2 1. THE DERIVATIVE OPERATOR

From

∂F

∂x

= tF we obtain

(1.4) Hn(x) = Hn−1(x),

for any n ≥ 1. Also, from

∂F

∂t

= (x − t)F we get the following recursive formula

(1.5) (n + 1)Hn+1(x) = xHn(x) − Hn−1(x),

for any n ≥ 1. The first Hermite polynomials are H0(x) = 1, H1(x) = x and

H2(x) =

1

2

(x2 − 1). From the expansion of F (0, t) = exp(−

t2

2

) in powers of t, we

get Hn(0) = 0 if n is odd and H2k(0) =

(−1)k

2kk!

for all k ≥ 1.

The relationship between Hermite polynomials and Gaussian random variables

is explained by the following result.

Lemma 1.3. Let X, Y be two random variables with joint Gaussian distribution

such that E(X) = E(Y ) = 0 and E(X2) = E(Y 2) = 1. Then for all n, m ≥ 0 we

have

E(Hn(X)Hm(Y )) =

0 if n = m,

1

n!

(E(XY

))n

if n = m.

Proof. For all s, t ∈ R we have

E exp(sX −

s2

2

) exp(tY −

t2

2

) = exp(stE(XY )).

Expanding both sides of this equality in power series of the variables s and t using

(1.3), and identifying the coeﬃcient of

sntm

yields

E(Hn(X)Hm(Y )) =

0 if n = m,

1

n!

(E(XY

))n

if n = m.

This completes the proof.

Suppose that H is infinite-dimensional and let {ei, i ≥ 1} be an orthonormal

basis of H. We will denote by Λ the set of all sequences a = (a1, a2, . . . ), ai ∈ N,

such that all the terms, except a finite number of them, vanish. For a ∈ Λ we set

a! =

∞

i=1

ai! and |a| =

∑∞

i=1

ai. For any multiindex a ∈ Λ we define

(1.6) Φa =

√

a!

∞

i=1

Hai (W (ei)).

The above product is well defined because H0(x) = 1 and ai = 0 only for a finite

number of indices.

Proposition 1.4. The family of random variables {Φa, a ∈ Λ} is a complete

orthonormal system in

L2(Ω,

F, P ).

Proof. For any a, b ∈ Λ we have, using the independence of the random

variables {W (ei), i ≥ 1} and Lemma 1.3,

E

∞

i=1

Hai (W (ei))Hbi (W (ei)) =

∞

i=1

E(Hai (W (ei))Hbi (W (ei)))

=

1

a!

if a = b,

0 if a = b.

This implies that the random variables {Φa, a ∈ Λ} are orthonormal. On the other

hand, suppose that F is an element in

L2(Ω,

F, P ) such that E(F Φa) = 0 for all