Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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, $$\sigma(X(\omega)) = \sigma(\emptyset, \{c,d\}, \Omega) = \{\emptyset, \{a,b\}, \{c,d\}, \Omega\}$$

share|cite|improve this answer
am I missing something, or are the values for X(w) switched in your answer. Shouldnt the RV take the value 3 when $\omega\in\left\{a,b\right\}$? – Justin May 22 '12 at 14:30
@Justin Yes. It was a typo. Fixed it. – user17762 May 22 '12 at 15:49
Thanks. I thought that was the case, but sometimes when I'm really looking at these concepts and dont fully understand them, I hesitate to think something is a typo. Thank you for the comment and help! – Justin May 22 '12 at 17:28
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? – Justin May 22 '12 at 20:08

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*}

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.