Interpretation of the technical requirement on a random variable

I found a slide where there is the definition of a random variable and after a technical requirement difficult to understand for me.

Can you explain it by using a counterexample please? What happens if X>x? What is an example in which the outcomes are not an event of the sample space?

I known that the sample space is the set of all possible outcomes of the random experiment, and that an event can be an elementary event if it coincides with a single outcome or a compound event if it is a group of two or more outcomes (in the slide it seems that it refers to a compound event constituted by all possible outcomes).

• Commented Jul 25, 2019 at 18:45
• A natural set of examples is afforded by the binomial asset price model: outcomes are paths along a binary tree whose branches represent changes in an asset price. The measurable random variables are those corresponding to events that do not require one to foresee the future! For details see Shreve
– whuber
Commented Jul 25, 2019 at 19:33
• Also relevant; stats.stackexchange.com/questions/199280/… Commented Feb 17, 2022 at 11:04

The kinds of questions I would like answers to are "What is the probability that the random variable has value between $$a$$ and $$b$$, that is, what is $$P\big(X \in (a,b)\big)$$?" or perhaps I want $$P\big(X \in (a,b]\big)$$ or $$P\big(X\in [a,b)\big)$$ or $$P\big(X\in [a,b]\big)$$. It turns out that all my questions can be answered if I know the function $$F_X(u)$$ (called the cumulative probability distribution function or CDF) of $$X$$ which is defined as $$F_X(u) = P(X \leq u) = P\big(\{\psi\colon X(\psi) \leq u\}\big).$$ But this means that I must restrict the name "random variable" to only those mappings $$X\colon \Omega \to \mathbb R$$ which have the property that $$\{\psi\colon X(\psi) \leq u\}$$ is an event for all choices of real number $$u$$. Recall that probabilities are defined only for those special subsets of $$\Omega$$ that are events, and that every subset of $$\Omega$$ is not necessarily an event, and such nonevents have no probability assigned to them. Thus the technical requirement is imposed so that we can be sure that we can talk sensibly about the CDF of $$X$$. For all choices of real number $$u$$, the set $$\{\psi\colon X(\psi) \leq u\}$$ whose probability is going to be defined as the value of $$F_X(u)$$ is guaranteed to be an event and so it does have a probability. In short, the technical requirement assures us that $$F_X(u)$$ is properly defined for all $$u$$.
So, are there mappings $$Y\colon \Omega\to \mathbb R$$ that are not random variables? Yes, and here is a contrived example. We take $$\Omega = \{\psi_1, \psi_2, \psi_3\}$$ and define the event space as the following collection of four subsets of $$\Omega$$: $$\big\{\emptyset, \{\psi_1\}, \{\psi_2, \psi_3\}, \Omega\big\}.$$ Note that we are explicitly denying that the singleton subsets $$\{\psi_2\}$$ and $$\{\psi_3\}$$ are events, and similarly that the doubleton subsets $$\{\psi_1, \psi_2\}$$ and $$\{\psi_1, \psi_3\}$$ are events. The singleton subset $$\{\psi_1\}$$ is an event and we can talk about the probability that the outcome $$\psi_1$$ has occurred, but the the singleton subsets $$\{\psi_2\}$$ and $$\{\psi_3\}$$ are not events and we are not allowed to talk about the probability that the outcome $$\psi_2$$ has occurred or the probability that the outcome $$\psi_3$$ has occurred; these occurrences don't have probabilities. We do have a probability associated with the event $$\{\psi_2, \psi_3\}$$ which occurs whenever the outcome is a member of the set $$\{\psi_2, \psi_3\}$$, that is, the outcome is either $$\psi_2$$ or $$\psi_3$$, but we can't subdivide this probability value into the probability that $$\psi_2$$ occurred and the probability that $$\psi_3$$ occurred: $$\{\psi_2\}$$ and $$\{\psi_3\}$$ are not events in the event space defined above. Thus, any mapping $$Y$$ that maps $$\psi_2$$ and $$\psi_3$$ onto two different real numbers is not a random variable defined on this sample space (and associated event space). Why not? Well, assume without loss of generality that $$Y(\psi_2) = x_2 < Y(\psi_3) = x_3$$, and consider the set $$\{\psi\colon Y(\psi) \leq x_2\}$$. This set is either $$\{\psi_1, \psi_2\}$$ or $$\{\psi_2\}$$ according as $$Y(\psi_1)$$ is smaller than or greater than $$x_2$$. But neither set is an event, and so $$F_Y(x_2)$$ is undefined. Ergo, $$Y$$ cannot be called a random variable.
• Sorry @DilipSarwate, I have an extra question about your example: if I map $\emptyset$ so that $X(\emptyset )=5$ and if I choose $u=1$, I have that the condition $X(\emptyset ) \leq u$ is not satisfied. What I missed? Commented Jul 23, 2019 at 8:24
• @GennaroArguzzi. You can't map $\emptyset$ to anything; the mapping is from elements of $\Omega$ (that is, $\psi_i, I = 1,2,3$) to real numbers. So, if $Y(\psi_2) \neq Y(\psi_3)$, $Y$ does not meet the technical condition for being called a random variable. Commented Jul 23, 2019 at 12:45
• I did a mistake in the previous question because the $\emptyset$ does not belong to $\Omega$. If I consider $\psi_1$ and I map it so that $X(\psi_1 )=5$, and if I choose again $u=1$, the condition is not satisfied. Which conceptual error am I making? Commented Jul 23, 2019 at 14:30
• @GennaroArguzzi Your latest query makes no sense. $\Omega$, the sample space is three elements $\{\psi_1,\psi_2,\psi_3\}$ and not what you have. It is the elements of $\Omega$ that are mapped to real numbers. The event space or $\sigma$-algebra, is a collection (set) whose elements are sets of outcomes, i.e. subsets of $\Omega$, The $\sigma$-algebra I defined has 4 elements (sets) in it. The 4 sets in the collection are $$\emptyset,\{\psi_1\},\{\psi_2,\psi_3\},\Omega.$$ Mapping is from elements of $\Omega$ to $\mathbb R$, not from elements of the $\sigma$-algebra to $\mathbb R$. Commented Jul 26, 2019 at 19:30