How Do Right and Left Filtrations Relate in Stochastic Calculus?

  • MHB
  • Thread starter gnob
  • Start date
In summary, the definitions of $\mathcal{F}_{t-}$ and $\mathcal{F}_{t+}$ from Karatzas and Shreve's book are equivalent to unions and intersections respectively. These unions and intersections can also be expressed using rational numbers and natural numbers. The proof involves showing that the unions and intersections are equal by using the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$ and that the intersection of a family of $\sigma$-fields is again a $\sigma$-field.
  • #1
gnob
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0
Hi! I am reading the book of Karatzas and Shreve (Brownian Motion and Stochastic Calculus - Ioannis Karatzas, Steven E. Shreve - Google Books). On page 4 we have the ff definitions:
$$
\mathcal{F}_{t-} := \sigma \left( \bigcup_{s<t} \mathcal{F}_s \right) \quad \text{and}\quad
\mathcal{F}_{t+} := \sigma \left( \bigcap_{\epsilon >0} \mathcal{F}_{t+\epsilon} \right),
$$
where $t,s \in \mathbb{R}$ and $t,s>0.$

My questions are:
(1) Is $\mathcal{F}_{t-} := \sigma \left( \bigcup_{s<t} \mathcal{F}_s \right)
= \sigma \left( \bigcup_{q<t} \mathcal{F}_q \right)
= \sigma \left( \bigcup_{n=n_k}^{\infty} \mathcal{F}_{t-\frac{1}{n}} \right)$ where $q \in \mathbb{Q} \cap (0,\infty)$ and $n_k \in \mathbb{Z}^+$ are chosen so that $t -\frac{1}{n_k} >0$?

(2) Is $\mathcal{F}_{t+} := \sigma \left( \bigcap_{\epsilon >0} \mathcal{F}_{t+\epsilon} \right)
=\sigma \left( \bigcap_{q >0} \mathcal{F}_{t+q} \right)
=\sigma \left( \bigcap_{n=1}^{\infty} \mathcal{F}_{t+\frac{1}{n}} \right)$ where $q \in \mathbb{Q} \cap (0,\infty)$ and $n \in \mathbb{Z}^+$?

My idea is to use the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$ but i am not sure how to write the proof. Any help would be appreciated. Thanks. :eek:
 
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  • #2
Check that the unions are the same using the fact that if $s<t$ then $\mathcal F_s\subset\mathcal F_t$. Do the same for intersections.
 
  • #3
girdav said:
Check that the unions are the same using the fact that if $s<t$ then $\mathcal F_s\subset\mathcal F_t$. Do the same for intersections.

Thanks for the suggestion. Please give your comment on the proof I produced. Thanks.

Recall that $\sigma\left(A\right)$ is the smallest $\sigma$-field that contains the set (or family of subsets) $A.$ Hence, for the first question regarding unions, it suffices to show that
$$
\bigcup_{s < t} \mathcal{F}_s =
\bigcup_{q < t} \mathcal{F}_q =
\bigcup_{\substack{n= n_k\\ \left(t-\frac{1}
{n_k}\right)>0}}^{\infty} \mathcal{F}_{t -
\frac{1}{n}}.
$$
To establish the first equality above, we consider an arbitrary $\sigma$-field
$\mathcal{F}_{s_0} \subseteq \bigcup_{ s <t} \mathcal{F}_s.$ Since $\mathbb{Q}$ is dense in $\mathbb{R},$ we can find $q_0 \in \mathbb{Q}$ such that
$s_0 < q_0 <t.$ Because $\{\mathcal{F}_t \}$ is an increasing sequence of $\sigma$-fields, we then have $\mathcal{F}_{s_0} \subseteq \mathcal{F}_{q_0} \subseteq \bigcup_{q <t} \mathcal{F}_q.$ Hence, $\bigcup_{s <t} \mathcal{F}_s \subseteq \bigcup_{q <t} \mathcal{F}_q.$ Conversely, let $\mathcal{F}_{q_0} \subseteq \bigcup_{q <t} \mathcal{F}_q.$ Note that $\mathbb{Q} \subseteq \mathbb{R}$ and so $\mathcal{F}_{q_0} \subseteq \bigcup_{s <t} \mathcal{F}_s,$ which implies that $\bigcup_{q <t} \mathcal{F}_q \subseteq \bigcup_{s <t} \mathcal{F}_s.$

Next, we establish that the extreme unions above are equal. First, observe that $t-\frac{1}{n} \in \mathbb{R}$ for every $n \in \mathbb{N}$ and so $\mathcal{F}_{t-\frac{1}{n}} \subseteq \bigcup_{s<t} \mathcal{F}_s,$ which shows that
$$
\bigcup_{\substack{n= n_k\\ \left(t-\frac{1}{n_k}\right)>0}}^{\infty}
\mathcal{F}_{t - \frac{1}{n}} \subseteq \bigcup_{s <t} \mathcal{F}_s.
$$
For the reverse inclusion, consider an arbitrary $\sigma$-field $\mathcal{F}_{s_0} \subseteq \bigcup_{s <t} \mathcal{F}_s$ and note that $s_0 < t.$ Let $\delta_{0}:= \frac{t - s_0}{2} \in\mathbb{R}_+.$ By the Archimedean property
of $\mathbb{R},$ we can find an integer $n_{k_0} \in \mathbb{N}$ for which $\frac{1}{n_{k_0}} < \delta_0$ so that $s_0 < t - \frac{1}{n_{k_0}} <t$ and
$$
\mathcal{F}_{s_0} \subseteq \mathcal{F}_{t-\frac{1}{n_{k_0}}}
\subseteq \bigcup_{\substack{n= n_k\\ \left(t-
\frac{1}{n_k}\right)>0}}^{\infty} \mathcal{F}_{t - \frac{1}{n}}.
$$
It follows by transitivity that all three unions above are equal, and consequently the last two equalities hold.
For the Second Question (on Intersections): We need to show that
$$
\mathcal{F}_{t+} := \sigma\left(\bigcap_{\epsilon > 0}
\mathcal{F}_{t + \epsilon} \right) = \bigcap_{\epsilon > 0}
\mathcal{F}_{t + \epsilon} = \bigcap_{q >0}
\mathcal{F}_{t + q} = \bigcap_{n=1}^{\infty} \mathcal{F}_{t + \frac{1}{n}}.
$$
It is a standard example in Measure Theory that the intersection of a family of $\sigma$-fields (defined on the same space) is again a $\sigma$-field. So, the first equality is trivial.

We now establish the equality $\bigcap_{\epsilon > 0} \mathcal{F}_{t + \epsilon}= \bigcap_{q >0} \mathcal{F}_{t + q}.$ If $A \in \bigcap_{\epsilon > 0} \mathcal{F}_{t + \epsilon}$ then $A \in \mathcal{F}_{t+\epsilon},\,\,\forall\, \epsilon >0.$ Thus, if we chose all $\epsilon$'s to be positive rationals ($\epsilon = q \in \mathbb{Q}_+$) we also have $A \in \mathcal{F}_{t+q},\,\,\forall\, q >0,$ which in turn implies that $A \in \bigcap_{q >0} \mathcal{F}_{t+q}.$ On the other hand, $B \in \bigcap_{q >0} \mathcal{F}_{t+q}$ implies that $B \in \mathcal{F}_{t + q},\,\,\forall\, q>0.$ Our goal is to show that $B \in\mathcal{F}_{t+\epsilon},\,\,\forall\,\epsilon >0.$ But for a fixed $\epsilon_0>0,$ since $\mathbb{Q}$ is dense in $\mathbb{R},$ we can find a sequence $\{ q^{(0)}_i\}_{i=1}^{\infty}$ such that $q^{(0)}_i \downarrow \epsilon_0.$ Hence,
$$
\mathcal{F}_{t+\epsilon_0} = \lim_{q^{(0)}_i \downarrow\epsilon}
\mathcal{F}_{t+q^{(0)}_i} = \left(\bigcap_{i=1}^{\infty}
\mathcal{F}_{t+ q^{(0)}_i}\right) \ni B.
$$
Consequently, we have $B\in \Big(\bigcap_{\epsilon >0} \mathcal{F}_{t+\epsilon}\Big).$

Lastly, we show that $\bigcap_{\epsilon > 0} \mathcal{F}_{t + \epsilon}= \bigcap_{n \in \mathbb{Z}_+} \mathcal{F}_{t + \frac{1}{n}}.$ But we can use similar argument as above to conclude that $\bigcap_{\epsilon > 0} \mathcal{F}_{t + \epsilon}\subseteq \bigcap_{n \in \mathbb{Z}_+} \mathcal{F}_{t + \frac{1}{n}}.$ For the reverse inclusion, a similar argument can be done by choosing a sequence $\{ \frac{1}{n_i(\epsilon)}\}_i^{\infty} \subseteq \mathbb{Q}$ such that $\frac{1}{n_i(\epsilon)} \downarrow \epsilon$ (Note: WLOG, we can take $\epsilon>0$ to be small since we are taking intersection and that $\{ \mathcal{F}_t\}$ is an increasing family).
 

FAQ: How Do Right and Left Filtrations Relate in Stochastic Calculus?

What is the purpose of right and left filtrations?

The purpose of right and left filtrations is to separate a given set into two subsets based on specific criteria. This allows for a more organized and efficient analysis of the set.

What is the difference between right and left filtrations?

The main difference between right and left filtrations is the direction in which they filter the set. Right filtrations start from the rightmost element and work towards the left, while left filtrations start from the leftmost element and work towards the right.

How are right and left filtrations used in scientific research?

Right and left filtrations are commonly used in scientific research to analyze data and identify patterns or trends. They can also be used to classify data into different categories based on specific criteria.

Can right and left filtrations be applied to any type of set?

Yes, right and left filtrations can be applied to any type of set, as long as there is a defined order or direction within the set. This can include numerical sets, alphabetical sets, and more.

Are there any limitations to using right and left filtrations?

One limitation of using right and left filtrations is that they can only divide a set into two subsets. This may not be sufficient for more complex data analysis, which may require the use of multiple filtrations or other methods.

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