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In optimal control theory, we often need a filtration do be right continuous. Consider a filtered probability space $(\Omega, \mathcal F, \mathbb P)$ equipped with a right continuous filtration $\mathcal F_t$. Let $y_t$ be an $\mathcal F_t$ measurable process. Define, $a_t := \mathbb{1}_{y_t = 0}$.

Let $\zeta_t := \int_0^t (1-a_s) ds$ and $\lambda(s) := \sup\{t \ge 0: \zeta_t \le s\}$. (Assume that $y_t = 1$ for all $t \ge T$ for some finite $T$ so that this sup is well-defined and finite for all $s$.)

Consider a filtration $\mathcal G_s := \mathcal F_{\lambda(s)}$. Is $\mathcal G_s$ a right continuous filtration in $s$?

I want to say it is because $\lambda(s)$ is right continuous. So, $\cap_{\epsilon >0} \mathcal G_{s+\epsilon} = \mathcal G_s$. But, continuity of filtrations seems sufficiently abstract for me to not have any confidence in this reasoning.

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  • $\begingroup$ Isn't $\mathcal G_s$ no more right-continuous than $\mathcal F_t$? $\endgroup$ – Mateusz Kwaśnicki Nov 25 at 9:42
  • $\begingroup$ In particular, if $y_t$ is identically $1$, then $\zeta_t = t$, $\lambda(s)=s$, and $\mathcal{G}_s =\mathcal{F}_s$. By the way, the Brownian motion $B_t$ doesn't seem to have been used anywhere. $\endgroup$ – Nate Eldredge Nov 25 at 10:24
  • $\begingroup$ Both of you are absolutely right. I changed the question to say $\mathcal F_t$ is rt cts. Also, I initially wanted to use the BM to define a stochastic integral but then decided against as I thought this was a cleaner way to ask the question and forgot to delete the brownian motion. Thanks. $\endgroup$ – avk255 Nov 25 at 11:22
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Just use the definitions.

Consider first the case of an arbitrary filtration $\mathcal F_t$ (not necessarily right-continuous). For every $s$, the random variable $\lambda(s)$ is a Markov time with respect to $\mathcal F_{t+}$, in the sense that $$\{\lambda(s) < t\} \in \mathcal F_t \quad \text{for every $t$.}$$ By definition, $E \in \mathcal F_{\lambda(s)+}$ if and only if $$E \in \mathcal F_\infty \quad \text{and} \quad E \cap \{\lambda(s) < t\} \in \mathcal F_t \quad \text{for every $t$.}$$ Similarly, $E \in \mathcal F_{\lambda(s+\varepsilon)+}$ if and only if $$E \in \mathcal F_\infty \quad \text{and} \quad E \cap \{\lambda(s + \varepsilon) < t\} \in \mathcal F_t \quad \text{for every $t$.}$$ It follows that if $E \in \bigcap_{\varepsilon > 0} \mathcal F_{\lambda(s+\varepsilon)+}$, then $E \in \mathcal F_\infty$ and $$E \cap \{\lambda(s) < t\} = \bigcup_n E \cap \{\lambda(s + \tfrac{1}{n}) < t\} \in \mathcal F_t$$ for every $t$, and thus $E \in \mathcal F_{\lambda(s)+}$ (here indeed we use right-continuity of $\lambda$). Of course, $\bigcap_{\varepsilon > 0} \mathcal F_{\lambda(s+\varepsilon)+}$ contains $\mathcal F_{\lambda(s)+}$, and consequently the two $\sigma$-algebras are equal.

If $\mathcal F_t$ is right-continuous, then $\mathcal F_{t+} = \mathcal{F}_t$, and consequently $\lambda(s)$ is a Markov time with respect to $\mathcal F_t$ (in the sense that $\{\lambda(s) \leqslant t\} \in \mathcal F_t$), and $\mathcal F_{\lambda(s)+} = \mathcal F_{\lambda(s)}$. Here $E \in \mathcal F_{\lambda(s)}$ if and only if $$E \in \mathcal F_\infty \quad \text{and} \quad E \cap \{\lambda(s) \leqslant t\} \in \mathcal F_t \quad \text{for every $t$.}$$ It follows that indeed $\mathcal G_s = \mathcal F_{\lambda(s)}$ is right-continuous.

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  • $\begingroup$ Thanks. I too was using the definition only but these objects aren't the sorts I am used to handling. $\endgroup$ – avk255 Nov 28 at 8:20
  • $\begingroup$ You're welcome. (No-one is, I suppose, unless one is accidentally teaching this to students at the very same time.) $\endgroup$ – Mateusz Kwaśnicki Nov 28 at 8:32

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