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This question arose from attending a seminar about stochastic processes using the book "Introduction to the Theory of Random Processes" by N.V. Krylov. I'm not gonna follow the notation of the book since it is a little outdated.

Let $X_t$ be an infinitely divisible cadlag process on $[0,\infty)$, i.e. a stochastically continuous time-homogeneous process with independent increment (also called a Levy process). For $0 \leq s < t < \infty$, define $\mathcal{F}_{s,t}^X$ as the completion of the $\sigma$-algebra generated by $X_r - X_s, r \in [s,t]$.

Here comes the statement in question: Since the increments of $X_t$ are independent, the $\sigma$-fields $\mathcal{F}_{0,t_1}^X, \mathcal{F}_{t_1,t_2}^X,\ldots,\mathcal{F}_{t_{n-1},t_n}^X$ are independent for any $0 < t_1 < \ldots < t_n$.

I'm not sure how to proceed on how to prove it, my plan was to show the independence of the $\sigma$-algebras generated by the first 2 increments, show they are independent, and to show by induction that all these $\sigma$-algebras are pairwise independent, and then extend the result to independence by using Proposition II.5.5 in Cinlar ("The sub-$\sigma$-algebras $\mathcal{F}_1,\mathcal{F}_2,\ldots$ of $\mathcal{H}$ are independent iff $\mathcal{F}_{\{1,\ldots,n\}}$ and $\mathcal{F}_{n+1}$ are independent.")

The other people in the seminar had trouble proving the argument, they defined some set (where some sets were independent, I missed the discussion since I was working on my approach), and then tried showing it is both a $\pi$- and a $\lambda$-system.

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  • $\begingroup$ Can you prove that the (uncompleted) $\sigma$-algebras generated by increments independent? $\endgroup$
    – zhoraster
    Commented Sep 11, 2015 at 3:40
  • $\begingroup$ I don't know how to proceed for the uncompleted $\sigma$-algebras either, is it very different in this case? $\endgroup$
    – Olorun
    Commented Sep 11, 2015 at 4:01
  • $\begingroup$ No, it follows straight from the definition, this is why I'm asking. $\endgroup$
    – zhoraster
    Commented Sep 11, 2015 at 4:04
  • $\begingroup$ Ah..but then the extension to the completion shouldn't be too complicated either - maybe I'm missing something? $\endgroup$
    – Olorun
    Commented Sep 11, 2015 at 4:36
  • $\begingroup$ No, it's not too complicated. Any progress? $\endgroup$
    – saz
    Commented Sep 12, 2015 at 6:57

1 Answer 1

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Here's my try on proving it, I might have some mistake in the notation..anyway, I'll be using several well-known results, which we will recall first.

  1. Let $\mathcal{A_1},\ldots,\mathcal{A_n}$ be $\pi$-systems that generate the $\sigma$-algebras $\mathcal{F_1},\ldots,\mathcal{F_n}$, respectively. If $\mathbb{P}(A_1 \cap\cdots \cap A_n) = \mathbb{P}(A_1) \cdots \mathbb{P}(A_n)$ for all $A_i \in \mathcal{A_i}, i=1,\ldots,n$, then $\mathcal{F_1},\ldots, \mathcal{F_n}$ are independent.

  2. If $\mathcal{F_1},\ldots,\mathcal{F_n}$ are independent, then $\overline{\mathcal{F}}_1,\ldots,\overline{\mathcal{F}}_n$, their respective completions, are independent.

So, the crux will be in finding the $\pi$-systems that are independent, which we will then use to generate the $\sigma$-algebras. This is mostly a problem with using the right notation for the process $X$.

For this purpose, let $$\mathcal{A}_k = \bigcup \sigma\left( X_{s_k^1}-X_{t_{k-1}},X_{s_k^2}-X_{t_{k-1}},\cdots,X_{s_k^{l_k}}-X_{t_{k-1}} \right),$$ where $s_k^i \in [t_{k-1},t_k], i=1,\ldots,l_k,$ and $l_k = 1,2,\ldots.$

This generates a $\pi$-system of independent $\mathcal{A}_k$.

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