# Tag Info

33

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 $\{1,2,3,4,5,6\}$. Let $X_2$ be the outcome of the second toss. As a sample space, we can take $\Omega = \{1,2,3,4,5,6\} \times \{1,2,3,4,5,6\}$, the set of ...

17

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 at time $t$. Single dice throw: Let's first consider the simple example of one dice throw: Before the throw, all you know is that the result will be "1 or ...

15

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 $t_n$. Suppose we decide to invest a certain percentage $\alpha_n$ of this money in a risky title which has value $S_n$ at time $t_n$ and put the remaining part $\... 15 Take the following simple model: a stochastic process$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$2$, it can jump to$c$or$d$if it was in$a$at time$1$, it can jump to$e$or$f$if it was in$b$at time$1$. In other words, there are four possible ... 14 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.e. $$\sigma(X) = \{ \{X \in B\}; B \in \mathcal{B}(\mathbb{R})\} \tag{1}$$ which is the smallest$\sigma$-algebra$\mathcal{F}$on$\Omega$such that$X: (\...

11

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 EVERY set in the intersection. So, if we think of each $F_t$ as the information contained in the system up to time $t$, the intersection $\cap_{\epsilon > 0} ... 7 I believe the desired result is not true. Let$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,y,z]$. Clearly$xy-z^n$is irreducible in$\mathbb{Q}[x,y,z]$, hence prime, and so$S$is a domain. Now, we let$f=\bar{x}$and$P=(\bar{x},\bar{z})$. To see ... 5 Since$S \leq T$, we have $$\{T \leq t\} = \{S \leq t\} \cap \{T \leq t\}.$$ Hence, $$F \cap \{T \leq t\} = \underbrace{(F \cap \{S \leq t\})}_{\in \mathcal{F}_t} \cap \underbrace{\{T \leq t\}}_{\in \mathcal{F}_t} \in \mathcal{F}_t$$ for any$F \in \mathcal{F}_S$and$t \geq 0$. 5 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. 5 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 information about the process up to time$t$. For instance, in your example, the set$\{ttt,tth,tht,thh\}$has some very useful information about the first toss: it ... 4 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 include the assumption:$1\in V_0$, as part of the definition of a filtration on an algebra$A$to make it a filtered algebra. If you do not include it as ... 4$\mathcal{F}_t\subset\tilde{\mathcal{F}}_{t+\delta}$for any$\delta>0$, so$ \bigcup_{t\geq 0}\mathcal{F}_t\subset \bigcup_{t\geq 0}\tilde{\mathcal{F}}_t$, and conversely the right-continuous filtration always contains the natural filtration. You don't need the rcll assumption. 4 For fixed$\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) \subseteq H$. Since$r \mapsto Y_r(\omega)$is right-continuous, there exists$\delta>0$such that $$\forall r \in [s,s+\delta] : |Y_r(\omega)-Y_{s}(\omega)| <... 4 According to the definition of the conditional expectation we need a random variable E[X\mid F_n](\omega) for which$$\int_A E[X\mid F_n](\omega)\ dP=\int_A X(\omega)\ dP$$for all A\in F_nand E[X\mid F_n] is constant over the sets [\frac k{2^n},\frac{k+1}{2n}) because E[X\mid F_n] has to be F_n measurable. For a k, let E[X\mid F_n](\omega)=... 4 You are right that the strong Markov property is not used in its full generality; somehow, it's an artifact of the definition of the (strong) Markov property by Karatzas & Shreve. They say that (X_t,\mathcal{F}_t)_t is Markov if$$\mathbb{E}(f(X_{s+t}) \mid \mathcal{F}_s) = \mathbb{E}(f(X_{s+t}) \mid X_s) \tag{1}$$and strong Markov if$$\mathbb{E}(... 4 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 question). Applying this for$S= \tau \wedge n$and$T=\tau \wedge (n+1)$gives immediately that$(\mathcal{F}_{n \wedge \tau})_{n \in \mathbb{N}}$is a filtration.... 4 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 fair coins. The sample space is$\Omega = \{ HH, HT, TH, TT\}$, and the terminal$\sigma$-algebra is$\mathcal{F} = 2^\Omega$. Then$\mathcal F_n$will be what ... 4 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 in$L^1$together with the continuity of the conditional expectation give $$\mathbb{E}\left[\mathbf{1}_AX\right]=\mathbb{E}\left[\mathbf{1}_AX_n\right]=\mathbb{E}... 4 Very good that you want to try out the definition on one of the standard examples! However, as people have pointed out in comments, it turns out that the group (\mathbb R, +) just does not serve as a very motivational example for this concept. A better example, or class of examples, would come from the upper triangular (unipotent) n \times n-matrices G =... 3 For any two R modules, A and B, there are some extensions M that fit into a long exact sequence of the form 0 \to A \to M \to B \to 0. Up to equivalence they are classified by the R module Ext_R^1(A,B), which can be non-trivial. There is always the trivial extension M = A \oplus B. Now, take the filtration 0 \subset A \subset M. The ... 3 \bigvee_n \mathcal{F}_n is the \sigma-algebra generated by \bigcup_n \mathcal{F}_n (i.e., smallest \sigma-algebra containing \bigcup_n \mathcal{F}_n). This is needed because in general, \bigcup_n \mathcal{F}_n might not be a \sigma-algebra. See here for an easy example. See here for an example where the sequence is nested, i.e. \mathcal{F}_1 \... 3 The first equality is indeed a direct consequence of Lévy's backward convergence theorem. For the second one the reasoning goes as follows: Choose h>0 sufficiently small such that t_{k-1} \leq t < t+h < t_{k}. Since f_j(X_{t_j}) is \mathcal{F}_{t+h} measurable for each j \leq k-1, we have$$\mathbb{E} \left( \prod_{j=1}^n f_j(X_{t_j}) \... 3 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 $$\sigma(X_{n \wedge T}; n \geq 0) \subseteq \sigma(\mathcal{F}_{n \wedge T}; n \geq 0).$$ Moreover, we have$\mathcal{F}_{n \wedge T} \subseteq \mathcal{F}_T$(as$...

3

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 $\mathcal{F}_{\tau}^X$-measurable for all $n \geq 1$, and therefore it just remains to show that $$\mathcal{F}_{\tau}^X \subseteq \sigma(X(\tau \wedge n); n \geq 1) =: \mathcal{H}. \tag{... 3 This is not true because \{\tau\ne t\}\notin\mathcal{F}_{\tau}\mid_{\{\tau=t\}} but \{\tau\ne t\}\in \mathcal{F}_t. You may show instead that$$ \mathcal{F}_{\tau}\mid_{\{\tau=t\}}=\mathcal{F}_t\mid_{\{\tau=t\}} .^1 $$^1 For any (\mathcal{F}_t)-stopping times \tau and \sigma, \mathcal{F}_{\sigma}\mid_{\{\sigma=\tau\}}=\mathcal{F}_{\tau}\mid_{\{... 3 The statement that the Markov property implies$$ \mathbb{P}(X_{t_n} \in B_{t_n} | X_{t_{n-1}} \in B_{t_{n-1}}, \dots, X_{t_1} \in B_{t_1}) = \mathbb{P}(X_{t_n} \in B_{t_n} | X_{t_{n-1}} \in B_{t_{n-1}}) $$is hopelessly false, as the example B_{t_{n-1}} = \mathbb{R} will show. The Markov property does not say anything about a window of space. The ... 3 No, the assertion does in general not hold true. Consider, for instance, a sequence of independent random variables (X_n)_{n \in \mathbb{N}} such that$$\mathbb{P}(X_n = -n^2+1) = \frac{1}{n^2} \qquad \mathbb{P}(X_n=1) = 1- \frac{1}{n^2}. It follows easily from the Borel-Cantelli lemma that $X_n \to 1$ almost surely (hence in probability). On the other ...

3

As you said, there is a map $\Omega^*_{B/A}\to\Omega^*_{P/A}$. This induces $F^i(\Omega^*_{B/A})\to \Omega^*_{P/A}$. Then $F^i(\Omega^*_{P/A})$ are the classes that can be written as a product of an element of the image of this map and an element of $\Omega^*_{P/A}$. Note that this is not all of $\Omega^*_{P/A}$ since $F^i(\Omega^*_{B/A})$ contains only ...

3

Suppose that $\left(X_t\right)_{t\geqslant 0}$ is a martingale with respect to the filtration $\left(\mathcal F_t\right)_{t\geqslant 0}$ where $\mathcal F_t=\mathcal F$ for all $t$. The condition of adaptedness implies that for all $t$, $X_t$ is $\mathcal F$-measurable. The condition $\mathbb E\left[X_t\mid\mathcal F_s\right]=X_s$ for $0\leqslant s\lt t$ ...

3

For the first point, notice that your probability space doesn't start with an inherent filtration. In this case, the natural filtration to work with is the one generated by the process $S_i$. That is $\mathcal{F}_i = \sigma(S_j : j \leq i)$. This is the smallest $\sigma$-algebra such that $S_j$ is $\mathcal{F}_i$-measurable for each $j \leq i$ and in ...

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