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I'm going over some materials on stochastic analysis, and stuck with a problem on Gaussian white noise:

Let $(\mathbb{R}^d,\mathcal{B},m)$ be the Borel measurable space on $\mathbb{R}^d$. A white noise is a mean zero Gaussian process $\{W(A):A\in\mathcal{B},m(A)<\infty\}$ with covariance $\mathbb{E}[W(A)W(B)]=m(A\cap B)$.

Fix positive numbers $\{c_k\}_{k\in\mathbb{N}}$ such that $\sum c_k^2=1$ but $\sum c_k=\infty$. Let $A=\bigcup_k A_k$ be a disjoint union of Borel sets with $m(A_k)=c_k^2$. Show that $\sum|W(A_k)|=\infty$ a.s.

Even though one can show that $\sum W(A_k)=W(A)$ a.s. by a standard result on random series, the problem above shows that $\sum W(A_k)$ doesn't converge absolutely and hence $W(\cdot,\omega)$ cannot be a signed measure.

Any thought will be appreciated.

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  • $\begingroup$ I am confused, is $W(A)$ a $N(0, m(A))$ rv? $\endgroup$ – Will M. Dec 13 '18 at 19:00
  • $\begingroup$ @WillM. Yes. And $W(A_k)$'s are independent. $\endgroup$ – SimonChan Dec 13 '18 at 20:16
  • $\begingroup$ I think this can be proved by Kolmogorov zero-one law and three series theorem. $\endgroup$ – SimonChan Dec 13 '18 at 21:16
  • $\begingroup$ I agree with your last comment, and do not think there is any easier method. $\endgroup$ – Mike Earnest Dec 13 '18 at 23:52

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