Natural filtration of martingales

Question

I don't quite understand what the natural filtration really is. Imagine e.g. a sequence of independent and identically distributed random $N(0,1)$ variables. What is their natural filtration, and how do I calculate, e.g, $$E(X_T | A_{t-1})$$ where $X_t$ is the $t$th variable and $A_{t-1}$ is the $t-1$th $\sigma$-algebra in the filtration?

According to the definition, the filtrations are given by $A_t = \sigma(X_t^{-1}(B), B \in A)$, but in this continuous case, I have no clue how to determine these. Are they all the same?

Answer 1

You need to be a bit more formal to get it. You should start with some probability space $(\Omega, \mathscr F, \mathsf P)$ and construct on it variables $X_t:\Omega\to\Bbb R$ such that they happen to be iid with a given distribution. Once you did that, it means that every measurable map $X_t$ pulls back Borel $\sigma$-algebra from $\Bbb R$ to a sub-$\sigma$-algebra $X^{-1}_t(\mathscr B(\Bbb R))\subseteq\mathscr F$. Each element of the natural filtration of $X$ is just a union of those $\sigma$-algebras (well, rather the $\sigma$-algebra generated by that union).

The conditional expectation you are talking about is $0$ since your variables are iid.

Answer 2

The "natural filtration" for a sequence $(X_n)_{n \in \mathbb N}$ is like this: $\mathcal A_n$ is the sigma-algebra generated by $X_1, X_2, \dots, X_n$.

But later you say the words "continuous case", so maybe you mean a stochastic process indexed by real numbers, like this: $(X_t)_{t \in [0,\infty)}$. Then you may take $\mathcal A_t$ to be the sigma-algebra generated by the set $\{X_s : s \le t\}$. That is an uncountable set of random variables, so it could be nasty; but if, for example, your process has continuous sample paths you can get the same sigma algebra with just countably many generators.