Demystifying definition of the Random Variable

Question

The intro on Random Variables says that it is a variable (in bold), whose value depends on a chance. IMO, it sounds like a random value generator, whose value depends on a chance, just as random variable does. What is more dismaying is that formal definition is absolutely different. It says that random variable X is a mapping, $$X: \Omega \to E$$

That is, I see two problems in the basic introduction:

1. Why do you insist that it is a variable if it is more like random value generator or, formally, a function, which is not a variable and not random at all and
2. How do you reconcile two definitions of the random variable: informal, which displays the RV as a random generator, and formal, which says about probability space to a measurable space mapping.

Answer 1

You are right, there is nothing random in the definition.

The point is that you can think of $\Omega$ as "states of the world": you don't know what is going to happen in the future, what $\omega$ is going to "materialise", so to speak.

So you don't know which values $X$ is going to take; but $X$ gives you a way to "translate" events from the real world (like "the dice will come up $6$) to a more "mathematical" object (usually $\mathbb R$)

Of course since $\omega$ is "random", so is $X(\omega)$.

In general though every measurable function is (by definition) a random variable (also a function like $f(x)=x-2$ if you take $\Omega = \mathbb R$ and the usual Borel sigma algebra) so there is no real difference, we just use this term in probability theory.

Another point is that we usually "forget" about $\Omega$. That is we say something like : Let $X$ be a normally distributed random variable... We do not specify $\Omega$, we don't know which kind of relation $X$ induces on the elements of $\Omega$; so what happens is that $X$ is basically "random", is the only random thing there is in our setting.

Answer 2

Consider a familiar example: Let $\Omega$ be the set of all outcomes of the toss of two dice: $$ \Omega = \left\{ \begin{array}{cccccc} 11 & 12 & 13 & 14 & 15 & 16 \\ 21 & 22 & 23 & 24 & 25 & 26 \\ 31 & 32 & 33 & 34 & 35 & 36 \\ 41 & 42 & 43 & 44 & 45 & 46 \\ 51 & 52 & 53 & 54 & 55 & 56 \\ 61 & 62 & 63 & 64 & 65 & 66 \end{array} \right\} $$ Let $X: \Omega \to \{2,3,4,5,6,7,8,9,10,11,12\}$ be defined by $X=$ the sum of the two numbers, so that, for example, if the member of $\Omega$ above is $53$ then $X=5+3=8$.

Each subset of $\Omega$ is assigned a probability by assigning to each one-element subset the probability $1/36$. That way we have \begin{align} \Pr(X=5) & = 4/36 \\ \Pr(X=6) & = 5/36 \\ \Pr(X=7) & = 6/36 \\ \Pr(X=8) & = 5/36 \\ \Pr(X=9) & = 4/36 \\ \Pr(X=10) & = 3/36 \\ & \qquad \text{etc.} \end{align}

The idea is that the "outcome" is one member of $\Omega$ chosen randomly, and the value of the random variable is determined by the outcome. More than one random variable can be defined on the space $\Omega$; for example, let $Y:\Omega\to\{1,2,3,4,5,6\}$ be the maximum of the two numbers, so that we would have \begin{align} \Pr(Y=6) & = 11/36 \\ \Pr(Y=5) & = 9/36 \\ \Pr(Y=4) & = 7/36 \\ & \qquad \text{etc.} \end{align}

Another example would be this:

$\Omega = $ the set of all voters in a state.

$X(\omega) =$ the income of the voter named $\omega$

$Y(\omega) = \begin{cases} 1 & \text{if $\omega$ votes “yes'' in the referendum}, \\ 0 & \text{otherwise}. \end{cases}$

If one voter $\omega\in\Omega$ is randomly chosen, then $X$ and $Y$ are random variables. If $E\subseteq\Omega$ is the set of voters whose incomes are more than $\$1{,}000{,}000$ per year, then $\Pr(E)$ is the probability that the chosen voter $\omega$ is in the set $E$, i.e. that the chosen voter has an income more than $\$1{,}000{,}000$ per year.