3
$\begingroup$

Let $B=(B_t)_{t\ge 0}$ be a Brownian motion on a probability space $(\Omega,\mathcal A,\operatorname{P})$, i.e. $B$ is a real-valued stochastic process with

  • $B_0=0$ almost surely
  • $B$ has independent and stationary increments
  • $B_t\sim\mathcal{N}_{0,\;t}$
  • $B$ is almost surely continuous

Let $\mathbb F=(\mathcal F_t)_{t\ge 0}$ be the filtration generated by $B$ and $$\mathcal F_t^+:=\bigcap_{s>t}\mathcal F_s\;\;\;\text{for all }t\ge 0\;.$$

**Claim:**$\;$ Let $t>0$ and $$B':=(B_{s+t}-B_t)_{s\ge 0}\;.$$ Then $B'$ is independent of $\mathcal F^+_t$.

Proof: $\;$ Since $B$ is almost surely right-continuous $$B_{s+t_n}-B_{t_n}\stackrel{n\to\infty}{\to}B'_s\;\;\;\text{almost surely}\tag{1}$$ for all $t_n\downarrow t$ and $s\ge 0$. And further, observing that $B'$ is a Brownian motion independent of $\mathcal F_t$, we see that $$\left(B_{s_1}',\ldots,B_{s_m}'\right)\stackrel{(1)}{=}\lim_{n\to\infty}\underbrace{\left(B_{s_1+t_n}-B_{t_n},\ldots,B_{s_m+t_n}-B_{t_n}\right)}_{=:X_n}\;\;\;\text{almost surely}\tag{2}$$ for all $s_1,\ldots,s_m\ge 0$. Again, it's easy to verify, that $$X_n\text{ is independent of }\mathcal F_{t_n}\supseteq\mathcal F_t\;\;\;\text{for all }n\in\mathbb{N}\;.$$ So, we're really close, but I don't find the right argument to conclude the independence of $\left(B_{s_1}',\ldots,B_{s_m}'\right)$ and $$\mathcal F_t^+=\bigcap_{n\in\mathbb{N}}\mathcal F_{t_n}\;.$$ So, how do I need to argue?

$\endgroup$
2
$\begingroup$

As $\mathcal{F}_{t_n} \supseteq \mathcal{F}_{t+}$, we know that $X_n$ is independent of $\mathcal{F}_{t+}$. Therefore,

$$(B_{t+s_1}-B_t, \ldots,B_{t+s_m}-B_{t}) = \lim_{n \to \infty} (B_{t_n+s}-B_{t_n},\ldots,B_{t_n+s_m}-B_{t_n}) = \lim_{n \to \infty} X_n$$

is also independent of $\mathcal{F}_{t+}$.

Lemma: Let $\mathcal{F}$ be a $\sigma$-algebra and $(X_n)_{n \in \mathbb{N}}$ such that $X_n$ and $\mathcal{F}$ are independent for each $n \in \mathbb{N}$ and $X_n \to X$ almost surely. Then $X$ and $\mathcal{F}$ are independent.

Proof. Fix $F \in \mathcal{F}$. For any $\xi, \eta \in \mathbb{R}$, we have by the dominated convergence theorem

$$\begin{align*} \mathbb{E} \exp(\imath \, (\xi X+\eta 1_F)) &= \lim_{n \to \infty} \mathbb{E}\exp(\imath \, (\xi X_n+\eta 1_F)) \\ &= \lim_{n \to \infty} \mathbb{E}\exp(\imath \, \xi X_n) \mathbb{E}\exp(\imath \, \eta 1_F) \\ &= \mathbb{E}\exp(\imath \, \xi X) \mathbb{E}\exp(\imath \, \eta 1_F). \end{align*}$$

By the uniqueness of the Fourier transform, this shows that the joint distribution $(X,1_F)$ equals the product of the distribution of $X$ and $1_F$; hence, $X$ and $1_F$ are independent (see also this question).

$\endgroup$
  • 2
    $\begingroup$ @0xbadf00d You should strip down your questions to the important things. If you don't know that the pointwise limit is again independent, then just ask "Why is the pointwise limit independent?" instead of adding all this Brownian motion stuff and so on. This would make it much more easier to answer your question. $\endgroup$ – saz Jul 5 '15 at 13:57
  • $\begingroup$ Let me try to understand your argumentation. $$\varphi_{1_F}(t):=\operatorname E\left[e^{it1_F}\right]\;\;\;\text{and}\;\;\;\varphi_X(u):=\operatorname E\left[e^{i\langle u,X\rangle}\right]\;\;\;\;\;\text{for }t\ge 0\;\text{and}\;u\in\mathbb{R}^d$$ are the characteristic functions of the random variable $X$ with values in $\mathbb{R}^d$ and the indicator function $1_F$, for $F\in\mathcal F$, respectively. A finite measure is uniquely determined by its characteristic function. Moreover, if $Y$ and $Z$ are independent r.v. with values in $\mathbb{R}^d$, then $\varphi_{X+Y}=\varphi_X\varphi_Y$. $\endgroup$ – 0xbadf00d Jul 5 '15 at 15:00
  • $\begingroup$ So far, so good. But it seems like you're using something like the reverse. I didn't thought that one could show that and I don't understand the proof you've linked to. $\endgroup$ – 0xbadf00d Jul 5 '15 at 15:00
  • $\begingroup$ @0xbadf00d $\varphi_{X+Y} = \varphi_X \varphi_Y$ does not imply independence, but $\varphi_{(X,Y)}(\xi,\eta) = \varphi_X(\xi) \varphi_Y(\eta)$ does. If you don't understand the proof, then I'm sorry, but I won't add another one to my answer. $\endgroup$ – saz Jul 5 '15 at 15:10
  • $\begingroup$ I think I almost got it. However, there is still one thing I don't understand. I've asked a new question. Maybe you can answer it: math.stackexchange.com/questions/1350411/… $\endgroup$ – 0xbadf00d Jul 5 '15 at 17:28
3
$\begingroup$

If $X_n\to X$ almost surely and $X_n$ is independent of $\mathcal G$, then $X$ is independent of $\mathcal G$. We use actually this.

Call $X$ the left hand side of (2). We can show that for each continuous and bounded function $f\colon\mathbf R^n\to\mathbf R$ and $E\in\mathcal F_t$, we have $\mathbb E\left[f(X)\mathbf 1(E)\right]=\mathbb E[f(X)]\mu(E)$ (this is true from your work when $X$ is replaced by $X_n$; then use dominated convergence). You can approximate the characteristic function of a closed set by continuous functions bounded by $1$, hence we have for each closed set $F\subset \mathbf R^n$ that $\mu(\{X\in F\}\cap E )=\mu(\{X\in F\})\mu(E)$. This extends by a monotone class argument to any Borel-measurable set instead of $F$.

For the second question, it suffices to show that $\mathcal F_t^+\supset\bigcap_{n\in\mathbb{N}}\mathcal F_{t_n}$; the other inclusion is obvious. Pick $s\gt t$ and $A\in \bigcap_{n\in\mathbb{N}}\mathcal F_{t_n}$. For some $n$, $t\lt t_n\lt s$, and by definition of $\mathcal F_s$, we have $A\in \mathcal F_{t_n} \subset \mathcal F_s$. Since $s$ is arbitrary, it follows that $A\in \mathcal F_t^+$.

$\endgroup$
  • $\begingroup$ Oh, you misunderstood me. Actually, I had only one question: How I need to argue, that $X$ and $\mathcal F_t^+$ are independent. I thought $\mathcal F_t^+=\bigcap_{n\in\mathbb{N}}\mathcal F_{t_n}$ might be crucial. $\endgroup$ – 0xbadf00d Jul 5 '15 at 13:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.