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I'm studying Martingales and below here filtrations. Given af probability space $(\Omega, F, P)$ I define a filter $(F_n)$ as a increasing sequence of $\sigma$-algebras of $F$, such that $F_t \subset F$ and $t_1 \leq t_2 \Longrightarrow F_{t_1} \subset F_{t_2}$. Here comes my question: How can the $F_t$'s be $\sigma$-algebras and subsets of $F$ without being exactly equal to $F$? I suppose that $F_t$'s being $\sigma$-algebras mean that they are $\sigma$-algebras with respect to the measure space $(\Omega, F)$. Can anyone explain why they are not necessarily equal to $F$ and give an example where this is obviously false?

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    $\begingroup$ Most trivial example: $\mathcal{F}_t := \{\emptyset,\Omega\}$ is the trivial $\sigma$-algebra and $\mathcal{F} := \mathcal{P}(\Omega)$ the power set. $\endgroup$ – saz May 13 '17 at 13:55
  • $\begingroup$ "$F_t$'s being $\sigma$-algebras mean that they are $\sigma$-algebras with respect to the measure space $(\Omega, F)$." Sorry but what do you mean by being a sigma-algebra with respect to $(\Omega, F)$? $\endgroup$ – Did Nov 7 '17 at 19:58
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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 2 or ... or 6". In set notation, this corresponds to $\Omega_1=\{1,2,\ldots,6\}$, i.e., to the full set of outcomes. Moreover, you can ask and answer the silly question "does nothing happen", which corresponds to the empty set. So the filtration is $\mathcal F_0=\{\Omega_1,\emptyset\}$.

  • After the throw, you get a single outcome $\omega_1 \in \Omega$. Now you can answer all kind of questions, for example: "is the result a 4" which corresponds to $\{4\}$, "is the result an odd number" corresponding to $\{1,3,5\}$, "is the result larger than 4" corresponding to $\{5,6\}$, and so on.

You see that all possible information after one throw -- all that can be asked and answered -- is contained in the power set $\mathcal P(\Omega_1)$ of $\Omega_1$.

Moreover, in a probability space, you cannot only ask whether a configuration might occur or not -- you can also consider the probability for a given configuration, for example $P(\{5,6\}) = 1/3$.

Two dice throw (ordered, i.e. thrown sequentially):

This is the example discussed in the answer by @GEdgar.

  • Before any throw, all you can ask and answer is that the result will be in $\Omega_2 = \Omega_1 \times \Omega_1$, so $ \mathcal F_0 = \{\emptyset,\Omega_2\} $.

  • After the first throw, you can only answer any of the questions related to the first throw we collected in the one-dice example, and nothing related to the second. So the filtration is $\mathcal F_1=\mathcal P(\Omega_1) \times \Omega_1$.

  • After two throws, you have the complete information, that is $\mathcal P (\Omega_2)$.

A few example configurations plus the corresponding question:

$\quad \quad \Big\{(1,5)\Big\} \in \mathcal F_2$ $\quad \longrightarrow$ first throw is a one, the second a five

$\quad \quad \Big\{(1,5),(5,1)\Big\} \in \mathcal F_2$ $\quad \longrightarrow$ one of of the two throws is a one, the other a five

$\quad \quad \Big\{\{1,3,5\}\times\{2,4,6\}\Big\} = \Big\{(1,2),(1,4),\ldots,(3,6),(5,6)\Big\} \in \mathcal F_2$ $\longrightarrow$ first throw odd, second even

Two dice throw (unordered, i.e. thrown at once):

Here, the set of outcomes is the set of ordered pairs $\Omega^+=\big\{(\omega_1,\omega_2)\ |\ 1\leq\omega_1\leq\omega_2\leq6\big\}$.

There is only one throw, and the filtration after it is -- just as in the one-dice throw example above -- given by $\mathcal P(\Omega^+)$. I won't go through this in detail. The only thing I want to stress is the following:

The unordered two-dice throw here can be reproduced by the ordered two-dice throw, if one restricts the sigma-algebra of the ordered example to contain all symmetric pairs. That is, for example, all elements of the form $\{(1,5)\}$ as in the previous example are disallowed and have to be replaced by $\{(1,5), (5,1)\}$. By this, the discrimination between "first" and "second" throw, basically disappears. In result, the unordered outcome set with the filtration being a power set is identical to the ordered outcome set with a filtration restricted to symmetric elements.

By this example, one can observe that the sigma-algebra is not a static thing. It depends on the set of outcomes $\Omega$ and on the questions one wants to ask.

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  • $\begingroup$ Thanks! I appreciate you answer. I have a short question, though. In the ordered two dice throw have I got it right that you mean $F_0 = \{ \emptyset, \Omega_2 \}$? $\endgroup$ – J. Goles Aug 22 '19 at 11:05
  • $\begingroup$ Sure, corrected the $0$ to $\emptyset$, thanks! $\endgroup$ – davidhigh Aug 22 '19 at 19:46
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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 all ordered pairs chosen from the set $\{1,2,3,4,5,6\}$. If $\omega \in \Omega$, then $\omega$ is an ordered pair, say $\omega = (\omega_1,\omega_2)$. Let the two random variables be $X_1(\omega) = \omega_1$ and $X_2(\omega) = \omega_2$. An "event" is a subset of $\Omega$.

[Note: a true probabilist thinks the first paragraph is quite natural, and the second paragraph is very artificial.]

The "times" that are relevant are: time $0$, before any tosses have been done, time $1$ after the first toss but before the second toss, and time $2$, after the second toss. For each time $t$, the sigma-algebra $\mathcal F_t$ is "the information known at time $t$". We have $\mathcal F_0 \subset \mathcal F_1 \subset \mathcal F_2$, with strict inclusion in all cases.

Now let's work out what these are.

$$ \mathcal F_0 = \{\varnothing, \Omega\} $$ since at time $0$ we have no information about which sample point is the true state of affairs. There are only two events such that we know whether it occurs: event $\Omega$ definitely occurs, and $\varnothing$ definitely does not occur. For any other event, we do not know whether or not it occurs. $$ \mathcal F_1 $$ consists of $2^6$ events: all sets of the form $A \times \{1,2,3,4,5,6\}$, where $A \subseteq \{1,2,3,4,5,6\}$. This is because, at time $1$, we know what the outcome $X_1$ was, that is, we know the first coordinate of $\omega$, but we do not know the second coordinate of $\omega$. The events in $\mathcal F_1$ are events that contain no information about the second toss $X_2$. Given an event $U$ in $\mathcal F_1$, at time $1$ we definitely know whether or not $U$ occurs. But given an event $V$ not in $\mathcal F_1$, at time $1$ we possibly do not know whether $V$ occurs. $$ \mathcal F_2 $$ consists of $2^{36}$ events: all subsets of $\Omega$. This is because, at time $2$ we know exactly what $\omega$ is, so, for any of the $2^{36}$ events $U$, we know whether or not $U$ has occurred.

[Probability language $$ \text{event $U$ occurs} $$ translates to set theory language $$ \omega \in U. $$ As you work with this, you will get more experience translating back and forth between them.]

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    $\begingroup$ You say all sets of the form $A\times \Omega$, although $\Omega$ is already a product. Is that right? The set $A$ has $2^6$ subsets, so F1 are sets holding triples? Why do we keep track of the later two components? $\endgroup$ – Nikolaj-K Jul 17 '18 at 0:36
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    $\begingroup$ Ah, okay. Can you still comment a bit on the intuition regard why subsets of $A$ are considered? E.g. say the first coin was flipped and I know it's value. That's then a single concrete value and all info about the first component. How would e.g. the subset $S=\{2,4,5\}$ of $A$ ever relate to that? I can understand that it's a formal definition and just so defined, but I have no intuition why arbitrary size subsets of all dice faces come into play. $\endgroup$ – Nikolaj-K Jul 17 '18 at 17:44
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    $\begingroup$ @GEdgar $F_1$ consists of $2^6$ events????? Shouldn't it be $6$??? When $\omega_1$ is known then the possible combinations are: $(\omega_1,1) , \dots, (\omega_1,6)$ . And similarly for $F_2$ shouldn't it be $6^2$ rather than $2^6$ ?? $\endgroup$ – Croma14 Nov 20 '18 at 11:26
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    $\begingroup$ @Croma14: If $A$ is any subset of $\{1,\dots,6\}$, then $A \times \{1,\dots,6\} \in \mathcal F_1$. That is, if we know $\omega_1$, then we also know whether or not $\omega \in A \times \{1,\dots,6\}$. There are $2^6$ such events. And every subset of $\{1,\dots,6\} \times \{1,\dots,6\}$ belongs to $\mathcal F_2$. There are $2^{36}$ such events. $\endgroup$ – GEdgar Nov 20 '18 at 12:09
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    $\begingroup$ @Nikolaj-K: the filtration contains any possible information of the process, that is, anything that can be answered. The subset $S=\{2,4,5\}$ corresponds to the information that answers the question is the first throw a two or a four or a five?. After one throw, you can answer this question, but you have no idea about the result of the second throw. This is what is stated by the element $\{2,4,5\} \times \Omega \in \mathcal F_1$. It says that you can reasonably ask the previous question about the first throw, but nothing specific related to the second throw (only that it's in 1 to 6). $\endgroup$ – davidhigh Aug 2 '19 at 20:51
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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 paths for the variable $X$: $\omega_1=0\to a\to c$, $\omega_2=0\to a\to d$, $\omega_3=0\to b\to e$ and $\omega_4=0\to b\to f$. They constitute our space of outcomes

$$\Omega=\{\omega_1,\omega_2,\omega_3,\omega_4\} \; .$$

Hence, $\Omega$ is the space of possible paths for $X$. On the other hand, you can define a filtration as follows

$$\begin{eqnarray} \mathcal{F}_t = &\{\emptyset,\Omega\} &, 0\leq t <1 ; \\ \mathcal{F}_t = &\{\emptyset,\{\omega_1,\omega_2\},\{\omega_3,\omega_4\},\Omega\} &, 1\leq t <2 ; \\ \mathcal{F}_t = & \mathcal{P}(\Omega) &, 2\leq t .\end{eqnarray}$$

with $\mathcal{P}(\Omega)$ the power set of $\Omega$. In this sense, the filtration is a reflection of the info you have at any time. In the beginning, you don't know which path the stochastic variable will follow, so your filter does not contain more than the events $\emptyset$ and $\Omega$, but in the next step, you can arrive at one of the values $a$ or $b$. Therefore you have two extra events in your set you can speak about. And finally at the final time, you have access to all possible events. As you can see $\mathcal{F}_1 \subset \mathcal{F}_2$ but $\mathcal{F}_1 \neq \mathcal{F}_2$.

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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 $j^{th}$ component, i.e.: $ X_j(x_1,x_2,\dots,x_n) = x_j $. Let $\mathcal{F}_j = $ sigma-algebra generated by $X_1, \dots, X_j$. $\mathcal{F}_j $ is the sigma-algebra of Borel sets that can be defined in terms of the first $j^{th}$ coordinates, in the sense that.: $$ \mathcal{F}_j = \{A \times\mathbb{R}^{n-j} | A \text{ is a Borel set of }\mathbb{R}^{j}\} $$

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