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I'm currently reading A First Look at Rigorous Probability Theory by J. S. Rosenthal, and I just got to the definition of random variable:

Definition 3.1.1 Given a probability triple $(\Omega,\mathcal F,\mathbf P)$, a random variable is a function $X$ from $\Omega$ to the real numbers $\mathbf R$, such that

$$\{\omega\in\Omega;X(\omega)\le x\}\in\mathcal F,\quad x\in\mathbf R\tag{3.1.2}$$

Equation (3.1.2) is a technical requirement, and states that the function $X$ must be measurable. It can also be written as $\{X\le x\}\in\mathcal F$, or $X^{-1}((-\infty,x])\in\mathcal F$, for all $x\in\mathbf R$.

Now, I understand this definition. What I don't understand is why we need it. I did some research, and I found this answer, which is pretty clear. Still, I have an issue: in the case used by the answer, it makes sense to use a map from the set of human beings to that of real numbers, since human beings are not mathematical objects which can be compared. However, I don't understand why random variables are necessary when, say, $\Omega=[0,1]$, or any subset of $\mathbf R$. In that case, the elements of the sample space are comparable, so do we still need random variables? If yes, why?

In case there's something I'm missing, the question would be: what is the correct interpretation of the given definition? Given some $\omega\in\Omega$, what does the number $X(\omega)$ represent, intuitively?

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  • $\begingroup$ Well, how would you prefer to define a random variable? The idea is that you want a function paired with a probability of different events happening. This offers a convenient way to do that. By referring to the sample space $\Omega$ you importing all the usefulness of measure theory, so that you can talk about the probability of the function doing this or that in a coherent way. $\endgroup$ Commented Sep 14, 2024 at 1:07
  • $\begingroup$ In short: random variables are not necessary when $\Omega$ is a subset of a vector space. In other cases they are necessary. Now let's move on. $\endgroup$ Commented Oct 1, 2024 at 6:33

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For a single random variable, your interpretation is right: a random variable gives a probability measure on the real numbers, and vice versa.

The problem begins when you have two random variables $X, Y.$ Probability is concerned with questions such as: are $X, Y$ independent? That is, if we know the value of $X,$ do we gain any knowledge on the value of $Y$? In order to model these types of questions, letting $\Omega \subseteq \mathbb R$ will not cut it, because two random variables following the same distribution may still have different types of interdependence.

Here is an example of a situation that we would like to model. Let $T_d$ be the (random) maximum temperature in a particular town at a future day $d \in \{1, \dots, 365\}$ of the year. For the sake of simplicity, assume the distribution of $T_d$ only depends on the season at day $d.$ Then the expected temperature is the same for all summer days, say $\mathbb E[T_d] = 29.$ However, let's say today is day $d$ in the middle of the summer. Then we know about $T_d,$ which might have been an unusually cold day, say $T_d = 20.$ In that case, it is reasonable to assume that $d+1$ has a high chance of being a cold day, too. Let's say that we expect the temperature at $d+1$ to be $23.$ Thus we want to express relationships such as $$\mathbb E[T_{d+1}\ |\ T_d = 20] = 23.$$ The space $(\Omega, \mathcal F, P)$ is the space of all outcomes. It encodes not only the distributions of each $T_d$ separately, but also joint expressions like the above. You can think of an element $\omega \in \Omega$ as a particular assignment of $T_d$ for every day $d.$ If $\omega$ is the "event that truly happens", then $T_d(\omega)$ is the real temperature at day $d.$

The mathematical construction of these types of dependencies is a bit subtle, but the intuition should be clear.

Finally, the fact that $X$ takes real values is mainly a matter of simplicity (for integration, etc.), but random variables taking values on more general spaces can be considered.

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An important thing to keep in mind is that we often refine $\Omega$ when we change the problem we're talking about. For example, $\{1, 2, \ldots, 6 \}$ with normalized counting measure is a fine model for the roll of a die, but if you want to roll two dice, you'd better look at ordered pairs $\widetilde{\Omega} = \{(1, 1), (1, 2), \ldots, (6, 6) \}$. As other commenters and answerers have noted, if you don't expand the probability space, you can't ask questions about things like independence.

Moreover, looking at ordered pairs is harmless even if you just want to talk about the first die. Indeed, if we imagine the bigger probability space $\widetilde{\Omega}$, the "value of the first die" is exactly the value of the random variable that takes the first coordinate and ignores the second. In general, probability calculations should be invariant under expanding the probability space. (And the smaller space can be recovered as the quotient of the larger one by an equivalence relation.)

Pumping this heuristic a little more, I think the best perspective is to imagine that there is one ridiculously enormous measure space $\Omega^!$ (that can't possibly really exist for set-theoretic reasons) which is big enough to discuss any probability problem. Then intuitively, $\Omega^!$ is just the "set of all possible worlds" and the measure of a subset of $\Omega^!$ is the probability of being in one of those possible worlds. A random variable is something we can measure in each of the possible worlds (e.g. what will the roll of the dice be in this world or that world), so that we can meaningfully ask "what is the probability that it is greater than 3," meaning, "what is the proportion of all possible worlds where this quantity is greater than 3?"

You might then imagine any given $\Omega$ in a problem as arising from an equivalence relation on $\Omega^!$. For example, if we care about the roll of this particular die in this particular casino this one time, we can say two possible worlds are equivalent if they have the same value of the die, and then we are content to work in the quotient of $\Omega^!$ by this equivalence relation. But of course the set of all equivalence classes is exactly in bijection with $\{1, 2, 3, 4, 5, 6\}$ with the expected measure.

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    $\begingroup$ (The reason we don't work with the enormous $\Omega^!$, in addition to it being set-theoretic nonsense, is that it's easier to compute when we pick a judiciously small $\Omega$ to model a given situation.) $\endgroup$ Commented Oct 2, 2024 at 6:26

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