Increasing sequence of sigma algebras

  • #1
Stephen Tashi
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Main Question or Discussion Point

The explanation of a continuous Markov process [itex] X(t) [/itex] defines an indexed collection of sigma algebras by [itex] \mathcal{F}_t = \sigma\{ X(s): s < t\} [/itex] and this collection is said to be increasing with respect to the index [itex] t [/itex].

I'm trying to understand why the notation used for set inclusion is used to express the relation of "increasing" for a collection of sigma algebras.

A straightforward approach is to think of a set of sigma algebras that are each a collection of subsets of the same set and to define the concept of sub-sigma algebra in terms of one collection of sets being a subset of another collection of sets.

However, don't [itex] \mathcal{F}_t [/itex] and [itex] \mathcal{F}_s [/itex] denote sigma algebras defined on different sets when [itex] s \ne t [/itex] ? I think of [itex] \mathcal{F}_t [/itex] as being a sigma algebra of subsets of (only) the set of all trajectories of the process up to time [itex] t [/itex]. [itex] \ [/itex] If [itex] s > t [/itex] then isn't [itex] \mathcal{F}_s [/itex] a sigma algebra of subsets of a different set of trajectories?
 

Answers and Replies

  • #2
andrewkirk
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Every ##\mathscr{F}_t## is a subset of the powerset of ##\Omega##, which is the set of all full trajectories, from the earliest time to the latest time.

The difference between ##\mathscr{F}_s## and ##\mathscr{F}_t## where ##s<t## is that the former groups together all full trajectories that have the same path up to time ##s## and the latter does the same in relation to the later time ##t##. So the latter is a finer subdivision because it has the extra info that emerges between times ##s## and ##t##.
 
  • #3
Stephen Tashi
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Every ##\mathscr{F}_t## is a subset of the powerset of ##\Omega##, which is the set of all full trajectories, from the earliest time to the latest time.
Ok
The difference between ##\mathscr{F}_s## and ##\mathscr{F}_t## where ##s<t## is that the former groups together all full trajectories that have the same path up to time ##s## and the latter does the same in relation to the later time ##t##.
What is the technical implementation of the concept of "groups together"?

So the latter is a finer subdivision because it has the extra info that emerges between times ##s## and ##t##.
I understand the concept of "extra info", but I don't see what mathematical object is being subdivided.
 
  • #4
andrewkirk
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Let ##t_0## be the earliest time and ##T## the latest time, and ##\Omega## be the set of all possible paths traced out by the process between these two. ##\Omega## is called the sample space. ##\mathscr{F}## (note the lack of subscript) is a subset of the powerset of ##\Omega##, that defines the set of all measurable subsets of ##\Omega##. We will denote paths in ##\Omega## by ##\omega##.

Denote the value of path ##\omega## at time ##t## by ##X(t,\omega)##.

Then for every time ##t## define an equivalence relation ##\sim_t## on ##\Omega## by:

$$\omega_1\sim_t\omega_2\textrm{ iff }\forall s\in[t_0,t]\ X(s,\omega_1)=X(s,\omega_2)$$

That is, the paths ##\omega_1,\omega_2## are equivalent by relation ##\sim_t## iff they are identical up to time ##t## (They may diverge after that).

Then, if we define ##\mathscr{C}_u## as the partition of ##\Omega## arising from the equivalence relation ##\sim_u##, we can define ##\mathscr{F}_t\equiv\bigcup{u=t_0}^t\mathscr{C}_u##. [That last sentence has been corrected, based on observations by Stephen Tashi]

The next para, italicised and enclosed in square brackets [], is wrong, so ignore it. I've left it in just to avoid the change causing confusion about what was originally written
[So if ##S\in\mathscr{F}_t## then all paths in ##S## are identical up to time ##t##. If ##S_1,S_2\in\mathscr{F}_t## and ##S_1\neq S_2## then every path in ##S_1## must differ from every path in ##S_2## for at least one time in the range ##[t_0,t]##.]

It is easy to deduce from this that ##\mathscr{F}_t## is a sigma algebra that is a subset (subalgebra) of ##\mathscr{F}## and that ##\mathscr{F}_s\subseteq\mathscr{F}_t## if ##s<t##.

The sequence of sigma algebras ##\{\mathscr{F}_t\}_{t\in[t_0,T]}## is called a filtration on the measure space ##(\Omega,\mathscr{F})##.
 
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  • #5
Stephen Tashi
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It is easy to deduce form this that ##\mathscr{F}_t## is a sigma algebra that is a subset (subalgebra) of ##\mathscr{F}## and that ##\mathscr{F}_s\subseteq\mathscr{F}_t## if ##s<t##.
If [itex] a \in \mathscr{F}_s [/itex] then [itex]a [/itex] can be written as an uncountable union of those members in [itex] \mathscr{F}_t [/itex] that have the same trajectory as [itex] a [/itex] on the interval [itex] [t_0,s] [/itex], but how do I see that [itex] a [/itex] can be expressed using the operations of a sigma algebra?
 
  • #6
andrewkirk
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how do I see that a a can be expressed using the operations of a sigma algebra?
Are you asking whether ##a\in\mathscr{F}_s## can be expressed in terms of elements of ##\mathscr{F}_t## using the operations that are closed within a sigma algebra? If so then it's easy, because ##a## is a member of ##\mathscr{F}_t## as well, since ##\mathscr{F}_s\subseteq\mathscr{F}_t##. So the expression needed is simply ##a=a##.
 
  • #7
Stephen Tashi
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If so then it's easy, because ##a## is a member of ##\mathscr{F}_t## as well,
The way I look at it intuitively is that [itex] \mathscr{F}_t [/itex] is generated by a collection of sets that are each "smaller than [itex] a [/itex]" or exclude [itex] a [/itex]. [itex] \ [/itex] Am I not visualizing this correctly? We did say that [itex] \mathscr{F}_t [/itex] is formed by taking equivalence classes of sets that make a finer partition of [itex] \mathscr{F} [/itex] than the partition used to define [itex] \mathscr{F}_s [/itex].

For example, pretending for the moment that we have a discrete index, an element of the partition used to define [itex] \mathscr{F}_s [/itex] could be as set like [itex] a = [/itex] all sets of the form [itex] \{1,5,7, ?, ?, ?, ... \} [/itex] where "[itex] ? [/itex]" can be an arbitrary value. An element of the partition used to define [itex] \mathscr{F}_t [/itex] could be something like [itex] b = [/itex] all sets of the form [itex]\{1,5,7,6,4,?,?,? ...\} [/itex]. We have [itex] b \subset a [/itex].
 
  • #8
andrewkirk
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For example, pretending for the moment that we have a discrete index, an element of the partition used to define ##\mathscr{F}_s## could be as set like ##a## = all sets of the form {1,5,7,?,?,?,...} where "?" can be an arbitrary value. An element of the partition used to define ## \mathscr{F}_t## could be something like ##b## = all sets of the form {1,5,7,6,4,?,?,?...}. We have ##b \subset a## .
Actually, this makes me realise that I didn't express it quite correctly in post 4, where I said we 'define ##\mathscr{F}_t## as the partition of ##\Omega## arising from the equivalence relation ##\sim_t##.' Most sigma-algebras are not partitions, because they contain overlapping sets, and in this case they certainly do.

What I should have written is that, if we use ##\mathscr{C}_u## to denote the collection of sets that is the partition of ##\Omega## defined by ##\sim_u##, then ##\mathscr{F}_t## is the sigma algebra generated by ##\bigcup_{u=t_0}^t \mathscr{C}_u##.

So every new value of ##t## adds a new bunch of sets to the growing sigma algebra. Because the ##t## index is continuous, we don't get a nice intuitive feel for this, that enables us to imagine the sigma algebra being 'constructed' as ##t## increases, like we do with discrete indices, because there is no 'next' or 'last' ##t##. But it's all perfectly well-defined and rigorous.

I hope that makes more sense. Sorry for misleading you.
 
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  • #9
Stephen Tashi
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What I should have written is that, if we use ##\mathscr{C}_u## to denote the collection of sets that is the partition of ##\Omega## defined by ##\sim_u##, then ##\mathscr{F}_t=\bigcup_{u=t_0}^t \mathscr{C}_u##.
Do we need to say [itex]\mathscr{F}_t = [/itex] the sigma algebra generated by [itex] \bigcup_{u=t_0}^t \mathscr{C}_u [/itex] ?

But anyway, I get the basic idea now. Thank you.
 
  • #10
andrewkirk
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