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Let $\Omega = \{0,1\}^\mathbb{N}$ and denote $\omega = (\omega_1,\omega_2,\ldots)$ as an element of $\Omega$. Now let $X_n: \Omega \rightarrow \mathbb{R}$ be defined by $X_n(\omega)= \omega_n$ and we put $\mathcal{F}_n = \sigma(X_1,\ldots,X_n)$.

My question now is how does $\mathcal{F}_1$ and $\mathcal{F}_2$ explicitly look like? I know that $\mathcal{F}_1 = \sigma(X_1) = \{X_1^{-1}(A)|A \in \mathcal{B}(\mathbb{R})\}$ but I don't know how to proceed further neither for $\mathcal{F}_2 = \sigma(X_1,X_2) = \sigma(X_1 \lor X_2)$. Any help is greatly appreciated!

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$X_1$ can only take two values: $0$ or $1$, so $\sigma(X_1)$ is generated by $$X_1^{-1}(\{0\}) = A_1 = \{\omega: \omega_1 = 0\}$$ and $$X_1^{-1}(\{1\}) = B_1 = \{\omega: \omega_1 = 1\}= A_1^c$$

So $\mathcal{F}_1 = \{\Omega, \emptyset, A_1, A_1^c\}$

Similarly, $\sigma(X_2)$ is generated by

$$X_2^{-1}(\{0\}) = A_2 = \{\omega: \omega_2 = 0\}$$ and $$X_2^{-1}(\{1\}) = B_2 = \{\omega: \omega_2 = 1\}= A_2^c$$

Then $\sigma(X_2) = \{\Omega, \emptyset, A_2, A_2^c\}$

So $\mathcal{F}_2 = \sigma(X_1, X_2)$ is generated by the $\sigma(X_1)$ and $\sigma(X_2)$ and is equal to

$$\{\Omega, \emptyset, A_1, A_1^c, A_2, A_2^c, A_1\cap A_2,A_1\cap A_2^c, A_1^c\cap A_2, A_1^c\cap A_2^c, A_1\cup A_2,A_1\cup A_2^c, A_1^c\cup A_2, A_1^c\cup A_2^c\}$$

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  • $\begingroup$ Thanks for the precise answer it is in fact the case that $X_i$ just takes 2 values which makes it easier. Thanks! $\endgroup$ – user155670 Nov 4 '14 at 14:57
  • $\begingroup$ @Rodel yes indeed, when $X$ take more values, it will be horrible to make the sigma algebra explicit. You are welcome! $\endgroup$ – Petite Etincelle Nov 4 '14 at 15:04
  • $\begingroup$ Isn't $(A_1\cap A_2^c) \cup (A_1^c \cap A_2)$ in $\mathcal{F}_2$? $\endgroup$ – YHH Oct 28 '18 at 5:28
  • $\begingroup$ @YHH it is $A_1^c\cup A_2^c$ $\endgroup$ – Petite Etincelle Oct 28 '18 at 20:31
  • $\begingroup$ Do you mean that $(A_1\cap A_2^c)\cup (A_1^c \cap A_2)=A_1^c \cup A_2^c$? Do you mind explaining it a little bit since I didn't see how the equality holds? Thanks a lot. $\endgroup$ – YHH Oct 29 '18 at 23:45
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Recall that $\mathcal F_1$ is the smallest sub-$\sigma$-algebra of $\mathcal F$ making $X_1$ measurable. Since $X_1$ can only take the values $0$ and $1$, it is enough to contain $\{X_1=0\}$ and $\{X_1=1\}$.

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$\mathcal{F}_n$ consists of cylinder sets on $\Omega$ formed by the permutation group of $\{\omega_1,...,\omega_n\}:=\pi(\omega_1,...,\omega_n)$ as the cylinder bases. I.e., you are forming a sub-$\sigma$-algebra on the restriction of $\Omega$ to the finite vectors $\omega'=(\omega_1,...,\omega_n)$ (a finite sample space, $\Omega'$). The sub-$\sigma$-algebra will be the Cartesian product of $2^{\Omega'}$ and $(\Omega')^c$. Therefore:

$$\mathcal{F}_n = 2^{\{\omega \in \pi(\omega_1,...,\omega_n)\} } \otimes \{\Omega \setminus \{\omega \in \pi(\omega_1,...,\omega_n)\}\}$$

Basically, $\mathcal{F}_n$ represents the $\sigma$-algebra on $\Omega$ than can be formed from only knowing the first $n$ elements of $\omega$

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