Let $\Omega = \{1,2,3,4,5,6,7,8\}$ be an outcome space. Let $X$ be a function defined on $\Omega$ given as $$X(\omega) = \begin{cases} 5 \quad \text{when $\omega = 1,8$} \\
\pi \quad \text{when $\omega = 2,3,4,5,6,7$}
\end{cases} $$
Find the smallest $\sigma$-algebra $F$ that makes $X$ into a random variable.
I know that the definition of a $\sigma$-algebra $\mathcal{F}$ is that:
- $\emptyset \in \mathcal{F}$
- If $A \in \mathcal{F}$, then $A^C \in \mathcal{F}$.
- If $A_1, A_2, \dots \in \mathcal{F}$, then $\cup_1^{\infty} A_i \in \mathcal{F}$
My initial thought is that the smallest $\sigma$-algebra would be $\{ \emptyset, (\{1\} \cup \{8\} ), (\{1\} \cup \{8\} )^C \}$. My reasoning is that the complement of $(\{1\} \cup \{8\} )$ would be all numbers $2,3,…,7$. Is this the correct answer?
Best Answer
In general, it is easy to check that the smallest sigma algebra which makes a function $X:\Omega\to\mathbb{R}$ a random variable is:
$\{X^{-1}(B): B\in\mathbb{B}(\mathbb{R})\}$
Where $\mathbb{B}(\mathbb{R})$ is the Borel sigma algebra on $\mathbb{R}$. In your case, if $B$ contains $5$ but not $\pi$ its inverse image is $\{1,8\}$. If it contains $\pi$ but not $5$ then the inverse image is $\{2,3,4,5,6,7\}$. If it contains both then the inverse image is $\Omega$, and if it contains none of them then the inverse image is the empty set. So the sigma algebra is:
$\{\emptyset, \{1,8\}, \{2,3,4,5,6,7\}, \Omega \}$