When is $F(x) = P(X \leq x) = P(X^2 \leq x^2)$?

Question

When is $F(x) = P(X \leq x) = P(X^2 \leq x^2)$ true, where $F$ is the cumulative distribution function, $P$ is a measure, $X$ is a r.v. and $x$ a real number? Why is the condition $P(X \leq 0) = 0$ apparently necessary for the equality $P(X \leq x) = P(X^2 \leq x^2)$ to be true? Of course, $F(x) = P(X \leq x)$ is true by definition.

Apparently, $x^2 \leq y^2 \iff x \leq y$, for $x, y \in \mathbb{R}$, only if $x, y >= 0$, but this does not really explain that necessary condition. If they are both negative, that $\iff$ does not hold. Here's a simple counter-example. Let $x = -1$ and $y=-2$, then $(-1)^2 \leq (-2)^2 \iff -1 \leq -2$ is not true.

Maybe a good answer would also explain the notation $X \leq x$ in $P(X \leq x)$ and provide a precise definition of a random variable, which is actually a function (so $X \leq x$ should not make much sense, given that a function cannot be smaller or equal to a number).

Answer 1

For a precise definition, one should use this: $$P(X\leq x) = P(\{\omega\in\Omega:X(\omega)\leq x\}).$$ This set is the inverse image of a Borel set through a measureable function, so it is measureable.

As for an example where the equality you stated fails, let $X$ be a discrete random variable taking values $-2$ and $1$, each with probability $1/2$. In this case, $P(X\leq 1) = 1$, but $P(X^2\leq 1^2) = 1/2$, precisely because the function $f(x) = x^2$ is no longer monotonically increasing if $x$ is allowed to take on negative values.

To explain why this fact about numbers shows up in this probability context, notice that the definition of $P(X\leq x)$ given above in actuality reduces to statements about numbers, i.e. $X(\omega)\leq x$ and you consider the values of the random variable $\omega$ by $\omega$.

Answer 2

A random variable is a measutrable function $X$ from a probability space $(\Omega, \mathcal F, P)$ into $\mathbb R$. $(X\leq x)$ is an abbreviation for $\{\omega \in \Omega: X(\omega) \leq x\}$.

Note that $P(X\leq 0)=P(X^{2}\leq 0)=P(X=0)$. If we know that $P(X=0)=0$ then we can conclude that $P(X \leq 0)=0$. Otherwise we cannot conclude that $P(X \leq 0)=0$ is necessary condition. A counter-example, is obtained by taking $X(\omega)=0$ for all $\omega$. In this case $P(X \leq 0)=1$!

$F(x)=P(X \leq x)$ is a right-continuous non-decreasing function on the real line with limits $+1$ at $\infty$ and $0$ at $-\infty$. Let $F(x-)$ denote $\lim_{y<x, y \to x} F(y)$. This limit always exists. $P(X^{2}\leq x^{2})=P(|X| \leq |x|)=P(-|x| \leq X \leq |x|)=F(|x|)-F(-|x|-)$. Hence a necessary and sufficient condition for $P(X \leq x)=P(X^{2} \leq x^{2})$ is $F(x)=F(|x|)-F(-|x|-)$.