# Help understanding filtered probability spaces

• Mar 4th 2011, 05:33 AM
tensorproduct
Help understanding filtered probability spaces
Hi

I am currently studying a course in financial economics for a professional actuarial qualification and I'm having trouble with some of the probability theory.

I'm having trouble understanding probability spaces and filtrations. Can anyone help? I figure that this level of mathematical theory won't appear on the exam but I'd feel more comfortable if I understood it.

From what I've read, a probability space is a triple (S, F, P)

• S is the space of all possible outcomes
• F is a collection of subsets of S, a sigma-algebra (i.e. closed under complement in S, and under countable (possibly infinite) unions and hence under intersection)
• P is a measure of the elements of F such that P:F -> [0,1] on the reals

For a discrete case, each s in S can be thought of as an event, a single outcome of running through an experiment or observing a share price move. Each element in F is a subset of S, a collection of events (possibly satisfying some condition, like every outcome in which the price increases by a certain amount). The probability measure P assigns a value between 0 and 1 to each element in F.

S and the null-set are elements of any sigma-algebra over S and have probabilities P(S) = 1 and P(null-set) = 0. Intuitively, the probability of anything at all happening is 1 and the probability of nothing happening is 0

I'm comfortable with everything above (though maybe I just think I understand it). My trouble is with filtrations.

A filtration {F_t}t>=0 is a collection of ordered sub-sigma algebras such that F_s is a subset of (or equal to) F_t if s <= t

Question:

• Does this mean that each F_t is also a subset of F? Hence that each F_t is also a sigma-algebra on S?

If t is thought of as the time, then each F_t is the history of the process up to t... This I don't get at all.

I'll use an example of a three-step binomial tree to illustrate my problem (this is exactly equivalent to tossing a coin three times or a one dimensional random walk).

• At each step, a value can randomly move up (u) or down (d).
• Thus, the state space S = {uuu, uud, udu, udd, duu, dud, ddu, ddd} - all possible outcomes of three steps.
• F could then be a collection of subsets of S. I think that F in a discrete state space is taken to be the power set of S: the set of all subsets of F, but I'm not sure.

How can the filtration {F_t} be understood as ths "history" of the process? My idea of what this means is outlined below, but even as I type it I don't think it makes sense.

Is F_2 a sigma algebra over a different state space, say S_2 = {uu, ud, du, dd}? In this case the state space after three steps would have to be reconstructed to include the elements of S_2 (and by extension) S_1 in order to allow F_2 to be a sub-sigma algebra of F (defined over S).
Thus, rewrite S = {u, d, uu, ud, du, dd, uuu, uud, udu, udd, duu, dud, ddu, ddd}
and construct F = pow(S).
Say the first step is up, and the second set is down. So do we construct F_2 as the smallest sigma-algebra over S which contains subset {ud}? This doesn't seem to make sense as such a collection would be (S, null, {ud}, S/{ud}), where S/{ud} is the complement.

I think that I'm rambling now, so I'll stop. Can anybody explain this to me, or tell me if I'm heading in completely the wrong direction?

Any help would be appreciated.

Many thanks
Barry
• Mar 5th 2011, 06:46 PM
Focus
Quote:

Originally Posted by tensorproduct
A filtration {F_t}t>=0 is a collection of ordered sub-sigma algebras such that F_s is a subset of (or equal to) F_t if s <= t

Question:

• Does this mean that each F_t is also a subset of F? Hence that each F_t is also a sigma-algebra on S?

Yes, $\displaystyle \mathcal{F}_t$ are increasing subsets of $\displaystyle \mathcal{F}$ such that each of them is a sigma algebra.

Quote:

If t is thought of as the time, then each F_t is the history of the process up to t... This I don't get at all.
This is not true in general. This is the case when you take $\displaystyle \mathcal{F}_t:=\sigma(X_s:0\leq s\leq t)$ which is exactly the information of the paths up to time t (in case the notion is not clear, it is the smallest sigma algebra generated by $\displaystyle X_s^{-1}(B)$ where B is a Borel set).

Quote:

I'll use an example of a three-step binomial tree to illustrate my problem (this is exactly equivalent to tossing a coin three times or a one dimensional random walk).

• At each step, a value can randomly move up (u) or down (d).
• Thus, the state space S = {uuu, uud, udu, udd, duu, dud, ddu, ddd} - all possible outcomes of three steps.
• F could then be a collection of subsets of S. I think that F in a discrete state space is taken to be the power set of S: the set of all subsets of F, but I'm not sure.

How can the filtration {F_t} be understood as ths "history" of the process? My idea of what this means is outlined below, but even as I type it I don't think it makes sense.

Is F_2 a sigma algebra over a different state space, say S_2 = {uu, ud, du, dd}? In this case the state space after three steps would have to be reconstructed to include the elements of S_2 (and by extension) S_1 in order to allow F_2 to be a sub-sigma algebra of F (defined over S).
Thus, rewrite S = {u, d, uu, ud, du, dd, uuu, uud, udu, udd, duu, dud, ddu, ddd}
and construct F = pow(S).
Say the first step is up, and the second set is down. So do we construct F_2 as the smallest sigma-algebra over S which contains subset {ud}? This doesn't seem to make sense as such a collection would be (S, null, {ud}, S/{ud}), where S/{ud} is the complement.

I think that I'm rambling now, so I'll stop. Can anybody explain this to me, or tell me if I'm heading in completely the wrong direction?

Any help would be appreciated.

Many thanks
Barry
You are getting confused between the value of the process and the state space. To define this process you want you need to consider $\displaystyle S=\{f: f:\{1,2,3\}\rightarrow \{u,d\}\}$.

I recommend reading Rogers & Williams if you need a reference for this stuff.
• Mar 6th 2011, 12:33 AM
tensorproduct
Hi Focus, thanks for the quick response.

Quote:

Originally Posted by Focus
You are getting confused between the value of the process and the state space. To define this process you want you need to consider $\displaystyle S=\{f: f:\{1,2,3\}\rightarrow \{u,d\}\}$.

You are definitely right in saying that I'm confused. I don't really know how to interpret that expression.
Is $\displaystyle f:\{1,2,3\}\rightarrow \{u,d\}$ a defined function?
Is there any way of listing explicitly the elements contained in $\displaystyle S$? How does it differ from a set of all possible outcomes?

Quote:

Originally Posted by Focus
I recommend reading Rogers & Williams if you need a reference for this stuff.

Is that this Rogers and Williams?
• Mar 6th 2011, 03:05 PM
Focus
Quote:

Originally Posted by tensorproduct
You are definitely right in saying that I'm confused. I don't really know how to interpret that expression.
Is $\displaystyle f:\{1,2,3\}\rightarrow \{u,d\}$ a defined function?
Is there any way of listing explicitly the elements contained in $\displaystyle S$? How does it differ from a set of all possible outcomes?

I mean the set of functions that map 1,2,3 to u,d.

This is essential all the things your process could be. This set will be the same as your set S with one added bonus that you can define the process X_n to be $\displaystyle X_n(f)=f(n)$.

A better example would be a simple random walk. Think about the space of functions $\displaystyle f:\{0,1,2\}\rightarrow \mathbb{Z}$ and X_n defined as before. What is F_1? Well it is X_1 either 1 or -1 so $\displaystyle \mathcal{F}_t=\{\{f:f(0)=0, f(1)=1\},\{f: f(0)=0, f(1)=-1\},\{f:f(0)=0, f(1)= \pm 1\}\}$.

Quote:

Is that this Rogers and Williams?
Yes, except you need volume 1 (not 2).
• Mar 8th 2011, 01:01 AM
tensorproduct
Quote:

Originally Posted by Focus
I mean the set of functions that map 1,2,3 to u,d.

This is essential all the things your process could be. This set will be the same as your set S with one added bonus that you can define the process X_n to be $\displaystyle X_n(f)=f(n)$.

Okay, but the elements of $\displaystyle S$ will be functions analagous to the elements I listed before: $\displaystyle {uuu, uud, ...}$
Where, for example, $\displaystyle uuu$ corresponds to a set of functions $\displaystyle \{f:f(1)=u,f(2)=u,f(3)=u\}$
Right? With $\displaystyle \mathcal{F}$ an algebra defined over these.

So, $\displaystyle \mathcal{F}_1$ would be the subset of $\displaystyle \mathcal{F}$ for which the outcome of the first move is known:
$\displaystyle \mathcal{F}_1=\{\{f:f(1)=u\},\{f:f(1)=d\},\{f:f(1) = u\text{ or }d\}\}$

Alternatively, in my previous notation:
$\displaystyle \mathcal{F}_1=\{\{uuu,uud,udu,udd\},\{duu,dud,ddu, ddd\},\emptyset,S\}$
(With the null-set there in order to make this an algebra.)

and
$\displaystyle \mathcal{F}_2=\{\{f:f(1)=u,f(2)=u\},\{f:f(1)=u,f(2 )=d\},...\}$

I'm not sure how $\displaystyle \mathcal{F}_2$ has "more" information than $\displaystyle \mathcal{F}_1$...

Quote:

Yes, except you need volume 1 (not 2).
Well, I've tracked that down. I may need to refresh a lot of the basic analysis in my head before i get to the meat of it.
• Mar 9th 2011, 09:26 AM
Focus
Quote:

Originally Posted by tensorproduct
I'm not sure how $\displaystyle \mathcal{F}_2$ has "more" information than $\displaystyle \mathcal{F}_1$...

The sigma algebra F_2 strictly contains F_1. This is why you have more information. The bigger the sigma algebra, the more questions you can ask. Think of a sigma algebra as the set of questions you can ask.

For example let $\displaystyle \Omega=\{1,2,3,4,5,6\}$ (roll of a dice) and let F be the discrete sigma algebra and $\displaystyle \mathcal{G}:=\{\emptyset,\{1,3,5\},\{2,4,6\},\Omeg a\}$. Now which sigma algebra gives you more information? With G, you can only know if you rolled an even or an odd number, whereas with F you know exactly which number you rolled.
• Mar 9th 2011, 03:06 PM
tensorproduct
Quote:

Originally Posted by Focus
The sigma algebra F_2 strictly contains F_1. This is why you have more information. The bigger the sigma algebra, the more questions you can ask. Think of a sigma algebra as the set of questions you can ask.

For example let $\displaystyle \Omega=\{1,2,3,4,5,6\}$ (roll of a dice) and let F be the discrete sigma algebra and $\displaystyle \mathcal{G}:=\{\emptyset,\{1,3,5\},\{2,4,6\},\Omeg a\}$. Now which sigma algebra gives you more information? With G, you can only know if you rolled an even or an odd number, whereas with F you know exactly which number you rolled.

Aha, now I get it - or at least I think I do. I'm still a long way from understanding the fully continuous case, but this is a good start.

Thanks, Focus, you've been a great help.
• Mar 14th 2011, 06:48 PM
Volga
Thanks Focus! I, too, have been struggling with F algebras, you just confirmed my intuitive understanding.
Interestingly, these concepts remind me of 'information sets' in game theory (extensive games with imperfect information), where the player sometimes knows and sometimes does not know his exact position the game. The more 'partitioned' his information sets are, the better he is informed of his place in the game (and consequences of his future moves). I wonder if you know what I am talking about, and if there are similarities with the above sigma algebra concepts.