Are all stochastic processes $\mathcal{F}_t$-measurable?

Question

I am currently reading about martingales, and the notion of a $\mathcal{F}_t$-measurable process has been introduced. It is stated that:

The filtration [of a process] $\mathcal{F}_t$ represents everything that can be known [about the past of a process] up to and including time $t$.

Some random variables will be known by time $t$. We say that $X_t$ is $\mathcal{F}_t$-measurable if the value of the process is known at time $t$, i.e. it belongs to $\mathcal{F}_t$.

This has caused some confusion for me, as I'd thought that the value of a process at time $t$ is included in the filtration $\mathcal{F}_t$. Does this not mean that all stochastic processes are $\mathcal{F}_t$-measurable?

If it is not always the case that the value of a stochastic process is known at time $t$, could someone give an example of this?

Answer 1

A priori, a filtration is just a one-parameter family of $\sigma$-fields that are ordered by inclusion, and a stochastic process is just a one-parameter family of random variables. On the face of it, there is no reason why they should be related to each other in any way.

In particular, you could take $\mathcal{F}_t = \{\emptyset, \Omega\}$ for every $t$. Obviously any nontrivial stochastic process will not be measurable (aka adapted) with respect to this filtration.

The condition that the process should be $\mathcal{F}_t$-measurable is a "compatibility" condition, asking that the process and the filtration should play nicely together. It is indeed very common to make this assumption, but it isn't inherent in the definitions of the objects themselves.

And even if we do assume we are working with a process $X_t$ and a filtration $\mathcal{F}_t$ such that $X_t$ is $\mathcal{F}_t$-measurable, yet if $Y_t$ is some completely different process there is no reason why it ought to be $\mathcal{F}_t$-measurable as well.

Answer 2

There is always "natural filtration" for a process $X=(X_t)_{t \in T}$ which is $(\mathcal F_t^X)_{t \in T}$ where $\mathcal F_t^X := \sigma( X_s : s \le t)$ (intuitivelly $\mathcal F_t^X$ is all information we know about process $X$ up to the time $t$). But there is nothing to stop asking whether some proces $X$ is adapted to another filtration $(\mathcal G_t)_{t \in T}$ (in other words whether $X_t$ is $\mathcal G_t$ measurable for all $t \in T$).

Just consider only one random variable, let's say $Y:[-1,1] \to [-1,1], Y(\omega) = \omega$ is uniformly distributed. Then $Y$ is $\sigma(Y)$ measurable, by definition. But you can then ask whether $Y$ is $\sigma(Y^2)$ measurable? And the answer will be no, because $Y^{-1}[[-1,0)] = [-1,0) \not \in \sigma(Y^2)$