ANNALES MATHÉMATIQUES BLAISE PASCAL

ALBERT BADRIKIAN

Transformation of gaussian measures

*Annales mathématiques Blaise Pascal*, tome S3 (1996), p. 13-58

<http://www.numdam.org/item?id=AMBP_1996__S3__13_0>

© Annales mathématiques Blaise Pascal, 1996, tous droits réservés.

L'accès aux archives de la revue « Annales mathématiques Blaise Pascal » (http://math.univ-bpclermont.fr/ambp/) implique l'accord avec les conditions générales d'utilisation (http://www.numdam.org/legal.php). Toute utilisation commerciale ou impression systématique est constitutive d'une infraction pénale. Toute copie ou impression de ce fichier doit contenir la présente mention de copyright.

NUMDAM

Article numérisé dans le cadre du programme
Numérisation de documents anciens mathématiques
http://www.numdam.org/
Transformation
of Gaussian measures
<figure><img src="image_1.png" /></figure>
Introduction

We shall be, in our lecture, mainly concerned by some particular cases of the following problem :

Let $(X, \mathcal{F}, \mu)$ be a measure space and $T: X \to X$ measurable. We denote by $T(\mu)$ or $\mu \circ T^{-1}$ the image of $\mu$ by $T$:

$$T(\mu)(A) = \mu \circ T^{-1}(A) = \mu(T^{-1}A), \quad \forall A \in \mathcal{F}.$$

When does $T(\mu) \ll \mu$ and how to compute the density?

**Example 1:** Let $X = \mathbb{R}^n$, $\mu = \lambda_n$ (the Lebesgue measure) and $T: X \to X$ a diffeomorphism. Then from the formula

$$\int f(T(x))|\det T'(x)|dx = \int f(y)dy,$$

we conclude that $T(\lambda_n)$ is absolutely continuous with respect to $\lambda_n$ and

$$T(\lambda_n)(dy) = |\det T'(T^{-1}y)|^{-1}dy = |\det (T^{-1})'(y)|dy.$$

**Example 2:** Let $(\Omega, \mathcal{F}, P)$ be the classical Wiener space, $\Omega = C_0([0, 1])$, $\mathcal{F}$ the Borel $\sigma$-field, $P$ the Wiener measure. Let $u: [0, 1] \times \Omega \to \mathbb{R}$ be a measurable and *adapted* stochastic process such that $\int_0^1 u_t^2(\omega)dt < \infty$ almost surely, and let $T: \Omega \to \Omega$ be defined by :

$$(T\omega)_t = \omega_t + \int_0^t u_s(\omega) \, ds.$$

Girsanov has proven that

$$T(P) \ll P.$$

On the other hand, let

$$\xi = \exp\left\{-\int_{0}^{1} u_{t} d\omega_{t} - \frac{1}{2} \int_{0}^{1} u_{t}^{2}(\omega) dt\right\}$$
then, if $\mathbb{E}(\xi) = 1$. $(T\omega)_t$ is a Brownian motion with respect to $(\Omega, \mathcal{F}, Q)$, where $\frac{dQ}{dP} = \xi$.
That is $Q \circ T^{-1} = P$.

(This fact was first proven by means of the Itô-calculus, but as we shall see, we can obtain this by analytic methods).

This has an application in Statistical Communication Theory:

Suppose we are receiving a signal corrupted by noise, and we wish to determine if there is indeed a signal or if we are just receiving noise.

If $x(t)$ is the received signal, $\xi(t)$ the noise and $s(t)$ the emitted signal :

$$x(t) = s(t) + \xi(t) \quad (A)$$

In general, we make an hypothesis on the noise : it is a white noise.

The "integrated" version of (A) is

$$X(t) = \int_{0}^{t} s(u) \, du + W_{t} = S_{t} + W_{t} \quad (A')$$

$(W$ is the standard Wiener process, $X(t) = \int_{0}^{t} x(s) \, ds$ is the cumulative received signal).

Now we ask the question : is there a signal corrupted by noise, or is there just a noise $(s(t) = 0, \forall t)$?

The hypotheses are :

$$H_0 : X_t = W_t$$

$$H_1 : X_t = \int_0^t s(u) \, du + W_t.$$

We consider the likelihood ratio

$$\frac{d\mu_w}{d\mu_x} = \exp\left(-\int_0^1 s(t) \, dW_t - \frac{1}{2} \int_0^1 s(t)^2 \, dt\right)$$

and we fix a threshold level for the type 1-error :

$$\text{if : } \frac{d\mu_w}{d\mu_x} (\omega) \le \lambda \quad \text{we reject } (H_0)$$

$$\text{if : } \frac{d\mu_w}{d\mu_x} (\omega) \ge \lambda \quad \text{we accept } (H_0).$$
**Some general considerations and examples.**

$$ \text{If } P \ll Q, \text{ then } T(P) \ll T(Q). \qquad (a) $$

Therefore, we do not lose very much if we suppose that $P$ and $Q$ are probabilities.

In the case where $Q$ is a probability, we can have an expression of $\frac{dT(P)}{dT(Q)}$ as conditional mathematical expectation.

**Remark :** From (a) we see that, if there exists a probability $Q$ such that

$$ P \ll Q \text{ and } T(Q) = P, \text{ then } T(P) \ll P. $$

The converse is true if moreover $\frac{dT(P)}{dP} > 0$. (The measures are equivalent). Therefore the following properties are equivalent :

$$ (i) : T(P) \sim P, $$

$$ (ii) : \exists Q \sim P \text{ such that } T(Q) = P. $$

Let us now consider an example which allows us to guess the situation in infinite dimensional space.

Let $\Omega = \mathbb{R}^n$ and $P = \gamma_n$ the canonical Gaussian measure with density :

$$ \frac{1}{(2\pi)^{n/2}} \exp \left( -\frac{\|x\|^2}{2} \right) $$

and let $T : \mathbb{R}^n \to \mathbb{R}^n$ be a diffeomorphism, then

$$ \begin{align*} \int_{\mathbb{R}^n} f(y)T(\gamma_n)(dy) &= \int_{\mathbb{R}^n} f(Tx) \gamma_n(dx) \\ &= \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n} f(Tx) \exp\left(-\frac{\|x\|^2}{2}\right) dx \\ &= \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n} f(Tx) \exp\left(-\frac{1}{2} \|T^{-1}Tx\|^2\right) dx \\ &= \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n} \frac{f(y)}{|\det T'(T^{-1}y)|} \exp\left(\frac{1}{2} \|y\|^2 - \frac{1}{2} \|T^{-1}y\|^2\right) \exp\left(-\frac{1}{2} \|y\|^2\right) dy. \end{align*} $$
Therefore :

$$ \frac{dT(\gamma_n)}{d\gamma_n}(y) = \frac{1}{|\det T'(T^{-1}y)|} \exp\left(\frac{1}{2} \|y\|^2 - \frac{1}{2} \|T^{-1}y\|^2\right) \\ = |\det(T^{-1})'(y)| \exp\left(\frac{1}{2} \|y\|^2 - \frac{1}{2} \|T^{-1}y\|^2\right). $$

Now if we write :

$$ T^{-1} = (I + S) \text{ with } S \text{ self adjoint,} $$

then :

$$ (T^{-1})'(y) = I + S'(y) $$

and we obtain :

$$ \frac{d(I + S)^{-1}(\gamma_n)}{d\gamma_n}(y) = |\det (I + S'(y))| \exp \{-(Sy, y)_{\mathbb{R}^n} - \frac{1}{2} \|S(y)\|^2\}. \quad (B) $$

This can be written as :

$$ |\det (I + S'(y))| \exp(-\operatorname{Trace} S'(y)) \exp \left\{ -(Sy, y)_{\mathbb{R}^n} + \operatorname{Trace} S'(y) - \frac{1}{2} \|S(y)\|^2 \right\}, $$

where $|\det (I + S'(y))| \exp(-\operatorname{Trace} S'(y))$ is the Carleman determinant.

**General remark :** If $T = \mathrm{Id}(\Omega)$, it is clear that $TP = P$ for every $P$. The idea is to perturb the identity operator.

The problem is :

"what does the word *perturbation* mean ?"
CHAPTER ONE

Anticipative stochastic integral

1 - Gaussian measures on Banach spaces

Let $E$ be a (real) separable Banach space, $E'$ its dual. A (Borelian) probability $\mu$
on $E$ is said to be "Gaussian centered" if for every $x' \in E'$, $\langle \cdot, x' \rangle_{E,E'} = x'(\cdot)$ is a
Gaussian centered (real) variable (eventually degenerated) under $\mu$. All what we shall say
is true whatever be the dimension of $E$ (finite or infinite).

If $x' \in E'$ we define $A : E' \to E$ by

$$
Ax' = \int_E \langle x, x' \rangle_{E,E'} x \, d\mu(x),
$$

(Bochner integral of a vector function). It is the **barycenter** of the measure $\langle \cdot, x' \rangle d\mu$.

A is injective if Supp $\mu = E$.

Let $x \in A(E')$ so $x = A(u')$ and let $y \in A(E')$ so $y = A(v')$, we shall put on
$A(E') \subset E$ the following scalar product :

$$
(x, y) \rightsquigarrow (x, y)_{\mu} := \int_{E} \langle u', z \rangle \langle v', z \rangle d\mu(z)
$$

(it does not depend on $u'$ and $v'$).

$A : E' \to E$ is continuous. (Since $\int_E \|x\|^2 d\mu(x) < \infty$ by Fernique's theorem).

Therefore, if $i$ denotes the canonical injection of $A(E')$ into $E$ :

$$
i : (A(E')), \|\cdot\|_{\mu}) \to (E, \|\cdot\|) \text{ is continuous.}
$$

Actually :

$$
\begin{align*}
\|Ax'\|_{E} &= \sup_{\|y'\| \le 1} \left| \int_{E} \langle x', x \rangle \langle y', x \rangle \, d\mu(x) \right| \\
&\le \sup_{\|y'\| \le 1} \left( \int_{E} |\langle x', x \rangle|^2 d\mu(x) \right)^{\frac{1}{2}} \left( \int_{E} |\langle y', x \rangle|^2 \, d\mu(x) \right)^{\frac{1}{2}} \\
&\le \left( \int |\langle x', x \rangle|^2 \, d\mu(x) \right)^{\frac{1}{2}} \left( \int \|x\|^2 \, d\mu(x) g \right)^{\frac{1}{2}};
\end{align*}
$$
hence,

$$ \|Ax'\|_E \leq C \|Ax'\|_{\mu} \quad (\text{where } C \text{ is a constant}). $$

Let $H_{\mu}$ be the completion of $A(E')$ with respect to $\|\cdot\|_{\mu}$. We have $\hat{i}: H_{\mu} \to E$. I say that $\hat{i}$ is injective (it will allow us to consider $H_{\mu}$ as a subspace of $E$).

$H_{\mu}$ is called the "reproducing kernel Hilbert space" (r.k.H.s.) of $\mu$.

**Example 1 : Finite dimension**

$$ E = \mathbb{R}^n, \quad \text{Supp } \mu = \mathbb{R}^n : $$

$$ Ax' = \int_E \langle x, x' \rangle x d\mu(x), $$

or :

$$ \langle Ax', y' \rangle = \int_E \langle x, x' \rangle \langle x, y' \rangle d\mu(x). $$

A is the covariance, it is invertible and

$$ (x, y)_{\mu} = \int_{E} \langle A^{-1}x, z \rangle \langle A^{-1}y, z \rangle d\mu(z) = \langle x, A^{-1}y \rangle, $$

and therefore :

$$ H_{\mu} = \mathbb{R}^n. $$

**Example 2 : Brownian motion, Wiener space.**

Let $T > 0$ and $\Omega = E = C([0, T], \mathbb{R})$ be the space of real continuous functions on $[0, T]$.

There exists an unique centered measure $\mu$ such that :

a) the support of $\mu$ is $C_0([0, T], \mathbb{R})$, the space of the continuous functions vanishing at 0,

b) $\forall t \in [0, T] : \omega \rightsquigarrow \omega_t$ has the variance $t$,

c) let $0 \le t_1 < t_2 < \dots < t_n \le T$, then : $\omega_{t_1}, \omega_{t_2} - \omega_{t_1}, \dots, \omega_{t_n} - \omega_{t_{n-1}}$ are independent.

We shall call $\mu$ the Wiener measure on $C([0, T], \mathbb{R})$; then $E'$ is the space of signed measures $\nu$ on $[0, T]$. We shall also denote :

$$ \omega_t = B(t, \omega) $$

and call $t \rightsquigarrow B(t, \cdot)$ : the "Brownian motion" on $[0, T]$.
For $\nu_1, \nu_2 \in E'$ let :

$$
\begin{aligned}
B(\nu_1, \nu_2) &= E [\langle \nu_1, B \rangle \langle \nu_2, B \rangle] \\
&= \int_{\Omega} \langle \nu_1, \omega \rangle \langle \nu_2, \omega \rangle d\mu(\omega).
\end{aligned}
$$

We have for $\nu \in E'$

$$
\langle \nu, B \rangle = \int_{[0,T]} B(t, \omega) \, d\nu(t) = \int_{0}^{T} \nu([t,T]) dB(t) \text{ (stochastic integral).}
$$

This fact can be verified as follows :

- it is true for $\nu = \delta_s$ (by definition of Brownian motion),

- by linearity this remains true if $\nu = \sum \alpha_i \delta_{t_i}$,

- then we apply a continuity argument.

Therefore

$$
B(\nu_1, \nu_2) = \int_{[0,T]} \nu_1([t,T]) \nu_2([t,T]) dt.
$$

Now let $\nu_1$ be a measure on $[0, T]$. We want to find the barycenter $m_{\nu_1}$ of the random variable on $\Omega: \omega \rightsquigarrow \langle\omega, \nu_1\rangle$. ($m_{\nu_1}$ is an element of $\Omega = C([0, T])$. It is defined by

$$
\nu \rightsquigarrow \langle m_{\nu_1}, \nu \rangle = \int_{[0,T]} m_{\nu_1}(t) \nu(dt) = B(\nu, \nu_1) = \int_{[0,T]} \nu_1([t,T]) \nu([t,T]) dt.
$$

By the generalized integration by parts this is equal to :

$$
\int_{[0,T]} J(\nu_1)(t) \, d\nu(t)
$$

where

$$
J(\nu_1)(t) = \int_0^t \nu_1([u, T]) \, du.
$$

$J(\nu_1)$ is then absolutely continuous. On the space

$$
\{J(\nu_1), \nu_1 \in \mathcal{M}([0, T])\}
$$
we put the norm

$$J(\nu_1) \rightsquigarrow \int_0^T \nu_1([t, T])^2 dt.$$

Its completion is the space of functions from $[0, T]$ into $\mathbb{R}$ absolutely continuous, null at zero, whose derivative belongs to $L^2([0, T], dt)$. It is the Cameron-Martin space.

Then the Cameron-Martin space is the reproducing kernel Hilbert space of the Wiener measure $\mu$.

**Definition :** We call an "abstract Wiener space" a triple $(H, E, \mu)$ where :

- $E$ is a separable Banach space, and $\mu$ is a centered Gaussian measure on $E$, whose topological support is $E$.

- $H$ is the r.k.H.s. associated to $\mu$.

Actually $H$ is dense in $E$. This can be proven as follows :

Let $i : H \to E$ be the canonical injection and $i^* : E' \to H$ its transpose (we identify $H$ to its dual).

Suppose that $\langle x', i(x) \rangle_{E,E'} = 0$ for every $x \in H$. This is equivalent in saying that :

$$(x \mid i^*(x'))_H = 0, \text{ for every } x \in H.$$

Therefore

$$i^*(x') = 0.$$

This means that

$$\|i^*(x')\|_H^2 = \int_E |\langle x', y \rangle_{E,E'}|^2 d\mu(y) = 0.$$

Therefore

$$\langle x', y \rangle = 0 \text{ almost surely,}$$

so this holds for all $y \in E$ since $\text{Supp } \mu = E$ and $x'$ is continuous.

The transpose $i^*$ from $i : H \to E$ is therefore injective and dense and we have :

$$E' \xrightarrow{i^*} H \xrightarrow{i} E \quad (i \text{ is the canonical injection}).$$

Every $x' \in E'$, defines a Gaussian centered random variable on $E'$, whose variance is

$$\|i^*(x')\|_H^2.$$
Now we give without proof some properties of an abstract Wiener space :

1) $H$ is separable, as a Hilbert space. Therefore it is a borelian subset of $E$,

2) $\mu(H) = 0$ or $1$ and $\mu(H) = 0 \Leftrightarrow \dim H = +\infty$ (therefore $\mu(H) = 1 \Leftrightarrow \dim H < \infty$),

3) $H$ is the intersection of the family of measurable subspaces of $E$, whose probability is equal to one,

4) the canonical injection $i: H \to E$ is compact,

5) for every Hilbert space $K$ and $u: E \to K$ linear continuous, $u \circ i: H \to K$ is Hilbert-Schmidt,

6) for every Hilbert space $K$ and $v: K \to E'$ linear continuous, $i^* \circ v: K \to H$ is Hilbert-Schmidt.

As a consequence of 5) and 6) we have :

7) let $K_1, K_2$ two Hilbert spaces ; $u_1: K_1 \to E'$ and $u_2: E \to K_2$ linear continuous then

$$K_1 \xrightarrow{u_1} E' \xrightarrow{i^*} H \xrightarrow{i} E \xrightarrow{u_2} K_2,$$

the composition $u_2 \circ i \circ i^* \circ u_1$ is nuclear (i.e. it possesses a trace).

## 2 - $L^2$-functionals on an abstract Wiener space

Let $(H, E, \mu)$ be an abstract Wiener space.

Suppose $(e_j)_{j \ge 1}$ is a sequence of elements of $E'$ such that $(i^*(e_j))_{j \ge 1}$ is an orthonormal basis in $H$. A function $f: E \to \mathbb{R}$ is said to be a polynomial in the $(e_j)$ if there exists an integer $n$ and a polynomial function $P$ on $\mathbb{R}^n$ such that

$$f(x) = P(e_1(x), \dots, e_n(x)), \quad \forall x \in E.$$

We denote $\deg f :\equiv \deg P$ ($P$ is not defined uniquely but the degree of $f$ is independent of the choice of $P$).

We denote by $\mathcal{P}((e_j))$ the set of polynomials and by $\mathcal{P}^n((e_j))$ the set of polynomials of degree $\le n$. It is easy to see that $\mathcal{P}((e_j))$ is contained in each $\mathcal{L}^p(E, \mu)$ $0 \le p < \infty$ (but clearly not in $\mathcal{L}^\infty(E, \mu)$). Moreover, $\mathcal{P}((e_j))$ is dense in $\mathcal{L}^p(E, \mu)$ for these $p$. Therefore, $\overline{\mathcal{P}((e_j))}_{L^p}$ is independent of the chosen orthonormal family $(e_j)$. The same is true for each $\mathcal{P}^n((e_j))$.
**Example:** If $n = 1$, $\mathcal{P}^1((e_j))$ is the family of affine continuous functions: an element of $\mathcal{P}^1((e_j))$ is a linear continuous function on $E$ plus a constant.

We have:

$$ \overline{\mathcal{P}^1}_{L^2(E, \mu)} \equiv H \oplus \mathbb{R} \quad (\text{see infra}). $$

We call $\overline{\mathcal{P}}^n L^2$ the set of generalized polynomials of degree at most $n$; $\overline{\mathcal{P}}^n L^2$ is a Hilbert space.

Let us now introduce the “Wiener chaos decomposition” (or “Wiener-Itô decomposition”). Let $C_0 = \overline{\mathcal{P}}^0 L^2$ the vector space of $(\mu$-equivalence classes of) constants. We define $C_n$ inductively as follows:

$C_n$ is the orthogonal complement in $\overline{\mathcal{P}}^n L^2$ of $(C_0 \oplus \dots \oplus C_{n-1})$.

(In other words $C_n$ is the set of generalized polynomials of degree $n$, orthogonal to all generalized polynomials of degree less than $n$).

It is clear that for every $n$:

$$ \overline{\mathcal{P}}^n L^2 = C_0 \oplus \dots \oplus C_n $$

and moreover

$$ L^2(E, \mu) = \bigoplus_{n=0}^{\infty} C_n . $$

The $C_n$ are called the “nth chaos” (or “chaos of order n”). $C_1$ is isomorphic to $H$. We have a description of elements of $C_n$ in term of Hermite polynomials.

We recall that the Hermite polynomials in one variable are defined by:

$$ H_n(t) = \frac{(-1)^n}{n!} \exp\left\{\frac{t^2}{2}\right\} \frac{d^n}{dt^n} \left(\exp\left\{-\frac{t^2}{2}\right\}\right), \quad n \in \mathbb{N}. $$

Then they satisfy:

$$ \sum_{n=0}^{\infty} \lambda^n H_n(t) = \exp\left\{-\frac{\lambda^2}{2} + \lambda t\right\} $$

$$ \frac{d}{dt} H_n(t) = H_{n-1}(t) $$

$$ \int_{\mathbb{R}} H_m(t) H_n(t) \frac{1}{\sqrt{2\pi}} \exp\left\{-\frac{t^2}{2}\right\} dt = \frac{1}{n!} \delta_{nm}. $$

Let $\alpha = (\alpha_1, \alpha_2, \dots) \in \mathbb{N}^\mathbb{N}$ such that $|\alpha| := \sum_{i=1}^\infty \alpha_i < \infty$. We set $\alpha! := \prod_{i=1}^\infty \alpha_i!$.
Now let $(e_n)_{n \ge 1}$ be a sequence of elements of $E'$ which is an orthonormal basis in $H$.
If $\alpha \in \mathbb{N}^\mathbb{N}$ let

$$H_\alpha(x) := \prod_{i=1}^{\infty} H_{\alpha_i} (e_i(x))$$

(This product is well defined). Then :

$\{\sqrt{\alpha!} H_\alpha(x), \alpha \in \mathbb{N}^\mathbb{N} \text{ and } |\alpha| < +\infty\}$ is an orthonormal basis in $L^2(E, \mu)$ and :
$\{\sqrt{\alpha!} H_\alpha(x), |\alpha| = n\}$ is an orthonormal basis in $C_n$.

In the case of the Wiener measure associated to Brownian motion, we have the following characterization of $C_n$ in terms of multiple stochastic integrals :

$F: C([0, T], \mathbb{R}) \to \mathbb{R}$ belongs to $L^2(P)$ where $P$ is the Wiener measure if and only if for each $n$ there exists $f_n \in L^2(\Delta_n, dt)$ where $\Delta_n = \{t \in \mathbb{R}^n, 0 \le t_1 \le t_2 \le \dots \le t_n \le T\}$ such that

$$F = \sum_n \int_{\Delta_n} f_n(t_1, \dots, t_n) dB(t_1) \dots dB(t_n) = \sum_n F_n.$$

Here

$F_0 = \mathbb{E}(F) \in C_0$ and $F_n \in C_n$.

### 3 - Measurable linear functionals and linear measurable operators

Let $(H, E, \mu)$ be an abstract Wiener space. Without loss of generality, we shall identify $H$ as a subspace of $E$ (i.e., $i(x) = x$).

A linear mapping $f: E \to \mathbb{R}$ is said to be a "linear measurable functional" if there exists a sequence of linear continuous functionals on $E$, converging to $f$, $\mu$-almost surely.

If $x \in H$, it defines a linear measurable functional $\tilde{x}(\cdot)$. Actually, if $x_n$ is a sequence of elements of $E' \subset H$ such that $x_n \to x$ in $H$, then $x_n(\cdot)$ converges to the random variable $\tilde{x}$ defined by $x$, in $L^2(E, \mu)$. Therefore, there exists a subsequence converging almost surely to $\tilde{x}$. Moreover,

$$\int_E |\tilde{x}(x)|^2 d\mu(x) < \infty.$$

The converse is true, shown by the following proposition.
If $h \in H$, the random variable $\tilde{h}$ on $E$ will be denoted by

$$x \rightsquigarrow (x, h)_H.$$

**Proposition :** Every linear measurable functional, $f$, has a restriction to $H$ which is continuous (for the Hilbertian topology). If we denote by $f_0$ this restriction we have

$$\|f\|_{L^2(E, \mu)} = \|f_0\|_H.$$

The converse is true.

**Proof :**

We have already noticed that the converse is true. Let $(x_n) \subset E' \subset H$ such that

$$x_n(x) \to f(x) \quad \forall x \in A, \text{ where } \mu(A) = 1.$$

Take $\mathcal{E}$ the linear subspace generated by $A$, we see that the above convergence holds for all $x \in \mathcal{E}$. Since $\mu(\mathcal{E}) = 1$, then $H \subset \mathcal{E}$ and therefore

$$x_n(x) \to f(x), \quad \forall x \in H.$$

Therefore the restriction of $f$ to $H$ is uniquely defined.

Now,

$$\int_E \exp \{i(x_n - x_m)(x)\} \mu(dx) = \exp \left\{-\frac{1}{2} \|x_n - x_m\|_H^2\right\} \to 1.$$

Therefore, $(x_n)$ converges in $H$, and

$$\int_E |x_n(x) - x_m(x)|^2 \mu(dx) = \|x_n - x_m\|_H^2 \xrightarrow[m,n \to \infty]{} 0.$$

Therefore $(x_n(\cdot))$ converges in $L^2(\mu)$. The limit is equal to $f$ almost surely, as we can see immediately.

— Q.E.D. —

Now let $K$ be a Hilbert space. As before we define a linear measurable function from $E$ to $K$, as the almost sure limit of a sequence of linear continuous functions from $E$ to $K$.

And, as before, if $A$ is a linear measurable function from $E$ into $K$, its restriction to $H$ is well defined and continuous from $H$ to $K$.

Let us remark that if $A$ is a linear measurable function from $E$ to $K$, we can define its transpose as a linear function from $K$ to $H$ since, for every $\varphi \in K$, $x \rightsquigarrow \langle Ax, \varphi \rangle_K$ is a linear measurable functional on $E$ therefore defined by an element of $H$. We have

$$\begin{aligned} \langle Ax, \varphi \rangle_K &= (\tilde{A}^*\varphi)(x), \quad \text{almost surely} \\ &= (x, A^*\varphi)_H \end{aligned}$$

where $A^*$ is the conjugate of the restriction of $A$ to $H$.
Now we can prove the following result :

**THEOREM** : If A is a linear measurable function from E to K. such that $\int \|Ax\|_K^2 d\mu(x) < \infty$, then its restriction to H is a Hilbert-Schmidt mapping B from H to K. Conversely if B is a Hilbert-Schmidt mapping from H to K, (we shall note $B \in L^2(H, K)$ or $B \in L_2(H, K)$), it possesses a linear measurable continuation on E, denoted by A.

Moreover, we have :

$$ \int_E \|Ax\|_K^2 d\mu(x) = \|B\|_{H-S}^2. $$

**Proof** :

Let $(\varphi_j)$ be an orthonormal basis of K.

We have :

$$ \|Ax\|_K^2 = \sum_j (Ax, \varphi_j)_K^2 \stackrel{a.s}{=} \sum_j (x, A^* \varphi_j)_H^2. $$

If we integrate term by term these equalities, we obtain :

$$ \begin{aligned} \int_E \|Ax\|_K^2 d\mu(x) &= \sum_j \int_E (x, A^* \varphi_j)_H^2 d\mu(x) \\ &= \sum_j \|A^* \varphi_j\|_H^2 = \sum_j \|B^* \varphi_j\|_H^2 = \|B^*\|_{H-S}^2. \end{aligned} $$

Conversely let $B \in L_2(H, K)$. We have for $x \in H$ :

$$ \begin{aligned} Bx &= \sum_j (Bx, \varphi_j)_K \varphi_j \\ &= \sum_j (x, B^* \varphi_j)_H \varphi_j. \end{aligned} $$

Now each term in the right-hand member possesses a linear measurable continuation to E, and the series converges in $L_2(E, \mu, K)$.

We have then defined a linear measurable extension of A to E.

— Q.E.D. —
## 4 - Derivatives of functionals on a Wiener space

Let $(E, H, \mu)$ be an abstract Wiener space and let $K$ be another Hilbert space. Let $f: E \to K$ be a function.

We say that $f$ possesses a Fréchet derivative in the direction of $H$, at the point $x_0 \in E$ if there exists an element denoted $f'(x_0)$ or $Df(x_0)$ or $\nabla f(x_0) \in L(H, K)$ such that $f(x_0 + h) - f(x_0) = f'(x_0) \cdot h + o(\|h\|_H)$, $\forall h \in H$.

Inductively we can define derivatives of all orders.

**Example 1:** Let $x_0 \in H \setminus i^*(E')$ and let $f$ be a measurable continuation of $h \mapsto (x_0, h)_H$ to $E$. ($f$ is not continuous).

Then $f$ is derivable at every $x$, and $f'(x_0) \in H$.

This example shows that a discontinuous function may have Fréchet derivatives in the direction of $H$.

**Definition 1:** Let us denote by $C^{2,1}(E, K)$ the set of functions $f: E \to K$ possessing the following properties:

- $f$ possesses $H$-derivatives at every point $x \in E$ and $f'(x)$ is Hilbert-Schmidt for every $x$,

- $f$ and $f'$ are continuous from $H$ to $K$ and to $L_2(H, K)$ respectively,

$$ - |||f|||_{2,1}^2 := \int_E \left[ \|f(x)\|_K^2 + \|f'(x)\|_{L_2(H,K)}^2 \right] \mu(dx) < \infty. $$

Then $C^{2,1}(E, K)$ is a vector space and $|||\cdot|||_{2,1}$ is a Hilbertian norm on this space.

**Definition 2:** Let $D^{2,1}(E, K)$ be the completion of $C^{2,1}(E, K)$ for the preceding norm; $D^{2,1}(E, K)$ is then a Hilbert space.

Clearly the elements of $D^{2,1}(E, K)$ are $\mu$-equivalence classes of functions.

**Convention:** Often we shall write $D^{2,1}(K)$ instead of $D^{2,1}(E, K)$. In the same manner we shall write $D^{2,1}$ instead of $D^{2,1}(E, \mathbb{R})$ or $D^{2,1}(\mathbb{R})$.

Now the map $f \mapsto f'$ from $C^{2,1}(E, K)$ into $L^2(E, \mu, L_2(H, K))$ is clearly continuous; therefore it possesses a unique continuous extension from $D^{2,1}(H, K)$ into $L^2(E, \mu, L_2(H, K))$. This extension is again denoted by $f'$, or $Df$, or $\nabla f$.

**Example 2:** Let $f$ be a polynomial function on $E$, with values in $\mathbb{R}$:

$$ f(x) = P(\langle f_1, x \rangle_{E',E}, \dots, \langle f_n, x \rangle_{E',E}), \quad f_1, \dots, f_n \in E'. $$
Then $f \in C^{2,1}$ and

$$f'(x) = \sum_{j=1}^{n} \frac{\partial P}{\partial y_j} (\langle f_1, x \rangle_{E',E}, \dots, \langle f_n, x \rangle_{E',E}) i^*(f_j).$$

The same result is true if $P$ is a $C^1(\mathbb{R}^n)$-function such that $P$ and the partial derivatives $\frac{\partial P}{\partial y_j}$ have polynomial growth.

In the same manner if $f$ is defined $(\mu$-almost everywhere) as

$$f(\cdot) = P(\tilde{h}_1(\cdot), \dots, \tilde{h}_n(\cdot)), \quad h_j \in H$$

with $P$ a polynomial function (or a $C^1(\mathbb{R}^n)$-function with polynomial growth together with its derivatives),

$$\nabla f = \sum_{j=1}^{n} \frac{\partial P}{\partial y_j} (\tilde{h}_1(\cdot), \dots, \tilde{h}_n(\cdot)) h_j.$$

**Example 2** : Let $\mu = \gamma_n$ the canonical Gaussian measure on $\mathbb{R}^n$, $\mathbb{D}^{2,1}$ is the Sobolev space $W^{2,1}(\gamma_n)$ of the distributions in $\mathbb{R}^n$ such that :

$$- f \in L^2(\mathbb{R}^n, \gamma_n),$$

- the distribution derivatives of $f$ belong to $L^2(\mathbb{R}^n, \gamma_n)$. The norm of $\mathbb{D}^{2,1}$ is the usual Hilbertian norm :

$$f \rightsquigarrow \left( \int_{\mathbb{R}^n} [|f(x)|^2 + \sum_{j=1}^n \left| \frac{\partial f}{\partial x_j}(x) \right|^2] d\gamma_n(x) \right)^{\frac{1}{2}}.$$

**Example 3** : If $f$ is a polynomial function with values in $K$ :

$$f(x) = \sum_{j=1}^{m} P_j(\langle f_1, x \rangle_{E',E}, \dots, \langle f_n, x \rangle_{E',E}) k_j \\ (k_j \in K, \ f_1, \dots, f_n \in E').$$

$$\nabla f(x) = \sum_j \sum_i \frac{\partial P_j}{\partial y_i} (\langle f_1, x \rangle_{E',E}, \dots, \langle f_n, x \rangle_{E',E}) f_i \otimes k_j.$$

(Analogous assertion for generalized polynomials, or "moderate" regular functions $P_j$).
**Example 4:** Characterization of the elements of $\mathbb{D}^{2,1}$ in the case of the Wiener measure.

If $E = C_0([0, T], \mathbb{R})$ and $\mu$ is the Wiener measure, we have seen that an element of $L^2(\mu)$ can be written as a series

$$F = \sum_{n=0}^{\infty} \sqrt{n!} \int_{\Delta_n} f_n(t_1, t_2, \dots, t_n) dB_{t_1}, \dots, dB_{t_n}$$

with

$$\sum_{n=0}^{\infty} n! \|f_n\|_{L^2(\Delta_n)}^2 < \infty.$$

Then $F$ belongs to $\mathbb{D}^{2,1}$ if and only if

$$\sum_{n=1}^{\infty} nn! \|f_n\|_{L^2(\Delta_n)}^2 < \infty$$

and in this case

$$\nabla F = \sum_{n=1}^{\infty} nJ(I_{n-1}(f_n^t))$$

where $f_n^t$ is the function defined on $\Delta_{n-1}^t = \{0 \le t_1 < t_2 < \dots < t_{n-1} < t\}$ by

$$f_n^t(t_1, t_2, \dots, t_{n-1}) = f_n^{\text{SYM}}(t_1, t_2, \dots, t_{n-1}, t),$$

$f_n^{\text{SYM}}$ being the symmetrisation of $f_n$.

The formula needs an explanation:

In the right member

$$(t, \omega) \rightsquigarrow I_{n-1}(f_n^t)(\omega) = g(t, \omega)$$

belongs to

$$L^2([0, T] \times \Omega, dt \otimes dP),$$

therefore for almost $\omega$,

$$t \rightsquigarrow g(t, \omega) \text{ is a } L^2([0, T], dt) \text{ function.}$$

$J(I_{n-1}(f_n^t))(\omega)$ is the indefinite integral null at zero of $I_{n-1}(f_n^t)(\omega)$:

$$J(I_{n-1}(f_n^t)) = \int_0^t I_{n-1}(f_n^s) ds.$$

Therefore $\nabla F(\omega)$ is an element of the Cameron-Martin space.
We now give several useful properties of $\mathbb{D}^{2,1}(E, K)$:

* The set of polynomial functions on $E$, with values in $K$ is dense in $\mathbb{D}^{2,1}(K)$.

* Therefore the algebraic sum of chaos $\sum C_n$ is dense in $\mathbb{D}^{2,1}$.

* The set of smooth functions on $E$ is dense in $C^{2,1}$ (a function is said to be "smooth" if it has the form :

$$x \rightsquigarrow f(\langle f_1, x \rangle_{E',E}, \dots, \langle f_n, x \rangle_{E',E})$$

with $f$ belonging to $C_b^\infty(\mathbb{R}^n)$; $f$ and its derivatives are bounded).

* Let $\varphi : \mathbb{R}^n \to \mathbb{R}$ be a function in $C_b^1(\mathbb{R}^n)$ and let $F^1, \dots, F^n \in \mathbb{D}^{2,1}$. Then $\varphi(F^1, \dots, F^n)$ is in $\mathbb{D}^{2,1}$ and

$$\nabla(\varphi(F^1, \dots, F^n)) = \sum_{i=1}^{n} \frac{\partial \varphi}{\partial y_i}(F^1, \dots, F^n) \nabla F^i.$$

This result is false if the above hypothesis is not satisfied. For instance on $\mathbb{R}$,

$$f = g = e^x \in \mathbb{D}^{2,1}, \text{ but } f \circ g \notin L^2(\mathbb{R}^n, \gamma_n).$$

**Remark:** The operator $\nabla$, called the "stochastic" gradient, or "stochastic" derivative, is very close to the ordinary gradient as we can see. The usual gradient at the point $x_0$ is an element of $E'$ (if the function takes its values in $\mathbb{R}$). The stochastic gradient is the composite of the ordinary gradient by the application $i^*$ from $E'$ to $H$.

In an analogous manner if $f: E \to K$ has an ordinary gradient, this gradient is a linear mapping of $E$ into $K$; $f': E \to K$.

The transpose $^t f'$ is a linear continuous mapping from $K$ into $E'$. Then the stochastic gradient is equal to $i^*(^t f') \in \mathcal{L}(K, H)$.

In his lectures at the EIPES in 1989, D. Nualart, in the case of usual Wiener space defined the stochastic derivative of the functional of the form :

$$F = f(W_{t_1}, \dots, W_{t_n}), \quad f \in C_b^\infty(\mathbb{R}^n) \quad (\text{or } f \text{ polynomial})$$

by

$$DF = \sum_{j=1}^{n} \frac{\partial F}{\partial y_j}(W_{t_1}, \dots, W_{t_n}) 1_{[0, t_j]}.$$
This definition is actually equivalent to ours, up to the notations.

Actually, let $h_j(t) = \int_0^t 1_{[0,t_j]} (s) \, ds$, $h_j$ belongs to the Cameron-Martin space and

$$W_{t_j} = \tilde{h}_j = \langle h_j, \cdot \rangle_{C-M}$$

The stochastic derivative of $F$ in our notations is therefore

$$\sum_{j=1}^{n} \frac{\partial F}{\partial y_j}(\tilde{h}_1, \dots, \tilde{h}_n) h_j.$$

There are actually equivalent since the Cameron-Martin space is isomorphic as Hilbert space to $L^2([0, T], dt)$. We shall have to consider $\nabla$ as an operator (densely defined) from $L^2(E, \mu, K)$ into $L^2(E, \mu, \mathcal{L}_2(H, K))$. It is a closed operator, naturally not continuous.

## 5 - Anticipative stochastic integral

**Definition:** The transpose of the operator $\nabla$ is called the **"Skorokhod integral"**, or the **"divergence operator"**.

The definition needs an explanation : on $L^2(E, \mu, K)$ ($K$ : Hilbert space) we have
defined the scalar product

$$(f,g) \rightsquigarrow \int_E \langle f(x), g(x) \rangle_K d\mu(x)$$

and on $L^2(E, \mu, \mathcal{L}_2(H, K))$ we have the pairing :

$$(F,G) \rightsquigarrow \int_E \langle F(x), G(x) \rangle_{\mathcal{L}_2(H,K)} d\mu(x) \\ = \int_E \operatorname{Trace} (G^*(x) \circ F(x)) d\mu(x).$$

Then $G \in L^2(E, \mu, \mathcal{L}_2(H, K))$ belongs to $\operatorname{dom}(\delta)$ if and only if the linear form on $\mathbb{D}^{2,1}(K)$ :

$F \rightsquigarrow \int_E \langle DF, G \rangle_{\mathcal{L}_2(H,K)}(x) d\mu(x)$ is continuous for the topology induced by $L^2(E, \mu, K)$.

We denote $\delta$ the Skorokhod integral and we have by definition, for every $F \in \mathbb{D}^{2,1}(K)$,

$$\int_E \langle F, \delta G \rangle_K d\mu = \int_E \langle \nabla F, G \rangle_{\mathcal{L}^2(H,K)} d\mu \quad \text{if } \delta(G) \text{ is defined.}$$
**Example 1:** Let $a \in H$, and $\varphi \in \mathbb{D}^{2,1}(K)$. Then $G := \varphi \otimes a$ is Skorokhod integrable and

$$ \delta(a \otimes \varphi) = \tilde{a}(\cdot) \varphi - \langle \nabla \varphi, a \rangle. $$

In particular, if $G: E \to H$ is such that $G(x) = a, \forall x$:

$$ \delta G = \tilde{a}(\cdot). $$

**Example 2:** $E = \mathbb{R}^n, \mu = \gamma_n$, $G: \mathbb{R}^n \to \mathbb{R}^n$.

Then

$$ \begin{aligned} \delta G(x) &= \langle x, G(x) \rangle_{\mathbb{R}^n} - \sum_{j=1}^{n} \frac{\partial G_j}{\partial x_j}(x) \\ &= \langle x, G \rangle - \operatorname{div} G(x). \end{aligned} $$

This formula can be written in another manner:

$$ \delta G = \langle \cdot, G \rangle - \operatorname{Trace}(\nabla G). $$

**Example 3:** If $G \in \mathbb{D}^{2,1}(E, \mu, \mathcal{L}^2(H, K))$, then it is $\delta$-integrable, and $\delta$ is continuous from $\mathbb{D}^{2,1}(\mathcal{L}_2(H, K))$ in $L^2(E, \mu, K)$.

**Example 4:** Let $F \in L^2(E, \mu, H)$ such that for every $h \in H: \nabla ((F, h)_H)$ exists. Then for every linear continuous operator $A: H \to H$ with **finite rank**, $A(F)$ is Skorokhod integrable.

More precisely, if $A = \sum_{j=1}^{n} \langle \cdot, a_j \rangle_H e_j$ (with $a_j$ and $e_j$ in $H$, $\langle e_j \rangle$ being orthonormal)

we have:

$$ A(F) = \sum_{j=1}^{n} \langle F, a_j \rangle_H e_j $$

$$ \delta(A(F)) = \sum_{j=1}^{n} \left[ \langle F, a_j \rangle \tilde{e}_j - \nabla_{e_j} (\langle F, a_j \rangle) \right]. $$

(see example 1).
This can be written in another manner :

Let $A^*$ be the transpose of $A$ : $A^* = \sum_{j=1}^{n} \langle \cdot, e_j \rangle_H a_j$ and let $\tilde{A}^*$ defined as :

$$ \tilde{A}^* = \sum_{j=1}^{n} a_j \tilde{e}_j . $$

Then

$$ \delta(A(F)) = \langle F, \tilde{A}^* \rangle_H - \sum_{j=1}^{n} \nabla_{e_j} (\langle F, a_j \rangle). $$

If we now suppose that $DF$ exists, we have :

$$ \sum_{j=1}^{n} \nabla_{e_j} (\langle F, a_j \rangle) = \text{Trace} (A \circ DF). $$

Therefore, we have :

$$ \delta(A(F)) = \langle F(\cdot), \tilde{A}^*(\cdot) \rangle_H - \text{Trace} (A \circ DF). $$

**Example 5** : The Skorokhod integral coincides with the ordinary Itô-Integral for adapted processes (see the above mentioned Nualart's Lecture Notes for a precise statement of this fact).

Now we give some properties of the Skorokhod integral :

a) Let $A: K \to K'$ be a linear continuous operator (K and K' Hilbert spaces) and let $F \in L^2(E, \mu, \mathcal{L}_2(H, K))$. If F is Skorokhod-integrable so is $A \circ F$ and we have

$$ \delta(A \circ F) = A(\delta F). $$

As a consequence we have :

- Let $F \in L^2(E, \mu, \mathcal{L}_2(H, K))$ such that $\delta(F)$ exists, then for every $k$ in $K$ we have $\langle \delta(F), k \rangle = \delta(F^*(k))$.
- Let $F \in L^2(E, \mu, \mathcal{L}_2(H, \mathcal{L}_2(H, K)))$ such that $\delta(F)$ exists, then

for every $h \in H$, $\delta(\check{F}(\cdot)(h))$ exists

and

$$ \delta(F)(h) = \delta(\check{F}(\cdot)(h)). $$

If $F \in \mathcal{L}^2(H, \mathcal{L}_2(H, K))$, $\check{F}$ denotes the operator of $\mathcal{L}^2(H, \mathcal{L}_2(H, K))$ such that :

$$ \check{F}(h)(h') = F(h')(h), \quad h, h' \in H. $$

b) Let $\varphi \in \mathbb{D}^{2,1}$, $F \in \mathcal{L}^2(E, \mu, H)$ such that $F$ is Skorokhod integrable. Suppose that $\varphi F \in L^2(E, \mu, H)$ and that $\delta(F)\varphi - \langle F, D\varphi \rangle_H$ belongs to $L^2(E, \mu)$, then $\varphi F$ is Skorokhod integrable and

$$ \delta(\varphi F) = \delta(F)\varphi - \langle F, D\varphi \rangle_H. $$

c) Let $A_n : H \to H$ a sequence of linear continuous operators such that $A_n \to \text{Id}_H$ in the simple convergence.

Let $F \in \mathbb{D}^{2,1}(\mathcal{L}_2(H, K))$, then $\delta(F \cdot A_n) \to \delta(F)$ in $L^2(E, \mu, K)$. In particular, if $(e_n)$ is an orthonormal basis of $H$, the sequence

$$ \left( \sum_{i=1}^{n} \tilde{e}_i F(e_i) - \nabla_{e_i} F(e_i) \right) $$

converges to $\delta(F)$.

d) Let $F, G$ in $\mathbb{D}^{2,1}(H)$ we have :

$$ \begin{aligned} \mathbb{E}(\delta(F)\delta(G)) &= \mathbb{E}\{\langle F, G \rangle_H\} + \mathbb{E}\{\langle DF, (DG)^* \rangle_{\mathcal{L}_2(H,H)}\} \\ &= \mathbb{E}\{\langle F, G \rangle_H\} + \mathbb{E}\{\text{Trace } DG(\cdot) \circ DF(\cdot)\}. \end{aligned} $$

More generally, if $F$ and $G$ belong to $\mathbb{D}^{2,1}(\mathcal{L}_2(H, K))$ we have :

$$ \mathbb{E}\{\langle \delta F, \delta G \rangle_K\} = \mathbb{E}\{\langle F, G \rangle_{\mathcal{L}_2(H,K)}\} + \mathbb{E}\{\langle DF, \check{DG} \rangle_{\mathcal{L}_2(H,\mathcal{L}_2(H,K))}\}. $$

e) The operator $\delta$, as an operator densely defined from $L^2(E, \mu, \mathcal{L}_2(H, K))$ into $L^2(\Omega, \mu, K)$ is closed.
We now briefly introduce the Ogawa integral.

Let $P: H \rightarrow H$ be an orthogonal projector with finite rank: $P(h) = \sum_{j=1}^{n} \langle h, e_j \rangle_H e_j$.

We denote $\tilde{P}$ the random variable with values in $H$:

$$ \tilde{P}(\cdot) := \sum_{j=1}^{n} \tilde{e}_j(\cdot) e_j. $$

Now let $F \in L^0(E, \mu, H)$ be a random variable with values in $H$. We shall say that $F$ is "Ogawa integrable", if there exists $G \in L^0(E, \mu)$ such that, for every increasing sequence $(P_n)$ of orthogonal projectors converging simply to $Id_H$, the sequence of real random variables $((F \cdot \tilde{P}_n)_H)_n$ converges to $G$ in probability.

We shall denote by $\mathring{\delta}(F)$ the Ogawa integral $G$ of $F$.

If $F \in L^2(E, \mu, H)$ is such that, for every $a \in H$:

$$ \langle F, a \rangle_H \tilde{a}(\cdot) \text{ belongs to } L^2(E, \mu), $$

we shall say that $F$ is "2-Ogawa integrable" when there exists $G \in L^2(E, \mu)$ such that

$$ \langle F, \tilde{P}_n \rangle_H \rightarrow G \text{ in quadratic mean.} $$

(The $P_n$ being as above).

**Example :** $(E, \mu) = (\mathbb{R}^n, \gamma_n)$. The Ogawa integral is equal to $\langle \cdot, F(\cdot) \rangle_{\mathbb{R}^n}$.

In this case, we have:

$$ \mathring{\delta}(F) = \delta(F) + \text{Trace}(\nabla F). $$

**Remark :** There exists elements of $\mathbb{D}^{2,1}(H)$ which do not possess an Ogawa integral (Rosinski).

For instance, in the case of the Brownian motion, the function : $\omega \rightarrow J(B(T - \cdot)(\omega))$ where $J$ denotes the indefinite integral null at zero, belongs to $\mathbb{D}^{2,1}(H)$ but is not Ogawa integrable.
Next we give a necessary and sufficient condition for Ogawa integrability :

Let $F \in \mathbb{D}^{2,1}(H)$; $F$ is Ogawa integrable if and only if, for almost every $x$ :

$$DF \in \mathcal{L}_1(H, H) \quad (\Longleftrightarrow DF \text{ is nuclear})$$

and we have :

$$\mathring{\delta}(F) = \delta(F) + \text{Trace}(DF).$$

**Sketch of the proof :**

Suppose $P: H \to H$ is an orthogonal projector with finite rank. We know that :

$$\delta(PF) = \langle F, \tilde{P} \rangle - \text{Trace}(DF).$$

Let $P_n \uparrow \text{Id}$. We know that

$$\delta(P_n F) \to \delta(F).$$

It is trivial that :

$$\langle F, \tilde{P}_n \rangle \to \mathring{\delta}(F)$$

(if $\mathring{\delta}(F)$ exists) and

$$\text{Trace}(D(P_n F)) \to \text{Trace}(DF)$$

— Q.E.D. —

# 6 - Extensions and remarks - Localization

Now we shall consider the case where $(E, H, \mu)$ is the Wiener space. If $F \in \mathbb{D}^{2,1}$, then $\nabla F$ is a random variable with values in the Cameron-Martin space. Therefore, if $t \in [0, T]$ we can speak of the value of $\nabla F(\omega)$ at $t$, denoted $\nabla_t F(\omega)$. Analogously, time derivative of $\nabla F(\omega)$ at time $t$ (defined for almost every $t$) makes sense. We shall denote it : $\mathring{\nabla}_t F(\omega)$. We have the equality :

$$\|\nabla F(\cdot)\|_{L^2(H)}^2 = \mathbb{E}\left(\int_0^t |\mathring{\nabla}_t F(\omega)|^2 dt\right).$$

**Lemma 1 :** Let $F \in \mathbb{D}^{2,1}$. Then $1_{\{F=0\}} \mathring{\nabla}_t F = 0$ almost everywhere on $[0, T] \times \Omega$.

For the proof see Nualart-Pardoux.

This results in a localization theorem : if $F$ is null (almost everywhere) on a set, so is its derivative. The derivation is a "local operator".
**Definition 1:** A random variable $F$ will be said to belong to $\mathbb{D}_{\text{loc}}^{2,1}$ if there exist

• a sequence of measurable sets of $E$, $E_k \uparrow E$ and

• a sequence $(F_k) \subset \mathbb{D}^{2,1}$ such that $F_{|E_k} = F_{k|E_k}$ a.s. $\forall k \in \mathbb{N}$.

Thanks to the preceding lemma we can define the derivation operator for an element of $\mathbb{D}_{\text{loc}}^{2,1}$.

**Definition 2:** Let $F \in \mathbb{D}_{\text{loc}}^{2,1}$ localized by the sequence $(E_k, F_k)$. $DF$ is the unique equivalence class of $dt \times dP$ a.e equal processes such that

$$DF_{|E_k} = DF_{k|E_k}, \quad \text{for all } k \text{ in } \mathbb{N}.$$

This generalized derivative has the usual properties of composition :

let $\varphi : \mathbb{R}^m \to \mathbb{R}$ of the class $C^1$; suppose $F = (F^1, ..., F^m)$ is a random vector whose components belong to $\mathbb{D}_{\text{loc}}^{2,1}$; then

$$\varphi(F) \in \mathbb{D}_{\text{loc}}^{2,1}$$

and

$$\nabla\varphi(F) = \sum_{i=1}^{m} \frac{\partial\varphi}{\partial x_i}(F) \cdot DF^i.$$

In the same manner we define $(\text{Dom } \delta)_{\text{loc}}$ as follows :
$F : E \to H$ belongs to $(\text{Dom } \delta)_{\text{loc}}$ if there exists a sequence $E_k \uparrow E$, and a sequence $F_k : E \to H$ such that $F_k \in (\text{Dom } \delta)$ for every $k$, such that

• $F = F_k$ on $E_k$

• $\delta(F_k) = \delta(F_\ell)|_{F_k}$ a.s. if $k < \ell$;

we shall say that $F$ is "localized" by $(E_k, F_k)$.

For sufficiently reasonable integrands on $(\text{Dom } \delta)$ Nualart-Pardoux have shown that $\delta$ is local.

**Definition 3:** Let $F \in (\text{Dom } \delta)_{\text{loc}}$ localized by $(E_k, F_k)$, $\delta(F)$ is defined as the unique equivalence class on random variables on $E$ such that

$$\delta(F)|_{E_k} := \delta(F_k)|_{E_k}, \quad \text{for all } k \text{ in } \mathbb{N}.$$

(Note that $\delta(F)$ may depend on the localizing sequence).
We shall need another notion of stochastic derivatives and Skorokhod integrals for some functions not necessarily belonging to $D^{2,1}$, nor Skorokhod integrable, introduced by Buckdahn:

Let $T: E \to E$ be a measurable mapping of the form:

$$x \rightsquigarrow x + Fx \quad \text{where } F \in D^{2,1}(H).$$

Let $\xi \in D^{2,1}$ and suppose that for every sequence of smooth random variables $(\xi_n) \in D^{2,1}$ converging to $\xi$ in $D^{2,1}$, the following limit exists and is independent of the approximating sequence chosen:

$$\lim_{n \to \infty} \nabla (\xi_n \circ T)$$

where the limit is taken in probability.

Let us remark that $\xi_n \circ T$ belongs to $D^{2,1}$ since the $\xi_n$ are smooth.

The common limit of the above sequences is denoted by $\tilde{\nabla}(\xi \circ T)$.

**Lemma 2:** Suppose that $T(\mu) \ll \mu$, then the limit exists and we have, $\mu$-almost surely:

$$\tilde{\nabla}(\xi \circ T) = (I_H + (\nabla F)^*)(((\nabla \xi) \circ T)) = (I_H + \nabla F)^*((\nabla \xi) \circ T)$$

(where $(\quad)^*$ denotes the adjoint of the bounded operator).

$$\text{Moreover, if } \xi \circ T \in D^{2,1}: \quad \tilde{\nabla}(\xi \circ T) = \nabla(\xi \circ T).$$

**Proof:**

We have, *since the* $(\xi_n)$ *are smooth*:

$$\nabla(\xi_n \circ T) = (I_H + \nabla F)^* ((\nabla \xi_n) \circ T).$$

Moreover, $\nabla \xi_n$ converges in probability, and since $T(\mu)$ is absolutely continuous with respect to $\mu$, $(\nabla \xi_n) \circ T$ converges in probability, so does $\nabla(\xi_n \circ T)$.

It now remains to prove that the limit does not depend upon the approximating sequence $(\xi_n)$.

Let $\xi_n \to \xi$ and $\eta_n \to \xi$ in $D^{2,1}$. Since the operator $\nabla$ is closed we have:

$$\lim_n \nabla(\xi_n \circ T) = \lim_n \nabla(\eta_n \circ T).$$

Therefore, $\tilde{\nabla}$ is well defined by what precedes. It is obvious that:

$$\tilde{\nabla} = \nabla \quad \text{if} \quad \xi \circ T \in D^{2,1}.$$

By duality, we can define a generalized Skorokhod integral of $\xi \circ T$, for $\xi \in D^{2,1}(H)$:

— Lemma 2 is proven. —
**Definition:** Let $(e_i)_{i \in \mathbb{N}}$ be a fixed orthonormal basis of $H$. We define

$$ \tilde{\delta}(\xi \circ T) := \sum_i (\langle \xi \circ T, e_i \rangle_H \tilde{e}_i - \tilde{\nabla}_{e_i} (\langle \xi \circ T, e_i \rangle_H), $$

if the limit of the right member is taken in probability.
($\tilde{\nabla}_{e_i}$ denotes the generalized derivative in the $e_i$-direction introduced just above).

**Lemma 3:** Suppose $T = I + F$ as above is such that $T(\mu) \ll \mu$. Then $\tilde{\delta}(\xi \circ T)$ exists and satisfies the following identity:

$$ (\delta(\xi)) \circ T = \tilde{\delta}(\xi \circ T) + \langle \xi \circ T, F \rangle_H + \text{Trace} \left( (\nabla \xi) \circ T \cdot \nabla F \right) \quad \mu\text{-almost surely.} $$

**Proof:**

Let $\xi^N = \sum_{i=1}^N \langle \xi, e_i \rangle_H e_i$, then

$$ \tilde{\delta}(\xi^N \circ T) = \sum_{i=1}^{N} \langle \xi \circ T, e_i \rangle_H \tilde{e}_i - \sum_{i=1}^{N} \tilde{\nabla}_{e_i} (\langle \xi \circ T, e_i \rangle_H). $$

But

$$ \tilde{e}_i \circ T = \tilde{e}_i + \langle F, e_i \rangle_H, $$

therefore:

$$ \begin{align*} \delta(\xi^N \circ T) &= \sum_{i=1}^{N} \left\{ \langle \xi \circ T, e_i \rangle_H [\tilde{e}_i \circ T - \langle F, e_i \rangle_H] - \langle (I_H + \nabla F)^* (\nabla(\langle \xi, e_i \rangle_H)) \circ T, e_i \rangle_H \right. \\ & \qquad \left. \text{ (by the preceding lemma)} \right. \\ &= \sum_{i=1}^{N} \left\{ \langle \xi \tilde{e}_i, e_i \rangle_H \circ T - \langle \xi \circ T, e_i \rangle_H \langle F, e_i \rangle_H - \langle (I_H + \nabla F)^* (\nabla(\langle \xi, e_i \rangle_H)) \circ T, e_i \rangle_H \right. \\ &= \sum_{i=1}^{N} \left[ \langle \xi, e_i \rangle_H \tilde{e}_i - \langle \nabla_{e_i} \xi, e_i \rangle_H \right] \circ T - \langle \xi^N \circ T, F \rangle_H - \text{Trace} (\nabla F^*, (\nabla \xi^N) \circ T). \end{align*} $$

Now $\xi^N \to \xi$ in $\mathbb{D}^{2,1}(H)$; then the right member of this last equality converges in $L^0(E, \mu)$. Hence the sum is convergent in $L^0(E, \mu)$ and

$$ \sum_{i=1}^{\infty} \langle \xi \circ T, e_i \rangle_H \tilde{e}_i - \tilde{\nabla}_{e_i} (\langle \xi \circ T, e_i \rangle_H) \text{ is convergent in } L^0(E, \mu). $$

— Lemma 3 is proven.—
CHAPTER TWO

Transformation of a Gaussian measure

Given an abstract Wiener space $(H, E, \mu)$ and $T: E \to E$ of the form :

$$Tx = x + F(x), \quad F: E \to H.$$

We shall examine when $T(\mu) \ll \mu$. We shall consider the following cases :

- $F$ is linear continuous from $E$ into $H$,

- $F$ is regular (i.e., possesses stochastic derivatives).

We shall give some expressions for the Radon-Nikodym density $\frac{dT(\mu)}{d\mu}$.

In the following chapter we shall study a family of flows : $T_t = I + F_t$ where $F_t : E \to H$, ($t \in [0, 1]$) and shall study the work of Cruzeiro, Buckdahn and Ustunel-Zakai on this subject. We shall only give the statements of the results and from time to time sketch of the proofs.

1 - Preliminary results on equivalence and orthogonality of product measures

Let $(E_k, \mathcal{B}_k)_{k \in \mathbb{N}}$ be a sequence of measurable spaces and for every $k$, let $\mu_k$ and $\nu_k$ be two probabilities on $(E_k, \mathcal{B}_k)$ such that $\mu_k \ll \nu_k$. Let us set $\rho_k = \frac{d\mu_k}{d\nu_k}$.

Let us consider the product measures :

$$\mu = \prod_{k=1}^{\infty} \mu_k$$

and

$$\nu = \prod_{k=1}^{\infty} \nu_k$$

and let

$$\alpha_k = \int_{E_k} \sqrt{\rho_k(x_k)} \nu_k(dx_k).$$

These notations having been fixed we have the following result of Kakutani :
**THEOREM 1 :** We have the dichotomy :

$$ \mu \ll \nu \text{ or } \mu \perp \nu. $$

a) $\mu \ll \nu \Leftrightarrow \prod \alpha_k$ converges; and in this case the density is equal to $\rho(x) = \prod_{k=1}^{\infty} \rho_k(x_k)$ (convergence in mean).

b) $\mu \perp \nu \Leftrightarrow \prod \alpha_n$ diverges to zero. (We cannot have divergence to infinity since $\alpha_k^2 \le 1$).

**Applications :** $E_k = \mathbb{R}$ for every $k$

$$ \nu_k(dx_k) = \frac{1}{\sigma_k \sqrt{2\pi}} \exp\left\{-\frac{(x_k - \gamma_k)^2}{2\sigma_k^2}\right\} dx_k $$

$$ \mu_k(dx_k) = \frac{1}{\lambda_k \sqrt{2\pi}} \exp\left\{-\frac{(x_k - \beta_k)^2}{2\lambda_k^2}\right\} dx_k. $$

Then

$$ \rho_k(x_k) = \frac{\sigma_k}{\lambda_k} \exp\left\{-\frac{1}{2\sigma_k^2 \lambda_k^2} \left[(x_k - \beta_k)^2 \sigma_k^2 - (x_k - \gamma_k)^2 \lambda_k^2\right]\right\} $$

and

$$ \alpha_k = \int_{\mathbb{R}} \sqrt{\rho_k(x_k)} \, d\nu_k(x_k) = \sqrt{\frac{2\lambda_k \sigma_k}{\lambda_k^2 + \sigma_k^2}} \exp\left\{-\frac{(\beta_k - \gamma_k)^2}{4(\lambda_k^2 + \sigma_k^2)}\right\}. $$

We now give some particular cases :

- Same covariance ($\lambda_k = \sigma_k$ for every $k$). $\mu$ and $\nu$ are equivalent if and only if

$$ \sum_k \frac{(\beta_k - \gamma_k)^2}{\sigma_k^2} < \infty $$

and the density is then equal to

$$ \exp\left\{\sum_{k=1}^{\infty} \frac{x_k(\beta_k - \gamma_k)}{\sigma_k^2} - \frac{\beta_k^2 - \gamma_k^2}{2\sigma_k^2}\right\}. $$

Otherwise, we have orthogonality of measures.
- Same mean $\beta_k = \gamma_k = 0$ for every $k$.

$\mu$ and $\nu$ are equivalent if and only if :

$$ \sum_{k=1}^{\infty} \frac{(\lambda_k - \sigma_k)^2}{\lambda_k \sigma_k} < \infty $$

and in this case the density is equal to :

$$ \frac{d\mu}{d\nu}(x) = \lim_{n \to \infty} \prod_{k=1}^{n} \frac{\sigma_k}{\lambda_k} \exp\left\{-\frac{x_k^2}{2} \left(\frac{\sigma_k^2 - \lambda_k^2}{\sigma_k^2 \lambda_k^2}\right)\right\}. $$

If this condition is not satisfied we have orthogonality.

## 2 - Affine transformations of Gaussian measures

Now let $(E, H, \mu)$ be an abstract Wiener space. If $(e_n)$ is an orthonormal basis of $H$, the random variables $\tilde{e}_n$ are independent Gaussian variables on $E$, with mean zero and variance one. The law of the sequence $(\tilde{e}_n)$ is therefore a product measure on $\mathbb{R}^n$:

$$ \gamma_{\mathbb{R}^n} = \bigotimes_{n=0}^{\infty} \gamma_n $$

where $\gamma_n = \gamma$ (Gaussian measure on $\mathbb{R}$) for every $n$.

Now we have a measurable (defined almost everywhere) map $\theta$ of $E$ into $\mathbb{R}^n$:

$$ x \rightsquigarrow (\tilde{e}_n(x))_n. $$

If the $e_n$ belong to $E'$, the $\tilde{e}_n$ are everywhere defined and $\theta$ is continuous from $E$ into $\mathbb{R}^n$.

It is clear now that the image of $\mu$ under $\theta$ is equal to $\gamma_{\mathbb{R}^n}$. We have $\theta(H) = \ell^2$ as we can see immediately (the $\tilde{e}_n(x)$ are defined in a unique way on $H$).

**Proposition 1** : Let $a \in E$ and $\tau_a(\mu)$ be the translate of $\mu$ by $a$. Then we have the dichotomy:

$$ \tau_a(\mu) \sim \mu \text{ or } \tau_a(\mu) \perp \mu, $$

$$ \tau_a(\mu) \sim \mu \text{ if and only if } a \in H \text{ and the density is equal to } \exp\{\tilde{a}(\cdot) - \frac{1}{2} \|a\|_H^2\}. $$
**Proof :**

$\tau_a(\mu)$ is a Gaussian (non centered if $a \neq 0$) measure with the same covariance than $\mu$.

Let $(e_n) \subset E'$ (orthonormal in $H$). It suffices to prove the same result for $\theta(\mu)$ and $\theta(\tau_a(\mu))$. But $\theta(\tau_a(\mu))$ is the product of Gaussian measures on $\mathbb{R}$ with variances one and mean $e_n(a)$. Therefore it suffices to apply the result of the previous paragraph.

— Q.E.D. —

Now let $T = I + F$ be a linear continuous transform of $E$ into $E$. Let us suppose that $F(E) \subset H$. In this case $F$ is continuous for the topology of $H$ by closed graph theorem.

Suppose moreover, that $T|_H = Id_H + F|_H$ is an **invertible operator**. Then $T: E \to E$ is also invertible and

$$T^{-1} = I - (T|_H)^{-1} \circ F.$$

**Proposition 2 :** Suppose $T = I + F$ with the above properties and that $F|_H$ is nuclear. Then $T^{-1}(\mu)$ and $\mu$ are equivalent and

$$\frac{dT^{-1}(\mu)}{d\mu}(x) = \exp\left\{-(Fx, x)_H - \frac{1}{2} \|Fx\|_H^2\right\} |\det T|.$$

**Proof :**

Let us explain what this formula means. Indeed, $F|_H$ being nuclear, admits the decomposition : $F|_H(x) = \sum_n \lambda_n (x, e_n)_H f_n$, $(e_n, f_n)$ orthonormal in $H$) and we can define $\langle F(x), x \rangle_H$ on $E$ by $\sum_n \lambda_n \tilde{e}_n(x) \tilde{f}_n(x)$, we set : $\det (I + F) = \prod_n (1 + \lambda_n)$. (This has sense since $\sum_n |\lambda_n| < \infty$).

• Let us suppose first that $F$ is symmetrical :

$$F(x) = \sum_n \lambda_n (x, e_n)_H e_n$$

where $e_n$ is an orthonormal basis composed of eigenvectors of $F$.

Let $\theta: E \to \mathbb{R}^N$ associated to these $e_n$. We have seen that : $\theta(\mu) = \gamma_N$ (product measure).
Now $\theta((I + F)^{-1}\mu)$ is the product of measures with densities :

$$ \frac{1}{\sqrt{2\pi}} (1 + \lambda_n) \exp\left\{-\frac{1}{2} (1 + \lambda_n)^2 x_n^2\right\}. $$

We have

$$ \frac{d((1 + \lambda_n)^{-1} \tilde{e}_n(\mu))}{d(\tilde{e}_n(\mu))} (x_n) = (1 + \lambda_n) \exp\left\{-\lambda_n x_n^2 - \frac{1}{2} \lambda_n^2 x_n^2\right\} $$

$$ \frac{d\left(\theta((I + F^{-1})(\mu))\right)}{d\theta(\mu)}(x) = \prod (1 + \lambda_n) \exp\left\{-(Fx, x)_H - \frac{1}{2} \|Fx\|_H^2\right\}. $$

* Now let us consider the general case (F non necessarily symmetrical)

$$ H \xrightarrow{i} E \xrightarrow{I+F} H \xrightarrow{i} E $$

$(I + F) \circ i$ is an operator from $H$ into $H$. There exists a unitary operator $U : H \to H$ "diagonalizing" $F \circ i$, therefore $(I + F) \circ i$. Let $\tilde{U}$ its extension to $E \to E$. We apply the result for $\tilde{U}(I + F) \tilde{U}^{-1}$.

— Q.E.D. —

Now we shall consider the case where $F|_H$ is not nuclear.

We know that in any case $F|_H$ is Hilbert-Schmidt.

* Suppose at first that rank (F) is finite.

Then the formula of Proposition 2 gives :

$$ \begin{aligned} & \prod_{i=1}^{n} (1 + \lambda_i) \exp\left\{-\sum_{i=1}^{n} \lambda_i x_i^2 - \frac{1}{2} \sum_{i=1}^{n} \lambda_i^2 x_i^2\right\} \\ & = \prod_{i=1}^{n} (1 + \lambda_i) e^{-\lambda_i} \exp\left\{-\left(\sum_{i=1}^{n} \lambda_i x_i^2 - \sum_{i=1}^{n} \lambda_i - \frac{1}{2} \|Fx\|_H^2\right)\right\}. \end{aligned} $$

* Now suppose F Hilbert-Schmidt with infinite rank :

$$ \prod_i (1 + \lambda_i) e^{-\lambda_i} \text{ converges since } \sum_i |\lambda_i|^2 < \infty. $$

The limit is called the "Carleman determinant".
Now we can prove that

$$ \lim_{n \to \infty} \exp\left\{-\left(\sum_{i=1}^{n} \lambda_i x_i^2 - \sum_{i=1}^{n} \lambda_i\right) - \frac{1}{2} \|Fx\|_H^2\right\} \text{ exists in } L^1(\mu) \text{ if } F \text{ is H-S.} $$

We denote it by :

$$ \exp\left\{-\left[\left(Fx, x\right)_H - \text{Trace } F\right]\right\} - \frac{1}{2} \|Fx\|_H^2 \right\}. $$

Therefore we have the following theorem :

**THEOREM 2 :** Let $T : E \to E$ linear continuous, such that $Tx = x+Fx$ with $F(E) \subset H$. Then $F|_H$ defines a Hilbert-Schmidt operator from $H$ into $H$. Suppose that $T|_H$ is invertible then $T : E \to E$ is invertible. Moreover, $T^{-1}(\mu)$ is absolutely continuous with respect to $\mu$ and we have

$$ \frac{d(T^{-1}(\mu))}{d\mu}(x) = \tilde{\Delta}(I+F) \exp\left\{-\left[\left(Fx, x\right)_H - \text{Trace } F\right]\right\} - \frac{1}{2} \|Fx\|_H^2 $$

with

$$ \tilde{\Delta}(I+F) = \prod_{1}^{\infty} (1+\lambda_i) e^{-\lambda_i}, $$

the $\lambda_i$ being the eigenvalues of $F$.

We have seen the affine case.

Now we may give the result for the general case announced in the beginning.

**THEOREM 3 :** Let $F \in D^{2,1}(H)$. Suppose that $(I+F)$ is invertible and that for every $x \in E$, the operator $I_H + \nabla F(x)$ from $H$ to $H$ is invertible, then $(I+F)^{-1}(\mu)$ is absolutely continuous with respect to $\mu$ and we have :

$$ \frac{d((I+F)^{-1}\mu)}{d\mu}(x) = \tilde{\Delta} (I_H + \nabla F(x)) \exp\left\{-\delta(F)(x) - \frac{1}{2} \|Fx\|_H^2\right\}. $$
CHAPTER THREE

# Transformation of Gaussian measures
under anticipative flows

Let $(\Omega, H, P)$ be an abstract Wiener space and let $T$ be an invertible transformation of $\Omega$ into $\Omega$ (the only interesting case will be of the form : $T := \text{Id} + F$ with $F \in \mathbb{D}^{2,1}(H)$).

**Definition** : A family of transformations $(T_t)_{t \in [0,1]}$ from $\Omega$ to $\Omega$ will be called an "interpolation" of the invertible transformation $T$ if

a) $T_0 = \text{Id}, T_1 = T,$

b) each $T_t$ is invertible,

c) for each $\omega$, $t \rightsquigarrow T_t\omega$ and $t \rightsquigarrow T_t^{-1}\omega$ are strongly continuous.

Moreover, if

d) for each $\omega$, $t \rightsquigarrow T_t\omega$ and $t \rightsquigarrow T_t^{-1}\omega$ are strongly continuously differentiable, the interpolation will be said to be "smooth".

**Example 1** : $T_t(\omega) = \omega + tA(\omega)$ where $A$ is a function from $\Omega$ to $H$, such that

$\omega \rightsquigarrow \omega + tA(\omega)$ is invertible for every $t$.

**Example 2** : Suppose $A : \Omega \to H$ is continuous and suppose that we have defined a family of transformations $(T_t)$ from $\Omega$ into $\Omega$ by :

$$T_t\omega = \omega + \int_0^t A(T_s\omega) \, ds \quad (\text{time homogeneous case})$$

$$\text{i.e.} \quad \begin{cases} \frac{dT_t}{dt}(\omega) = A(T_t\omega) \\ T_0(\omega) = \omega \end{cases}$$

we have then :

$$\frac{dT_t}{dt}(T_t^{-1}(\omega)) = A(\omega).$$
**Example 3**: $T_t(\omega) = \omega + \int_0^t \sum(s, T_s(\omega)) \, ds$.

If $\sum(r, \omega)$ is continuous on $[0, 1] \times \Omega$ into $\Omega$ or into $H$ and satisfies a global Lipschitz condition :

$$|\sum(t, \omega_1) - \sum(t, \omega_2)| \le L\|\omega_1 - \omega_2\|\Omega$$

We can consider $T_t(\omega)$ as the solution of the ordinary differential equation

$$\begin{cases} \frac{dT_t}{dt}(\omega) = \sum (t, T_t(\omega)) \\ T_0(\omega) = \omega \end{cases}$$

on the Banach space $\Omega$.

If for every $t \in [0, 1]$, $\sum (t, \cdot)$ is Fréchet differentiable, with Fréchet differential denoted by $\partial \sum (t, \omega)$, and if we assume that $\partial \sum (t, \omega)$ is bounded continuous on $[0, 1] \times \Omega$, then the equation

$$T_t \omega = \omega + \int_0^t \sum (r, T_r(\omega)) \, dr$$

has a unique solution.

Moreover, $\omega \rightsquigarrow T_t(\omega)$ is Fréchet differentiable and $\partial T_t(\omega)$ is continuous, invertible on $[0, 1] \times \Omega$, and satisfies the differential equation :

$$\frac{d}{dt} (\partial T_t \omega) = (\partial \sum (t, \cdot) \circ T_t(\omega)) \cdot \partial T_t(\omega).$$

Its inverse $\partial^{-1} T_t \omega$ satisfies :

$$\frac{d}{dt} (\partial^{-1} T_t \omega) = -\partial^{-1} T_t(\omega) \cdot (\partial \sum (t, \cdot) \circ T_t(\omega)).$$

Consequently, by the global inverse theorem, $T_t(\omega)$ is a $C_1$-diffeomorphism. Therefore, we have an interpolation of $T$ defined by

$$T(\omega) = \omega + \int_0^1 \sum (r, T_r \omega) \, dr.$$

Later on we shall come back to this example. Now let us return to the general situation.
**THEOREM 1**: Let $T$ be a transformation from $\Omega$ to $\Omega$ and $(T_t, t \in [0, 1])$ be an interpolation of $T$. Let us assume moreover that

(a) $T_t(P) \ll P$, $\forall t \in [0, 1]$ and let $X_t(\omega) = \frac{dT_t(P)}{dP}(\omega)$,

(b) $G_t = T_t^{-1} - I \in \mathbb{D}^{2,1}(H)$ and $\frac{dT_t^{-1}}{dt} \in H$,

(c) $\frac{dT_t^{-1}}{dt}$ as a function from $[0, 1] \times \Omega$ into $H$ is almost surely continuous in $(t, \omega)$ (for $dt \otimes dP$) and $\nabla T_t^{-1}(\omega)$ will be assumed to possess a continuous extension $[0, 1] \times \Omega$,

(d) $\frac{dT_s^{-1}}{ds} \circ T_s \in \mathbb{D}^{2,1}(H)$.

Then

$$X_t(\omega) = \exp\left\{-\int_0^t \left(\delta\left[\frac{dT_s}{ds} \circ T_s\right] \circ T_s^{-1}(\omega) \right) ds\right\} \quad (1)$$

This implies that the measures $T_t(P), T_t^{-1}(P)$ and $P$ are equivalent. Moreover

$$X_t = \exp\left\{ -\int_0^t \tilde{\delta}\left[\frac{dG_s}{ds}\right] ds - \frac{1}{2} \langle G_t, G_t \rangle_H - \int_0^t \text{Trace} \left[ \left( \nabla\left[\frac{dG_s}{ds} \circ T_s\right] \circ T_s^{-1} \right) \cdot \nabla G_s \right] ds \right\} \quad (2)$$

where $\tilde{\delta}$ was defined precedently by :

$$\tilde{\delta}(\xi \circ T) = (\delta\xi) \circ T - \langle \xi \circ T, F \rangle_H - \text{Trace}((\nabla\xi) \circ T \cdot \nabla F).$$

Moreover, if $\frac{dG_s}{ds}$ and $G_s$ are in $\mathbb{D}^{2,1}(H)$, then the formula (2) becomes :

$$X_t = \exp\left\{ -\delta(G_t) - \frac{1}{2} \langle G_t, G_t \rangle_H - \int_0^t \text{Trace} \left[ \left( \nabla\left[\frac{dG_s}{ds} \circ T_s\right] \circ T_s^{-1} \right) \cdot \nabla G_s \right] ds \right\}. \quad (3)$$
**Proof of (1):**

We have:

$$ 
\begin{aligned}
0 &= \frac{1}{\epsilon} [T_{t+\epsilon}^{-1} \circ T_{t+\epsilon} - T_t^{-1} \circ T_t] \\
&= \frac{1}{\epsilon} [T_{t+\epsilon}^{-1} \circ T_{t+\epsilon} - T_{t+\epsilon}^{-1} \circ T_t] + \frac{1}{\epsilon} [T_{t+\epsilon}^{-1} \circ T_t - T_t^{-1} \circ T_t].
\end{aligned}
$$

Therefore by (c)

$$ [(\nabla T_t^{-1}) \circ T_t(\omega)] \cdot \frac{dT_t}{dt}(\omega) + \frac{dT_t^{-1}}{dt} \circ T_t \omega = 0 \quad (4) $$

Let now $a: \Omega \to \mathbb{R}$ smooth and let $h \in H$. By (d) we have:

$$ 
\begin{aligned}
\langle (\nabla a) \circ T_t(\omega), h \rangle_H &= \lim_{\epsilon \to 0} \frac{\partial}{\partial \epsilon} a (T_t \omega + \epsilon h) \\
&= \lim_{\epsilon \to 0} \frac{\partial}{\partial \epsilon} [(a \circ T_t) (T_t^{-1} (T_t \omega + \epsilon h))] \\
&= \lim_{\epsilon \to 0} \frac{\partial}{\partial \epsilon} [(a \circ T_t) (\omega + \epsilon (\nabla T_t^{-1}) \circ (T_t \omega) \cdot h + o(\epsilon))] \\
&= \langle \nabla (a \circ T_t), (\nabla T_t^{-1}) \circ T_t(\omega) \cdot h \rangle_H.
\end{aligned}
$$

Now if we set $h = \frac{d}{dt} T_t(\omega)$, comparing with (4), we obtain:

$$ \langle (\nabla a) \circ T_t \omega, \frac{d}{dt} T_t \omega \rangle_H = - \langle \nabla (a \circ T_t) (\omega), \frac{dT_t^{-1}}{dt} \circ T_t (\omega) \rangle_H. $$

But the left-hand member of this equality is equal to $\frac{d}{dt} (a \circ T_t)(\omega)$. Therefore we obtain:

$$ 
\begin{aligned}
\mathbb{E}\{a \circ T_t \omega - a(\omega)\} &= \mathbb{E} \left( \int_0^t \frac{d}{ds} (a \circ T_s \omega) \, ds \right) \\
&= -\mathbb{E} \left( \int_0^t \langle \nabla (a \circ T_s) (\omega), \frac{dT_s^{-1}}{ds} \circ T_s \omega \rangle \, ds \right).
\end{aligned}
$$

But from condition (d), $(\frac{dT_s^{-1}}{ds} \circ T_s \in \mathbb{D}^{2,1}(H))$, and integrating by parts we obtain:

$$ \mathbb{E}\{a \circ T_t(\omega) - a(\omega)\} = - \int_0^t \mathbb{E}\{(a \circ T_s \omega) \delta[\frac{dT_s^{-1}}{ds} \circ T_s](\omega)\} ds $$
and

$$ \mathbb{E} \{a(\omega) \cdot (X_t(\omega) - 1)\} = -\mathbb{E} \left( \int_0^t a(\omega) X_s(\omega) \left( \delta\left[ \frac{dT_s^{-1}}{ds} \circ T_s \right] \right) \circ T_s^{-1} \omega \, ds \right). $$

Since this last inequality is true for smooth functions we have :

$$ X_t(\omega) = 1 - \int_0^t X_s(\omega) \left( \delta\left[ \frac{dT_s^{-1}}{ds} \circ T_s \right] \right) \circ T_s^{-1} \omega \, ds. $$

Finally, since $X_t$ is P-almost surely positive, $T_t P$ and $P$ are equivalent.

On the other hand, if $a: \Omega \to \mathbb{R}$ is smooth, then :

$$ \mathbb{E} \{a \circ T_t^{-1} X_t\} = \mathbb{E}a. $$

Hence if $B$ is a Borelian subset of $\Omega$, then

$$ P(B) = 0 \iff \mathbb{E}\{1_B \circ T_t^{-1} X_t\} = 0 \iff 1_B \circ T_t^{-1} = 0, \text{ a.s.} $$

Therefore, $T_t^{-1}(P)$ and $P$ are equivalent.

**Proof of (2) :**

We start from

$$ (\delta\xi) \circ T = \tilde{\delta}(\xi \circ T) + \langle \xi \circ T, F \rangle_H + \text{Trace}((\nabla\xi) \circ T \cdot \nabla F) $$

with

$$ \xi = \frac{dT_s^{-1}}{ds} \circ T_s, \quad T = T_s^{-1}, \quad F = T - \text{Id} = G_s $$

and

$$ \frac{dG_s}{ds} = \frac{dT_s^{-1}}{ds}. $$

Then

$$ \delta\left[\frac{dT_s^{-1}}{ds} \circ T_s\right] \circ T_s^{-1} = \tilde{\delta}\left(\frac{dG_s}{ds}\right) + \langle \frac{dG_s}{ds}, G_s \rangle + \text{Trace} \left( \left( \nabla\left[\frac{dG_s}{ds} \circ T_s\right] \right) \circ T_s^{-1} \cdot \nabla G_s \right) $$

and we integrate from 0 to $t$.
**Proof of (3):**

It is immediate from (2) since $\tilde{\delta} = \delta$ under this hypothesis.

We have expressed the density $X_s$ in terms of $\frac{dT_s^{-1}}{dt}$. (The next result will give an expression of $X_t$ in terms of $\frac{dT_s}{ds}$).

— Q.E.D. —

**Corollary:** *Under the assumptions and conditions of the theorem 1 let us replace $T$, $T_t$, $T_s$ and $X_t$ by $T^{-1}$, $T_t^{-1}$, $T_s^{-1}$, $\frac{dT_t^{-1}(P)}{dP} = Y_t$. Then we have :*

$$X_t(\omega) = \frac{dT_t(P)}{dP}(\omega) = \exp\left\{\int_0^t \left(\delta\left[\frac{dT_s}{ds} \circ T_s^{-1}(\cdot)\right]\right) \circ T_s T_t^{-1}(\omega) \, ds\right\}$$

and

$$X_t(\omega) = \exp\left\{-\delta(G_t)(\omega) - \frac{1}{2} \langle G_t, G_t \rangle_H(\omega) + \int_0^t \text{Trace} \left[ \left(\nabla\left[\frac{dT_s}{ds} \circ T_s^{-1}\right] \circ T_s T_t^{-1}(\omega)\right) \cdot \nabla\left(G_t - G_s (T_s T_t^{-1})(\omega)\right) \right] ds\right\}.$$

**Proof:**

By Theorem 1:

$$Y_t(\omega) = \exp\left\{-\int_0^t \left(\delta\left[\frac{dT_s}{ds} \circ T_s^{-1}\right]\right) \circ T_s(\omega) \, ds\right\}. \quad (A)$$

On the other hand, if $a$ is a smooth functional :

$$\begin{align*}
\mathbb{E}\{a(\omega) Y_t^{-1} (T_t^{-1}\omega)\} &= \mathbb{E}\{a(T_t T_t^{-1}\omega) Y_t^{-1} (T_t^{-1}(\omega))\} \\
&= \mathbb{E}\{a(T_t(\omega)) Y_t^{-1}(\omega) Y_t(\omega)\} \\
&= \mathbb{E}\{a(\omega) X_t(\omega)\}.
\end{align*}$$

Therefore :

$$X_t(\omega) = Y_t^{-1} (T_t^{-1}(\omega)) = \exp\left\{\int_0^t \left(\delta\left[\frac{dT_s}{ds} \circ T_s^{-1}(\cdot)\right]\right) \circ T_s \circ T_t^{-1}(\omega) \, ds\right\},$$

— which proves the first formula. —
To prove the second formula let us start from

$$T_s \omega = \omega + F_s(\omega)$$

which implies

$$T_s T_t^{-1} \omega = T_t^{-1} \omega + F_s (T_t^{-1} \omega),$$

and if $s = t$

$$\omega = T_t^{-1} \omega + F_t (T_t^{-1} \omega).$$

Therefore

$$T_s T_t^{-1} \omega = \omega + F_s (T_t^{-1} \omega) - F_t (T_t^{-1} \omega).$$

Now

$$G_t(\omega) = T_t^{-1}(\omega) - \omega = -F_t(T_t^{-1}\omega).$$

Therefore:

$$T_s T_t^{-1} \omega = \omega + G_t(\omega) - G_s (T_s T_t^{-1} \omega).$$

In the formula

$$X_t(\omega) = \exp\left\{\int_0^t \left(\delta\left[\frac{dT_s}{ds} \circ T_s^{-1}\right]\right) \circ T_s T_t^{-1} \omega \, ds\right\},$$

let us apply the formula given $\delta$ in terms of $\tilde{\delta}$. We obtain:

$$X_t(\omega) = \exp\left\{ \begin{aligned} & \int_0^t \left( \tilde{\delta}\left[\frac{dT_s}{ds} \circ T_t^{-1}\right](\omega) \right. \\ & \qquad \left. + \langle \frac{dT_s}{ds} \circ T_t^{-1}(\omega), G_t(\omega) - G_s(T_s T_t^{-1}\omega) \rangle_H \right. \\ & \qquad \left. + \operatorname{Trace} \left[ \left( \nabla\left[\frac{dT_s}{ds} \circ T_s^{-1}\right] \circ T_s T_t^{-1}(\omega) \right) \cdot \nabla \left( G_t - G_s(T_s T_t^{-1}) \right)(\omega) \right] \right) ds \end{aligned} \right\}$$

Now we integrate with respect to $s$, by using:

$$\frac{d}{ds}(T_s \circ T_t^{-1}(\omega)) = -\frac{d}{ds}(G_s(T_s T_t^{-1}\omega)) = \frac{d}{ds}(G_t(\omega) - G_s(T_s T_t^{-1}\omega)).$$

— We obtain the second formula.—
Now we give an integral equation satisfied by $X_t$.

**THEOREM 2 :** Let $T : \Omega \to \Omega$ and $T_t : \Omega \to \Omega$ ($t \in [0, 1]$) be an interpolation of $T$. Assume that for each $t \in [0, 1]$, $T_t(P) \ll P$ and that $X_s = [\frac{dT_s}{ds} \circ T_s^{-1}] \in \mathbb{D}_{\text{loc}}^{2,1}(H)$ (this condition is satisfied if $\frac{dT_s}{ds} \circ T_s^{-1} \in \mathbb{D}^{2,1}(H)$ and $X_s \in \mathbb{D}_{\text{loc}}^{2,1}$), then $X_t$ satisfies :

$$X_t = 1 + \int_0^t \delta[X_s \frac{dT_s}{ds} \circ T_s^{-1}] \, ds.$$

**Proof :**

Let $a$ be a smooth functional. Then

$$
\begin{align*}
\mathbb{E}\{X_t(\omega)a(\omega)\} &= \mathbb{E}\{a(T_t(\omega))\} \\
&= \mathbb{E}\{a(\omega) + \int_0^t \frac{da(T_s(\omega))}{ds} \, ds\} \\
&= \mathbb{E}\{a(\omega) + \int_0^t \langle(\nabla a) \circ T_s \omega, \frac{d}{ds} T_s(\omega)\rangle \, ds\} \\
&= \mathbb{E}\{a(\omega)\} + \int_0^t \mathbb{E}\{X_s(\omega)\langle\nabla a)(\omega), [\frac{dT_s}{ds} \circ T_s^{-1}(\omega)]\rangle \, ds \\
&= \mathbb{E}\{a(\omega)\} + \int_0^t \mathbb{E}\{a(\omega) \delta[X_s \frac{dT_s}{ds} \circ T_s^{-1}](\omega)\} \, ds
\end{align*}
$$

— Q.E.D. —

**Applications of these formulas.**

* In the example (1) : $T_t(\omega) = \omega + t A(\omega)$,

$$X_t(\omega) = \exp\left\{\int_0^t \left(\delta[A(T_s^{-1}(\cdot))]\right) \circ T_s T_t^{-1}(\omega) \, ds\right\}$$

(this result was obtained by Bell).

* In the example (2) : $T_t(\omega) = \omega + \int_0^t A(T_s(\omega)) \, ds$

$$\frac{dT_s}{ds}(T_s^{-1}(\omega)) = A(\omega)$$
and

$$X_t(\omega) = \exp\left\{\int_0^t \left(\delta(A)\right) \circ T_s T_t^{-1}(\omega) \, ds\right\}.$$

• We shall now study the example three :

$$T_t(\omega) = \omega + \int_0^t \sum(r, T_r(\omega)) \, dr. \qquad (B)$$

We have given some hypotheses insuring that $T_t\omega$ is a solution of the ODE with values in the Banach space $\Omega$

$$\left| \frac{dT_t}{dt}(\omega) \right| = \sum(t, T_t(\omega)) \\ T_0(\omega) = \omega$$

and that $\omega \rightsquigarrow T_t(\omega)$ and $\omega \rightsquigarrow T_t^{-1}(\omega)$ are Fréchet differentiable (in $\omega$). Then :

$$I_H + \nabla \int_0^t \sum(s, T_s\omega) \, ds$$

is invertible and satisfies the hypotheses of Ramer's theorem

As a consequence the probabilities

$$T_t P, P \text{ and } T_t^{-1} P \text{ are equivalent.}$$

Now in (B) we replace $\omega$ by $T_s^{-1}\omega$ :

$$T_t T_s^{-1}(\omega) = T_s^{-1}(\omega) + \int_0^t \sum(r, T_r T_s^{-1}(\omega)) \, dr.$$

Setting : $T_t T_s^{-1}(\omega) = \varphi_{s,t}(\omega)$ and $T_s T_t^{-1}(\omega) = \psi_{s,t}(\omega)$, $t \ge s$, we have :

$$\psi_{s,t} \circ \varphi_{s,t} = \varphi_{s,t} \circ \psi_{s,t} = \text{Id}$$

and :

$$\varphi_{s,t}(\omega) = \omega + \int_s^t \sum(r, \varphi_{s,r}(\omega)) \, dr$$

$$\psi_{s,t}(\omega) = \omega - \int_s^t \sum(r, \psi_{r,t}(\omega)) \, dr.$$

Note that $\varphi_{(1-s)t,t}$, $s \in [0,1]$ is, for $t$ fixed, an interpolation of $T_t$ and naturally $(T_t)_{t \in [0,1]}$ is an interpolation of $T_1 : \varphi_{s,t}$ is a "two-parameter" interpolation of $T$.
• Now we shall specialize the example in the case $\Omega = C_0[0, 1]$, with the Wiener measure and we shall use the following notations in this case :

If $U$, $U_1$ and $U_2$ are random functions with values in $H$; if $H$ is the Cameron-Martin space, then

$$ 
\begin{align*}
U(\omega)(\cdot) &= \int_{0}^{\cdot} \dot{u}(\theta, \omega) \, d\theta \\
\delta(U) &= \int_{0}^{1} \dot{u}(\theta, \omega) \delta_{\theta}(W) \\
\langle U_1, U_2 \rangle_{H} &= \int_{0}^{1} \dot{u}_{1}(\theta, \omega) \dot{u}_{2}(\theta, \omega) \, d\theta.
\end{align*}
$$

But if $H$ is the $L^2[0, 1]$ space

$$ 
\begin{align*}
U(\omega)(\cdot) &= u(\cdot, \omega) \\
\delta U &= \int_{0}^{1} u(\theta, \omega) \delta_{\theta}(W) \\
\langle U_1, U_2 \rangle_{H} &= \int_{0}^{1} u_{1}(\theta, \omega) u_{2}(\theta, \omega) \, d\theta \\
(T_t \omega)(\cdot) &= \omega(\cdot) + \int_{0}^{t} \rho(r, \cdot) \sigma(r, T_r \omega) \, dr \tag{C}
\end{align*}
$$

where $\rho$ is a smooth function on $[0, 1]^2$ and $\sigma : [0, 1] \times \Omega \to \mathbb{R}$ is assumed to satisfy Lipschitzian and differentiability conditions.

In terms of $\varphi_{s,t}$ and $\psi_{s,t}$, ($s \le t$) we have :

$$ 
\begin{align*}
\varphi_{s,t}(\omega)(\cdot) &= \omega(\cdot) + \int_{s}^{t} \rho(r, \cdot) \sigma(r, \varphi_{s,r}(\omega)) \, dr \\
\psi_{s,t}(\omega)(\cdot) &= \omega(\cdot) - \int_{s}^{t} \rho(r, \cdot) \sigma(r, \psi_{r,t}(\omega)) \, dr.
\end{align*}
$$

We consider these equations as ODE in Banach space (the first in $t$ with $s$ fixed; the second in $s$ for $t$ fixed), we have existence and unicity of solutions with

$$ \varphi_{s,s}(\omega) = \omega, \quad \psi_{t,t}(\omega) = \omega \quad \text{and} \quad \varphi_{s,t} \circ \psi_{s,t}(\omega) = \omega. $$

Then $\psi_{s,t}(\omega)$ and $\varphi_{s,t}(\omega)$ are Fréchet differentiable in $\omega \in C_0([0, 1])$.
Consequently, $\partial\varphi_{s,t}$ and $\partial\psi_{s,t}$ restricted to $H$ are invertible, and by Ramer's theorem: $\varphi_{s,t}(P)$, $\psi_{s,t}(P)$ and $P$ are equivalent.

Set

$$L_{s,t}(\omega) = \frac{d\varphi_{s,t}(P)}{dP}$$

and

$$\Lambda_{s,t} = \frac{d\psi_{s,t}(P)}{dP}.$$

Now let us fix $t$ in the equation:

$$T_t \omega (\cdot) = \omega(\cdot) + \int_0^t \rho(r, \cdot) \sigma(r, T_r \omega) dr.$$

Let $s = t - \lambda$ and $\lambda \in [0, t]$ be the interpolation parameters.

Now let us recall that (cf (3))

$$X_t = \exp\left\{ -\delta(G_t) - \frac{1}{2} \langle G_t, G_t \rangle_H \right. \\ \left. - \int_0^t \operatorname{Trace} \left( \nabla \left[ \frac{dG_s}{ds} \circ T_s \right] \circ T_s^{-1} \bullet \nabla G_s \right) ds \right\} \quad (D)$$

where $G_t = T_t^{-1} - \text{Id}$, and apply the result for $T_t$ satisfying the relation:

$$T_t \omega (\cdot) = \omega(\cdot) + \int_0^t \rho(r, \cdot) \sigma(r, T_r \omega) dr.$$

Then we obtain an expression for $X_t$:

$$X_t = \exp\left\{ \int_0^1 \left[ \int_0^t \frac{\partial \rho}{\partial \theta}(r, \theta) \sigma(r, \psi_{0,r}) dr \right] \delta_{\theta}(W) \right. \\ - \frac{1}{2} \int_0^1 \left[ \int_0^t \frac{\partial \rho(r, \theta)}{\partial \theta} \sigma(r, \psi_{0,r}) dr \right]^2 d\theta \\ \left. - \int_0^t \int_0^t \int_0^t \left[ \int_0^{\lambda} \frac{\partial \rho(r, \eta)}{\partial \eta} D_{\theta} \sigma(r, \psi_{0,r}) dr \right] \circ \frac{\partial \rho(\lambda, \theta)}{\partial \theta} (D_{\eta} \sigma(\lambda, \cdot)) \circ \psi_{0,\lambda} d\lambda d\theta d\eta \right\}$$

We can obtain another formula for the Radon-Nikodym density using the relation:

$$\delta(aU) = a\delta U - \langle \nabla a, U \rangle_H$$
in the expression :

$$X_t(\omega) = \exp \left\{ \int_0^t \left( \delta\left[ \frac{dT_s}{ds} \circ T_s^{-1} \right] \right) \circ T_s T_t^{-1}(\omega) \, ds \right\}.$$

We then obtain :

$$L_{s,t} = \exp \left\{ \int_s^t \sigma(r, \psi_{r,t}) \left[ \delta\rho(r, \cdot) - \int_s^r \sigma(u, \psi_{u,t}) \langle \rho(r, \cdot), \rho(u, \cdot) \rangle_H du \right] dr \right. \\ \left. - \int_s^t \langle (\nabla\sigma)(r, \psi_{r,t}), \rho(r, \cdot) \rangle_H dr \right\}.$$

## REFERENCES

* BUCKDAHN, R., *Anticipative Girsanov transformations and Skorokhod stochastic differential equations*. Memoirs of the A.M.S. n°:533, 1994.

* CRUZEIRO, A.B., *Equations differentielles sur l'espace de Wiener et formules de Cameron-Martin non-linéaires*. J.F.A., 54, pp.206-227, 1983.

* NUALART, D., *Noncausal stochastic integrals and calculus*. L.N.M., 1516, pp.80-129, 1988.

* NUALART, D. and PARDOUX, E., *Stochastic calculus with anticipating integrands*. Probability Theory and Related Fields. 78, pp.535-581, 1988.

* NUALART, D. and ZAKAI, M., *On the relation between the Stratonovich and Ogawa integrals*. Ann.Proba. 17, pp.1536-1540, 1989.

* USTUNEL, A.S. and ZAKAI, M., *Transformation of Wiener measure under anticipative flows*. Probability Theory and Related Fields 93, pp. 91-136, 1992.