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The definition from my lecture notes of a random variable is as follows:

$\textbf{Definition} \hspace{2mm} \text{A random variable is a map X}:\Omega \rightarrow \mathbb{R} \text{ such that for any} A \in \mathscr{B}(\mathbb{R}),$ $$X^{-1}(A) = \{\omega \in \Omega:X(\omega)\in A\} \in \mathscr{F}.$$

Surely, for example, if we take X to be the number of even numbers in two rolls of a die and we take some arbitrary interval in the set of Borel $\sigma$-algebras of $\mathbb{R}$, such as $(-1,2)$, not every member of this set can be mapped back to a sample point? What would $X^{-1}(-1,2)$ represent?

Edit: Also, what is $X(\emptyset)$?

nonuser
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python_learner
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1 Answers1

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There are two misconceptions about set theory here. This is unrelated to probability theory.

First, the preimage of a set $S$ by a function $f:E\to F$ is the set

$$f^{-1}(S)=\{x |x\in E \wedge f(x)\in S\}$$

That is, not every element $s$ of $S$ needs to have a preimage $x$ such that $f(x)=s$. But those which do yield all such $x$ in the preimage of $S$.

Second point, $X$ is a function from the set $\Omega$ to $\Bbb R$. Unless $\Omega$ contains the empty set as an element, $X(\emptyset)$ is meaningless. For instance, since you consider two rolls of a die, $\Omega$ is a set of pairs of integers $(i,j)$ such that $1\leq i\leq 6$ and $1\leq j\leq 6$. Thus $\emptyset\notin \Omega$. Of course $\emptyset\subset\Omega$, but it's not related to the existence of $X(\emptyset)$.

Another problem: you didn't define $\mathscr F$. This has to be a $\sigma$-algebra on $\Omega$. For a countable set $\Omega$, you may take $\mathscr F$ to be the set of all subsets of $\Omega$.

Jean-Claude Arbaut
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