# Natural filtration and Kolmogorov existence theorem

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.

• 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.
– Ian
Commented Mar 5, 2017 at 1:02
• 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).
– Ian
Commented Mar 5, 2017 at 1:05
• @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. Commented Mar 5, 2017 at 1:16
• requiring to be seperable, one can use the special case here: en.wikipedia.org/wiki/… Commented Mar 5, 2017 at 1:17
• @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 Commented Mar 7, 2017 at 14:23

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 $X^*$.

• Sorry could you specify what $\mathscr G$ is? Commented Mar 8, 2017 at 17:13
• 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 $\Omega$ Commented Mar 8, 2017 at 17:26
• $X_s_j$ 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. Commented Mar 9, 2017 at 9:17