# Increasing sequence of sigma algebras

Tags:
1. Oct 30, 2015

### Stephen Tashi

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

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

2. Oct 30, 2015

### andrewkirk

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. Oct 30, 2015

### Stephen Tashi

Ok
What is the technical implementation of the concept of "groups together"?

I understand the concept of "extra info", but I don't see what mathematical object is being subdivided.

4. Oct 30, 2015

### andrewkirk

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 $(\Omega,\mathscr{F})$.

Last edited: Oct 31, 2015
5. Oct 31, 2015

### Stephen Tashi

If $a \in \mathscr{F}_s$ then $a$ can be written as an uncountable union of those members in $\mathscr{F}_t$ that have the same trajectory as $a$ on the interval $[t_0,s]$, but how do I see that $a$ can be expressed using the operations of a sigma algebra?

6. Oct 31, 2015

### andrewkirk

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. Oct 31, 2015

### Stephen Tashi

The way I look at it intuitively is that $\mathscr{F}_t$ is generated by a collection of sets that are each "smaller than $a$" or exclude $a$. $\$ Am I not visualizing this correctly? We did say that $\mathscr{F}_t$ is formed by taking equivalence classes of sets that make a finer partition of $\mathscr{F}$ than the partition used to define $\mathscr{F}_s$.

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 $b \subset a$.

8. Oct 31, 2015

### andrewkirk

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.

Last edited: Nov 1, 2015
9. Nov 1, 2015

### Stephen Tashi

Do we need to say $\mathscr{F}_t =$ the sigma algebra generated by $\bigcup_{u=t_0}^t \mathscr{C}_u$ ?

But anyway, I get the basic idea now. Thank you.

10. Nov 1, 2015