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Consider a stochastic process $(X_t)_{t \in T}$ where $T\subseteq[0,\infty)$. The natural filtration $\mathscr F^X=(\mathscr F_t^X)_{t \in T}$ is defined by $$\mathscr F_t^X=\sigma\{X_s^{-1}(B)|0\lt s\le t,B\in\mathscr B(\Bbb R) \}$$

In a book I have read, the author wants to define a mapping $$\psi:\omega\mapsto (X_t)_{t\in T}(\omega):=(\omega(t))_{t\in T}$$ from the underlying probability space $(\Omega,\mathscr A, P)$ to a measure space $(\Bbb R^T, \mathscr S_{cyl})$, the space of sample paths with the sigma-algebra generated by the cylinder-sets. Thus, $$\Bbb R^T=\{f:T\to \Bbb R\}\qquad\mathscr S_{cyl}=\sigma \{f∈\Bbb R^T:\forall i=1,2,...,n,\ f(t_i)\in B_i\}$$ and $$X_t:(\Bbb R^T, \mathscr S_{cyl})\to(\mathbb R,\mathcal B(\mathbb R)),\quad X_t(f)=f(t)$$ is the coordinate mapping in order to use the Kolmogorov Existence Theorem to construct a probability distribution between the time instance $0\le t_1\le\cdots\le t_n\le T$. Equivalently they use a separable process and define the natural filtration $\mathscr G=(\mathscr G_t)$ as the sigma algebras $\mathscr G_t$ generated by the following cylinder sets $$A=\{\omega \in \Omega:\forall i=1,2,...,n,\ X_{s_i}(\omega) \in B_i\}$$

for $0\le s_1\le.....\le s_n \le t$.

My questions are:

1.Where is the natural filtration $\mathscr G$ defined? On $(\Bbb R^T, \mathscr S_{cyl})$? Where is the equivalence to the usual definition $\mathscr F^X$ given above?
2. If I want to calculate the expectation value $E[X^*_t]$ with $$X^*_t=\sup_{0\le t\le T} X_t$$ am I calculating it on $(\Bbb R^T, \mathscr S_{cyl})$ or $(\Omega,\mathscr A, P)$? I am confused between this pair of spaces and how to use them.

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  • $\begingroup$ If I'm understanding the setting correctly, the filtration consists of $\sigma$-algebras of subsets of $\Omega$ and expectation values are Lebesgue integrals over $\Omega$ against the measure $P$. The mapping $\psi$ simply maps $\omega$ values to sample functions. $\endgroup$ – Ian Mar 5 '17 at 1:02
  • $\begingroup$ The way I usually think about this is that a stochastic process a priori has a signature of $\Omega \times T \to V$ where $V$ is some set of values of the process, usually $\mathbb{R}^n$. Then the mapping $\psi$ is the curried version of this function, i.e. it has signature $\Omega \to (T \to V)$. The remainder of this discussion amounts to making sense of the idea that $\psi$ is a "path-valued random variable" (i.e. it is a measurable function with values in a space of paths, with respect to some sigma-algebra on the space of paths). $\endgroup$ – Ian Mar 5 '17 at 1:05
  • $\begingroup$ @Ian Thanks! If I am understanding it correctly, the second definition of the natural filtration, i.e. $\sigma(A)$ is more a natural filtration on the space of paths? Because it is actually a cylinder-set, which is the generator of the sigma algebra of linear function(which is actually the sample path). Can it somehow related it to the usual definition? The point I don't understand is just, in the first definition, they take all uncountable time instants while the second only finite time instants are considered. The both can somehow not be the same. $\endgroup$ – quallenjäger Mar 5 '17 at 1:16
  • $\begingroup$ requiring to be seperable, one can use the special case here: en.wikipedia.org/wiki/… $\endgroup$ – quallenjäger Mar 5 '17 at 1:17
  • $\begingroup$ @quallenjäger : A few remark on your notations, unless mistaken to be consistent with the first part of your post $\sigma_{cyl}=\sigma \{f\in \Omega :f(t_i)(ω)\in B_i\ for\ i=1,2,...,n\}$ should be written : $\sigma_{cyl}=\sigma \{f\in \Bbb R^T:f(t_i)\in B_i\ for\ i=1,2,...,n\}$ if your cylindrical sigma field is over $\Bbb R^T$. You need the canonical process $\psi$ to bridge both, if you wanted to write things with respect to a sigma field over $\Omega$: $\sigma_{cyl}=\sigma \{\omega \in \Omega :X_{t_i}(\psi(\omega)) \in B_i\ for\ i=1,2,...,n\}$. Best regards $\endgroup$ – TheBridge Mar 7 '17 at 14:23
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Now that did performed a great editing work on your post (I added 2 minor changes to your post on notations).

I am able to give you an answer to the first question, of course $\mathcal{G}$ lives on $\Omega$, does it needs the filtered space $(\Bbb R^T, \mathscr S_{cyl})$ to be defined ? The answer is indeed !

To show you this and answer the last part of your first question let me write a plain generator set $A\in \mathcal{G} $ again departing from your definition and noting that there is an abuse of notation here :

$$A=\{\omega \in \Omega:\forall i=1,2,...,n, s_i \in [0,t]\cap T, \ X_{s_i}(\omega) \in B_i\}$$

Note that $\ X_{s_i}(\omega)$ has no meaning unless you identify (which is done implicitly) $\omega$ with $\psi(\omega)$, so getting back to the definition we get : $$A=\{\omega \in \Omega:\forall i=1,2,...,n, s_i \in [0,t]\cap T,\ X_{s_i}(\psi(\omega))\in B_i\}$$

Ok now we get a clearer picture, to get $\mathscr F_t^X$, you have to use a measure theoretical result but on a collection of set that generates this filtration. But let's suppose that we know that it is generated by sets B (let's call them cylindrical sets ) which can be defined like this :

$$B=\{\omega \in \Omega:\forall i=1,2,...,n, s_i \in [0,t]\cap T,\ \omega \in X^{-1}_{s_i}(B_i)\}$$ here again $X_{s_i}$ is assimilated to the product $X_{s_i}\ (\psi(.)) $ so $X_{s_i}^{-1}(.)= \psi^{-1}(X_{s_i}^{-1}(.))$

So if we want to write things without any implicit notations we get :

$$B=\{\omega \in \Omega:\forall i=1,2,...,n, s_i \in [0,t]\cap T,\ \omega \in \psi^{-1}(X^{-1}_{s_i}(B_i))\}$$

Now compare sets $A$ and $B$ those are exactly the same sets, given the looked for equivalence, as long as you know that the sets of the form $B $ (or $A$) generate the natural filtration (I still need to find the proper reference for this but it is usaly proven with Monotone Class Theorem argument unless mistkan which is a bit heavy for me so I pass).

For the second question, you certainly calculate this over $\Omega$ but to achieve any calculus of this type, you (always) use implicitly the transfer theorem (this a french appellation I don't know the correct name for this in english so please edit this if you know the correct denomination in english) to get for example the image measure form $\Omega$ to $\Bbb R$ measure with respect to Lebesgue measure to express the law of $X^*$.

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  • $\begingroup$ Sorry could you specify what $\mathscr G$ is? $\endgroup$ – quallenjäger Mar 8 '17 at 17:13
  • $\begingroup$ So far I understand, is the Random Variable $X_{s_j}$ defined on the space $\Bbb R^T$? The existence of the space $\Bbb R^T$ depends on the mapping of $\psi$. The set $B$ is defined on the space $\Omega$. I thought I need a filtration on $\Bbb R^T$, but if I take $B$ as generator, I would have a filtration instead on $\Omega$ $\endgroup$ – quallenjäger Mar 8 '17 at 17:26
  • $\begingroup$ $X_s_j$ formally is not a RV (random variable) in your context (because $\Bbb R^T$ is not $\Omega$ it is only a regular space, a RV goes from an abstract space to a state space). It is only a measurable function from $\Bbb R^T$ to $\Bbb R$, it becomes a RV once you get the abuse of notation. This abuse makes $X_s_j$ a RV from $\Omega \to \Bbb R$. Once again this not rigorous but hey that the way it is so get used to it ! I understand all the puzzlement it generates, once you get used to this you won't even think of it as an abuse, that's why it is used. $\endgroup$ – TheBridge Mar 9 '17 at 9:17

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