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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

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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 $Z = f(X_1,\ldots, X_n)$.

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    $\begingroup$ Yes, I finally figure out that the goal of the exercise was to intuitively understand this result. Thank you. $\endgroup$
    – Td_21
    Oct 13, 2022 at 11:54

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