Intuition of sub sigma-algebra definition

Question

I am having trouble understanding the sub σ-algebra definition on Wikipedia. I understand the following:

Let $X$ be a set, and let $A,B$ be σ-algebras on $X$. Then $B$ is said to be a sub-σ-algebra of $A$ iff $\mathcal B \subseteq \mathcal A$.

For example, let $X=[0,1]$ and let $A$ be the σ-algebra generated by the sets $[0,1/4), [1/4,1/2), [1/2,3/4), [3/4,1]$.

Then the σ-algebra $B$ generated by the sets $[0,1/2), [1/2,1]$ is a sub σ-algebra of $A$. Is this correct?

And now regarding the wikipedia definition that I have trouble with:

Formally, since you need to use subsets of Ω, this is codified as the σ-algebra

${\displaystyle {\mathcal {G}}_{n}=\{A\times \{H,T\}^{\infty }:A\subset \{H,T\}^{n}\}.}$

Observe that then ${\displaystyle {\mathcal {G}}_{1}\subset {\mathcal {G}}_{2}\subset {\mathcal {G}}_{3}\subset \cdots \subset {\mathcal {G}}_{\infty },}$ where ${\displaystyle {\mathcal {G}}_{\infty }}$ is the smallest σ-algebra containing all the others.

I don't understand the last observation. My reasoning is that ${\mathcal {G}}_{1}$ (for example a sequence of H, T starting with H) has more elements than ${\mathcal {G}}_{2}$ (a sequence of H, T starting with HH).

Answer 1

I had the same question after looking at the wikipedia page, and this is what I figured out after looking at the comments posted to the question. Please feel free to correct me if I am wrong.

$$ G_1 = \{\Omega, \phi, \{H,...\}, \{T,...\} \} $$

$$ G_2 = \{\Omega, \phi,\{H,H,...\},\{H,T,...\},\{T,H,...\},\{T,T,...\},\{H,...\}, \{T,...\}, ...\}$$

You get $\{H,...\}$ and $\{T,...\}$ in $G_2$ since these are the unions of $\{H,H,...\},\{H,T,...\}$ and $ \{T,H,...\},\{T,T,...\}$ respectively. This follow from the definition of sigma-algebras, which states that sigma-algebras must contain all possible unions of elements.

$\{H, ... \}$ etc. are sets of all possible sequences with the first n coin toss results.

Answer 2

If the question is about intuition, I think the best way to see this is by looking at a simpler example.

Consider just two coins thrown sequentially (instead of infinite). Write the set of possible outcomes of a single coin throw as $\Omega_{1}=\left\{ H,T\right\}$ and both coin throws as $\Omega_2 = \left\{ HH, HT, TH, TT \right\} $ . The information before both throws, after one throw, and after both throws, respectively, are modeled as the following filtration:

$$ \begin{align} \mathcal{F}_{0} & =\left\{ \emptyset,\Omega_{2}\right\} \\ \mathcal{F}_{1} &=2^{\Omega_{1}}\times\Omega_{1}\\ & =\left\{ \emptyset,\left\{ H\right\} ,\left\{ T\right\} ,\left\{ H,T\right\} \right\} \times\left\{ H,T\right\} \\ &=\left\{ \emptyset,\left\{ HH,HT\right\} ,\left\{ TH,TT\right\} ,\left\{ HH,HT,TH,TT\right\} \right\} \\ \mathcal{F}_{2} & =2^{\Omega_{2}} \\ & = \left\{ \begin{array}{c} \emptyset,\\ \left\{ HH\right\} ,\left\{ HT\right\} ,\left\{ TH\right\} ,\left\{ TT\right\} \\ \left\{ HH,HT\right\} ,\left\{ HH,TH\right\} ,\left\{ HH,TT\right\} ,\left\{ HT,TH\right\} ,\left\{ HT,TT\right\} ,\left\{ TH,TT\right\} \\ \left\{ HH,HT,TH\right\} ,\left\{ HH,HT,TT\right\} ,\left\{ HH,TH,TT\right\} ,\left\{ HT,TH, TT\right\} \\ \left\{ HH,HT,TH,TT\right\} =\Omega_{2} \end{array}\right\} \end{align} $$

Before any throw, all you can ask and answer is that the result will be in $\Omega_{2}=\Omega_{1}\times\Omega_{1}$, so $\mathcal{F}_{0}=\left\{ \emptyset,\Omega_{2}\right\}$. After the first throw, you can now answer any of the questions related to the first throw, but nothing related to the second (e.g. “Is the first throw heads?”). So the relevant sub sigma field is $\mathcal{F}_{1}=2^{\Omega_{1}}\times\Omega_{1}$, written out explicitly above. After both throws, you can answer any questions (e.g. "Did you get at least one tails?"), so you have the power set of $\Omega_2$

Check that at each t, $\mathcal{F}_{t}$ is a sigma-field. Also, once you write it out, it's clear that $\mathcal{F}_{0}\subset\mathcal{F}_{1}\subset\mathcal{F}_{2}$. Thus we have a filtration.

I think once you see this, it's much easier to understand your example, and in fact it does hold that:

$$\mathcal{G_1} \subset \mathcal{G_2} \subset \mathcal{G_3} \subset \cdots$$