Let $X_t : \Omega \to E, \ t \geq 0$ be continuous-time stochastic process with (Polish) state space $E$ and canonical filtration $\mathcal{F}_t := \sigma(X_u \ | \ u \leq t)$. Let $Y_t : \Omega \to \mathbb{R}$ be a real-valued process that is adpated to $\mathcal{F}_t$.

From measure theory it is known that if a $\sigma$-Algebra $\mathcal{A}$ is generated by a real-valued map $h$, i.e. $\mathcal{A} = \sigma(h)$ then every real-valued map $f$ that is $\mathcal{A}$-measurable can be represented as measurable function of $h$, i.e. $f = g \circ h$ for some measurable map $g$.

Applying this to the stochastic process $X_t$ it follows that for each $t$ there exists a measurable map $g_t : E^{[0,t]} \to \mathbb{R}$ such that $Y_t = g_t((X_u)_{u \leq t})$.

I want to know, if there is some possibility to have only one map $g(t, (X_u)_{u \leq t})$ that is jointly measurable in $t \in \mathbb{R}$ and the sample paths $X_u$ up to time $t$ such that $Y_t = g(t, (X_u)_{u \leq t})$ for all $t$. Do I need $X_t$ to be jointly measurable and $Y_t$ to be progressively measurable? How can I formally choose a suitable domain for such a function $g$? Is it some fibre space related to $X$, e.g. a subspace of $[0, \infty) \times E^{[0, \infty)}$ that consists of all elements of the form $(t, (X_u)_{u \leq t})$ (i.e. some sort of measurable upper-diagonal of $[0, \infty) \times E^{[0, \infty)}$)?

If MSE is the appropriate forum for this kind of question, I can leave it here or otherwise move this thread to MO.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.