Let $\left(\Omega\ ,\mathcal{F}\ ,\mathbb{P} \right)$ be a probability space with filtration $F=\left\{ \mathcal{F_t} \right\}_{t\geqslant0}$ , and $X=\left\{ X_t \right\}_{t\geqslant0}$ is an adapted stochastic process w.r.t. $F$.

Definition 1: A process $X$ is called stochastically continuous , if for all $t\geqslant0$ , $X_s\overset{\mathbb{P}}{\rightarrow} X_t$ as $s\rightarrow t$.
Definition 2: An adapted process $X$ is called a process with independent increments w.r.t. $F=\left\{ \mathcal{F_t} \right\}_{t\geqslant0}$ , if for all $t>s\geqslant0$ , $X_t-X_s$ is independent of $\mathcal{F_s}$.

Proposition: If a stochastically continuous process $X$ has independent increments w.r.t. $F=\left\{ \mathcal{F_t} \right\}_{t\geqslant0}$ , then $X$ has independent increments w.r.t. $F_+:=\left\{ \mathcal{F_{t+}} \right\}_{t\geqslant0}$ , where $\mathcal{F_{t+}}:=\bigcap_{r>t}{\mathcal{F_r}} $ .

Now we want to prove that if $t>s\geqslant0$ , then $X_t-X_s$ is independent of $\mathcal{F_{s+}}=\bigcap_{n\in\mathbb{N}}{\mathcal{ F_{s+1/n} }} $ . The condition tell us that $X_t-X_{s+1/n}\overset{\mathbb{P}}{\rightarrow}X_t-X_s$ as $n\rightarrow\infty$. In fact, $\overset{\mathbb{P}}{\rightarrow}$ can be replaced by $\overset{a.s.}{\rightarrow}$ because of independent increments.
Note that $X_t-X_{s+1/n}$ is independent of ${\mathcal{ F_{s+1/n} }}$ , it seems that $$X_t-X_s=\lim_{n\rightarrow\infty}\left(X_t-X_{s+1/n}\right) \text{ is independent of }\ \mathcal{F_{s+}}=\bigcap_{n\in\mathbb{N}}{\mathcal{ F_{s+1/n} }}\ ,$$ but I don't know how to get the result.

  • $\begingroup$ Personal input? An approach seems natural, did you try to make it work? $\endgroup$ – Did Jun 29 '16 at 9:17
  • $\begingroup$ @Did I have tried but failed. I don't konw how to keep the independence with the limit in probability. $\endgroup$ – Wang Ke Jun 29 '16 at 9:26
  • $\begingroup$ Please include whichever tries you made, in the question. $\endgroup$ – Did Jun 29 '16 at 9:28
  • $\begingroup$ @Did I added my thinking. $\endgroup$ – Wang Ke Jun 29 '16 at 9:53
  • $\begingroup$ And now we see that $X_t-X_{s+1/n}$ is independent of $\mathcal F_{s+}$ and that $X_t-X_{s+1/n}\to X_t-X_s$ almost surely hence the question is: "Assume that $(Y_n)$ is independent of $\mathcal G$ and that $Y_n\to Y$ almost surely then $Y$ is independent of $\mathcal G$." Any idea for this? $\endgroup$ – Did Jun 29 '16 at 10:56

Thanks for @Did's help. Now I have got the answer.

Because $X_t-X_{s+1/n}$ is independent of ${\mathcal{ F_{s+1/n} }}$ and $\mathcal{ F_{s+} }\subset\mathcal{F_{s+1/n}}\ $ , $X_t-X_{s+1/n}\ $ is independent of ${\mathcal{ F_{s+} }}$ .
Now the question is to prove "Assume that $\left\{Y_n\right\}$ is independent of $\mathcal{G}$ and that $Y_n\overset{a.s.}{\rightarrow}Y$ , then $Y$ is independent of $\mathcal{G}$."


  • " $Y$ is independent of $\mathcal{G}$ " is equivalent to " $\forall A\in\mathcal{G}$ , $Y$ is independent of $I_A$ ".
  • $Y_n\overset{a.s.}{\rightarrow}Y$ , hence $\exp\left(it_1Y_n\right) \overset{a.s.}{\rightarrow} \exp\left(it_1Y\right)$ and $\exp\left(it_1Y_n+it_2I_A\right) \overset{a.s.}{\rightarrow} \exp\left(it_1Y+it_2I_A\right)$ .
  • By dominated convergence theorem, we have that $\mathbb{E}\left[\exp\left(it_1Y_n\right)\right] \rightarrow \mathbb{E}\left[\exp\left(it_1Y\right)\right]$ and that $\mathbb{E}\left[\exp\left(it_1Y_n+it_2I_A\right)\right] \rightarrow \mathbb{E}\left[\exp\left(it_1Y+it_2I_A\right)\right]$ .
  • For each $n$, $Y_n$ is independent of $I_A$, hence $$\mathbb{E}\left[\exp\left(it_1Y_n+it_2I_A\right)\right]=\mathbb{E}\left[\exp\left(it_1Y_n\right)\right]\mathbb{E}\left[\exp\left(it_2I_A\right)\right], $$
    then we have that
    $$\begin{align*} &\mathbb{E}\left[\exp\left(it_1Y+it_2I_A\right)\right] = \lim_{n\rightarrow\infty}{\mathbb{E}\left[\exp\left(it_1Y_n+it_2I_A\right)\right]} \\ &\qquad\qquad\qquad\qquad\quad =\lim_{n\rightarrow\infty}{\mathbb{E}\left[\exp\left(it_1Y_n\right)\right]\mathbb{E}\left[\exp\left(it_2I_A\right)\right]} \\ &\qquad\qquad\qquad\qquad\quad =\mathbb{E}\left[\exp\left(it_1Y\right)\right]\mathbb{E}\left[\exp\left(it_2I_A\right)\right]. \end{align*}$$
    It is concluded that $Y$ is independent of $I_A$.
  • $\begingroup$ The fact that, for every $n$, $\sigma\left(Y_n\right)$ is independent of $\mathcal{G}$, does not imply that $\sigma\left(Y_n,n\in\mathbb{N}\right)$ is independent of $\mathcal{G}$. $\endgroup$ – Did Jun 29 '16 at 11:48
  • $\begingroup$ @Did I have edited. $\endgroup$ – Wang Ke Jun 29 '16 at 12:36

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