# Right continuous filtration

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.

• Isn't $\mathcal G_s$ no more right-continuous than $\mathcal F_t$? – Mateusz Kwaśnicki Nov 25 '19 at 9:42
• 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. – Nate Eldredge Nov 25 '19 at 10:24
• 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. – avk255 Nov 25 '19 at 11:22

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.