# Generated Sigma Algebras

I am not a mathematician, rather I pick up on topics on the go, when I need something for the topic I am studying in the given time. So I am sorry if this is trivial to most of you and apologies for any conceptual mistakes I might make in the description - I'll try to be as precise as possible.

At the moment, I am studying Probability Theory, from this course: https://www.youtube.com/playlist?list=PL5B3KLQNAC5jT6yjV1199ji1zUy1YUp6P , [for the purposes of understanding Stochastic Calculus for Finance (Vol. II - S. Shreve)], and I stumbled upon sigma algebras.

While I do understand the concept; if we have a set which is a collection of subsets of Omega (i.e. if we have a collection of events) denoted by F, then F is a sigma-algebra if it satisfies the following three conditions;

1. Omega belongs in F,
2. F is closed under complements,
3. F is closed under countable Unions

So far so good and I also understand the properties that derive from the definition as well as how they are derived. In addition I know that we have the trivial sigma algebra, the smallest sigma algebra on Omega and the Discrete Sigma Algebra, which is the power set of Omega, being the largest sigma algebra on Omega.

My problem is with generated sigma algebras. I do understand the definition; Let A be an arbitrary collection of subsets of Omega, then sigma(A) is the generated sigma algebra, generated from A and is the smallest sigma algebra containing A. Further, we can find the smallest sigma algebra by intersecting all sigma algebras containing A, as the intersection of sigma algebras is also a sigma algebra.

The last part is the one I don't understand and confuses me. I get that we have the power set of Omega that definitely contains the collection A - But what exactly do we mean by intersecting all the sigma algebras containing A to find the smallest one containing A? Does it mean that if we have a sigma algebra containing the collection A and another collection of subsets, B (which is a sigma algebra containing A, but i get that it is not the smallest) and intersect it with the power set of Omega, we generate sigma(A), which is indeed the smallest and more refined to answer the questions that we need in our problem? But, where exactly does the bigger sigma algebra (on the collections A and B) come from?

If anyone could provide a more intuitive explanation or even better give an example (finite, like a die roll), I would be very grateful.

The point is not to take a $$\sigma$$-algebra $$B$$ containing $$A$$ and intersect it with the powerset of $$\Omega$$. (Note that if $$B$$ is any collection of subsets of $$\Omega$$, and $$P(\Omega)$$ is the powerset of $$\Omega$$, then $$B\cap P(\Omega)=B$$. So intersecting with the powerset of $$\Omega$$ doesn't do much.) The point is that if we have two $$\sigma$$-algebras $$B_1$$ and $$B_2$$ that contain $$A$$, then $$B_1\cap B_2$$ is also a $$\sigma$$-algebra containing $$A$$ (exercise). Moreover, $$B_1\cap B_2$$ is going to be smaller than $$B_1$$ and $$B_2$$ (unless one of $$B_1$$ or $$B_2$$ contains the other).

So $$\sigma(A)$$ takes this idea to its extreme: we intersect all $$\sigma$$-algebras containing $$A$$. In symbols, let $$\mathscr{B}$$ be the set of $$\sigma$$-algebras on $$\Omega$$ that contain $$A$$. Then $$\sigma(A)=\bigcap_{B\in\mathscr{B}}B$$. Then $$\sigma(A)$$ is a $$\sigma$$-algebra containing $$A$$ (exercise), and if $$B$$ is a $$\sigma$$-algebra containing $$A$$ then $$\sigma(A)\subseteq B$$ by definition. So it makes sense to call $$\sigma(A)$$ the smallest $$\sigma$$-algebra containing $$A$$, or the $$\sigma$$-algebra generated by $$A$$.

Now you ask where these bigger $$\sigma$$-algebras come from, and that very much depends on the particular example. In general, the collection $$\mathscr{B}$$ above could be quite complicated. The most we can say in general is that there is always at least one $$\sigma$$-algebra in $$\mathscr{B}$$, namely, the powerset of $$\Omega$$.

The construction of $$\sigma(A)$$ described above is good for a definition, but rather difficult to put into practice because it might be difficult or at least very time consuming to calculate $$\mathscr{B}$$. Given a particular $$A$$, if one wants to get a more explicit description of $$\sigma(A)$$ then this usually involves calculating families of sets that must be in any $$\sigma$$-algebra containing $$A$$ until you come up with a family that is itself a $$\sigma$$-algebra. Indeed, if you can come up with a collection $$B$$ which is a $$\sigma$$-algebra containing $$A$$ and must be contained in any $$\sigma$$-algebra that contains $$A$$, then it follows that $$\sigma(A)=B$$.

When $$\Omega$$ is finite the brute force idea is a little more reasonable because you can just start closing $$A$$ under intersections and complements until you get an algebra.

• Would the downvoter care to provide criticism? I am happy to fix/edit, etc. – halrankard2 Sep 1 '20 at 13:00
• Hey, this description is exactly what I have been looking for! I agree on the downvoter providing an explanation for the downvote to avoid any mistakes! Many thanks! One thing;The first is, in the case that we don't have any other sigma algebra containing an arbitrary collection of events, then the smallest sigma algebra containing that collection is the power set of the Omega. Is that right? – Konstantinos Bampalis Sep 1 '20 at 13:06
• Yes that is correct, it could be very possible that $\sigma(A)$ is the full powerset. An example would be if $\Omega$ is finite and $A$ is the set of all singletons. – halrankard2 Sep 1 '20 at 13:10
• Btw, if you find my answer helpful you can indicate that with an upvote. You can upvote multiple answers. You can also mark the green check to indicate the "accepted" answer which perhaps might be the one you find the most helpful. Given your specific interests, that is probably Kavi's answer in this case. Though people often wait a bit to see other possible answers, etc. Welcome to M.SE! – halrankard2 Sep 1 '20 at 13:11
• Thanks! I have been trying to upvote your answer for a while.. Does it show – Konstantinos Bampalis Sep 1 '20 at 13:13

Consider, for example, the real line and the collection of all one-point sets $$\{\{x\}: x \in \mathbb R\}$$. One sigma algebra containing these sets is the power set. Another one is the family of all countable sets and their complements. There are many others too. Now take the sets that are common to all such sigma algebras. That gives the sigma algebra generated by our family. In this case this turns out to be exactly countable sets and their complements. Reason: Any sigma algebra that contains singletons must contain all countable sets (since they are countable unions of singletons). It must also contain their complements. Hence the smallest sigma algebra containig all singeltons is exactly the family of all countable sets and their complements.

• So in a less abstract way if my sample space consists of the outcomes of a die roll Omega = {1,2,3,4,5,6}, and lets say i the collection of events A={{1,2},{3,4},{5,6},}, then the sigma algebra of A is F={null, omega, {1,2}, {3,4}, {5,6}, {3,4,5,6},{1,2,3,4}}. If I had then a collection B={{1,2},{3,4}}, how could i use the fact that intersections give the smallest sigma algebra? I know that the collection A contains the collection B and also the power set of Omega contains B..? Could you please help on that? – Konstantinos Bampalis Sep 1 '20 at 12:15
• In the first case you missed $\{1,2,5,6\}$. In the second case note that if a sigma algebra contains $\{1,2\}$ and $\{3,4\}$ it would contain their union $\{1,2,3,4\}$ and hence its complement $\{5,6\}$. So there is no difference between the sigma algebra generated by these two collections! @KonstantinosBampalis – Kavi Rama Murthy Sep 1 '20 at 12:21
• Oh yes your are right, quick typo for {1,2,5,6} thanks! Ok I do understand, so if this is not too much to ask then, could you be kind enough to provide a simple example on the intersection results on this simple die roll sample space? Appreciated :) – Konstantinos Bampalis Sep 1 '20 at 12:26
• Any sigma algebra which contains $B$ must contain your $F$. So the sets in $F$ belong to the intersection of these sigma algebras also. On other hand intersection of sigma algebras containing $B$ is contained in each particular sigma algebra containing $B$. In particular it is contained in $F$. @KonstantinosBampalis – Kavi Rama Murthy Sep 1 '20 at 12:32
• thanks Kavi ! all the best :) – Konstantinos Bampalis Sep 1 '20 at 12:45