# Understanding the relationship between filtration $\mathcal{F}_t$ and an observed trajectory $O_{t}$

Introduction: I understand the filtration $$\mathcal{F}_t$$ to model all knowledge of a stochastic process $$\{X_t:t=0,1,2,\dots,T\}$$ up to time $$t$$ which in this case is discrete-time (due to my basic level of understanding this makes it easier). It does this via a growing set of sigma algebras $$\mathcal{F}_t \subset \mathcal{F}_{t+1}$$. This means that as time passes, we are able to distinguish more events, i.e. the observer is becoming smarter. For example at $$t=1$$ in a heads (H) and tails (T) coin toss game that lasts 2 rounds, we can't distinguish between the events $$\{HH,HT\}$$ as well as $$\{HH,HT\}$$ such that $$\mathcal{F}_1 = \{ \Omega,\varnothing,\{HH,HT\},\{TH,TT\}\}$$ but at $$t=2=T$$ we can distinguish between all events $$\mathcal{F}_2 = \{ \Omega,\varnothing,\{HH\},\{HT\},\{TH\},\{TT\}\} = 2^\Omega$$. To me, the concepts of distingushing was important in comprehending filtrations.

Question: I have a problem with the two ideas/notation. The first conditions on the filtration $$P(X_t|\mathcal{F}_t) \mbox{ and } \mathbb{E}[X_t|\mathcal{F}_t]$$ while the other conditions on an observed realisation of the stochastic process $$O_t = \{X_\tau(\omega):\tau=1,2,\dots,t\}$$ such that one would consider $$P(X_t|O_t) \mbox{ and } \mathbb{E}[X_t|O_t].$$ In the coint-toss example, we might have $$O_1 = \{H\}$$ or $$O_1 = \{T\}$$. I am much more comfortable conditioning on observations. A course on Hidden Markov Models would make use of observations, etc. Is there a connection between the two?

My attempt: I would like to show effort through showing my attempt at understanding the relationship between the two. Firstly, $$O_t$$ should tell us where we are both in state and time/index. The index part is important as it means we know to use $$\mathcal{F}_t$$ to denote our available information. Such a statement is new to me; I would have always referred to $$O_t$$ as our available information or history but now know $$\mathcal{F}_t$$ to be the available information. Secondly, $$O_t$$ tells us what subset of $$\mathcal{F}_t$$ would be permissible. In other words, what part of $$\mathcal{F}_t$$ contains information pertaining to the realisation $$O_t$$. Let's denote that as $$\mathcal{H}_{O_n}$$ such that $$\mathcal{H}_{O_t}\subseteq \mathcal{F}_t$$. In the coin toss example, if we have $$O_1= \{H\}$$ then we have $$\mathcal{H}_{O_1} = \{ \{HT,HH\} \} \subset \mathcal{F}_1$$. I think we could also say that if $$\mathcal{H}_{O_t}\not\subseteq \mathcal{F}_t$$ then we are not observing the stochastic process $$\{X_t:t=1,2,\dots,T\}$$ anymore but perhaps another $$\{Y_t:t=1,2,\dots,T\}$$, i.e. of one observed a dice roll of six as the second outcome $$O_2=\{H,6\}$$. This leads to the following conclusion: $$P(X_t|\mathcal{F}_t)$$ is a generalisation of $$P(X_t|O_t)$$ which says put an observation here to condition on BUT the observation must be able to locate a subset of $$\mathcal{F}_t$$. Hence, the filtration tells us what informative observations would look like but also what the minimum requirements for $$O_t$$ would be. For example, $$O_2=\{H,6\}$$ does not suffice nor does $$O_2 = \{H\}$$. The former contains no relevant information at the second observation while the latter does not have enough. Something like $$O_2=\{H,T,6\}$$ is informative enough but the inclusion of 6 is redundant. Hopefully this make sense and is in the ballpark of correct intuition.

If $$\mathcal F_t$$ is the filtration generated by $$(X_1,...,X_t)$$, then we actually have that $$\mathbb{E}[X_t|\mathcal F_t] = X_t$$ because the information in $$\mathcal F_t$$ is sufficient to determine $$X_t$$.
In general, if $$Z$$ is a random variable, you can think of $$\mathbb{E}[Z|\mathcal F_t]$$ as a function mapping the sets in $$\mathcal F_t$$ to estimates of $$Z$$. $$Z$$ might be $$X_T$$ for some $$T > t$$, or it might be a different stochastic process, or any other random variable you are interested in. Since $$\mathcal F_t$$ is generated by $$(X_1,...,X_t)$$, there is actually a function $$f$$ such that $$\mathbb{E}[Z|\mathcal F_t] = f(X_1,...,X_t).$$ Note that since $$X_1,...,X_t$$ are all random variables, $$\mathbb{E}[Z|\mathcal F_t]$$ is also a random variable. The connection between conditioning on $$\mathcal F_t$$ and conditioning on an observed realization $$O_t = (X_1(\omega),...,X_t(\omega))$$ is that you are just plugging that observed realization into the function $$f$$, i.e. $$\mathbb{E}[Z|O_t] = f(X_1(\omega),...,X_t(\omega)).$$
It looks like you were also interested in the case where your observation $$O_t$$ contains some redundant information. As you said, the observation must be able to identify a subset of $$\mathcal F_t$$. Since each trajectory corresponds to a unique subset of $$\mathcal F_t$$, that means it must also be able to identify $$(X_1(\omega),...,X_t(\omega))$$, in which case you again just plug that into the function $$f$$ and ignore any extra information in $$O_t$$.