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Let $\{X_t\}, t\in [0,\infty)$ be a collection of random variables and $$C_1 = \{ E \mid E = \{ (X_{t_1}, X_{t_2}, X_{t_3}, \cdots, X_{t_n}) \in B\},n \in \mathbb{N},B\in \mathcal{B}(\mathbb{R}^n)\}$$ where $t_i \in [0,\infty)~\forall i$.

$C_1$ can be shown to be an algebra (field) basically by "embedding" the Borel sets in higher dimensions.

Now, if the definition is changed slightly to:

$$C_2 = \{ E \mid E = \{ (X_{t_1}, X_{t_2}, X_{t_3}, \cdots, X_{t_n}) \in B\},\\n \in \mathbb{N},B\in\mathcal{B}(\mathbb{R}^n), 0 \leq t_1 < t_2 < t_3 \cdots < t_{n-1} < t_n\}\}.$$

Is $C_2$ still an algebra? The issue is that the Borel sets some how have to be interspersed and I am not sure how to do this.

Thanks, Phanindra

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I don't understand your definition of $C_1$ and $C_2$. A set $E$ is in $C_1$ if ...? –  martini May 4 '12 at 6:09
    
@martini : A set $E$ is in $C$ if there exists a $n \in \mathbb{N}$, a Borel set $B \in \mathbb{B}(\mathbb{R})^n$ such that $(X_{t_1},X_{t_2},\cdots, X_{t_n})^{-1}(B) = E$ for some $t_1,t_2,\cdots, t_n$. For $C_2$ we have an additional order constraint. –  jpv May 4 '12 at 12:53

1 Answer 1

up vote 1 down vote accepted

It seems that $C_2$ is indeed an algebra. The proof is not hard, but quite boring.

Take $E_1$ and $E_2$ two elements of $C_2$. We have to check that their intersection still is in $C_2$. $E_1$ has the form $E_1=\{(X_{s_1},\ldots,X_{s_m})\in B_1\}$, and $E_2=\{(X_{t_1},\ldots,X_{t_n})\in B_2\}$, where $B_1\in\mathcal B(\Bbb R^m)$, $B_2\in\mathcal B(\Bbb R^n)$, $m$ and $n$ are integers and $0\leq s_1<\ldots<s_m$, $0\leq t_1<\ldots<t_n$. Let $r_1,\ldots,r_N\geq 0$ such that $r_1<r_2<\ldots<r_N$ such that $r_k$ is one of the $s_i$ or $t_j$, and all the $t_j$ and $r_j$ are in this list (I denoted $N$ because we don't know how many repetition we have between the $s_j$ and $t_j$). Let $I_1:=\{k\in\{1,\ldots,N\}\mid \exists i\in \{1,\ldots,m\},r_k=s_i\}\}=\{a_1,\ldots,a_{N_1}\}$ and $I_2:=\{k\in\{1,\ldots,N\}\mid \exists i\in \{1,\ldots,n\},r_k=t_i\}\}=\{b_1,\ldots,b_{N_2}\}$ (ordered). We define $$B:=\{(x_1,\ldots,x_N)\in\Bbb R^n\mid (x_{a_1},\ldots,x_{a_{N_1}})\in B_1\mbox{ and }(x_{b_1},\ldots,x_{b_{N_2}})\in B_2\}.$$ It's a Borel subset of $\mathcal B(\Bbb R^N)$ and $$E_1\cap E_2=\{(X_{r_1},\ldots,X_{r_N})\in B\}.$$

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