Skorokhod Representation Theorem

• $(\Omega, \mathcal{F}, P) := ([0, 1], \mathcal{B}([0, 1]), P)$,
  • $\mathcal{B}([0, 1])$ borrel sigma algebra over $[0, 1]$,
  • $P$ is Lebesgue measure over $[0, 1]$
  • the triplet is probability space
• $F: \mathbb{R} \rightarrow [0, 1]$
  • non decreasing
  • $\lim_{x \rightarrow \infty}F(x) = 1$,
  • $\lim_{x \rightarrow -\infty}F(x) = 0$,
  • right continuous
• (a)

\[\begin{eqnarray} \omega \in \Omega, \quad A^{+}(\omega) & := & \{ z \in \mathbb{R} \mid F(z) > \omega \} \nonumber \\ \omega \in \Omega, \quad X^{+}(\omega) & := & \inf A(\omega) \nonumber \\ & = & \sup\{ y \in \mathbb{R} \mid F(y) \le \omega \} \nonumber \\ \omega \in \Omega, \quad A^{-}(\omega) & := & \{ z \in \mathbb{R} \mid F(z) \ge \omega \} \nonumber \\ \omega \in \Omega, \quad X^{-}(\omega) & := & \inf A^{-}(\omega) \nonumber \\ & = & \sup\{ y \in \mathbb{R} \mid F(y) < \omega \} \nonumber \end{eqnarray}\]

• (b) $X^{-1}$ has distribution function $F$,
• (c) $X^{+}$ has distribution function $F$, and

\[P(X^{+} = X^{-1}) = 1\]

proof

(a)

We first remember that

\[\begin{eqnarray} (z > X^{-}(\omega)) & \Rightarrow & (F(z) \ge \omega) . \nonumber \end{eqnarray}\]

Indeed, suppose that $F(z) < \omega$ for some $\omega$. There exists $z^{\prime} \in A^{-}(\omega)$ such that $X^{-}(\omega) < z^{\prime} < z$. By monotoncity, we have $F(X^{-}(\omega)) \le F(z^{\prime}) \le F(z) < \omega$. However $z^{\prime} \in A^{-}(\omega)$ contradicts $F(z^{\prime}) < \omega$.

We can take a sequence \(\{z_{n}\}_{n \in \mathbb{N}} \subset A^{-}(\omega)\) which converges to $X^{-}(\omega)$. By $F(z^{n}) \ge \omega$ and right continuity of $F$, we have

\[\begin{eqnarray} F(X^{-}(\omega)) \ge \omega . \nonumber \end{eqnarray}\]

Combining above, we have

\[(X^{-}(\omega) \le c) \Rightarrow ( \omega \le F(X^{-}(\omega)) \le F(c) ) .\]

Thus,

\[(\omega \le F(c)) \Leftrightarrow (X^{-}(\omega) \le c)\]

so that

\[P(X^{-} \le c) = F(c) .\]

$\Box$

Reference