Properties of Cumulative Distribution Function

Theorem

Let $\left({\Omega, \Sigma, \Pr}\right)$ be a probability space.

Let $X$ be a random variable on $\left({\Omega, \Sigma, \Pr}\right)$.

Let $F \left({X}\right)$ be the cumulative distribution function of $X$, that is:

$\forall x \in \R: F \left({X}\right) = \Pr \left({X \le x}\right)$

Then the following conditions apply to $F \left({X}\right)$:

Bounds of CDF

$0 \le F \left({X}\right) \le 1$

CDF is Increasing

$x_1 < x_2 \implies F \left({x_1}\right) \le F \left({x_2}\right)$

That is, $F$ is an increasing mapping.

Limits of CDF

$\displaystyle \lim_{x \to -\infty} F \left({x}\right) = 0, \lim_{x \to \infty} F \left({x}\right) = 1$

Proof

Proof of Bounds of CDF

This follows directly from the definition of $\Pr$.

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle S \in \Sigma\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle \varnothing \subseteq S \subseteq \Omega\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle 0 \le \Pr \left({S}\right) \le 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Probability Measure is Monotone

$\blacksquare$

Proof that CDF is Increasing

Suppose $x, y \in \R: x \le y$.

Let $X \left({\omega}\right) \le x$.

Then $X \left({\omega}\right) \le y$, and so:

$\left\{{\omega \in \Omega: X \left({\omega}\right) \le x}\right\} \subseteq \left\{{\omega \in \Omega: X \left({\omega}\right) \le y}\right\}$

Hence the result.

$\blacksquare$

Proof of Limits of CDF

As $x \to -\infty$, $\left({-\infty \, . \, . \, x}\right] \to \varnothing$.

So $X^{-1} \left({\left({-\infty \, . \, . \, x}\right]}\right) \to \varnothing$ and so $F \left({x}\right) \to 0$.

Similarly, as $x \to +\infty$, $\left({-\infty \, . \, . \, x}\right] \to \R$.

So $X^{-1} \left({\left({-\infty \, . \, . \, x}\right]}\right) \to \Omega$ and so $F \left({x}\right) \to 1$.

$\blacksquare$