Tell me more ×
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.
up vote 3 down vote favorite

Let's consider a probability space $(\Omega, \mathcal{F}, P)$ corresponding to experiments on throwing a dice and defined in the following way: $\Omega = \{1, 2, 3, 4, 5, 6\}$, $\mathcal{F} = \{\Omega, \varnothing, \{1, 3, 5\}, \{2, 4, 6\}\}$. So, in this $\sigma$-algebra we only have 4 events: something happened, nothing happened, odd number rolled, even number rolled. Can anyone give me an example of random real variable defined for this probability space?

share|improve this question
3  
A side remark: "dice" is the plural form, and the singular is "die". (The teacher in my first probability course at university said "we roll a die" with a heavy Indian accent a gazillion times, so I won't forget that!) – Hans Lundmark Jun 19 '11 at 20:37
1  
A minor remark: the events $\Omega$ and $\varnothing$ do not describe the fact that something happened vs nothing happened. Rather they correspond to something certain happened vs something impossible happened (like, the result of the throw of the dice is an integer vs the result of the throw of the die is $7$). But something always happens. – Did Jun 27 '11 at 10:18

2 Answers

up vote 6 down vote accepted

Yes, $X(\omega)=0$ if $\omega \in \{1,3,5\}$, and $X(\omega)=1$ if $\omega \in \{2,4,6\}$.

share|improve this answer
1  
As a trivial example, you can take $X$ to be a constant random variable, for example $X(\omega)=1$ for all $\omega \in \Omega$. – Shai Covo Jun 19 '11 at 18:09
Thank you. Any other examples (of variables having other "structures")? – Leo Jun 19 '11 at 18:13
You can generalize as follows: $X(\omega)=a$ if $\omega \in \{1,3,5\}$, and $X(\omega)=b$ if $\omega \in \{2,4,6\}$. – Shai Covo Jun 19 '11 at 18:18
Note that the requirement is that for any $x \in \mathbb{R}$ fixed, $\{\omega: X(\omega) \leq x\} \in \mathcal{F}$. – Shai Covo Jun 19 '11 at 18:32
Or equivalently, since $\Omega$ is finite, for any $x \in \mathbb{R}$ fixed, $\{\omega:X(\omega) = x\} \in \mathcal{F}$. – Shai Covo Jun 19 '11 at 19:06
up vote 4 down vote

The random variables $X$ on this probability space are exactly the functions $X:\Omega\to\mathbb{R}$ such that there exists $a$ and $b$ with $X(\omega)=a$ if $\omega\in\{1,3,5\}$ and $X(\omega)=b$ otherwise. That is, every such $X$ is a random variable and every random variable $X$ is like that.

share|improve this answer
3  
Or in other words, $X$ has to be ${\mathcal F}$-measurable which means it can't depend on any other information than whether the result was even or odd. – Marek Jun 19 '11 at 18:22

Your Answer

 
discard

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.