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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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    I want to use the definition given in my first comment. Thanks.2012-05-02

1 Answers 1

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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, 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.

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

Thus

$\bigcup_m \bigcup_{s_1<\ldots

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