Inverse Transform Method
Inverse transform method is also known as inversion method. The method generates random vairable with cumulative distribution $F$ from an uniformly distributed random variable.
One-dimensional case
- \((\Omega, \mathcal{F}, P)\),
- probability space
- \(F: \mathbb{R} \rightarrow [0, 1]\),
- 1-dim c.d.f.
$F$ is monotonically increasing so that this definition is well-defined.
Proposition1.
\(\begin{eqnarray} F^{-1}(u) \le y & \Leftrightarrow & u \le F(y), \end{eqnarray}\)
proof.
($\Rightarrow$) There is a sequence \(y_{n}\) such that \(y_{n} \searrow y^{*} := F^{-1}(y)\), \(F(y_{n}) \ge u\). Since $F$ is right continuous, \(F(y^{*}) \ge u\). \(F\) is monotone so that \(F(y^{*}) \le F(y)\).
($\Leftarrow$) \(u \le F(y)\) implies \(y \in \{x \in \mathbb{R} \mid F(x) \ge u\}\). Hence by definition of \(F^{-1}\), \(F^{-1}(u) \le y\).
Thereom2. Inversion method
Let $U$ be uniformly distributed random variable on $(0, 1)$ (i.e. \(U:\Omega \rightarrow [0, 1]\)). Then $F^{-1}(U)$ is a random variable with distribution $F$.
proof.
\(\begin{eqnarray} P(\{\omega \mid F^{-1}(U(\omega)) \le x\}) & = & P(\{\omega \mid U(\omega) \le F(x)\}) \nonumber & = & F(x) \nonumber \end{eqnarray}\)
Multi-dimensional case
- \((\Omega, \mathcal{F}, P)\),
- probability space
- \(X: \Omega \rightarrow \mathbb{R}^{n}\),
- $n$-dim r.v.
- \(\mathcal{B}(\mathbb{R}^{n})\),
- borel sets
- \(\mathcal{I}_{n}\),
- \(J_{n}: \mathbb{R}^{n} \rightarrow \mathcal{I}_{n}\),
- \(J_{n}(x) := (-\infty, x_{1}] \times \cdots \times (-\infty, x_{n}]\),
Cumulative distribution function of $X$ is defined as
\[x \in \mathbb{R}^{n}, \ F(x) := P(X^{-1}(J_{n}(x))) .\]\(\mathcal{I}_{n}\) is a $\pi$-system so that the RHS of the above equation uniquly determines probability measure over \((\mathbb{R}^{n}, \mathcal{B}(\mathbb{R}^{n}))\).
TODO: Validate the statements from here. $F$ is monotonically increasing so that we can define
\[\begin{eqnarray} \mathcal{A} & := & \{ I \in \mathcal{I}_{n} \mid \exists x \in \mathbb{R}^{n}, \ I = J_{n}(x), \ F(x) \ge u \} \\ F^{-1}(u) & := & \inf_{I \in \mathcal{A}} I \nonumber \\ & = & \bigcap_{I \in \mathcal{A}} I . \nonumber \end{eqnarray}\]where inf is taken as infimum of family of sets. The proposition1 is easily exntend to multi dimensional case.
Proposition3.
\[F^{-1}(u) \subseteq J_{n}(y) \Leftrightarrow u \le F(y)\]proof.
($\Rightarrow$) We can construct a sequence of sets converging to \(F^{-1}(u)\). Let \(\phi_{i}(J_{n}(x)) := x_{i}\) and
\[z_{i} := \inf \{ \phi_{i}(I) \in \mathbb{R} \mid I \in \mathcal{A} \} .\]We define \(z := (z_{1}, \ldots, z_{n})\). Then we observe
\[F^{-1}(u) = J_{n}(z) .\]Indeed, $\subseteq$ is obvious so that we will show $\supseteq$. To show that, it suffices to prove
\[\forall I \in \mathcal{A}, \ J_{n}(z) \subseteq I.\]Suppose that \(\exists I \in \mathcal{A}\) such that
\[\exists z^{\prime} := (z_{1}^{\prime}, \ldots, z_{n}^{\prime}) \in J_{n}(z), \ z^{\prime} \notin I .\]This implies that \(\exists z^{\prime}_{i} \in \mathbb{R}\) s.t. \(\phi_{i}(I) < z_{i}^{\prime} \le z_{i}\).
($\Leftarrow$) $u \le F(y)$ implies \(J_{n}(y) \in \mathcal{A}\). That is \(F^{-1}(u) \subseteq J_{n}(y)\).
Theorem4. Inversion Method
Let $U$ be uniformly distributed random variable on $(0,1)$ (i.e. \(U:\Omega \rightarrow (0,1)\)). Then \(F^{-1}(U)\) is a random variable with distribution $F$.
proof.
Let $X := F^{-1}(U)$. Then by proposition3,
\[\begin{eqnarray} P(X^{-1}(J_{n}(x))) & = & P(\{\omega \mid F^{-1}(U(\omega)) \subseteq J_{n}(x) \}) \nonumber \\ & = & P(\{\omega \mid U(\omega) \le F(x) \}) \nonumber \\ & = & F(x) \nonumber \end{eqnarray}\]Example1 uniform distribution over unit ball
\[\begin{eqnarray} h: (-1, 1) \rightarrow \mathbb{R}, \quad h(x) & := & \mathrm{tanh}^{-1}(x) \nonumber \\ C_{a, b} & := & (a, b)^{n} \subseteq \mathbb{R}^{n} \nonumber \end{eqnarray} .\] \[\begin{eqnarray} \phi_{0}: C_{0, 1} \rightarrow C_{-1, 1}, \quad \phi_{0}(x) & := & 2x - 1 \nonumber \\ \phi_{1}: C_{-1, 1} \rightarrow \mathbb{R}, \quad \phi_{1}(x_{1}, \ldots, x_{n}) & := & ( h(x_{1}), \cdots h(x_{n}) ) \nonumber \\ \phi_{2}: \mathbb{R} \rightarrow B_{0, 1}, \quad \phi_{2}(x) & := & \frac{ x }{ \sqrt{ 1 + |x|^{2} } } \nonumber \\ \phi: C_{0, 1} \rightarrow B_{0, 1}, \quad \phi(x_{1}, \ldots, x_{n}) & := & \phi_{2}(\phi_{1}(\phi_{0}(x_{1}, \ldots, x_{n}))) . . \end{eqnarray} .\]$\phi$ is bijection. $\phi^{-1}: real$ is continuous.