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Let $X=(X_t)$ be a stochastic process and we define the raw filtration by $F=(\mathcal{F}_t)$, where $\mathcal{F}_t:=\sigma (X_s;s\le t)$ Now I want to prove that $\sigma (\mathcal{C})=\mathcal{F}_t$, where $\mathcal{C}:=\{\prod_{k=0}^nf_k(X_{t_k});t_n\le t,f_k:\mathbb{R}\to \mathbb{R} \mbox{ bounded and measurable}\}$ I was able to prove one inclusion: since $\mathcal{F}_t$ is generated by $X^{-1}_s(B)$ where $B\in \mathcal{B}(\mathbb{R})$, we take $n=0$, $f_0=\mathbf1_B$ and $t_0=t_n=s$. The other inclusion is bothering me, i.e. $\sigma (\mathcal{C})\subset \mathcal{F}_t$?

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up vote 1 down vote accepted

Let $t \ge 0$, $n \in \mathbb N$, $f_k\colon \mathbb R \to \mathbb R$ measurable and bounded, $t_k \le t$ for $k \le n$. Then, as multiplication is measurable, and the $X_{t_k}$ are $\mathcal F_t$-measurable, $\prod_k f_k \circ X_{t_k}$ is $\mathcal F_t$-measurable. As $\prod_k f_k \circ X_{t_k} \in \mathcal C$ was arbitrary, and $\sigma(\mathcal C)$ is the smallest $\sigma$-algebra which makes these functions measurable, we have $\sigma(\mathcal C)\subseteq \mathcal F_t$.

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@ martini: Just a small additional question: The proof still works if we restrict $f_k$ to be positive, i.e. $f_k:\mathbb{R}\to[0,\infty)$, right? – user20869 Jul 12 '12 at 9:40
@hulik, yes. ${}$ – martini Jul 12 '12 at 10:05

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