomial chaos. The figures 22 and 23 are presented in several papersand slides related to multivariate orthogonal polynomials, including [12], for example.

3 Polynomial chaos

3.1 Introduction

The polynomial chaos, introduced by Wiener [16], uses Hermite polynomials as the basis andinvolves independent Gaussian random variables.

Denote [4, 13] the set of multi-indices with finite number of nonzero components as

J = {α = (αi)i≥1, αi ∈ {0, 1, 2, ...}, |α| <∞} , (209)

where

|α| =∞∑i=1

αi. (210)

If α ∈ J , then there is only a finite number of nonzero components (otherwise |α| would beinfinite).

Let {Xi}i≥0 be an infinite set of independent standard normal random variables and let f bethe associated multivariate tensor product normal distribution function. Assume that g(X) is arandom variable with finite variance. We are interested in the decomposition of g(X) onto Heαi

,the Hermite polynomial of degree αi ≥ 0.

The expectation of g(X) is

E(g(X)) =

∫g(x)f(x)dx, (211)

and its variance is:

V (g(X)) = E((g(X)− µ)2

), (212)

where µ = E(g(X)).For any α ∈ J , the Wick polynomial is

Ψα(X) =∞∏i=1

Heαi(Xi). (213)

The degree of the polynomial Tα is |α|. Notice that, since there is only a finite number of nonzerocomponents in α, the right hand side of the expression 213 has a finite number of factors.

We consider the inner product

〈g,Ψα〉 =

∫g(x)Ψα(x)f(x)dx. (214)

39

We use the norm

‖Ψα‖2 =

∫Ψα(x)2f(x)dx. (215)

The following theorem is due to Cameron and Martin [5].

Theorem 3.1. ( Cameron-Martin) Let {xi}∞i=1 be an infinite set of independent standard nor-mal random variables. Let g(X) be a random variable with finite variance. Then g(X) has thedecomposition

g(X) =∑α∈J

aαΨα(X) (216)

where X = (ξ1, ξ2, ...) and

aα =〈g,Ψα〉‖Ψα‖2

. (217)

Moreover,

E(g(X)) = aα0 , (218)

V (g(X)) =∑α∈J

a2α‖Ψα‖2, (219)

where α0 = (0, 0, ...).

The previous theorem is due to [5] and will not be proved here.Moreover, the decomposition 216 converges in the L2

w sense.

3.2 Truncated decomposition

In the decomposition 216, there is an infinite number of random variables xi and an unrestrictedpolynomial degree |α|. In order to use such an expansion in practice, we have to perform a doubletruncation. This is why we keep only p independent normal random variables and only the orderd multivariate polynomials. Define the truncated index set [9]

Jp,d = {α = (α1, α2, ..., αp), |α| ≤ d} . (220)

The finite decomposition is

g(X) ≈∑

α∈Jp,d

aαΨα(X), (221)

where X = (X1, X2, ..., Xp) are independent standard normal random variables. There is a one-to-one mapping from the multi-indices of α of the set Jp,d and the indices k = 1, 2, ..., P p

d definedin the section 2.4. Therefore, the decomposition 216 can be written:

g(X) ≈P pd∑

k=1

akΨk(X), (222)

where Ψk is defined by the equation 164.

40

Proposition 3.2. ( Expectation and variance of the truncated PC expansion) The truncatedexpansion 222 is such that

E(g(X)) = a1 (223)

V (g(X)) =

P pd∑

k=2

a2kV (Ψk(X)). (224)

Since the expansion 222 involves only a finite number of terms, it is relatively easy to provethe previous proposition.

Proof. The expectation of g(X) is

E(g(X)) = E

P pd∑

k=1

akΨk(X)

(225)

=

P pd∑

k=1

E(akΨk(X)) (226)

=

P pd∑

k=1

akE(Ψk(X)), (227)

since the expectation of a sum is the sum of expectations. Then, the equation 223 is a straight-forward consequence of the proposition 2.16.

The variance of g(X) is

V (g(X)) = V

P pd∑

k=1

akΨk(X)

(228)

=

P pd∑

k=1

V (akΨk(X)) +

P pd∑

k,`=1k 6=`

Cov(akΨk(X), a`Ψ`(X)) (229)

=

P pd∑

k=1

a2kV (Ψk(X)) +

P pd∑

k,`=1k 6=`

aka`Cov(Ψk(X),Ψ`(X)). (230)

However, the proposition 2.18 states that the covariances are zero. Moreover, the variance of Ψ1

is zero, since this is a constant. This leads to the equation 224.

In the following proposition, we prove the equation 217 in the case of the truncated decom-position.

Proposition 3.3. ( Coefficients of the truncated decomposition) The truncated expansion 222 issuch that

ak =〈g,Ψk〉‖Ψk‖2

. (231)

41

Proof. Indeed,

〈g,Ψk〉 =

∫I

g(x)Ψk(x)w(x)dx (232)

=

P pd∑

`=1

a`

∫I

Ψ`(x)Ψk(x)w(x)dx (233)

=

P pd∑

`=1

a` 〈Ψ`,Ψk〉 (234)

= ak 〈Ψk,Ψk〉 (235)

= ak‖Ψk‖2, (236)

where the equation 235 is implied by the orthogonality of the functions {Ψk}k≥1.

An immediate consequence of the equation 3.3 is that the decomposition 222 is unique. Indeed,assume that ak and bk are real numbers and consider the decompositions

g(X) ≈P pd∑

k=1

akΨk(X), (237)

and

g(X) ≈P pd∑

k=1

bkΨk(X). (238)

The equation 231 is satisfied both by the coefficients ak and bk, so that

ak = bk. (239)

The following decomposition shows how the coefficients be be expressed in termes of expecta-tions and variances.

Proposition 3.4. ( Coefficients of the truncated decomposition (2)) The truncated expansion 222is such that

a1 = E(g(X)) (240)

and

ak =E (g(X)Ψk(X))

V (Ψk(X))(241)

for k > 1.

Proof. By definition of the expectation, we have

E (g(X)Ψk(X)) =

∫I

g(x)Ψk(x)f(x)dx (242)

=1∫

Iw(x)dx

∫I

g(x)Ψk(x)w(x)dx (243)

=〈g,Ψk〉∫Iw(x)dx

(244)

=ak‖Ψk‖2∫Iw(x)dx

, (245)

42

where the last equation is implied by 231. Hence,

E (g(X)Ψk(X)) =ak∫

Iw(x)dx

∫I

Ψk(x)2w(x)dx (246)

= ak

∫I

Ψk(x)2f(x)dx (247)

= akE(Ψk(X)2). (248)

This implies:

ak =E (g(X)Ψk(X))

E(Ψk(X)2). (249)

For k = 1, we have Ψ1(X) = 1, so that the previous equation implies 240. For k > 1, we have

ak =E (g(X)Ψk(X))

V (Ψk(X)) + E(Ψk(X))2. (250)

However, we know from proposition 2.16 that, for k > 1, we have E(Ψk(X)) = 0, which concludesthe proof.

3.3 Univariate decomposition examples

In this section, we present several examples of decomposition of univariate random variables onHermite polynomials.

Assume that X is a standard univariate normal random variable (i.e. p = 1) and consider therandom variable g(X) where g is a square integrable function. Its truncated polynomial chaosdecomposition is

g(X) ≈P 1d∑

k=1

akΨk(X), (251)

where Ψk are the univariate Hermite polynomials. The number of univariate polynomials is givenby the proposition 2.6, which states that, for p = 1, we have

P 1d =

(1 + dd

)(252)

=(1 + d)!

d!1!(253)

= 1 + d. (254)

Moreover, the equation 164 states that the functions Ψk are defined in terms of the Hermitepolynomials as

Ψk(x) = Heα(k)(x), (255)

for any x ∈ R and k ≥ 1, where

α(k) = k − 1. (256)

43

This implies :

Ψk(x) = Hek−1(x), (257)

for any x ∈ R and k ≥ 1. The equation 251 simplifies to:

g(X) ≈ a1He0(X) + a2He1(X) + ...+ ad+1Hed(X), (258)

where d ≥ 1 is an integer and {Hei}i=0,1,...,d are Hermite polynomials.The proposition 3.4 states that the coefficients are

a1 = E(g(X)) (259)

and

ak =E (g(X)Hek−1(X))

V (Hek−1(X))(260)

=E (g(X)Hek−1(X))

(k − 1)!(261)

for k > 1. In general, this requires to compute the integral

E (g(X)Hek−1(X)) =

∫Rg(x)Hek−1(x)f(x)dx, (262)

for k ≥ 1, where f is the Gaussian probability distribution function defined in the equation 48.In this section, we consider several specific examples of functions g and present the associated

coefficients ai, for i = 1, 2, ..., d+ 1.

Example (Constant) Assume that

g(x) = c, (263)

for some real constant c. A possible exact decomposition is:

g(X) = cHe0(X). (264)

Since the polynomial chaos decomposition is unique, this is the decomposition. But it is interestingto see how the coefficients can be computed in this particular case. We have

a1 = E(c) (265)

= c. (266)

Moreover,

ak =E (cHek−1(X))

(k − 1)!(267)

= cE (Hek−1(X))

(k − 1)!(268)

However, the proposition 1.9 states that E (Hek−1(X)) = 0 for k > 1. This implies

ak = 0 (269)

for k > 1.

44

Example (Standard normal random variable) Assume that

g(x) = x, (270)

for any x ∈ R. The exact polynomial chaos decomposition is

g(X) = He1(X), (271)

since He1(X) = X. Again, it is interesting to see how the coefficients can be computed in thisparticular case. We have

a1 = E(X) (272)

= 0, (273)

since, by assumption, X has a mean equal to zero. Moreover,

ak =E (XHek−1(X))

(k − 1)!(274)

=E (He1(X)Hek−1(X))

(k − 1)!, (275)

since the first Hermite polynomial is He1(x) = x. The equation 39 then implies ak = 0, for k > 2.Moreover,

a2 =E (He1(X)2)

V (He1(X))(276)

=V (He1(X))

0!(277)

=0!

1(278)

= 1, (279)

where the previous equation comes from the equation 40. This immediately leads to the equation271.

Example (The square function) Consider the function

g(x) = x2, (280)

for any x ∈ R. Consider the random variable g(X) = X, where X is a standard normal randomvariable. From the table 6, we see that

x2 = x2 − 1 + 1 (281)

= He2(x) +He0(x). (282)

Therefore, the exact polynomial chaos decomposition of g is

g(x) = He0(x) +He2(x). (283)

45

k a0 a1 a2 a3 a4 a5 a6 a7 a8 a9x0 1x1 1x2 1 1x3 3 1x4 3 6 1x5 15 10 1x6 15 45 15 1x7 105 105 21 1x8 105 420 210 28 1x9 945 1260 378 36 1

Figure 24: Coefficients ak in the decomposition of xn into Hermite polynomials, for n = 0, 1, ..., 9.

Example (Decomposition of xn) We can compute the decomposition of xn, where n is a positiveinteger. For example, we have

x5 = 15He1(x) + 10He3(x) +He5(x). (284)

In the figure 24, we present the coefficients ak in the decomposition:

xn =∑k≥0

akHek(x). (285)

Example (Transformation) TODO : X normal µ, σ and the standardization of X.

Example (Some approximate decompositions) We compute the approximate polynomial chaosdecomposition of several functions, where X is a standard normal random variable. In order toperform the numerical integration involved in the integral 262, we have used the intg function,with an absolute tolerance equal to 1.e-8, and numerical bounds equal to -10 and +10. Wecompute the coefficients up to the degree 14.

These numerical experiments are consistent with the following exact integrals, computed fromWolfram Alpha [14].

∫ ∞−∞

cos(x)Hen(x)w(x)dx = { (286)

3.4 Generalized polynomial chaos

In 2002, Xiu and Karniadakis [18, 19] introduced the generalized polynomial chaos. They noticethat the Hermite polynomials are quite effective to solve some stochastic differential equationsinvolving Gaussian as well as non Gaussian random variables, e.g. log-normal random variables.However, they emphasize that, for general non-Gaussian random inputs, the convergence rate isnot fast. This is why they introduce the generalized polynomial chaos, where each probabilitydistribution function is associated to a family of orthogonal polynomials.

46

cos(x) sin(x)a0 0.6065307a1 0.6065307a2 -0.3032653a3 -0.1010884a4 0.0252721a5 0.0050544a6 -0.0008424a7 -0.0001203a8 0.0000150a9 0.0000017a10 -0.0000002a11 -1.519D-08a12 1.266D-09a13 9.740D-11a14 -6.957D-12

Figure 25: Coefficients ak in the decomposition of several functions into Hermite polynomials.

Distribution Polynomial SupportContinuous Normal Hermite (−∞,∞)Distribution Gamma (Exponential) Laguerre [0,∞)

Beta Jacobi [a, b]Uniform Legendre [a, b]

Discrete Poisson Chalier {0, 1, 2, ...}Distribution Binomial Krawtchouk {0, 1, 2, ..., N}

Negative Binomial Meixner {0, 1, 2, ...}Hypergeometric Hahn {0, 1, 2, ..., N}

Figure 26: Map from a distribution function to the associated orthogonal polynomials [12].

The involved orthogonal polynomials are constructed from limit relationships from the hyper-geometric orthogonal polynomials. This leads to a classification which allows to create a tree ofrelations between the polynomials, called the Askey scheme. The mapping from the distributionto the corresponding orthogonal polynomial is presented in the figure 26.

The finite decomposition is

g(X) ≈∑

α∈Jp,d

aαΨα(X), (287)

where X = (X1, X2, ..., Xp) are independent random variables and Ψk is the tensor product oforthogonal polynomials φ, so that

Ψk(x) =

p∏i=1

φα(k)i

(xi) (288)

for k = 1, 2, ..., P pd .

47

Example Assume that X1 is a standard normal random variable, X2 is a uniform random vari-able in the interval [−1, 1], and assume that X1 and X2 are independent. Therefore, the gener-alized polynomial chaos associated with the random variable X = (X1, X2) involves the Hermiteorthogonal polynomials (for X1) and the Legendre orthogonal polynomials (for X2).

Example (Transformation) TODO : X U(0,1) and G(X)=exp(X).

3.5 Transformations

TODO

3.6 Notes and references

The book [11] and the thesis [10] present spectral methods, including polynomial chaos.In the figure 24, we present the coefficients ak in the decomposition of xn into Hermite poly-

nomials, for n = 0, 1, ..., 9. This table is similar to the table 22.12 in the chapter 22 of [1]. Butour table presents the coefficients for the probabilist’s Hermite polynomials Hen, while the table22.12 presents the coefficients of the physicist’s Hermite polynomials Hn.

4 Thanks

TODO

A Gaussian integrals

A.1 Gaussian integral

Proposition A.1. ( Gaussian integral)∫ +∞

−∞exp(−x2/2)dx =

√2π. (289)

Proof. We have,(∫ +∞

−∞exp(−x2)dx

)2

=

(∫ +∞

−∞exp(−x2)dx

)(∫ +∞

−∞exp(−x2)dx

)(290)

=

(∫ +∞

−∞exp(−x2)dx

)(∫ +∞

−∞exp(−y2)dy

)(291)

=

∫ +∞

−∞

∫ +∞

−∞exp

(−x2 − y2

)dxdy. (292)

Let us consider the polar change of variable:

x = r cos(θ), (293)

y = r sin(θ) (294)

48

where r ∈ R and θ ∈ [0, 2π]. Its Jacobian is J(r, θ) = r (see, e.g. [6]). Hence,(∫ +∞

−∞exp(−x2)dx

)2

=

∫ +∞

0

∫ 2π

0

exp(−r2

)rdrdθ (295)

= 2π

∫ +∞

0

r exp(−r2

)dr. (296)

Consider the change of variable s = −r2. We have ds = 2rdr, which implies(∫ +∞

−∞exp(−x2)dx

)2

= π

∫ 0

−∞exp (−s) ds (297)

= π(exp(0)− 0) (298)

= π. (299)

Hence, ∫R

exp(−x2)dx =√π. (300)

Now consider the change of variable y =√

2x. We have dx = dy/√

2, which implies∫R

exp(−x2)dx =1√2

∫R

exp(−y2/2)dy. (301)

We plug the equation 300 into the previous equality and get 289.

The previous proof does not take into account the improper integrals which appear during thecomputation.

A.2 Weighted integral of xn

Proposition A.2. For any odd n,∫ +∞

−∞xn exp(−x2/2)dx = 0, (302)

Proof. If n is odd, the function xn is antysymmetric, since (−x)n = −xn for any x ∈ R. Onthe other hand, the function exp(−x2/2) is symmetric, since exp(−(−x)2/2) = exp(−x2/2). Weconsider the change of variable y = −x and get:∫ 0

−∞xn exp(−x2/2)dx =

∫ +∞

0

(−y)n exp(−(−y)2/2)dy (303)

= −∫ +∞

0

yn exp(−y2/2)dy. (304)

This implies∫ +∞

−∞xn exp(−x2/2)dx =

∫ 0

−∞xn exp(−x2/2)dx+

∫ +∞

0

xn exp(−x2/2)dx (305)

= 0. (306)

and concludes the proof.

49

Proposition A.3. For any even n, we have∫ +∞

−∞xn exp(−x2/2)dx = 1 · 3 · 5 · ... · (n− 1)

√2π. (307)

Proof. Let us denote by In the integral

In =

∫ +∞

−∞xn exp(−x2/2)dx. (308)

We are going to prove the equality

In+2 = (n+ 1)In. (309)

Integrating by part, we get

In =

[xn+1

n+ 1exp(−x2/2)

]+∞−∞

+

∫ +∞

−∞

xn+1

n+ 1x exp(−x2/2)dx. (310)

On the other hand,

limx→+∞

xn exp(−x2/2) = 0, (311)

for any integer n. Hence, [xn+1

n+ 1exp(−x2/2)

]+∞−∞

= 0. (312)

This implies

In =

∫ +∞

−∞

xn+2

n+ 1exp(−x2/2)dx (313)

=1

n+ 1In+2. (314)

Moreover, we from from proposition A.1 that I0 =√

2π, which concludes the proof.

Consider the Gaussian weight function w(x) = exp(−x2/2). The consequence of the proposi-tions A.2 and A.3 is that a monomial xn is in L2

w(R), for any integer n.

50

References

[1] Milton Abramowitz and Irene A. Stegun. Handbook of Mathematical Functions with Formu-las, Graphs, and Mathematical Tables. Dover, New York, ninth dover printing, tenth gpoprinting edition, 1964.

[2] George E. Andrews, Richard Askey, and Ranjan Roy. Special functions, volume 71 of En-cyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge,1999.

[3] G. Arfken. Mathematical Methods for Physicists, 3rd ed. Elsevier Academic Press, 2005.

[4] M. Branicki and A.J Majda. Fundamental limitations of polynomial chaos for uncertaintyquantification in systems with intermittent instabilities, 2012.

[5] R. H. Cameron and W. T. Martin. The orthogonal development of non-linear functionals inseries of fourier-hermite functionals. Annals of Mathematics, 48(2):385–392, April 1947.

[6] Philip J. Davis and Philip Rabinowitz. Methods of Numerical Integration. Academic Press,New York, 1984.

[7] William Feller. An Introduction to Probability Theory and Its Applications - Volume I. JohnWiley and Sons, third edition edition, 1968.

[8] Nicholas J. Higham. Accuracy and Stability of Numerical Algorithms. Society for Industrialand Applied Mathematics, Philadelphia, PA, USA, second edition, 2002.

[9] Thomas Y. Hou, Wuan Luo, Boris Rozovskii, and Hao min Zhou. Wiener chaos expansionsand numerical solutions of randomly forced equations of fluid mechanics. J. Comput. Phys,216:687–706, 2006.

[10] O.P. Le Maıtre. Methodes spectrales pour la propagation d’incertitudes parametriques dansles modeles numeriques, memoire d’habilitation a diriger des recherches. Springer, 2005.

[11] O.P. Le Maıtre and O.M. Knio. Spectral methods for uncertainty quantification. Springer,2010.

[12] Didier Lucor. Introduction to stochastic spectral methods, November 2008.

[13] Wuan Luo. Wiener Chaos Expansion and Numerical Solutions of Stochastic Stochastic Par-tial Differential Equations. PhD thesis, California Institute of Technology, May 2006.

[14] Wolfram Research. Wolfram alpha. http://www.wolframalpha.com.

[15] Gabor Szego. Orthogonal polynomials. American Mathematical Society, 1975.

[16] Norbert Wiener. The homogeneous chaos. Amer. J. Math., 60(4):897–936, 1938.

[17] Wikipedia. Hermite polynomials — wikipedia, the free encyclopedia, 2013. [Online; accessed26-January-2013].

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[18] Dongbin Xiu and George Em Karniadakis. The wiener–askey polynomial chaos for stochasticdifferential equations. SIAM J. Sci. Comput., 24(2):619–644, February 2002.

[19] Dongbin Xiu and George Em Karniadakis. Modeling uncertainty in flow simulations viageneralized polynomial chaos. J. Comput. Phys., 187(1):137–167, May 2003.

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