Centered stochastic process with independent increments is a martingale

Question

A process with independent increments, $\{\xi_t, t \in T\}$ is defined as a stochastic process, such that its finite distributions, $$(\xi_{t_1}, \xi_{t_2} - \xi_{t_1}, ..., \xi_{t_n} - \xi_{t_{n-1}})$$ are independent. My lecturer "proved", that for such a process to be central [$E(\xi_t - \xi_s) = 0] \forall s, t$] is equivalent for it to be a martingale. The argument is, that since the process has independent increments, any increment will be independent with the natural filtration: $$E(\xi_t - \xi_s | \mathscr{F}_s) = E(\xi_t-\xi_s)$$ The natural filtration for a process is defined as $$\mathscr{F}_s = \sigma\{\mathscr{F}_{\xi_r}, r \le s\}$$ What I don't see, is why will it be independent? I more or less understand that it will be independent if the filtration at any given moment had been generated by a finite number of $\sigma$-algebras, but in general it isn't. I don't even see how it can be - since we only have that finite distributions are independent, why should we have independence with a family generated by infinitely many $\sigma$-algebras.

Answer 1

Your question boils down, essentially, to the following result:

Suppose $\{X_i\}_{i\in I}$ is a collection of random variables, and let $\mathscr F$ be the $\sigma$-field generated by this collection. Suppose $Y$ is a random variable such that $Y$ and $X_i$ are independent for all $i\in I$. Then $Y$ is independent of $\mathscr F$.

Applying the above result with $\{X_i\}_{i\in I} = \{\xi_r\}_{r\le s}$ and $Y=\xi_t - \xi_s$ will give you what you need.

How do we prove this result? To start with, define $$ \mathscr A:= \Big\{ \{X_{i_1}\in E_1,X_{i_2}\in E_2,\ldots,X_{i_n}\in E_n\}\,:\,n\in\mathbb N, i_1,\ldots,i_n \in I, E_j={i_k}\text{ Borel}\Big\}.$$ In words, $\mathscr A$ is the collection of events obtained by observing finitely many of the $X_i$. Also define $$ \mathscr G:= \Big\{ A \in \mathscr F: \text{$A$ is independent of $Y$}\Big\}.$$

Since you know the result for finite collections, one has $\mathscr A \subset \mathscr G \subseteq \mathscr F$, and it is clear that $\sigma(\mathscr A) = \mathscr F$. The idea now is to use the $\pi$-$\lambda$ Theorem. Specifically, we will show

• $\mathscr G$ is a Dynkin system (i.e. contains the empty set, and is closed under complements and countable disjoint unions), and
• $\mathscr A$ is a $\pi$-system (i.e. is non-empty and is closed under finite intersections),

from which it follows by the $\pi$-$\lambda$ Theorem that $\mathscr F = \sigma(\mathscr A) \subseteq \mathscr G$, and hence $\mathscr G = \mathscr F$, which shows that every element of $\mathscr F$ is independent of $Y$, as required.

Let us now prove these two points. We know $\emptyset\in\mathscr G$ since $\emptyset$ is independent of every event (since $P(\emptyset\cap B) = P(\emptyset) = 0 = P(\emptyset)P(B)$). Suppose $A_j\in \mathscr G$ for $j\in\mathbb N$, and the $\{A_j\}$ are disjoint. Then for any $Y$-measurable event $B$, we have $$P(A_1^c\cap B) = P(B) - P(A_1\cap B) = P(B) - P(A_1)P(B) = \big(1-P(A_1)\big)P(B) = P(A_1^c)P(B),$$ and $$P\left(\left\{\bigcup_jA_j\right\}\cap B\right) = P\left(\bigcup_j(A_j\cap B)\right) = \sum_jP(A_j\cap B)=\sum_jP(A_j)P(B)=P\left(\bigcup_jA_j\right)P(B), $$ so $A_1^c$ and $\bigcup_jA_j$ are both independent of $B$. Since $B$ was arbitrary, both events are independent of $Y$ and are hence elements of $\mathscr G$. This shows $\mathscr G$ is a Dynkin system, as claimed.

For the second point, non-emptiness is trivial, and closure under finite intersections is not much more difficult: suppose $A,B\in\mathscr A$, so $A=\{X_{i_1}\in E_1,\ldots,X_{i_n}\in E_n\}$ and $B=\{X_{j_1}\in F_1,\ldots,X_{j_m}\in F_m\}$ for some $n,m\in\mathbb N$, $i_1,\ldots,i_n,j_1,\ldots,j_m\in I$, and Borel sets $E_1,\ldots,E_n,F_1,\ldots,F_m$. Then

$$ A\cap B = \{X_{i_1}\in E_1,\ldots, X_{i_n}\in E_n, X_{j_1}\in F_1,\ldots,X_{j_m}\in F_m\} \in \mathscr A,$$

where repeated indices can be interpreted appropriately (i.e. if $i_k = j_\ell$, then we interpret $X_{i_k}\in E_k,X_{j_\ell}\in F_\ell$ as $X_{i_k} \in E_k\cap F_\ell$). This shows $\mathscr A$ is a $\pi$-system, and completes the proof.