# Independent of generated $\sigma$-algebra if independent of random variables

Let $$(X_t)_{t\geq 0}$$ be a stochastic process and let $$\mathcal F_t=\sigma(X_s:0\leq s\leq t)$$ be the generated filtration. If $$Y$$ is a random variable independent of $$X_s$$ for all $$0\leq s\leq t$$, must $$Y$$ be independent of $$\mathcal F_t$$?

Edit: Answer is no. Suppose that $$Y$$ is a random variable independent of $$(X_{t_1},\dots,X_{t_n})$$ for all $$n>0$$ and $$0\leq X_{t_1}<\dots, must $$Y$$ be independent of $$\mathcal F_t$$?

• Consider independent r.v.s $(Z_i)_{i=0}^{\infty}$ such that $\mathbb{P}(Z_i = 1) = \mathbb{P}(Z_i = -1) = 1/2$. Set $X_t = Z_{\lfloor t \rfloor}$ and let $Y = Z_0 Z_1$. Then $X_s$ and $Y$ are independent for any $s \geq 0$, but $Y$ is not independent from $\mathcal{F}_t$ for $t \geq 1$. – Sangchul Lee Feb 7 '19 at 3:58
• @SangchulLee Thanks. Seems like the problem was that $Y$ only needed to be independent of a bounded number of $X_{t_i}$. What happens if we relax the requirement to independent of any finite collection of $X_{t_i}$? – Empty Set Feb 7 '19 at 14:17
• Then $$\forall A_1,\cdots,A_n, B \in \mathcal{B}, \quad \mathbb{P}\left(\{Y\in B\}\cap\bigcap_{i=1}^{n}\{X_{t_i}\in A_i\}\right)=\mathbb{P}(Y\in B)\prod_{i=1}^{n} \mathbb{P}(X_{t_i} \in A_i),$$ and notice that sets of the form $\bigcap_{i=1}^{n}\{X_{t_i}\in A_i\}$ for $A_i$'s in $\mathcal{B}$ and $t_i$'s in $[0, t]$ generates $\mathcal{F}_t$. So $Y$ must be independent of $\mathcal{F}_t$. – Sangchul Lee Feb 7 '19 at 19:24
• @Sangchul Lee We only have $P\left([Y\in B]\bigcap\cap_{i=1}^{n}[X_{t_{i}}\in A_{i}]\right)=P([Y\in B])P\left(\cap_{i=1}^{n}[X_{t_{i}}\in A_{i}]\right)$. It is unclear why $P\left(\cap_{i=1}^{n}[X_{t_{i}}\in A_{i}]\right)=\prod_{i=1}^{n}P\left([X_{t_{i}}\in A_{i}]\right).$ – Danny Pak-Keung Chan May 8 '19 at 21:25
• Moreover, to argue that: $\mathcal{C}$ and $\mathcal{G}$ are independent $\Rightarrow$ $\sigma(\mathcal{C})$ and $\mathcal{G}$ are independent, it is crucial that $\mathcal{C}$ is a $\pi$-class. In this case, Dynkin's $\pi$-$\lambda$ theorem is invoked. Although in our case, the generator is a $\pi$-class, however, this important fact should be mentioned explicitly. – Danny Pak-Keung Chan May 8 '19 at 21:29