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?

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    $\begingroup$ 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 $X_0$, ..., $X_n$). $\endgroup$ – Did Mar 7 '15 at 9:46
  • $\begingroup$ 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 $\Omega$. $\endgroup$ – saz Mar 7 '15 at 10:01
  • $\begingroup$ 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? $\endgroup$ – user48547 Aug 10 '16 at 20:46
  • $\begingroup$ @oeda Post this as a question on the site then. $\endgroup$ – Did Aug 10 '16 at 23:12
  • $\begingroup$ OK, it will also help me to make sense of the question. $\endgroup$ – 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!

  • $\begingroup$ 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. $\endgroup$ – user415535 Sep 5 '17 at 11:05
  • $\begingroup$ Mmm I don't think so, that sounds to me as a Martingale!! $\endgroup$ – Max Sep 5 '17 at 12:40
  • $\begingroup$ 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? $\endgroup$ – user415535 Sep 5 '17 at 13:10
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    $\begingroup$ 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 $X_n$ ? $\endgroup$ – Max Sep 5 '17 at 13:14
  • $\begingroup$ 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 $X_{n}$? $\endgroup$ – 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.


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