I currently have an understanding of what a filtration is and will illustrate this through an example. From this example, I will convey what my idea of natural filtration is. My question is similar to this one which has not yet been answered. Before reading one, what I seek to know is:

  • Is my interpretation of filtration correct? I am looking to build intuition. (Example 1)
  • How does a filtration differ from a natural filtration? (Example 2)
  • Does filtration differ from process history?

Example 1: Let's consider the generic binomial model of stock prices moving over $N$ periods where it can go up $U$ or down $D$ with probability $p$ and $1-p$, respectively. If the process is stock price $S_n$ then denote the stochastic process as $\{S_n:n=0,1,2,\dots,N\}$. For convenience, define the following Cartesian product $$ \mathcal{C}_k := \underbrace{\{U,D\} \times \{U,D\} \times \dots \times \{U,D\}}_{k\mbox{ times}} $$ from which the set of outcomes is $\Omega = \mathcal{C}_N $. The first filtration is trivial in that an elementary event occurs or does not such that $$ \mathcal{F}_{0} = \{ \varnothing,\Omega \}. $$ In the next trading period/iteration, the first stock movement will be revealed. The process will be $\mathcal{F}_1$-adapted in that the future is unknown and cannot be revealed as this is not information that the process has. At this point $\mathcal{F}_1$ is the information of the process and is given as $$ \mathcal{F}_1 = \{U\}\times \mathcal{C}_{N-1}\cup\{D\}\times \mathcal{C}_{N-1} \cup\mathcal{F}_0 $$ which satisfies $\mathcal{F}_0 \subseteq \mathcal{F}_1 \subseteq \dots \subseteq \mathcal{F}_{N-1} \subseteq \mathcal{F}_N$. Inductive reasoning deduces the following: $$ \mathcal{F}_n = \mathcal{F}_{n-1} \cup \bigcup_{\sigma \in \mathcal{C}_n} \sigma \times \mathcal{C}_{N-n}, \quad n = 1,2,\dots,N . $$ From the above, can we interpret $\mathcal{F}_n$ as follows. Lets ignore the inclusion of $\mathcal{F}_{n-1}$ first and focus on the part that is new. This new part is the uncertain behaviour we can expect over $n+1,n+2,\dots,N$ (no information part) for every possible realised path/history up to $n$ (the information part). Including $\mathcal{F}_{n-1}$ is a bit unclear to me, but I think it is there to denote that the $n^{th}$ trading period did not occur. Am I right in saying so?

Example 2: Now we get to the natural filtration. Some texts describe is as the minimal filtration. This gives me the idea that the filtration from example 1 is then a natural filtration. Planet Math says that a stochastic process generates this minimal filtration. This is maybe why it is difficult to imagine examples that aren't natural filtrations because we are thinking of how the stochastic process evolves and building an example of the filtration from that. Such reasoning will inevitably give a natural filtration. Perhaps an example of a filtration that is not a natural filtration can be built using a Hidden Markov Model with underlying states $x_n$ and observations $y_n$? The natural filtration will pertain to $x_n$ but the non-natural filtration might describe $(x_n,y_n)$.

Lastly, it seems to me that a filtration is different from process histories. We can consider each $\sigma \in \mathcal{C}_n$ to be a process history? Hence, a filtration considers all uncertain futures for each possible realised process history.

Edit: The following perhaps appears to be a bit funky: $$ \mathcal{F}_1 = \{U\}\times \mathcal{C}_{N-1}\cup\{D\}\times \mathcal{C}_{N-1} \cup\mathcal{F}_0 $$ $$ \mathcal{F}_n = \mathcal{F}_{n-1} \cup \bigcup_{\sigma \in \mathcal{C}_n} \sigma \times \mathcal{C}_{N-n}, \quad n = 1,2,\dots,N . $$ The convention is unconventional with respect to $\mathcal{C}_n$. The reason behind this is that it is how I went about reasoning on how one might write a general program to output filtrations for discrete state stochastic processes evolving over discrete points. To see how it works, consider that a trajectory has been observed at $n=3$ such that $\sigma_3 = \{U,U,D\}$. The stochastic process is of length $N=4$ such that $\mathcal{C}_{4-3} = \{ U,D\}$. Hence, $\mathcal{F}_4 = \mathcal{F}_3 \cup \sigma_{3}\times \{U\} \cup \sigma_{3}\times\{D\} = \mathcal{F}_3 \cup \{U,U,D,U\} \cup \{U,U,D,D\}$. Furthermore, it seems that I have mistaken an observed trajectory $\sigma_n$ to be the process history which is actually $\mathcal{F}_n$.

Update if anyone with an "applied" background as opposed to one using rigorous mathematics, please look at this gem of a paper. I found it after asking this question and I really feel that it has expanded my understanding.


2 Answers 2

  • I find the notation

$$ \mathcal{F}_1 = \{U\}\times \mathcal{C}_{N-1}\cup\{D\}\times \mathcal{C}_{N-1} \cup\mathcal{F}_0 $$ quite odd, in particular because of the $\times \mathcal{C}_{N-1}$. Better is in my opinion $$ \mathcal{F}_1 = \{\varnothing,U,D,\Omega\} $$ where $U$ and $D$ represent the events that $S_1$ has gone up, resp. down.

  • A typical example of a filtration that is larger than the natural one is the filtration of a multi-dimensional Brownian motion. Every one-dimensional component of such a BM is a BM with respect to its natural filtration and with respect to the larger filtration.

  • A standard treatment of a filtration for a discrete binomial process can be obtained as follows:

  1. Fix a maturity $N$ and let $\Omega$ be the set that consists of all "paths" $\{U,D\}^N$ i.e. all $N$-tuples with values in $\{U,D\}$. The final $\sigma$-algebra ${\cal F}={\cal F}_N$ on this finite set of paths is clearly the set of all subsets of $\Omega$ i.e. the power set ${\cal P}(\Omega)$.

  2. Since the above applies to all maturities $n$ strictly before $N$ we have a $\sigma$-algebra ${\cal F}_n={\cal P}(\{U,D\}^n)$ for each $n$. It is somewhat of a formal pseudo problem that a priory the elements of ${\cal F}_n$ are not subsets of $\Omega=\{U,D\}^N$.

  3. A way around this is to consider an event $A\in{\cal F}_N$ and take the maximal index $n$ that is needed to uniquely describe it. For example $$ A=\{S_1=U,S_2-S_1=D\}. $$ This event belongs to ${\cal F}_2$.

Why do you want to write a program to generate the discrete filtration $({\cal F}_n)_{n=1,...,N}$ ? If this is that important then the simple approach in 2. is the route that I would take.

  • $\begingroup$ Thank you for your answer. In regard to process histories, I perceive them as the trajectory or path produced by the process up to index $n$. For example, we might observe a realisation $\sigma_3 = \{U,U,D\}$. If $N=4$ then the $\sigma$ algebra results from the uncertainty in the fourth stock price change such that $\mathcal{F}_4 = \mathcal{F}_3 \cup \{U,U,D,U\} \cup \{U,U,D,D\}$. This is the same as $\mathcal{F}_4 = \mathcal{F}_3 \cup \{\sigma \times \{U,D\}\}$ hence the funky notation. Perhaps process history is not the same as observed history? Is observed history a realised trajectory? $\endgroup$ Jul 31, 2022 at 15:41
  • $\begingroup$ The funky notation stems from wanting to write a program to produce discrete filtrations. This is why creating combinations from Cartesian products seemed to make sense. However, I realise it does not fit the convention. $\endgroup$ Jul 31, 2022 at 15:45
  • $\begingroup$ I have updated my question to include the above discussed comments. $\endgroup$ Jul 31, 2022 at 15:57
  • 1
    $\begingroup$ I suggest to stick to the example ${\cal F}_1$ . There is not much insight provided by adding ${\cal F}_2$ etc. At time $1$ we know what has happened to date, and that is a process history. It is either $U$ or $D$. The filtration has to include both and has to be a $\sigma$-algebra because it wants to model all possible process histories up to time $1$. Don't overthink it by trying to understand just words. $\endgroup$
    – Kurt G.
    Jul 31, 2022 at 16:35
  • $\begingroup$ I think this video has confused me: youtu.be/bsxYi0hkb4I?t=1041 $\endgroup$ Jul 31, 2022 at 17:12

I am submitting this answer to verify what I have learned from the answer of Kurt G. along with chapter 8 of Mathematical Techniques in Finance:Tools for Incomplete Markets (2nd ed.). I have included a figure from this book an will be using it in my explanation.

What was wrong: my original concepts of the filtration $\mathcal{F}_n$ was that it was a collection of all the distinguishable elementary events up to the $n^{th}$ trading period. If we look at the figure below, we see that the distinguishable elementary events at each epoch is:

  • $n=0:\quad \mathcal{E}_0 = \{ \Omega,\varnothing \}$ where $\Omega = \{UUU,UUD,UDU,UDD,DUU,DUD,DDU,DDD \}$
  • $ n = 1: \quad \mathcal{E}_1 = \{ \{UUU,UUD,UDU,UDD\},\{DUU,DUD,DDU,DDD \} \} $
  • $n=2:\quad \mathcal{E}_2=\{\{UUU,UUD\},\{UDU,UDD\},\{DUU,DUD\},\{DDU,DDD\} \} $
  • $ n= 3: \quad \mathcal{E}_3 = \{ \{UUU\},\{UUD\},\{UDU\},\{UDD\},\{DUU\},\{DUD\},\{DDU\},\{DDD\} \} $

Figure from book

So we see that at epoch $n$, an elementary event consists of a known path which I called $\sigma$ in my question. This was probably a bad idea due to the clash with $\sigma$-algebras. Let's rather call a known path $\phi$. The unknown part in my question was given by $\mathcal{C}_{N-n}$ such that the known part was $\phi_n$. Concatenating a known part to each of its future unknowns and then repeating for all possible known $n$-length paths $\phi_n \in \mathcal{C}_n$ is what my original formula did.

If one looks at the figure, one can see that there was merit to this approach. At $n=1$ there are two known paths: $\{ \{U\},\{D\}\}$ which leads to the first branching. To each of the known paths, one add the remaining unknowns from $\mathcal{C}_{3-1}$.

How do we fix it: knowing the indistinguishable elementary events at each epoch is great. This seems like all we need to know. I think this is a common mistake. In fact, this video did exactly this (hence my confusion).

What is missing? We need to know all events that can answer questions up to epoch $n$. We can answer questions about all events distinguishable up to epoch $n$. So if $\mathcal{P}(\cdot)$ denotes the power-set of a given set then the filtration at $n$ follows as $$ \mathcal{F}_n = \mathcal{P}\left( \mathcal{E}_n\right) .$$ Finally, $\mathcal{F}_N = \mathcal{P}\left( \Omega\right)$. Additional intuition in regard to the empty set $\varnothing$ in $\mathcal{E}_0$, this simply tells us whether we can distinguish the process from something else, i.e. we know whether we are looking at the process or not. That is intuition that was provided by this question.


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