# Tag Info

• 15k
Accepted

### 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 $t_n$. ...
• 3,985

### 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....
• 114k

### 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 ...
• 151
Accepted

### 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 ...
• 9,427
Accepted

• 114k
Accepted

• 10.3k
Accepted

### 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 ...
• 7,035
Accepted

### 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 ...
• 1,590

### 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 $\mathcal{F}_{\tau}^X$-...
• 114k

• 114k

### 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 ...
• 20.4k
Accepted

### 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 ...
• 114k
Accepted

### Conditional expectations of $X$ with respect to a filtration converges to $X$

Let $X_n=\mathbb{E}\left[X\mid\mathcal{F}_n\right]$. This is a closed martingale so it converges in $L^1$ and almost surely to $\tilde{X}$. For all $n\geq 1$ and $A\in\mathcal{F}_n$, the convergence ...
• 813
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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 ...
• 21.7k
Accepted

### Filtered objects and associated graded objects

For any two $R$ modules, $A$ and $B$, there are some extensions $M$ that fit into a long exact sequence of the form $0 \to A \to M \to B \to 0$. Up to equivalence they are classified by the $R$ module ...
• 19.8k
Accepted

• 114k

### 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 \...
• 114k
The text gives you a hint for this: "For any given $\omega$ find an $\alpha \ldots$". To extend this hint a little: Let $\omega\in\Omega$, with $\omega=(\omega_1,\omega_2,\ldots)$ and with infinitely ...
This is a variation on @GEdgard example, that can be applied to real valued random variables. The probability plays no role. Let $X_j$ be the projection of $\mathbb{R}^n$ onto the the $j^{th}$ ...