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I'm looking for an example of a stochastic process, such that the natural filtration and the completed natural filtration aren't right-continuous.

I defined the process $(Z_t)_{t \geq 0}$ as $Z_t = t \cdot X$ where $X$ is a a.s. not constant random variable. For $t=0$ I get for the natural filtration $\mathcal{F}_0$:

$\mathcal{F}_0 = \sigma(Z_0) = \sigma(0) = \{\emptyset, \Omega\}$.

For $t>0$ I get

$\mathcal{F}_t = \sigma(Z_s, s \leq t) = \sigma(s \cdot X, s \leq t) = \sigma(X).$

(Is the last step correct, and if so, why exactly?)

Now I get $\mathcal{F}^+_0 = \bigcap_{\epsilon >0} \mathcal{F}_{\epsilon} = \sigma(X) \neq \{\emptyset, \Omega\} = \mathcal{F}_0$.

So the natural filtration isn't right-continuous.

Now I add the null sets and have a look at the filtration $\hat{\mathcal{F}}_t = \sigma(\mathcal{F}_t \cup \mathcal{N}_{\mathbb{P}})$. Now I get

$\hat{\mathcal{F}}_0 = \sigma(\{\emptyset,\Omega\} \cup \mathcal{N}_{\mathbb{P}})$ and

$\hat{\mathcal{F}}^+_0 = \bigcap_{\epsilon > 0} \sigma(\mathcal{F}_\epsilon \cup \mathcal{N}_{\mathbb{P}}) = \bigcap_{\epsilon > 0} \sigma(\sigma(X) \cup \mathcal{N}_{\mathbb{P}}) = \sigma(\sigma(X) \cup \mathcal{N}_{\mathbb{P}})$.

Since $X$ is a.s. not constant, $\sigma(X) \neq \{\emptyset, \Omega\}$ and so I can find a $B \in \sigma(X), B \neq \Omega, B \neq \emptyset$ with $\mathbb{P}(B) \neq 0$ (Can I assume that and if so, why?). So $(B \cup N) \in \hat{\mathcal{F}}^+_0$ for $N \in \mathcal{N}_{\mathbb{P}}$, but $(B \cup N) \notin \hat{\mathcal{F}_0}$ and therefore the completed natural filtration is also not right-continuous.

I have some doubts about my example and the proof; what do I have to change? Or does someone has another and better example?

Thanks a lot in advance.

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    $\begingroup$ What do you mean by "$\{B,N\} \in \hat{\mathcal{F}}_0^+$" for two sets $B$ and $N$? $\endgroup$
    – saz
    Jun 10, 2015 at 16:36
  • $\begingroup$ Sorry for the wrong notation; I meant $(B \cup N) \in hat{\mathcal{F}}^+_0$ for $B \in \sigma(X)$ and $N \in \mathcal{N}_{\mathbb{P}}$. But I don't think that this is correct since it isn't an element of the sigma algebra. $\sigma(X) \cup \mathcal{N}_{\mathbb{P}}$ is an element of $\hat{\mathcal{F}}^+_0$, but not of $\hat{\mathcal{F}}_0$. Is that correct? And what about the approach itself and the rest? Thanks again! $\endgroup$
    – Max93
    Jun 10, 2015 at 20:25

1 Answer 1

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Yes, the idea of your proof is correct and I don't think that there much easier examples. Concerning your questions:

  1. $\sigma(s \cdot X; s \leq t)$ is the smallest $\sigma$-algebra such that $\omega \mapsto s \cdot X(\omega)$ is measurable for any $s \leq t$. Since $X$ is measurable wrt to $\sigma(X)$, we know that $s \cdot X$ is measurable wrt to $\sigma(X)$ for any $s \leq t$. Consequently, $\sigma(s X; s \leq t) \subseteq \sigma(X)$. On the other hand, $$X = \frac{1}{s} (sX), \qquad s \neq 0,$$ is measurable with respect to $\sigma(s X_s)$ for any $s \neq 0$ and therefore $\sigma(X) \subseteq \sigma(s X; s \leq t)$.
  2. One can show that any $\hat{A} \in \hat{\mathcal{F}}_t$ can be written as $$\hat{A} = A \cup N$$ where $A \in \mathcal{F}_t$ and $N$ is a subset of a $\mathbb{P}$-null set. Since $\mathcal{F}_0 = \{\emptyset,\Omega\}$, this implies in particular $\mathbb{P}(\hat{A}) \in \{0,1\}$ for any $\hat{A} \in \hat{\mathcal{F}}_0$. As you already figured out, since $X$ is not constant almost surely, there exists a set $B \in \sigma(X)$ such that $\mathbb{P}(B) \notin \{0,1\}$. By the above considerations, this implies $B \notin \hat{\mathcal{F}}_0$. On the other hand, $B \in \sigma(X) \subseteq \hat{\mathcal{F}}_0^+$. Therefore, we have shown that $\hat{\mathcal{F}}_0 \subsetneq \hat{\mathcal{F}}_0^+$.
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  • $\begingroup$ Thanks again, I finally understand each step of the proof. $\endgroup$
    – Max93
    Jun 12, 2015 at 13:24
  • $\begingroup$ @Susan Glad I could help you. $\endgroup$
    – saz
    Jun 12, 2015 at 16:15

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