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Assume that $X,\{Y_\alpha\}$ (where $\alpha \in A$, $A$ can be uncountable) are random variables. If $X$ and $Y_\alpha$ are independent for all $\alpha \in A$, i.e., $\sigma(X)$ and $\sigma(Y_\alpha)$ are independent for all $\alpha \in A$, then it need not be true (I was unable to prove so!) that $\sigma(X)$ and $\sigma(\{Y_\alpha\}, \alpha\in A)$ are also independent.

Now, let $X=B(t)-B(s)$ (where $B$ is the standard brownian motion and $t>s$) and $Y_\alpha= B(\alpha)$ with $A = [0,s]$. $X$ is independent of $Y_\alpha$ for all $\alpha \in A$ from the properties of Brownian Motion.

However, it seems that $X$ is also independent of $\sigma(\{Y_\alpha\}, \alpha\in A)$, as the textbook I am following (Introduction to Stochastic Integration by Hui-Hsiung Kuo page 18) uses this result. I am wondering if the properties of Gaussian Random Variables make this plausible. However, I would like to see a proof of this result.

Any help is much appreciated.

Thanks, Phanindra

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You probably need $t \gt s$ – Henry May 1 '12 at 6:36
The question is empty since the fact that $B(t)-B(s)$ is independent of $\sigma(B(u),u\in[0,s])$, for every $t\geqslant s$, is part of the definition of standard Brownian motion. – Did May 1 '12 at 7:36
@Didier : The book referred in the question defines a Brownian Motion as a stochastic process sastisfying the three properties: 1) $B(0)=0~a.s.$ 2) $B(t)-B(s) \sim N(0,t-s), t > s$ 3) all increments are independent i.e. if $0 \leq t_1 < t_2 < \cdots <t_n$ then $B(t_1), B(t_2) - B(t_1), \cdots B(t_n) - B(t_{n-1})$ are independent. I am not sure how the sigma algebras defined in the question become independent by this definition. – jpv May 1 '12 at 7:42
And since the sigma-algebra $\sigma(B(u),u\in[0,s])$ is the smallest one such that $B(u)$ is measurable, for every $u\in[0,s]$, it is generated by the collection of vectors $(B(t_1),B(t_2)-B(t_1),\ldots,B(t_n)-B(t_{n-1}))$ for every $n$ and every $t_1\leqslant\cdots\leqslant t_n\leqslant s$, hence all is well. You could check the definition of $\sigma(Y_a,a\in A)$ when $A$ is infinite. – Did May 1 '12 at 7:52
@Didier: Thanks for your reply. However, I am still unable to resolve the issue. Even if I assume that the $\sigma(B(u), u \in [0,s])$ is the same as the $\sigma$-algebra generated by all the vectors of the form $(B(t_1), B(t_2)-B(t_1),\cdots,B(t_n)-B(t_{n-1}))$ I still need to prove that $B(t)-B(s)$ is independent of this new $\sigma$-algebra. I surely know that $B(t)-B(s)$ is independent of the $\sigma$-algebra generated by each of these vectors individually but I want to prove that $B(t)-B(s)$ is independent of the $\sigma$-algebra generated by all of these vectors put together. – jpv May 2 '12 at 7:30

1 Answer 1

Let $X_0,\ldots,X_n$ random variables. Then

$$\sigma(X_j; j=0,\ldots,n) = \sigma(X_0,X_j-X_{j-1};j=1,\ldots,n) \qquad (1)$$

(follows straight from the definition).

Now let $0:=s_0<s_1<\ldots<s_m = s <t$, then we know that $B_{t}-B_s,B_{s_m}-B_{s_{m-1}},\ldots,B_{s_1}-B_{s_0}$ are independent, i.e.

$$\newcommand{\Perp}{\perp \! \! \! \perp} \sigma(B_{s_j}-B_{s_{j-1}}; j=1,\ldots,m\} \Perp \sigma(B_t-B_s) \\ \stackrel{(1)}{\Rightarrow} \sigma(B_{s_j};j=0,\ldots,m) \Perp \sigma(B_t-B_s)$$


$$\bigcup_m \bigcup_{s_1<\ldots<s_m=s} \sigma(B_{s_j};j=0,\ldots,m) \Perp \sigma(B_t-B_s)$$

Since the left hand side is a generator of $\mathcal{F}_s$ and closed under finite intersections we conclude $\mathcal{F}_s \Perp B_t-B_s$.

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