This is just a short question, and may be to basic, but:
is there a way to construct a sequece of independent wiener processes on a given probability spaces?
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$\begingroup$ As you know, not every probability space supports a wiener process. $\endgroup$– AkiraCommented May 3 at 15:15
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$\begingroup$ As long as you have a single normal random variable, you can construct a sequence of independent Wiener processes. $\endgroup$– mikeCommented May 3 at 15:17
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$\begingroup$ @mike that sounds promissing. Can you give more details please? $\endgroup$– PerelmanCommented May 3 at 15:18
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$\begingroup$ it's pretty standard, the base two expansion of a uniform yields iid bernoulli's which can be put back together into infinitely many independent uniforms, which can be transformed into normals with inverse cumulant transforms & then made into as many wiener processes as you like via, say, the fourier expansion. $\endgroup$– mikeCommented May 3 at 15:23
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$\begingroup$ Do you have a response to the answer below? $\endgroup$– Iosif PinelisCommented May 5 at 13:15
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1 Answer
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$\newcommand\om\omega\newcommand\Om\Omega\newcommand\F{\mathcal F}$Let $(\Om,\F,P)$ be a probability space.
A necessary and sufficient condition for $(\Om,\F,P)$ to support a sequence of independent Wiener processes is that $(\Om,\F,P)$ be non-atomic.
Indeed, as noted in the comment by mike, it is enough to show that a necessary and sufficient condition for $(\Om,\F,P)$ to support a random variable $U$ (r.v.) uniformly distributed on $[0,1]$ is that $(\Om,\F,P)$ be non-atomic.
The necessity is clear: If $A$ is an atom, then for for some real $c$ we will have $P(U=c)\ge P(A)>0$.
To prove the sufficiency, note that, by the (proof of) Sierpiński's theorem on non-atomic measures, there a nondecreasing family $(A_t)_{t\in[0,1]}$ in $\F$ such that $P(A_t)=t$ for all $t\in[0,1]$; without loss of generality, $A_1=\Om$.
For each $\om\in\Om$, let now $$U(\om):=\inf\{t\in[0,1]\colon\om\in A_t\}.$$ Then for each $s\in[0,1)$ $$U^{-1}([0,s])=\bigcap_{n=1}^\infty A_{\min(1,s+1/n)},$$ so that $U$ is $\F$-measurable and $$P(U^{-1}([0,s]))=\lim_{n\to\infty}P(A_{\min(1,s+1/n)})=s.$$ So, $U$ is a r.v. on $(\Om,\F,P)$ that is uniformly distributed on $[0,1]$. $\quad\Box$
answered May 3 at 16:30