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I have a quesition about an isonormal Gaussian process and measurability.

Let $\mathcal{H}$ be a real separable Hilbert space with inner product $\langle \cdot,\cdot \rangle$ and norm $\|\cdot\|=\langle\cdot,\cdot\rangle^{1/2}$. Let $X=\{X_{h}:h \in \mathcal{H}\}$ be an isonormal Gaussian process over $\mathcal{H}$ i.e. $X$ is a collection of jointly Gaussian random variables defined on a probability space $(\Omega,\mathcal{F},P)$ and such that $E \left[X_{h}X_{g} \right]=\langle h,g\rangle$ for all $h,g \in \mathcal{H}$, $E\left[X_{h} \right]=0$ for all $h \in \mathcal{H}$.

We shall assume $\mathcal{F}=\sigma[X]$.

My question

Let $(h_{j})_{j =1}^{\infty}$ be an orthonormal basis of $\mathcal{H}$ and $\mathcal{F}_{m}\,(m\in\mathbb{N})$ be the $\sigma$-algebra generated by $X_{h_{1}},\ldots,X_{h_{m}}$. In this case can we deduce $\mathcal{F}=\sigma \left[ \bigcup_{m=1}^{\infty} \mathcal{F}_{m} \right]$?

My attempt:

I think it is enough to show that $\forall h \in \mathcal{H}$, $X_{h}$ is $\sigma \left[ \bigcup_{m=1}^{\infty} \mathcal{F}_{m} \right]$-measurable.

Since $h=\lim_{n \to \infty} \hat{h}_{n}$ in $\mathcal{H}\quad(\hat{h}_{n}=\sum_{j=1}^{n}\langle h,h_{j} \rangle h_{j})$,

\begin{align*} E \left[\left|X_{h}-X_{\hat{h}_{n}} \right|^{2} \right]=\|h-\hat{h}_{n}\|^{2}\to0\quad(n \to \infty) \end{align*}

Therefore $X_{h}=\lim_{k \to \infty}X_{\hat{h}_{n_{k}}}$ $P$-a.s. but not in pointwise. If we could prove $X_{h}(\omega)=\lim_{k \to \infty}X_{\hat{h}_{n_{k}}}(\omega)$ for all $\omega \in \Omega$, $X_{h}$ is $\sigma \left[ \bigcup_{m=1}^{\infty} \mathcal{F}_{m} \right]$-measurable.

(since $X_{\hat{h}_{n}}$ is $\sigma \left[ \bigcup_{m=1}^{\infty} \mathcal{F}_{m} \right]$-measurable.)

Thank you in advance.

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No.

Take a non-zero $g\in\mathcal{H}$ such that $g$ is not an element of the orthonormal basis $(h_j)_{j=1}^\infty$. Let $B$ be a Bernoulli random variable with parameter $1/2$, independent of $X$. Define, for $h\in\mathcal{H}$, $$ \tilde{X}_h = \begin{cases}X_g\mathbf{1}(B = 0)\qquad &\text{if $g=h$}\\ X_h \qquad &\text{otherwise} \end{cases} $$

$\tilde{X}_g$ is not measurable with respect to $\sigma(\tilde{X}_{h_1},\tilde{X}_{h_2},\cdots)=\sigma(X_{h_1},X_{h_2},\cdots)$, since it depends on the outcome of $B$! (Prove this if it's not obvious.)

Now instead choose the parameter of $B$ to zero (then $B=0$, a.s.). This doesn't change the measurability of $\tilde{X}_g$. Now $\tilde{X}$ is an isonormal process, but $\tilde{X}_g$ still isn't measurable with respect to $\sigma(\tilde{X}_{h_1},\tilde{X}_{h_2},\cdots)$.

This is why it is typical to work with a complete $\sigma$-algebra.

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  • $\begingroup$ Thanks for your reply. I hardly understand. I try to understand your answer. $\endgroup$ – sharpe Mar 23 '15 at 2:58
  • $\begingroup$ @sharpe You're welcome. If you don't understand anything, let me know and I'll add detail to the answer. $\endgroup$ – Ben Derrett Mar 23 '15 at 9:09

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