# Sigma-algebra in the coin toss modelling

Let $$X_n$$ be the result of the $$n$$-th flip of a coin: $$X_n(\omega) := \omega_n$$ where $$\omega = (\omega_1, \omega_2, ...)$$ and $$\forall i, \omega_i \in \{0,1\}$$. $$\mathcal{G}_n$$ is the $$\sigma$$-field up to time $$n$$: $$\mathcal{G}_n := \sigma(X_1, \dots, X_n)$$.

I am trying to determine all random variables that are $$\mathcal{G}_n$$-measurable.

I wrote, for small values of $$n$$: for $$n=1$$, $$\mathcal{G}_1 = \big\{ \emptyset, \{(0, \omega_2, \omega_3, \dots )\}, \{(1, \omega_2, \omega_3, \dots ) \}, \{(0, \omega_2, \omega_3, \dots ), (1, \omega_2, \omega_3, \dots )\} \big\}$$ (the set of all sequences starting by either 0 or 1 and their union that represents all the possible sequences); for $$n=2$$, $$\mathcal{G}_2 = \big\{ \emptyset, \{(0, 1, \omega_3, \cdots)\}, \{(0, 0, \omega_3, \cdots)\}, \{(1, 0, \omega_3, \cdots)\}, \{ (1, 1, \omega_3, \cdots) \}, \cdots, \{(0, 1, \omega_3, \cdots), (0, 0, \omega_3, \cdots), (1, 0, \omega_3, \cdots), (1, 1, \omega_3, \cdots) \} \big\}$$.

We can remark that $$\sigma(X_2) = \big\{ \emptyset, \{(\omega_1, 1, \omega_3, \cdots)\}, \{(\omega_1, 0, \omega_3, \cdots)\}, \cdots, \{(\omega_1, 1, \omega_3, \cdots), (\omega_1, 0, \omega_3, \cdots) \} \big\} \not \subset \mathcal{G}_1$$ but, since $$\forall \omega_2 \in \{0,1\}$$, $$X_2(\omega) \times (1 - X_2(\omega)) = \omega_2 (1 - \omega_2) \equiv 0$$, $$\sigma \big( X_2 (1-X_2) \big) \subset \mathcal{G}_1$$.

So, to find $$\mathcal{G_n}$$-measurable r.v., I am led to determine all r.v. that are built thanks to the r.v. $$(X_n)_{n \in \mathbb{N}}$$ that remove the information beyond time $$n$$ using the binary property of the $$X_n$$.

What do you guys think? Thanks

A random variable $$Z$$ is $$\mathcal G_n$$-measurable iff you can write it as a function of $$X_1,\ldots, X_n$$, i.e. if there exists a function $$f : \{0,1\}^n \to \mathbb R$$ such that $$Z = f(X_1,\ldots, X_n)$$.

• Yes, I finally figure out that the goal of the exercise was to intuitively understand this result. Thank you. Oct 13, 2022 at 11:54