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Question

Let $\Omega=C([0,2])$ (set of continious funtions) and $X$ a stochastic process on $\mathbb R$ such that $$X_t(\omega):=\omega(t)$$ with the natural filtration $\mathbb F:=(\mathcal F_t)_{t\in [0,T]}$ given by $\mathcal F_t:=\sigma(X_s:s\le t)$.

I would like to prove that $\mathbb F$ is not right-continuous.


Proof Attempt

Let $t>0$ and $S_+^t:=\{\omega \in \Omega:X_t(\omega)>0\}=X^{-1}_t((0,\infty))$, then $S_+^t \in \mathcal F_{0+}$, where $\mathcal F_{0+}:=\bigcap_{\varepsilon>0}\mathcal F_{0+\varepsilon}$, but $S_+^t$ is not $\mathcal F_0$-measurable since $t>0$.


Is this proof correct?

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  • $\begingroup$ Why is $S_+^t \in \mathcal F_{0+}$? $\endgroup$ Mar 2, 2023 at 18:13
  • $\begingroup$ Oh, I think I found a mistake in my proof since you asked me. $t$ can't be just bigger than 0 since for $t=1$ $S_+^t \not \in \mathcal F_{0+}$, right? So I would make the following change: $S_+:=\{\omega \in \Omega:X_t(\omega)>0, \forall t>0\}$, now it should be in $\mathcal F_{0+}$, right? $\endgroup$
    – scholar
    Mar 2, 2023 at 18:22
  • $\begingroup$ I think that's on the right track, but that $S_+$ still isn't in $\mathcal F_{0+}$ because it is not in $\mathcal F_t$ for any fixed $t > 0$. $\endgroup$ Mar 2, 2023 at 18:27
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    $\begingroup$ Ok, I am sure that $S_+^t \in \mathcal F_t$ because this is explained with the definition of the chosen filtration. So then $\tilde S:=\cap_{t>0} S_+^t$ should be in $\mathcal F_{0+}$, is that better? $\endgroup$
    – scholar
    Mar 2, 2023 at 18:31
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    $\begingroup$ Right, it's possible for the intersection (or union) of two sets not in $\mathcal F$ to be in $\mathcal F$. For example, if $\mathcal F$ is any $\sigma$-algebra and $A$ is any set, then $A \cap A^c \in \mathcal F$, regardless of whether or not $A$ is in $\mathcal F$. $\endgroup$ Mar 2, 2023 at 19:51

1 Answer 1

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As discussed in the comments, if we define $\bar S := \bigcap_{t > 0} S^t_+$, then $\bar S \in \mathcal F_{0+}$. The last step is to verify that $\bar S \not \in \mathcal F_0$. Since $\mathcal F_0 = \sigma(X_0)$, it is enough to show that there exist $\omega_1, \omega_2 \in \Omega = C([0,2])$ such that $\omega_1(0) = \omega_2(0)$ and $\omega_1 \in \bar S$ but $\omega_2 \not \in \bar S$. So we can just let $\omega_1(t) := t$ and $\omega_2(t) := -t$.

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  • $\begingroup$ I got it now! Thank you for your time and help, I really appreciate it! $\endgroup$
    – scholar
    Mar 2, 2023 at 23:16

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