# (PTE) 1.2. Distributions

In this subsection, we define random variables and distribution functions.

## Random variables

In measure theory, a function on a measurable space $A$ onto another measurable space $B$ is *measurable* if its inverse image of measurable sets are measurable sets. If $A$ is a probability space and $B$ is a Borel measurable space of $\mathbb{R}$, then we call this function a random variable.

We also say $X$ is *$\mathcal{ F}$- measurable* or write $X \in \mathcal{ F}$. That is, random variables are Borel measurable real-valued functions defined on a probability space.

(2) An indicator function $\mathbf{1}_A$, $A \in \mathcal{ F}$ is a random variable.

In undergraduate statistics, we used random variables as if they were values. Now we can understand true meaning behind these notations: \(P(-1\le X \le 1) = P(X^{-1}([-1, 1])) = P(\{\omega\in\Omega:~ X(\omega) \in [-1,1]\})\) and such. Random variables are defined as a measurable function so that their inverse images can be measured by the probability measure $P$.

An important fact is that every random variable *induces a probability measure on $(\mathbb{ R}, \mathcal{ B}(\mathbb{ R}))$*. (note that it is not on $(\Omega, \mathcal{ F})$.)

It is not difficult to show that such $P_X$ is a probability measure.

## Distribution functions

A distribution function of $X$ is defined in terms of probability.

$\Leftrightarrow$ $F(x) := P(X\le x) = P_X(-\infty, x],~ \forall x \in \mathbb{ R}$.

We already saw in section 1.1 that a distribution function uniquely determines a distribution (or a random variable). The following theorem implies that every function that satisfies some conditions can be regarded as a distribution function of some random variable.

## expand proof

($\Rightarrow$) is trivial.

($\Leftarrow$) we proof this by construction. Let $\Omega = [0,1]$, $\mathcal{ F} = \mathcal{B}([0,1])$, $P=\lambda$ (Lebesgue measure). Let $X(\omega) = \sup\{y:~ F(y)<\omega\}$. Then $X$ is a random variable. To show $P(X\le x) = P(\{\omega:~ 0\le\omega\le F(x)\}),~ \forall x$, it is equivalent to show that $\{\omega:~ X(\omega) \le x\} = \{\omega: \omega \le F(x)\},~ \forall x$.

First given $x$, $\omega_0 \in \{\omega:~ \omega \le F(x)\}$, we get $x \notin \{y:~ F(y) < \omega_0\}$. Thus $X(\omega_0) \le x$ and we get $\{\omega:~ X(\omega) \le x\} \supset \{\omega: \omega \le F(x)\}$. Next, suppose $\omega_0 \notin \{\omega: \omega \le F(x)\}$, then $\omega_0 > F(x)$. Since $F$ is right-continuous, there exists $\epsilon > 0$ such that $F(x) \le F(x+\epsilon) < \omega_0$. Hence $x<x+\epsilon\le X(\omega_0)$ and $\{\omega:~ X(\omega) \le x\} \subset \{\omega: \omega \le F(x)\}$. Finally we get $F(x) = \lambda(0, F(x)] = P(\{\omega:~ X(\omega) \le x\}) = P(X \le x)$.

*Acknowledgement*

This post is based on the textbook *Probability: Theory and Examples*, 5th edition (Durrett, 2019) and the lecture at Seoul National University, Republic of Korea (instructor: Prof. Johan Lim).