# Is a random variable bijective?

Given a probability space $(\Omega, \mathit{F}, \mathbb{P})$, a random variable is defined as a function $X: \Omega \rightarrow \mathbb{R}$ such that the set $\{ \omega: X(\omega) \leq c \}$ is $\mathit{F}$-measurable.

I'm wondering whether the random variable $X$ has to be bijective (i.e. one-to-one and onto). If so, why? If not, how is it that many textbooks directly refer to the function $X^{-1}$.

-
Most properties of random variables make it only (sensible to consider it) well-defined up to a.e.-equivalence so it wouldn't really make sense to ask about bijectivity anyway. – Quinn Culver Sep 26 '12 at 1:51

Usually they're not bijective. Suppose, for example, that $X_1,X_2,X_3,\ldots$ is an infinite sequence of random variables, each with the same distribution, and $\Pr(X_1=0)=\Pr(X_1=1)=1/2$. Now consider just the first three. We have $$(X_1,X_2,X_3)=\begin{cases} (0,0,0) & \text{with probability }1/8 \\ (0,0,1) & \text{with probability }1/8 \\ (0,1,0) & \text{with probability }1/8 \\ (1,0,0) & \text{with probability }1/8 \\ (0,1,1) & \text{with probability }1/8 \\ (1,0,1) & \text{with probability }1/8 \\ (1,1,0) & \text{with probability }1/8 \\ (1,1,1) & \text{with probability }1/8 \end{cases}$$ Here you have eight disjoint subsets of $\Omega$, and in four of them, one has $X_1(\omega)=0$. So $X_1$ is certainly not one-to-one! Nowhere near it. And if one considers the first four random variables in this sequence, one further splits each of these eight subsets in half, getting 16 subsets, and so on. So there are infinitely many $\omega\in\Omega$ such that $X_1(\omega)=0$. Very far from bijective!

The set $X_1^{-1}(\{0\})$ is precisely $\{\omega\in\Omega : X_1(\omega)=0\}$. It's an infinite set.

Generally, $f^{-1}(A)$ is defined as $\{x : f(x)\in A\}$. So the notation $f^{-1}$ does not refer to an inverse function in this context.

Sometimes in set theory one uses a notation with square brackets, writing $f^{-1}[A]$, and similarly $f[A]= \{f(x): x\in\text{domain}\}$. The reason for that is that set theorists often deal with sets that are simultaneously a member and a subset of the domain, so that $f(A)$ differs from $f[A]$.

-

$X^{-1}$ is not the inverse. It is the inverse image.

-

No, it doesn't have to be bijective - take a constant function.

-