# Definition of previsible processes?

Definition from my textbook:

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

Okay, but that surely cannot mean that I can really predict the next outcome $X_{n+1}$ if I know $X_0, \ldots, X_n$ (regardless if $X$ is also adapted to $\mathbb{F}$), right?

• No you cannot, rather this means that one can predict the next outcome $X_{n+1}$ if one knows $\mathcal{F}_n$ (and there is usually more information in this than just $X_0$, ..., $X_n$). – Did Mar 7 '15 at 9:46
• I agree with @Did. An easy example for a filtration $\mathcal{F}_n$ which contains more information than $X_0,\ldots,X_n$ is $\mathcal{F}_n := \mathcal{A}$ where $\mathcal{A}$ denotes the $\sigma$-algebra on the probability space $\Omega$. – saz Mar 7 '15 at 10:01
• Somehow I'm interested in the notion @Did mentioned. How can conditioning on the filtration and conditioning on the sample path differ? I guess conditioning on $X_{1} = x_{1},X_2 = x_2, \ldots, X_n = x_n$ is the same thing with conditioning on $\mathcal{F}_n = \sigma(X_1,\ldots,X_n)$, no? What is the alternative filtration which can cause the case mentioned above? – user48547 Aug 10 '16 at 20:46
• @oeda Post this as a question on the site then. – Did Aug 10 '16 at 23:12
• OK, it will also help me to make sense of the question. – user48547 Aug 11 '16 at 13:30

It is not clear -mathematically speaking- what do you mean by predict, but in a certain way, you can predict $X_n$ by knowing $\mathcal{F}_{n-1}$ (which may be the same as knowing $X_0,X_1,...,X_{n-1}$ if $\mathcal{F}_n$ is the natural filtration).

The thing is, in applications, a previsible process has to do with a strategy (i.e. a decision you make taking into account the previous information, for example a gambling strategy) rather than with a system you are analysing.

See for example Probablity with Martingales , by Williams.

I'm sorry if this answer arrives late!

• Am I right to think that a for a previsable process, one can in some sense estimate $X_{n}$ without having to taking avarages of values with some subsets. One can estimate it just as the immediate predecessor. – user415535 Sep 5 '17 at 11:05
• Mmm I don't think so, that sounds to me as a Martingale!! – Max Sep 5 '17 at 12:40
• hmm if we have a martingale then one takes the avarages of $X_{n}$ over the sets in $F_{n-1}$ which in turn have the same distriution as $X_{n-1}$. In this case this we dont take avarages but get the possible values and thier probabilites just by looking at the predecessor $X_{n-1}$ to get the value of $X_{n}$. Or have I slipped somewhere? – user415535 Sep 5 '17 at 13:10
• You are right about the averages. On the other hand, take the trivial previsible (deterministic) process $X_n =n$. In which sense you estimate $X_{n+1}$ as $X_n$ ? – Max Sep 5 '17 at 13:14
• Right, I was thinking that $C_{n}$ was a martingale aswell...So it more like "if we know what event in $\mathcal{F}_{n-1}$ did occur then we know the value of $X_{n}$? – user415535 Sep 5 '17 at 13:22

Technically, you can always predict in terms of the best available information but you will NOT necessarily know the value of $$X_{n+1}$$ if you know the values of $$X_0, \ldots, X_n$$. This is easy to show since based on the definition of a previsible process, if you are at $$(n-1)^{th}$$ index of the filtration (i.e. $$\mathcal{F}_{n-1}$$), you already know the values of $$X_0, \ldots, X_n$$ but you don't get to know the value of $$X_{n+1}$$. You will know $$X_{n+1}$$ when you are at $$n^{th}$$ index of filtration (i.e. $$\mathcal{F}_{n}$$). I think in the question there seems to be a confusion regarding what subscript $$n$$ means in $$X_n$$. It doesn't mean when you know $$X_0, \ldots, X_n$$ you are at index $$n$$ in the filtration. In fact, you know those values when you are at $$\mathcal{F}_{n-1}$$ which doesn't tell what $$X_{n+1}$$ is.