Sub-sigma-algebras: infinite coin flips example on wikipedia

Question

On the wiki there is this section about sub sigma-algebras for the first $n$ coin flips

Specifically, they say that after the first $n$n coin flips we can describe the observed information in terms of the sub sigma-algebra

$$\tag{1} \mathcal{G}_n = \{A\times \{H,T\}^\infty:A\subset\{H,T\}^n\} $$ Then they say that $$ \mathcal{G}_1\subset \mathcal{G}_2 \subset \mathcal{G}_3 \subset \dots \subset \mathcal{G}_\infty $$ I don't see why this subset relation between the sets is true (intuitively, we "know more" after more coin flips)

Perhaps the wiki should say "the sub sigma-algebra for the first $n$ flips is the sigma algebra generated by sets of the form $\mathcal{G}_n$?

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Specifically, according to (1), $\mathcal{G}_1$ should contain all sequences that start with either an $H$ or a $T$.

Similarly, $\mathcal{G}_2$ should contain all sequences that start with an element of the set $\{\{H,T\},\{H,H\},\{T,H\},\{T,T\}\}$

but all these sequences in $\mathcal{G}_2$ start with either an $H$ or a $T$, so shouldn't they be in $\mathcal{G}_1$ as well?

Am I just reading the notation wrong and the sigma algebra is actually a sigma algebra that contains sets of the form $\mathcal{G}_n$. (i.e., perhaps $\mathcal{G}_1$ should be a set that has 4 elements: the empty set, the set of all infinite sequences, a set of all sequences starting with $H$, and a set of all sequences starting with $T$. Then $\mathcal{G}_2$ would be the above set, along with the set of all sequences starting with $(H,T)$ and so on, plus complements and unions)?

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Thanks

Answer 1

$\mathcal G_2$ is a set of subsets of $\{H,T\}^\infty,$ not a subset $\{H,T\}^\infty.$ The set of all sequences that begin with $(H,H)$ or $(H,T)$ or, etc is a subset of $\{H,T\}^\infty$ (in fact it is just $\{H,T\}^\infty).$

$\mathcal G_2$ is a set of sixteen subsets of $\{H,T\}^\infty,$ one corresponding to each element of the sigma algebra for two coin flips $\mathcal P(\{H,T\}^2).$ For instance, $A=\{(H,H),(H,T)\}$ is an element of the sigma algebra for two coin flips. It corresponds to an element of $\mathcal G_2:$ the set of all sequences in $\{H,T\}^\infty$ that start with either $(H,H)$ or $(H,T).$ (Note this is a subset of $\{H,T\}^\infty$).

Also notice that this is equal to an element of $\mathcal G_1:$ it is the set of all sequences that start with $H.$ From this example, you should be able to see that $\mathcal G_1\subset \mathcal G_2.$

Answer 2

But each $\mathcal G_n$, as you defined it, is already a $\sigma$-algebra, there is no need to generate anything.

Note that there is no sequence in $\mathcal G_n$. There are only sets of sequences. So for example, in $\mathcal G_1$, there are four sets: the set of sequences starting with $H$, the set of sequences starting with $T$, the empty set, and the set of all sequences. All four of these sets are in $\mathcal G_2$ (just take $A$ of the form $A'\times \{H,T\}$ for $A'\subseteq \{H,T\}$), and hence $\mathcal G_1\subseteq \mathcal G_2$.

Answer 3

The definition is $\mathcal G_n = \{A{\times}\{H,T\}^\infty: A\subseteq\{H,T\}^n\}$

For compactness, let's use $X=\{H\}, Y=\{T\}, Z=\{H,T\}, W=Z^\infty$ and type $XYZ$ to mean te cartesian product $X\times Y\times Z$ and so forth. ie $XW=\{H\}\times\{H,T\}^\infty$.

$$\mathcal G_n=\{AW: A\subseteq Z^n\}$$

So since $\mathcal P(Z^1)=\{\emptyset, X, Y, Z\}$ therefore $\mathcal G_1 =\{\emptyset, XW, YW, W\}$ since $\emptyset W=\emptyset, ZW=W$.

Likewise $\mathcal G_2=\{\emptyset, X^2W, XYW, XW, YXW, Y^2W, YW, ZXW, ZYW, (XY\cup YX)W,..., W\}$

Thus $\mathcal G_1\subset \mathcal G_2$ as all events in $\mathcal G_1$ are in $\mathcal G_2$ but $\mathcal G_2$ contains more events.

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While it is true that all events in $G_2$ are subsets of some event in $\mathcal G_1$, that is a quite different claim; that $\forall G_2\in \mathcal G_2\exists G_1\in\mathcal G_1:G_2\subseteq G_1$.