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Let $\mathcal I:=\left\{I\subseteq[0,\infty):I\text{ is finite}\right\}$ and $(X_t)_{t\ge 0}$ and $(Y_t)_{t\ge 0}$ be stochastic processes on a common probability space taking values in some measurable spaces. Clearly, if

  1. $X$ and $Y$ are independent, i.e. $\sigma(X)=\sigma(X_t,t\ge 0)$ and $\sigma(Y)=\sigma(Y_t,t\ge 0)$ are independent,

then

  1. $\mathcal A_I:=\sigma(X_t,t\in I)\subseteq\sigma(X)$ and $\mathcal B_J:=\sigma(Y_t,t\in J)\subseteq\sigma(Y)$ are independent for all $I,J\in\mathcal I$.

How can we show, that 2. implies 1. too?

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3 Answers 3

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Let $$\mathcal A:=\bigcup_{I\in\mathcal I}\mathcal A_I\;\;\;\text{and}\;\;\;\mathcal B:=\bigcup_{I\in\mathcal I}\mathcal B_I\;.$$

Claim 1 $\;$ $\mathcal A$ is $\cap$-stable.

Proof: $\;$ Let $A,B\in\mathcal A$ $\Rightarrow$ $\exists I,J\in\mathcal I$ with $A\in\mathcal A_I$ and $B\in\mathcal A_J$ $\Rightarrow$ $$A\cap B\in\mathcal A_I\cup\mathcal A_J\subseteq\mathcal A_{I\cup J}\subseteq\mathcal A\;,$$ since $A_I,\mathcal A_J\subseteq\mathcal A_{I\cup J}$.

Clearly, we obtain the $\cap$-stability of $\mathcal B$ by the same argumentation.

Claim 2 $\;$ $\sigma(\mathcal A)=\sigma(X)$.

Proof: $\;$ By definition, $\mathcal A_I\subseteq\sigma(X)$ for all $i\in\mathcal I$ $\Rightarrow$ $$\sigma(\mathcal A)\subseteq\sigma(X)$$ by monotonicity of the $\sigma$-operator. On the contrary, $\left\{t\right\}\in\mathcal I$ for all $t\ge 0$ and thereby $$\bigcup_{t\ge 0}\mathcal A_{\left\{t\right\}}\subseteq\mathcal A\;.$$ Thus, $$\sigma(X)\subseteq\sigma(\mathcal A)$$ by monotonicity of the $\sigma$-operator.

Clearly, we obtain $\sigma(B)=\sigma(Y)$ by the same argumentation. By assumption (2.), $\mathcal A$ and $\mathcal B$ are independent. Using the well-known fact

Theorem $\;$ Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $I$ be a set and $\mathcal E_i\subseteq\mathcal A$ for all $i\in I$. If $\mathcal E_i\cup\left\{\emptyset\right\}$ is $\cap$-stable for all $i\in I$, then $(\mathcal E_i)_{i\in I}$ is $\operatorname P$-independent if and only if $\left(\sigma\left(\mathcal E_i\right)\right)_{i\in I}$ is $\operatorname P$-independent.

we can conclude the independence of $X$ and $Y$ by combining claim 1 and 2.

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a. If $A\in\sigma(X)$ and $B\in\sigma(Y)$ then there is a countable set $L\subset[0,\infty)$ such that $A\in\sigma(X_t:t\in L)$ and $B\in\sigma(Y_t: t\in L)$.

b. The collection $\tilde{\mathcal A}_L:=\cup_{J\subset L, J\text{ finite}}\sigma(X_t:t\in J)$ is an algebra of sets, and $\sigma(\tilde{\mathcal A}_L)=\sigma(X_t:t\in L)$. One can define $\tilde{\mathcal B}_L:=\cup_{J\subset L, J\text{ finite}}\sigma(Y_t:t\in J)$ analogously.

c. Fix $A\in\sigma(X)$ and $B\in\sigma(Y)$, and let $L$ be as in point a. above. Using b. one can show that given $\epsilon>0$ there exists $J_\epsilon$ (a finite subset of $L$) and sets $A_\epsilon\in \sigma(X_t:t\in J_\epsilon)$ and $B_\epsilon\in \sigma(Y_t:t\in J_\epsilon)$ with $P(A\Delta A_\epsilon)<\epsilon$ and $P(B\Delta B_\epsilon)<\epsilon$. By hypothesis $A_\epsilon$ and $B_\epsilon$ are independent. Clearly $P(A)$, $P(B)$, $P(A\cap B)$ differ from $P(A_\epsilon)$, $P(B_\epsilon)$, $P(A_\epsilon\cap B_\epsilon)$ (respectively) by at most $\epsilon$, $\epsilon$, $2\epsilon$ (respectively). It follows that $P(A\cap B)$ differs from $P(A)P(B)$ by at most $2\epsilon+\epsilon^2<3\epsilon$. As $\epsilon>0$ was arbitrary, $P(A\cap B)=P(A)P(B)$, so $A$ and $B$ are independent.

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  • $\begingroup$ I began writing an answer before your answer appeared. What do you think about my solution? $\endgroup$
    – 0xbadf00d
    Feb 14, 2016 at 17:55
  • $\begingroup$ Yours is fine, even preferable. Your well-known fact follows by an easy application of the monotone class theorem, as in @Blackbird's answer. $\endgroup$ Feb 14, 2016 at 18:11
  • $\begingroup$ Thanks for the feedback. $\endgroup$
    – 0xbadf00d
    Feb 14, 2016 at 18:13
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Let $\Omega_X$ and $\Omega_Y$ be the measurable spaces in which $X$ and $Y$ take their values.

Let $\mu_X$ and $\mu_Y$ be the measures on $\Omega_X$ and $\Omega_Y$ induced by $X$ and $Y$, that is, $\mu_X(E):=\mathbb{P}(X\in E)$ for measurable $E\subseteq\Omega_X$ and $\mu_Y(F):=\mathbb{P}(Y\in F)$ for measurable $F\subseteq\Omega_Y$. Similarly, let $\mu_{X,Y}$ be the measure induced on $\Omega_X\times\Omega_Y$ by the pair $(X,Y)$. That $X$ and $Y$ are independent is equivalent to saying that $\mu_{X,Y}=\mu_X\times\mu_Y$ (why?).

Now, in order for two probability measures to be the same, it is sufficient that they agree on a $\pi$-system generating the $\sigma$-algebra of measurable sets. So, it is enough to show that the family of sets $E\times F$ such that $\{X\in E\}\in\mathcal{A}_I$ and $\{Y\in F\}\in\mathcal{B}_J$ for some $I,J\in\mathcal{I}$ is a $\pi$-system generating the $\sigma$-algebra on $\Omega_X\times\Omega_Y$. This is the case if the original $\sigma$-algebras on $\Omega_X$ and $\Omega_Y$ were product $\sigma$-algebras.

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