50 votes

Example of filtration in probability theory

Another simple example. The natural filtration when we model tossing a die twice in a row. Here is how it works. Let $X_1$ be the outcome of the first toss. So the values of $X_1$ are in the set $\...
GEdgar's user avatar
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29 votes
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Example of filtration in probability theory

Let me first state an interpretation for the meaning of a filtration: A filtration $\mathcal F_t$ contains any information that could be possibly asked and answered for the considered random process ...
davidhigh's user avatar
  • 418
23 votes

What is meant by a filtration "contains the information" until time $t$?

In order to understand the intuition behind filtrations, it's a good idea to start with a very particular case: the $\sigma$-algebra generated by a single random variable $X:\Omega \to \mathbb{R}$, i....
saz's user avatar
  • 120k
20 votes

Example of filtration in probability theory

Take the following simple model: a stochastic process $X$ that starts at some value $0$. From that value, it can jump at time $1$ to either the value $a$, either the different value $b$. And at time $...
Raskolnikov's user avatar
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19 votes
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What exactly is a 'predictable process'?

The classical interpretation of $\sigma$-algebras is information. A nice example comes from quantitative finance. Suppose we have a certain amount of money $V_n$ at a certain (discrete) time $t_n$. ...
Lonidard's user avatar
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14 votes

What is the intuition behind right-continuous filtration?

The idea is that you gain no additional information by taking an infinitesimal step forward in time. Remember that an $\mathit{intersection}$ means that we are taking only the elements contained in ...
Charles Beer's user avatar
12 votes
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Stochastic processes - Why do we need filtration?

You know the probability of every event in every $\mathcal{F}_t$, but the idea is that at time $t$ you know specifically which event you are in. For an easy example, you can think of flipping two ...
user6247850's user avatar
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7 votes
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Intersection multiplicity does not go down after restriction to closed subvariety: proof using filtrations

I believe the desired result is not true. Let $S=\mathbb{Q}[x,y,z]\big/(xy-z^n)$, with $n\geqslant 2$. For convenience we refer to the coset $p+(xy-z^n)\in S$ as $\bar{p}$, for each $p\in\mathbb{Q}[x,...
Atticus Stonestrom's user avatar
6 votes
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Show that the hitting time of an open set for a right-continuous process is a stopping time

For fixed $\omega \in \{\tau<t\}$ there exists $\tau(\omega) \leq s < t$ such that $Y_s(\omega) \in H$. As $H$ is open, there exists some $\epsilon>0$ such that $B(Y_s(\omega),\epsilon) \...
saz's user avatar
  • 120k
6 votes
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adapted process, translation between measurable and information?

It's not that $\mathcal{F}_t$ itself "contains information about the process up to time $t$" in the way you seem to be assuming. Rather, elements of $\mathcal{F}_t$ are allowed to depend on ...
Eric Wofsey's user avatar
5 votes
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Stopping Times, the $\inf$ is not a stopping time

Your way of writing the event is wrong. Note that $\inf \limits_n \frac{1}{n} = 0$, while $\frac{1}{n} \le 0$ never holds.
Dominik's user avatar
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5 votes
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Simple question regarding stopping times.

Since $S \leq T$, we have $$\{T \leq t\} = \{S \leq t\} \cap \{T \leq t\}.$$ Hence, $$F \cap \{T \leq t\} = \underbrace{(F \cap \{S \leq t\})}_{\in \mathcal{F}_t} \cap \underbrace{\{T \leq t\}}_{\...
saz's user avatar
  • 120k
5 votes

Why does a filtration of a group consist of normal subgroups, and not any subgroups?

I have to say that I'm a working group theorist (in geometric group theory) and I don't agree with Wikipedia's definition here. If $G$ is a group and someone says "filtration of $G$", I would expect ...
Jim Belk's user avatar
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5 votes
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Example/meaning of filtration on a group $(\mathbb{R},+)$

Very good that you want to try out the definition on one of the standard examples! However, as people have pointed out in comments, it turns out that the group $(\mathbb R, +)$ just does not serve as ...
Torsten Schoeneberg's user avatar
5 votes
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Meaning of measurableness

Here is an intuitive answer which may address your concern about $\sigma$-algebras and information. Assume that only $3$ mutually exclusive events may happen at time $T$. Let these be denoted by $\...
Holden's user avatar
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5 votes
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What is the correct filtration?

$X_t$ and $Y_t$ will be Brownian motions under their natural filtrations; i.e. under $$\mathcal{F}_t^{X} = \sigma(X_s\, : \, s \leq t) \qquad \text{ and } \qquad \mathcal{F}_t^{Y} = \sigma(Y_s\, : \, ...
Jose Avilez's user avatar
5 votes
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Is $Z_n = X_n Y_n$ a martingale?

I was somewhat surprised that the answer is negative. (However, see the answer by John Dawkins, who notes that $Z_n$ is a Martingale with respect to some filtration.) Let $\{\xi_n\}$ and $\{\eta_n\}$...
Yuval Peres's user avatar
4 votes
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union of natural filtration vs union of right continuous filtration

$\mathcal{F}_t\subset\tilde{\mathcal{F}}_{t+\delta}$ for any $\delta>0$, so $ \bigcup_{t\geq 0}\mathcal{F}_t\subset \bigcup_{t\geq 0}\tilde{\mathcal{F}}_t$, and conversely the right-continuous ...
snarfblaat's user avatar
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4 votes
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Exercise on algebra filtrations

I think that it is a matter of definition: I have looked at both M. Sweedler's book on Hopf algebras (p.230, 1969 edition) and E.Abe's book on Hopf algebras (p.20, 1977 edition) and they both seem to ...
KonKan's user avatar
  • 7,344
4 votes

Sigma algebra generated by the stopped process.

Yes, the two $\sigma$-algebras coincide. For any discrete stopping time $\tau: \Omega \to \mathbb{N}_0$ the random variable $X_{\tau}$ is $\mathcal{F}_{\tau}$-measurable, and this implies that $$\...
saz's user avatar
  • 120k
4 votes

For a discrete stopping time $\tau$, $\mathcal{F}_\tau^X = \sigma(X(t\wedge\tau):t\ge 0).$

Let $(X_n)_{n \in \mathbb{N}}$ be a stochastic process and $\tau: \Omega \to \mathbb{N}$ an $\mathcal{F}^X$-stopping time. You have already shown that $X(\tau \wedge n)$ is $\mathcal{F}_{\tau}^X$-...
saz's user avatar
  • 120k
4 votes

Conditional expectation as a random variable new

According to the definition of the conditional expectation we need a random variable $E[X\mid F_n](\omega)$ for which $$\int_A E[X\mid F_n](\omega)\ dP=\int_A X(\omega)\ dP$$ for all $A\in F_n$and $E[...
zoli's user avatar
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4 votes
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Where is the Strong Markov property(SM) being used in the proof that augmented filtration of a Strong Markov process is right continuous?

You are right that the strong Markov property is not used in its full generality; somehow, it's an artifact of the definition of the (strong) Markov property by Karatzas & Shreve. They say that $(...
saz's user avatar
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4 votes
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Convergence in probability of conditional expectation

No, the assertion does in general not hold true. Consider, for instance, a sequence of independent random variables $(X_n)_{n \in \mathbb{N}}$ such that $$\mathbb{P}(X_n = -n^2+1) = \frac{1}{n^2} \...
saz's user avatar
  • 120k
4 votes
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Stopping time clarification

The idea is that $F_t$ consists of all events that depend on your stochastic process only up to time $t$. So to say that $\{\omega\in\Omega:\tau(\omega)\leq t\}\in F_t$ means that given any ...
Eric Wofsey's user avatar
4 votes
Accepted

Tensor product commutes with associated graded

One can prove that $\phi_k$ is an isomorphism by noting that it maps a basis of the left hand side to a basis of the right hand side. The following bookkeeping device will be useful to find suitable ...
Nils Matthes's user avatar
  • 4,454
4 votes

Help understanding the definition of a "filtration" in probability theory

Sigma algebras are often thought of as containing "information". Conditioning on a larger sigma algebra corresponds to "knowing more" about the values of random variables (more things are measurable ...
kccu's user avatar
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4 votes
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Uniformly integrable martingale problem: typo (wrong filtration)?

No, it's not a typo; $(\mathcal{F}_{n \wedge \tau})_{n \in \mathbb{N}}$ is a filtration. It holds for any two stopping times $S \leq T$ that $\mathcal{F}_S \subseteq \mathcal{F}_T$ (see e.g. this ...
saz's user avatar
  • 120k
4 votes
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A counterxample of a non-adapted process?

Consider a real-valued Brownian motion $(B_t)_{t \geq 0}$ in its natural filtration, $\left(\mathcal{F}_t\right)_{t \geq0}$. Then, if we define a new process $(X_t)_{t \geq 0} = (B_{t+1})_{t \geq 0}$, ...
spetrevski's user avatar
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