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Let $(\Omega, \mathscr{A}, P)$ be a probability space, $(E, \mathscr{E})$ a measurable space and $X_t : \Omega \to E$, $t \geq 0$ a measurable stochastic process, i.e. the map $X : [0, \infty) \times \Omega \to E$ is measurable where the domain is equipped with the product $\sigma$-algebra.

Let $Y_t : \Omega \to \mathbb{R}$, $t \geq 0$ be a stochastic process that is adapted to the canonical filtration $\mathscr{F}_t := \sigma(X_u \ | \ u \leq t)$ of $X_t$.

Q1: Is it true that $Y$ is also measurable (or has a measurable modification)? Surely, if I know that $Y$ is progressively measurable w.r.t. $\mathscr{F}_t$ then we are done. But is adaptedness enough?

Q2: If the answer to Q1 is no, what is the minimal requirement for $X$ and $Y$ to assure that $Y$ is measurable?

In my applications I know more about $X$, namely that $t \mapsto X_t$ is a.s. cadlag with piecewise constant sample paths. But from $Y_t$ I only know that it is adapted to $X_t$.

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I will give a short answer for general filtrations.

ad 1) Adaptedness is not enough. Let $f: [0,\infty) \to \mathbb{R}$ be a non-measurable function. Set $Y = f$ for all $\omega\in\Omega$. For all $t\in[0,\infty)$ we have that $Y_t$ is a deterministic value, hence a random variable and $\mathcal{F}_t$-measurable. So $(Y_t)$ is an adapted process. Measurability fails though, because $f$ is not measurable.

ad 2) Right-continuity (or left-continuity) and adaptedness imply progressiveness. To see this, approximate by piecewise constant processes, which are progressive, and take the limit. Note, this is not a minimal requirement. Maybe some expert can help out.

If you take the counterexample from 1) it is adapted to $\sigma(X_u \ | \ u \leq t)$, since it’s deterministic, but not measurable. So you need stronger assumptions.

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  • $\begingroup$ Thanks, nice counterexample. My process $Y_t$ is neither right- nor left-continuous, but I can restrict it to be regulated, i.e. both right and left limits exist. I think, as for right-continuous processes it can be shown in the same way, that regulated adapted processes are also progressively measurable. $\endgroup$ – yadaddy Dec 21 '14 at 18:01
  • $\begingroup$ I’m not quite sure you can write a general regulated (LLRL) process as a limit of progressive processes. Can you state your idea? $\endgroup$ – dlrlc Dec 22 '14 at 15:10
  • $\begingroup$ A regulated function $f(t)$ has on each compact interval $[0,T]$ only a finite number of discontinuities which are all jumps. Just an idea: approximate the regulated function by step functions defined between the jumps of $f$. Or alternatively, approximate the cadlag version $f(t+)$ and modify the values at points of jump of $f$. $\endgroup$ – yadaddy Dec 23 '14 at 9:07
  • $\begingroup$ This will surely work for deterministic jump times. For random jump times you have piecewise constant approximation on random intervals, showing that this is progressively measurable might not be that easy. $\endgroup$ – dlrlc Dec 23 '14 at 10:12
  • $\begingroup$ I see. In my case, if $Y_t$ jumps then the cadlag $X_t$ also jumps. And since $Y_t$ is adapted to $X_t$ it might work to show that the regulated $Y_t$ is progressively measurable. As an example for a regulated process $Y_t$ that is adapted to $X_t$ take $Y_t = g(X_{t−},X_t)$ for some measurable function $g : E \times E \to \mathbb{R}$. This $Y_t$ seems to be progressively measurable. $\endgroup$ – yadaddy Dec 24 '14 at 11:20

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