Confusion about probability space associated with infinite coin flips

Question

Let $\Omega = \{ \omega = (\omega_1, \omega_2, \ldots) : \omega_j = 1 \text{ or } 0 \}$. For each positive integer $n$, let $\Omega_n = \{ \omega = (\omega_1, \ldots, \omega_n) : \omega_j = 1 \text{ or } 0 \}$. We can consider $\Omega_n$ as a probability space with $\sigma$-algebra $ 2^{\Omega_n}$ and probability induced by $\mathbb{P}_n(\omega) = 2^{-n}$. We define $F_n$ to be the collection of all subsets $A$ of $\Omega$ such that there is an $E \in 2^{\Omega_n}$ with \begin{equation} A = \{(\omega_1, \omega_2, \ldots) : (\omega_1, \ldots, \omega_n) \in E\}. \tag 1 \end{equation} $F_n$ is a finite $\sigma$-algebra (containing $2^{2^n}$ subsets) and $F_1 \subset F_2 \subset F_3 \subset \cdots$ (i.e., an ascending sequence of $\sigma$-algebras). If $A$ is of the form $(1),$ we let $\mathbb{P}(A) = \mathbb{P}_n(E_n)$. This gives a function $\mathbb{P}$ on on \begin{equation} F^{0} = \bigcup_{j=1}^{\infty} F_j \end{equation} This is followed by the proposition that $F^{0}$ is an algebra but not a $\sigma$-algebra. All of the above and most of the proof makes sense to me. But in the proof for the proposition I just mentioned states: $ \Omega \in F_0$ since $\Omega \in F^1$. I get a bit confused because $F_1$ should just consist of the first flip being tails and the first flip being heads. Does this mean that $F_1$ consists of each as the first element in two infinite sequences where such as $(1, \omega_2, \ldots)$ and $(0, \omega_2, \ldots)$, where $\omega_j, j > 1$ are all just not given? But then doesn't $F_1$ trivially contain all possible sequences? Maybe it's because I'm a bit rusty on measure theory or I'm missing something but I'm a bit confused.

Answer 1

Note that $F_n$ is not a subset of $\Omega$, it is a set of subsets of $\Omega$. So it doesn't make sense to ask about $F_1$ containing all possible infinite sequences.

By definition, $F_1$ consists of four subsets of $\Omega$ (corresponding to the four subsets of $\Omega_1$). These four subsets are:

$\Omega$ (corresponding to $E=\Omega_1$)

$\emptyset$ (corresponding to $E=\emptyset$)

all sequences in $\Omega$ that start with $0$ (corresponding to $E=\{(0)\}$)

all sequences in $\Omega$ that start with $1$ (corresponding to $E=\{(1)\}$)

In general, $\Omega_n$ has size $2^n$ hence has $2^{2^n}$ subsets. These $2^{2^n}$ subsets of $\Omega_n$ correspond to $2^{2^n}$ subsets of $\Omega$, which together make up the collection $F_n$. The correspondence is: start with a subset $E$ of $\Omega_n$ (so this is some collection of sequences of $0$'s and $1$'s of length $n$). Then define the subset $A_E$ of $\Omega$ to consist of all infinite sequences whose first $n$ entries are a sequence in $E$. Now $F_n$ is the collection $\{A_E:E\subseteq \Omega_n\}$.

Answer 2

Say instead of $n=\infty$, you had $n=2$. I believe that $$F_0=\{\emptyset,\{\{0,0\},\{0,1\},\{1,0\},\{1,1\}\}\}$$ and $$F_1=\{\emptyset,\{\{0,0\},\{0,1\}\},\{\{1,0\},\{1,1\}\},\{\{0,0\},\{0,1\},\{1,0\},\{1,1\}\}\}.$$ This means that, after the first flip, you know whether heds or tails turned up in that first flip, something you did not know before the flip. Compare to $F_2$ $$F_2=\{\emptyset,\{\{0,0\}\},\{\{0,1\}\},\{\{1,0\}\},\{\{1,1\}\},\{\{0,0\},\{0,1\}\},\{\{1,0\},\{1,1\}\},\{\{0,0\},\{0,1\},\{1,0\},\{1,1\}\}\},$$

where you can know the full history of flips. For $n=\infty$, the number of terms in each $F_i$ is the same.

Answer 3

This is to complement the answers others have posted to your question.

There are many probability spaces where one can define precisely the random variables that model the crossing of a coin. Surely, the product space $\{0,1\}^\mathbb{N}$ with the product $\sigma$-algebra is one.

Here is another one which can also be considered canonical:

Consider the unit interval in the real line with the Borel $\sigma$-algebra and the Lebesgue measure $\lambda$ on it, that is $([0,1],\mathscr{B}[0,1],\lambda)$. Notice that in the this space, the identity function $\theta(x)=x$ is a uniformly distributed $U[0,1]$ random variable.

Recall that every $x\in[0,1]$ has a unique binary expansion $$x=\sum_{n\geq1}r_n/2^n$$ where $r_n\in\{0,1\}$, and $\sum_{n\geq1}r_n=\infty$ for $x>0$. For each $n\in\mathbb{N}$, the $n$--th bit map $x\mapsto r_n(x)$ defines a measurable function from $([0,1],\mathscr{B}([0,1]))$ to $(\{0,1\},2^{\{0,1\}}))$, where $2^{\{0,1\}}$ is the collection of all subsets of $\{0,1\}$.

We will see that

1. The map $\beta:[0,1]\rightarrow\{0,1\}^{\mathbb{N}}$ given by $x\mapsto(r_n(x))$ is measurable, that is, it is a random variable taking values in $\{0,1\}$.
2. $\{r_n:n\in\mathbb{N}\}$ is an i.i.d. sequence of Bernoulli random variables.

Lemma 1: Suppose $\theta$ is a uniformly 0-1 distributed random variable defined in some probability space $(\Omega,\mathscr{F},\mathbb{P})$. Define $\{X_n=r_n\circ\theta\}$. Then, $\{X_n\}$ is an i.i.d. Bernoulli sequence with rate $p=\tfrac12$. Conversely, if $(Y_n)$ is an i.i.d. Bernoulli sequence with rate $p=\tfrac12$, then $\theta=\sum_{n\geq1}2^{-n}Y_n\sim U[0,1]$.

Here is a short proof:

Suppose that $\theta\sim U(0,1)$. For any $N\in\mathbb{N}$ and $k_1,\ldots,k_N\in\{0,1\}$, $$\begin{align} \bigcap^N_{j=1}\{x\in(0,1]:r_j(x)=k_j\}&=&(\sum^N_{j=1}\tfrac{k_j}{2^j}, \sum^N_{j=1}\tfrac{k_j}{2^j}+\tfrac{1}{2^N}]\tag{1}\label{one}\\ \{x\in(0,1]: r_N(x)=0\}&=&\bigcup^{2^{N-1}-1}_{j=0}(\tfrac{2j}{2^N},\tfrac{2j+1}{2^N}]\tag{2}\label{two}\\ \{x\in(0,1]:r_N(x)=1\}&=&\bigcup^{2^{N-1}-1}_{j=0} (\tfrac{2j+1}{2^N},\tfrac{2(j+1)}{2^N}]\tag{3}\label{three} \end{align} $$ It follows immediately that $x\mapsto (r_n(x):n\in\mathbb{N})$ is measurable, and that $ \mathbb{P}[\bigcap^N_{j=1}\{X_j=k_j\}]=\tfrac{1}{2^N}=\prod^N_{j=1}\mathbb{P}[X_j=k_j]$. Hence $\{X_n\}$ is an i.i.d. Bernoulli($\tfrac12$) sequence.

Conversely, suppose $\{Y_n:n\geq1\}$ is a Bernoulli sequence with rate $\tfrac12$. Let $\widetilde{\theta}$ be a $U(0,1)$-distributed random variable defined in some probability space $(\Omega,\mathscr{F},\mathbb{P})$ (for examples $\widetilde{\theta}(t)=t$ on $([0,1],\mathscr{B}([0,1]),\lambda)$). Then, the first part shows that the sequence of bits $\{\widetilde{Y}_n\}\stackrel{law}{=}\{Y_n\}$. Therefore, $$ \theta:=\sum_{n\geq1}2^{-n}Y_n\stackrel{law}{=} \sum_{n\geq1}2^{-n}\widetilde{Y}_n=\widetilde{\theta} $$ since $\theta$ is a measurable function of $\{Y_n\}$.