$\Omega = \{ {\omega_1, \omega_2,\omega_3} \}$ where each state is equally probable.

Two random variables exist $\widetilde{x}$ and $\widetilde{y}$ that are functions of these states:

$\widetilde{x}(\omega_i)=a_i$ where $a_1 \neq a_2 \neq a_3$


$\widetilde{y}(\omega_1) = b_1$ and $\widetilde{y}(\omega_2) = \widetilde{y}(\omega_3)=b_2.$

The question is, what is the $\sigma$-field generated by $\widetilde{y}?$

I think the answer is $F = \{\emptyset, \{\omega_1\}, \{\omega_2, \omega_3\}, \{\omega_1, \omega_2, \omega_3 \} \}$ for the following reasons:

  1. $A \in F \subseteq \Omega $
  2. $A \in F \implies A^c \in F$
  3. The intersection of any number of the elements of $F$ is an element of $F$.
  4. The union of any number of the elements of $F$ is an element of $F$
  5. $\omega_2$ and $\omega_3$ are indistinguishable from one another so we don't need to include $\{\omega_2\}$ nor $\{\omega_3\}$ in $F$.

Is that correct?

Also, since the question is about $\widetilde{y}$ any information about $\widetilde{x}$ is irrelevant, right?

Furthermore, the probability of each state of the world is also irrelevant when considering the $\sigma$-field, right?


1 Answer 1


You're right, but you might appreciate knowing how to find this sigma field using the definition:

The sigma-field generated by a random variable $X:\Omega\to\mathbb{R}$ consists of all the inverse images $X^{-1}(B)$ of the Borel sets $B\subset \mathbb{R}.$

Because $y$ has only two possible values $b_1$ and $b_2,$ there are exactly four kinds of Borel sets $B$ relevant to $y:$

  1. $b_1\in B$ and $b_2\in B.$ In this case, $y^{-1}(B) = \{\omega\in\Omega\mid y(\omega)\in B\}= \Omega.$

  2. $b_1\in B$ but $b_2\notin B.$ Now $y^{-1}(B) = \{\omega\in\Omega\mid y(\omega)\in B\}=\{\omega_1\}.$

  3. $b_1\notin B$ yet $b_2\in B.$ Now $y^{-1}(B) = \{\omega\in\Omega\mid y(\omega)\in B\}=\{\omega_2,\omega_3\}.$

  4. $b_1\notin B$ and $b_2\notin B.$ Clearly $y^{-1}(B) = \emptyset.$

That's it--we have listed precisely the elements you gave for $\mathfrak F.$

(Implicitly, we have used the facts that the Borel sets form a sigma field ; every real number is an element of some Borel set; and any two distinct real numbers can be separated by a Borel set in the sense that one of them is inside the set and the other is outside it.)

Some things to observe and remember:

  • You don't have to demonstrate the properties $(1)-(4)$ (in your question) of a sigma field . Because the Borel sets of $\mathbb R$ form a sigma field , necessarily the collection of their inverse images under $y$ forms a sigma field . That's proven using basic set theory and you only have to prove it once, not every time you deal with a random variable.

  • The sigma field for $y$ is generated by the inverse images of any pi-system that generates the Borel sets of $\mathbb R.$ A standard pi system consists of the sets of the form $(-\infty, a]$ that are used to define distribution functions. Although this observation wouldn't have simplified this exercise, it greatly simplifies the considerations involving more complicated random variables.

  • Sigma fields are logically prior to probabilities: you can't define a probability until you have a sigma field. Think of it this way: the sigma field is a declaration (by you, the modeler) of what events you may assign probabilities to. You can't make those assignments until you know what these events are! (The need for this comes to the fore in complex situations where there are infinitely many random variables to analyze: that is, for stochastic processes on infinite index sets.)


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