10 1. THE DERIVATIVE OPERATOR

where the kernels fn are symmetric functions from

L2(T n).

The derivative DtF

can be easily computed using this expression.

Proposition 1.17. Let F ∈

D1,2

be a random variable with a development of

the form (1.19). Then, we have

(1.20) DtF =

∞

n=1

nIn−1(fn(·, t)).

Proof. Suppose that F = In(fn), where fn is a symmetric and elementary

function of the form (1.18). Then

DtF =

n

j=1

k

i1,...,in=1

ai1···in W (Ai1 ) ··· 1Aij (t) ··· W (Ain ) = nIn−1(fn(·, t)).

The result follow easily.

Suppose that F is a random variable in the space Dk,2 with a Wiener chaos

expansion F = E(F ) +

∑∞

n=1

In(fn). Then, applying Proposition 1.17 k times we

obtain that

DkF

is a random field parametrized by T

k

given by

Dt1,...,tk

k

F =

∞

n=k

n(n − 1) ··· (n − k + 1)In−k(fn(· , t1, . . . , tk)).

Notice that the L2 norm of this expression is given by (1.15). As a consequence, if

F belongs to D∞,2 = ∩kDk,2, then

(1.21) fn =

1

n!

E(DnF

)

for every n ≥ 0 (cf. Stroock [48]).

Stroock’s formula (1.21) is a useful tool to compute Wiener chaos expansions.

For example, suppose that W = {Wt, t ≥ 0} is a Brownian motion and F = W1

3.

Then,

f1(t1) = E(Dt1 W1

3)

= 3E(W1

2)1[0,1](t1)

= 31[0,1](t1),

f2(t1, t2) =

1

2

E(Dt1,t2

2

W1

3

) = 3E(W1)1[0,1](t1 ∨ t2) = 0,

f3(t1, t2, t3) =

1

6

E(Dt1,t2,t3

3

W1

3

) = 1[0,1](t1 ∨ t2 ∨ t3),

and we obtain the Wiener chaos expansion

W1

3

= 3W1 + 6

1

0

t1

0

t2

0

dWt1 dWt2 dWt3 .

Let A ∈ B. We will denote by FA the σ-field (completed with respect to the

probability P ) generated by the random variables {W (B), B ⊂ A, µ(B) ∞}. We

will make use of the following technical result.

Proposition 1.18. Let A ∈ B and suppose that F ∈ D1,2 is FA-measurable.

Then DtF is zero almost everywhere in Ac × Ω.

Proof. Suppose that Fn ∈ S converges to F in the norm of the space

D1,2,

where

Fn = fn(W (h1

n

), . . . , W (hkn

n

),