3
$\begingroup$

consider a probablity space $(\Omega,\mathcal{F}, \mathcal{P})$ and a filtration $(\mathcal{F}^0_t)$. In general $(\mathcal{F}^0_t)$ doesn't satisfy the usual conditions (it is not both complete at any $t$ and right continuous). To overcome this problem one usualy set for all $t$ $$\mathcal{F}_t= \mathcal{F}^0_{t+}V \mathcal{N}$$ where $\mathcal{N}$ are the negligible sets for $\mathbb{P}$. The new filtration $(\mathcal{F}_t)$ satisfies the usual conditions and it is called the usual augmentation of $(\mathcal{F}_t^0)$. On the other hand one could also consider the filtration $(\mathcal{G}_t)$ defined by $$\mathcal{G}_t = \cap_{\epsilon>0} (\mathcal{F}_{t+\epsilon}^0 V \mathcal{N})$$ It is also right continuous and complete. Obviously for all $t$ $\mathcal{F}^0_t \subset \mathcal{F}_t \subset \mathcal{G}_t$, so $(\mathcal{G}_t)$ does not seem optimal. It has however some good properties. My question is the following : do you have an example of what we loose by using $(\mathcal{G}_t)$ instead of $(\mathcal{F}_t)$. For instance a Theorem which would fail ?

$\endgroup$
4
$\begingroup$

They're the same, $\mathcal G_t=\mathcal F_t$.

Indeed, suppose $A\in\mathcal G_t$. So in particular $A\in\bigcap_{n=1}^\infty(\mathcal F_{t+1/n}\vee\mathcal N)$.

Note that for any $\sigma$-algebra $\mathcal M$, the $\sigma$-algebra $\mathcal M\vee\mathcal N$ consists of all sets whose symmetric difference with a set in $\mathcal M$ is null, i.e., sets that are almost equal to an element of $\mathcal M$.

Thus $ A $ is almost equal to some $ B_n \in\mathcal F^0_{t+1/n} $ for each $ n $.

But then $ A $ is almost equal to $B:=\bigcup_{n=1}^\infty\bigcap_{m\ge n} B_m=\bigcup_{n=N}^\infty\bigcap_{m\ge n} B_m\in \mathcal F^0_{t+1/N}$ for each $ N $, hence this set $B\in \mathcal F^0_{t+} $. Thus, $A\in\mathcal F^0_{t+}\vee\mathcal N$.

$\endgroup$
  • $\begingroup$ Question by @user58269: Why do we have $B:=\bigcup_{n=1}^\infty\bigcap_{m\ge n} B_m=\bigcup_{n=N}^\infty\bigcap_{m\ge n} B_m\in \mathcal F^0_{t+1/N}$ ? $$B:=\left(\bigcup_{n=1}^{N-1}\bigcap_{m\ge n} B_m\right) \bigcup \left(\bigcup_{n=N}^\infty\bigcap_{m\ge n} B_m\right)$$ $\bigcup_{n=N}^\infty\bigcap_{m\ge n} B_m$ is in $\mathcal{F}^0_{t+1/n}$ but $\bigcup_{n=N}^\infty\bigcap_{m\ge n} B_m$ is in $\mathcal{F}^{0}_{t+1}$ taking their union we obtain an element of $\mathcal{F}^{0}_{t+1}$ $\endgroup$ – Bjørn Kjos-Hanssen Sep 17 '14 at 14:30
  • $\begingroup$ @user58269 why don't you formulate this latest comment as a question for math.stackexchange.com you could ask why are these two expressions for $ B $ equal $\endgroup$ – Bjørn Kjos-Hanssen Sep 17 '14 at 14:35
0
$\begingroup$

They are indeed the same, i.e. $\sigma(\cap_{s>t}\mathcal{F}_s,\mathcal{N})=\cap_{s>t}\sigma(\mathcal{F}_s,\mathcal{N})$.

Let $(\Omega,\overline{\mathcal{F}},\overline{P})$ be the completion of the original measure space so $\overline{\mathcal{F}}=\sigma(\mathcal{F},\mathcal{N})$where $\mathcal{N}:=\{N\subseteq \Omega| \exists A\in \mathcal{F},s.t. N\subseteq A,P(A)=0\}$.

***First let's show $\sigma(\cap_{s>t}\mathcal{F}_s,\mathcal{N})\subseteq\cap_{s>t}\sigma(\mathcal{F}_s,\mathcal{N})$. For any $E\in\sigma(\cap_{s>t}\mathcal{F}_s,\mathcal{N})$. Notice $\sigma(\cap_{s>t}\mathcal{F}_s,\mathcal{N})$ is the completion of $\cap_{s>t}\mathcal{F}_s$ relative to $\overline{\mathcal{F}}$ so there exist $A,B\in\cap_{s>t}\mathcal{F}_s$ such that $$A\subseteq E\subseteq B \mbox{ and }\overline{P}(B\setminus A)=0 $$ Because $A,B\in\cap_{s>t}\mathcal{F}_s$ so $A,B\in\mathcal{F}_s$ for each $s>t$. Thus (1) with $A,B\in\mathcal{F}_s$ implies that $E\in\sigma(\mathcal{F}_s,\mathcal{N})$ for each $s>t$ (by the property of $\sigma(\mathcal{F}_s,\mathcal{N})$ as the relative completion of $\mathcal{F}_s$ with respect to $\overline{\mathcal{F}}$) So $E\in \cap_{s>t}\sigma(\mathcal{F}_s,\mathcal{N})$.

***On the other hand, let's show $\sigma(\cap_{s>t}\mathcal{F}_s,\mathcal{N})\supseteq\cap_{s>t}\sigma(\mathcal{F}_s,\mathcal{N})$.

For any $E\in\cap_{s>t}\sigma(\mathcal{F}_s,\mathcal{N})$, $E\in\sigma(\mathcal{F}_s,\mathcal{N})$ for each $s>t$. Let $\{s_k\}_{k=1}^\infty$ be a sequence that decreases to $t$ but larger than $t$ (i.e. $t<\dots\le s_3 \le s_2 \le s_1$ and $s_k\to t$ as $k\to \infty$) so $E\in\sigma(\mathcal{F}_{s_k},\mathcal{N})$. Thus, for each $k=1,2,\dots$, there exist $A_k,B_k\in\mathcal{F}_{s_k}$ such that $A_k\subseteq E \subseteq B_k$ and $P(B_k\setminus A_k)=0$. It is easy to see that we also have $\cap_{k\ge m}A_k \subseteq E\subseteq \cup_{k\ge m}B_k$ for $m=1,2,\dots$, and hence $\cup_{m\ge 1}\cap_{k\ge m}A_k \subseteq E\subseteq \cap_{m\ge1}\cup_{k\ge m}B_k$. Define $A=\cup_{m\ge 1}\cap_{k\ge m}A_k$ and $B=\cap_{m\ge1}\cup_{k\ge m}B_k$.

Claim 1: $A\in\cap_{s>t}\mathcal{F}_s$. To see why, notice that $\cap_{k\ge m}A_k$ is non-decreasing in $m$ so $A=\cup_{m\ge 1}\cap_{k\ge m}A_k=\cup_{m\ge 2}\cap_{k\ge m}A_k=\cup_{m\ge 3}\cap_{k\ge m}A_k=\dots$. Let $\epsilon>0$ be arbitrary. Since $s_k$ decreases to $t$, there exists $M>0$ such that $t<s_k<t+\epsilon$ for all $k\ge M$. So $A=\cup_{m\ge M}\cap_{k\ge m}A_k\in\mathcal{F}_{t+\epsilon}$ because in $\cup_{m\ge M}\cap_{k\ge m}A_k$, all $k$ are greater than or equal to $M$ and thus $A_k\in\mathcal{F}_{s_k}\subseteq\mathcal{F}_{t+\epsilon}$. Since $\epsilon>0$ is arbitrary, so $A\in \cap_{\epsilon>0}\mathcal{F}_{t+\epsilon}=\cap_{s>t}\mathcal{F}_s$.

Claim 2: $B\in\cap_{s>t}\mathcal{F}_s$. It can be proved using the fact $\cup_{k\ge m}B_k$ is non-increasing in $m$ and similar argument in the proof of claim 1.

Claim 3: $P(B\setminus A)=0$. To show this claim, notice that $$\begin{aligned}B\setminus A&=(\cap_{m\ge1}\cup_{k\ge m}B_k)\setminus (\cup_{m\ge 1}\cap_{k\ge m}A_k)=(\cap_{m\ge1}\cup_{k\ge m}B_k)\cap (\cap_{m\ge 1}\cup_{k\ge m}A_k^c)\\&=\cap_{m\ge 1}[(\cup_{k\ge m}B_k)\cap(\cup_{k\ge m}A_k^c)]=\cap_{m\ge 1}\cup_{k\ge m}(B_k\cap A_k^c)\\&=\cap_{m\ge 1}\cup_{k\ge m}(B_k\setminus A_k)\subseteq\cup_{k\ge 1}(B_k\setminus A_k).\end{aligned}$$ So $0\le P(B\setminus A)\le P (\cup_{k\ge 1}(B_k\setminus A_k))\le \sum_{k=1}^\infty P(B_k\setminus A_k)=0.$

Now using the fact that $A\subseteq E \subseteq B$ and claims 1,2,3, we conclude $E\in \sigma(\cap_{s>t}\mathcal{F}_s,\mathcal{N})$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.