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Is there are standard probability space $(\Omega, \mathcal{A}, P)$ and a process $X_t : \Omega \to \{ -1, 1 \}$, $t \in [0,1]$ such that $X_t$ is uniformly distributed on $\{ -1, 1 \}$ and all the $X_t$ are independent (or uncorrelated)?

By Kolmogorov one can construct such a process but it is not clear to me whether $\Omega$ can be chosen to be standard.

I can show at least that such a process is not measurable (i.e. $X : \Omega \times [0,1] \to \{ -1, 1 \}$ is not measurable) but one can find a version with measurable sample paths $X(\omega) : [0,1] \to \{ -1, 1 \}$ for almost all $\omega$.

EDIT: The Kolmogorov construction gives $\Omega = \{ -1, 1 \}^{[0,1]}$ with the product $\sigma$-algebra and product measure (since all $X_t$ are independent). But $\Omega$ has cardinality $2^{c} > c$. Thus $\Omega$ is never standard Borel thus not a standard probability space (when a standard probability space is defined to be the completion of a standard Borel space equipped with some Borel probability measure). So we need to show that any other choice for $\Omega$ has cardinality $> c$.

yada
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  • By "all the $X_t$ are independent" do you mean that for any finite $E\subset[0,1]$, the set of random variables ${X_t:t\in E}$ is independent, or that $X$ has independent increments? – Math1000 Nov 20 '15 at 09:46
  • Yes, every finite subset is independent. I do not mean independent increments (the state space ${ -1, 1 }$ need not carry any linear structure). – yada Nov 20 '15 at 09:47
  • The only example I can think of is where $\Omega={H,T}\times[0,1]$, but there are some issues. See here: http://mathoverflow.net/questions/173037/uncountable-family-of-random-variables – Math1000 Nov 20 '15 at 12:21
  • How do you define the process $X$ on $\Omega = { H, T } \times [0,1]$? Do you mean something like $X_t((H,\omega)) := H$ if $\omega = t$? I think one has to show (if it is true) that on a standard probability space a process can have a measurable version (which would contradict such an existence for the iid process above)). – yada Nov 20 '15 at 13:49
  • As I said, there are some issues :) – Math1000 Nov 20 '15 at 13:49

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