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Let $X = (X_t)_{t \geq 0}$ be a continuous, real-valued process defined on some probability space $(\Omega,\mathcal{F},P)$, and let $\mathbb{F}^X = (\mathcal{F}_{t}^X)_{t \geq 0}$ be the filtration generated by $X$. Suppose that $\alpha = (\alpha_t)_{t \geq 0}$ is real-valued and $\mathbb{F}^X$-progressive. Can we write

  • $\alpha_t(\omega) = \tilde{\alpha}(t,X(\omega))$ for some product measurable function $\tilde{\alpha} : \mathbb{R}_+ \times C(\mathbb{R}_+;\mathbb{R}) \rightarrow \mathbb{R}$

or

  • $\alpha_t(\omega) = \tilde{\alpha}(t,X_{t \land \cdot}(\omega))$ for some product measurable function $\tilde{\alpha} : \mathbb{R}_+ \times C(\mathbb{R}_+;\mathbb{R}) \rightarrow \mathbb{R}$ ?

Here $C(\mathbb{R}_+;\mathbb{R})$ denotes the space of continuous functions from $\mathbb{R}_+$ to $\mathbb{R}$.

The answer does not seem to be a simple application of the Doob-Dynkin factorization lemma + functional monotone class argument as the progressive $\sigma$-algebra (from my understanding) is not generated by a random variable nor does it have "nice" elementary generators.

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  • $\begingroup$ Here $X(\omega)$ stands for the function $t \mapsto X_t(\omega)$ ? $\endgroup$
    – rubikscube09
    Commented Apr 22, 2021 at 14:37
  • $\begingroup$ @rubikscube09 yes, exactly. $\endgroup$
    – vaoy
    Commented Apr 22, 2021 at 14:53

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