5
$\begingroup$

My book has the following definition for the range of a random variable.

Since a random variable is defined on a probability space, we can calculate these probabilities given the probabilities of the sample points. Let $a$ be any number in the range of a random variable $X$. Then the set $\{\omega \in \Omega :X(\omega)=a\}$ is an event in the sample space (simply because it is a subset of $\Omega$).

I think I'm misunderstanding the provided notation but to me this expression:

$$\{\omega \in \Omega :X(\omega)=a\}$$

seems to be saying "For each sample point $\omega$ in the sample space $\Omega$, compute $X(\omega)$ and store the result $a$ in the set". If the above expression is saying what I think it says how can we say that a set of $\{X(\omega_1), X(\omega_2),\ldots,X(\omega_{|\Omega|})\}$ is an event in the sample space? Aren't events a subset $\{ \omega_1, \omega_2, \ldots\}$ of $\Omega$ ?

$\endgroup$
  • 3
    $\begingroup$ The thing you quote is certainly not an attempt to give a definition of the range. $\endgroup$ – Michael Hardy May 16 '17 at 2:05
6
$\begingroup$

$\{\omega \in \Omega :X(\omega)=a\}$ means the sets of points of $\Omega$ such that every point is mapped to the value of $a$ by the random variable $X: \Omega \mapsto \mathbb R$.

$\{X(\omega_1), X(\omega_2),\ldots,X(\omega_{|\Omega|})\}$ is better to be noted as $X(\Omega)$, as $\omega_{|\Omega|}$ sometimes does not make sense if $\Omega$ is not finite. But you are right that it is not an event, because it is a subset of $\mathbb R$.

But some times we write $X$ directly as a presentation/notation. For example, we write $u<X\leq v$ as a short form to represent the event of $\{\omega \in \Omega: u < X(\omega) \le v\}$. We always focus on subsets of $\Omega$ as events, because the probability is defined over the measurable subsets of $\Omega$, which is in a $\sigma$-algebra.

$\endgroup$
4
$\begingroup$

The first part of the constructor indicates from where the elements are drawn.   The second part of the constructor indicates the criteria for acceptance.


Thus $\{\omega \in \Omega : X(\omega)=a\}$ reads: "the set of outcomes in the sample space, $\Omega$, whose $X$-measure equals $a$".

That is, examine each outcome from the sample space, but only accept it if the $X$ value equal $a$, else discard it.


The same way, a circle is $\{(x,y)\in\Bbb R^2: x^2+y^2=c^2\}$; read as-"the set of points in the real plane whose sum of squares equals $c^2$".

$\endgroup$
3
$\begingroup$

Events are measurable subsets of $\Omega$ and $\Omega$ is the domain, not the range, of the random variable. What is defined in the quoted passage is a subset of $\Omega,$ i.e. a subset of the set of all inputs to $X$ not of the set of all outputs.

For example, suppose you throw a pair of dice and each gives you a number in the set $\{1,2,3,4,5,6\}.$ Your probability space is the set $$ \Omega = \left\{ \begin{array}{cccccc} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\(5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array} \right\}. $$ Let $X$ be the sum of the two numbers in the pair.

Then the event $X=10$ is the set $$ \{\omega\in\Omega : X(\omega) = 10\} \quad = \quad \{ (4,6),\ (5,5),\ (6,4) \}. $$ That is a set of three of the $36$ points in $\Omega.$

You refer to "the following definition for the range of a random variable". But the statement you quote is certainly not an attempt to define the range. The range is the set $\{X(\omega) : \omega\in\Omega\}.$

The notation $\{\omega \in \Omega :X(\omega)=a\}$ certainly does not mean "for each $\omega\in\Omega$ compute $X(\omega)$ and store the result in $a$." In the first place, the expression as a whole should be read as a noun. It names a particular set. It does not instruct you to do something (with that set or otherwise). The object called $a$ is pre-supposed to be already defined before the definition can make sense; it's not telling you to store anything in $a$ and it's not saying "Let $a=\text{something}.$" Rather, it identifies certain elements of $\Omega$ and excludes others.

$\endgroup$
  • $\begingroup$ Very nice example, well explained! (+1) $\endgroup$ – Yujie Zha May 16 '17 at 2:23
  • $\begingroup$ @YujieZha : Thank you. $\endgroup$ – Michael Hardy May 16 '17 at 2:26
2
$\begingroup$

In the question, the expression $$\{\omega \in \Omega :X(\omega)=a\}$$ is interpreted as follows:

For each sample point $\omega$ in the sample space $\Omega$, compute $X(\omega)$ and store the result $a$ in the set.

One fundamental misunderstanding here is the notion that the expression on the right-hand side of the colon contributes any values to the set at all. The only values that can be placed in the set are the values named $\omega$ on the left side of the colon. The formula $\omega \in \Omega$ says that every value $\omega$ that is a member of the set $\{\omega \in \Omega :X(\omega)=a\}$ must also be a member of $\Omega.$ That is practically the definition of a subset.

What the expression on the right-hand side does is to select the subset of $\Omega$ as follows:

For each sample point $\omega$ in the sample space $\Omega$, compute $X(\omega),$ compare it to the value $a$, and if the two values match then that value of $\omega$ belongs to the set $\{\omega \in \Omega :X(\omega)=a\}.$

In this procedure, $a$ is value that was already selected from the range of $X$ before we even started to write $\{\omega \in \Omega :X(\omega)=a\}$; during the entire definition of an particular subset $\{\omega \in \Omega :X(\omega)=a\}$ of $\Omega$ only that one single value of $a$ is ever used.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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