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Given a stochastic process $\{X_t: t\in R^+\}$, which takes value in $R$, there is always a natural filtration $(\mathcal F^X_t)$ induced by $X_t$, i.e. $\mathcal F_t^X = \sigma(\{X_s^{-1}(A): s\le t, A \text{ is Borel-measurable}\})$.

Now, consider the inverse situation: Given a filtration $(\Omega, \mathcal F_t)$, (we may consider the special case: $\Omega = R^{R^+}$ first) is it true that there is always a stochastic process $(X_t)$ which is adapted to $\mathcal F_t$, and that $\mathcal F_t = \mathcal F^X_t$ for all $t\in R^+$? And if it is not true, can we add some conditions (hopefully, only some weak conditions) to make it true?

Olorun
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Jay.H
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  • Not sure if I am missing a point but for each $t$ you could define a r.v. $Z_t$ such that $\sigma(Z_t) = \mathcal{F}_t$. If you stack these random variables you get a process that generates the given filtration. – Calculon Jan 19 '16 at 07:23
  • I think this question is equivalent to: do every $\sigma$-algebra have a random variable that induces it. Sounds like yes if the $\sigma$-algebra have cardinality no greater than $\mathbb{R}$, though I may be wrong. – Shengjia Zhao Jan 19 '16 at 07:53
  • @ShengjiaZhao Indeed, you can define the countable sum of indicator functions over the elements of the sigma-algebra as your random variable. I think that should work. – Calculon Jan 19 '16 at 09:10
  • Good point, I think you guys are right, the question can be reduced to a problem for each single t. But, @Calculon, I don't understand your second point, the cadinality of a sigma- algebra can be more than "countable many", how would construct the rv in such cases? – Jay.H Jan 19 '16 at 13:38
  • @Jay.H You should take what I am about to write with a grain of salt as I am not 100% sure of its truthfulness. Suppose you start with a certain sigma algebra $\mathcal{F}$ and a r.v. $X$ that is measurable wrt $\mathcal{F}$. Then you can find a sequence of simple functions (sums of indicators) that converges to $X$ pointwise. Clearly, you cannot exhaust all the elements of $\mathcal{F}$ this way. So it seems to me that you can't find a r.v. that would do what you need. – Calculon Jan 19 '16 at 13:51
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    For fixed $t$ there does not necessarily exist $Z_t$ such that $\sigma(Z_t) = \mathcal{F}_t$; see this question: http://math.stackexchange.com/q/267584/ (Note that this implies in particular that the there does, in general, not exist a process which generates the filtration.) – saz Jan 19 '16 at 15:18
  • Thanks saz. So that discussion relies on cardinality. I guess that was also what in @Shengjia's mind when he commented about the cardinality. I would like to push this a little further, and posted a related question where cardinality will not be helpful. http://math.stackexchange.com/questions/1618337/extending-borel-sigma-algebra-a-little-bit – Jay.H Jan 19 '16 at 16:06

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