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Simple question here. I am trying to enumerate the sigma field generated by the random variable: $X(\omega)=2+1_{\left\{a,b\right\}}(\omega)$ where $\Omega=\left\{a,b,c,d\right\}$.

I think what is confusing me is that I am used to looking for the pre-images under a function in $\mathbb{R}$ or similar. I know the simple discrete space should make this exercise easier, but it just isn't clicking in my head.

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First note that $X(\omega) = \begin{cases} 2 & \omega \in \{c,d\}\\ 3 & \omega \in \{a,b\}\end{cases}$

Now consider $\{\omega \in \Omega : X(\omega) \leq \alpha\}.$

  1. For $\alpha <2$, you get $\emptyset$.
  2. For $\alpha \in [2,3)$, you get $\{c,d\}$.
  3. For $\alpha \geq 3$, you get $\{a,b,c,d\} = \Omega$.

You want these set to be in your $\sigma$-algebra. Hence, $\sigma(X(\omega)) = \sigma(\emptyset, \{c,d\}, \Omega) = \{\emptyset, \{a,b\}, \{c,d\}, \Omega\}$

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    If I were to extend this to determine the sigma-field for (X,Y) with Y defined similarly (but not exactly the same), would it simply be a list of all the events in $\sigma(X), \sigma(Y)$ listed as ordered pairs, such as $\sigma(X,Y)=$ {{a,b},{c,d}} etc?2012-05-22
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I will include another pretty easy approach since I think the reply of Marvis had a typo in the function (not that it would make any major difference in this particular case, but I think this detail made you question the final result).

Note that $X$ takes only two values in $\mathbb{R}$: $2$ and $3$. For this reason, the preimage of any Borel set $B\subset \mathbb{R}$ is equals the preimage of either the singleton $\{2\}$ or $\{3\}$. The function $X$ in fact is $X(w)=3$ if $w\in\{a,b\}$ and $X(w)=2$ if $w\in\{c,d\}$. So \begin{align*} \sigma(X)=\sigma(X^{-1}\{2\},X^{-1}\{3\})=\sigma(\{a,b\},\{c,d\})=\{\emptyset,\{a,b\},\{c,d\},\Omega\}. \end{align*}