Basic Struggle with Adapted Stochastic Process and Predictable Stochastic Processes

I am going through Klenke's Probability Theory. Here are the definitions he presents for adapted

Definition 9.10 A stochastic process $$X = (X_t,t \in I)$$ is called adapted to the filtration $$\mathbb{F}$$ if $$X_t$$ is $$\mathcal{F}_t$$-measurable for all $$t \in I$$.

and for predictable

Definition 9.12 A stochastic process $$X = (X_n,n\in \mathbb{N}_0)$$ is called predictable with respect to the filtration $$\mathbb{F} = (\mathcal{F}_n,n \in \mathbb{N}_0)$$ if $$X_0$$ is constant and if, for every $$n \in \mathbb{N}$$ $$X_n$$ is $$\mathcal{F}_{n-1}$$-measurable.

The definition of predictable makes sense to me if $$\mathbb{F} = \sigma(X)$$, but when this is not the case, I do not know how to predict future values in $$X_n$$.

I constructed a concrete example. Let $$(\Omega,\mathcal{F},\mathbb{P}) = ([0,1),\mathcal{B}(\Omega),\mu)$$. Then, take the following process. Let $$X_0(\omega) = 0$$. Then, for $$i \geq 1$$, let $$\omega_j$$ be the jth digit in the decimal expansion of $$\omega$$ and $$X_i(\omega) = \begin{cases} 1 & \omega_j \text{ even} \\ 0 & \omega_j \text{ odd}.\end{cases}$$

If we let $$\mathbb{F} = \mathcal{F}$$ (giving out the original $$\sigma$$-algebra at each time step), then each $$X_i$$ is $$\mathcal{F}_0$$-measurable. Now we start the process and find that $$X_0 =0$$. How do I then predict the value of $$X_1$$?

The point you're missing is that the $$\ t^\text{th}\$$ term $$\ \mathcal{F}_t\$$ of the filtration is meant to represent all the knowledge available to you at the instant $$\ t\$$, in the sense that, for every event $$\ A\in \mathcal{F}_t\$$, either it is known that $$\ A\$$ has already occurred, or it is known that $$\ \Omega\setminus A\$$ has already occurred (i.e. it is known at that instant whether $$\ \omega\in A\$$ or $$\ \omega\in\Omega\setminus A\$$ ).
If $$\ \left(X_n, n\in\mathbb{N}_0\right)\$$ is predictable with respect to a filtration $$\ \left(\mathcal{F}_n, n\in\mathbb{N}_0\right)\$$, and $$\ v=X_n(\omega)\$$ is the realised value of $$\ X_n\$$, then at time $$\ n-1\$$, even though you might not—and usually won't—know $$\ \omega\$$, you do know that $$\ \omega\in X_n^{-1}\left(\{v\}\right)\$$, because $$\ X_n\$$ is $$\ \mathcal{F}_{n-1}$$-measurable, and so $$\ X_n^{-1}\left(\{v\}\right)\in\mathcal{F}_{n-1}\$$, and hence you know that $$\ X_n(\omega)= v\$$
In your example, by making $$\ \mathcal{F}_0=\mathcal{F}=\mathcal{B}\left([0,1)\right)\$$ you are specifying that $$\ \omega\$$ is already completely known at the start of the process, because all singleton events $$\ \{\omega\}\$$ belong to $$\ \mathcal{B}\left([0,1)\right)\$$.
• Ok, I think I am understanding most of what you are saying. In the general case (your first paragraph), how is it that we know whether $\omega$ is in a given $\mathcal{F}_t$-measurable set or not? Also, just to double check, is there only one value of $\omega$ that is chosen at the start of the process and is plugged into every $X_n$? – sandaga Feb 5 '20 at 13:32
• In practice (at least, in the limited contact I've had with it) the information is acquired by learning the values of some sets of random variables. In that case, if $\ \mathcal{S}_t\$ is the set of random variables whose values you've learned by time $\ t\$, then $\ \mathcal{F}_t\$ would be the $\ \sigma$-algebra generated by $\ \left\{X^{-1}(B)\,|\,X\in\mathcal{S}_t, B\in\mathcal{B}\right\}\$, where $\ \mathcal{B}\$ is the Borel $\ \sigma$-algebra on the ranges of the $\ X$s in $\ \mathcal{S}_t\$. – lonza leggiera Feb 5 '20 at 14:07