> [!definition] > > Let $X$ be a [[Random Variable|random variable]] on a [[Probability|probability]] space $(\Omega, \cf, \bp)$, then the **cumulative distribution function** $F_X: \real \to [0, 1]$ is given by > $ > F_X(r) = \bp\bracs{X \le r} > $ > and its [[Probability Distribution|probability distribution]] is the measure $\mu_X$ on $\cb(\real)$. > > Since all probability measures on $\real$ are bounded on bounded sets, $\mu_X$ is a [[Lebesgue-Stieltjes Measure]], and $F_X$ is its associated Stieltjes function. > [!theorem] > > Let $F$ be a CDF, then there exists a [[Random Variable|random variable]] $X: [0, 1] \to \real$ on the probability space $([0, 1], \cb([0, 1]), m)$ such that $F_X = F$, where $m$ is the [[Lebesgue Measure|Lebesgue measure]]. > > *Proof*. Define $G: [0, 1] \to \ol\real$ be defined by > $ > G(p) = \inf\bracs{x: F(x) \ge p} > $ > > ### Measurable > > $G$ is measurable. > > *Proof*. Since the set on the right shrink as $p$ increases, $G$ is non-decreasing. If $G(p) \le r$, then $G(q) \le r$ for all $q \le p$. This allows assigning > $ > G^{-1}([-\infty, r]) = [0, p] \text{ or }[0, p) \quad p = \sup_{G(q) \le r}q > $ > In both cases, $G$ sends measurable sets back to measurable sets. > > ### Right Inverse > > $F(r) \ge q$ if and only if $r \ge G(q)$. > > *Proof*. Suppose that $F(r) \ge q$, then $r \in \bracs{x: F(x) \ge q}$ and $r \ge G(q)$. > > Suppose that $r \ge G(q)$, then for any $\varepsilon > 0$ there exists $s < r + \varepsilon$ such that $F(s) \ge q$. Using this fact, the right continuity of $F$ and its monotonicity, $F(r) \ge q$ as well. > > ### Correspondence > > $F$ is the cumulative distribution function of $G$. > > *Proof*. > $ > \begin{align*} > \bp\bracs{G \le r} &= \bp\bracs{x: r \ge G(x)} \\ > &= \bp\bracs{q: F(r) \ge q} \\ > &= \bp[0, F(r)] = m([0, F(r)]) = F(r) > \end{align*} > $ > [!theorem] > > Let $\seq{F_n}$ be a sequence of CDFs, then there exists a sequence of *independent* random variables such that each $X_n$ has CDF $F_n$. > > In other words, there exists an *independent* coupling for any countable sequence of probability distributions. > > *Proof*. Let $A_n = \bigcup_{0 \le k < 2^n}\left(\frac{k}{2^n}, \frac{k + 1}{2^n}\right]$, and $R_n = \chi_{A_n}$. Enumerate the prime numbers as $\seq{p_n}$, and let > $ > U_n = \sum_{i \in \nat}\frac{R_{p_n^i}}{2^i} > $ > Then $\seq{U_n}$ are each uniform $[0, 1]$ random variables that are mutually independent. Apply the above technique and use the function $G$ on each $U_n$, then we have the desired independent sequence.