Isonormal Gaussian process

normal distributionstochastic-analysisstochastic-processes

I am reading Nualart's book The Malliavin Calculus and Related Topics and there is some issues that I am stuck wtih.

Let $H$ be a real separable Hilbert Space with. A stochastic process $W=\{W(h)\;;h\in H\}$ defined in a complete probability space $(\Omega,\mathcal{F},P)$ is an isonormal Gaussian process if $W$ is a centered Gaussian family of random variables such that $E(W(h)W(g))=\langle h,g\rangle$ for all $g,h\in H.$

The map $h\mapsto W(h)$ is linear. Now Nualart said

  1. The mapping $h\to W(h)$ provides a linear isometry of $H$ onoto a closed subspace of $L^2$ that will denote by $\mathcal{H}_1.$

Question $1$: Why the map provides a linear isometry ?

Denote, now, by $\mathcal{G}$ the $\sigma-$alebra generated by the random variables $\{W(h)\;; h\in H\}.$

  1. The set $\{\exp(W(h)),h\in H\}$ form a total subset of $L^2(\Omega,\mathcal{G},P)$

The proof goes as follow, we take $X\in L^2$ such that $E(Xe^{W(h)})=0$ for all $h\in H.$

By linearity of $h\to W(h)$ we have $$E\big( X\exp(\sum_{i=1}^n t_iW(h_i))\big)=0.\quad (3)$$

This equation says that Laplace transform of the signed measure $E\big(X\mathbb{1}_{B}(W(h_1),W(h_2),\ldots,W(h_n))\big)$ is identically zero on $\Bbb{R}^n.$

Then $E(X\mathbb{1}_{G})=0$ for all $G\in\mathcal{G}$ so that $X=0.$

Quesstion $2$: I don't understand why linearity of $h\to W(h)$ gives us equation $(3)$ and cannot write down that is the Laplace transform of of the signed measure $E\big(X\mathbb{1}_{B}(W(h_1),W(h_2),\ldots,W(h_n))\big).$

Best Answer

First question: $\|W(g)\|^{2}=\langle W(g), W(g) \rangle =\langle g, g \rangle =\|g\|^{2}$ so $\|W(g_1)-W(g_2)\|=\|W(g_1-g_2)\|=\|g_1-g_2||$. This proves that $g \to W(g)$ is an isometry. To prove (3) all you have to do is write $\sum_{i=1}^{n} t_iW(h_{i})$ as $W(h)$ with $h =\sum_{i=1}^{n} t_i h_{i}$. For the last part you need the following fact: suppose $\mu(B)=EXI_B(Z)$ for Borel sets $B$ in $\mathbb R^{n}$ where $Z$ is a random vector with $n$ components. Then, for any non-negative measurable function $f$ we have $\int f\, d\mu =EXf(Z)$. To prove this last equation you just have to note that it holds when $f$ is an indiactor function (by definition), hence for all simple functions $f$, hence for all non-negative measurable functions $f$.

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