Lévy's Inversion Formula

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Theorem

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Let $X$ be a real-valued random variable on $\struct {\Omega, \Sigma, \Pr}$.

Let $ F_X$ be the distribution function of $X$.

Let $\phi : \R \to \C$ be the characteristic function of $X$.


Then for all $a < b$ such that:

$\map \Pr {X \in \set {a,b} } = 0$

we have:

\(\ds \map {F_X} b - \map {F_X} a\) \(=\) \(\ds \map \Pr {a < X \le b}\)
\(\ds \) \(=\) \(\ds \lim _{T \to \infty} \frac 1 {2 \pi} \int^T _{-T} \dfrac {e^{-ita} - e^{-itb} }{it} \map \phi t \rd t\)


Integrable Characteristic Function

Let $X$ be a real-valued random variable.

Let $P_X$ be the probability distribution of $X$.

Let $\phi : \R \to \C$ be the characteristic function of $X$.


Suppose that $\phi$ is Lebesgue integrable, i.e.:

$\ds \int _\R \cmod {\map \phi t} \rd t < + \infty$


Then $P_X$ is absolutely continuous with respect to the Lebesgue measure.

Moreover, the Radon-Nikodym derivative is given by:

$\ds \map g x := \dfrac 1 {2 \pi} \int_\R \map \phi t e^{- i t x} \rd t$


Proof

In fact, the first equality holds for all $a < b$, as:

\(\ds \map {F_X} b - \map {F_X} a\) \(=\) \(\ds \map \Pr {X \le b} - \map \Pr {X \le a}\) Definition of $F_X$
\(\ds \) \(=\) \(\ds \map \Pr {\set {X \le b} \setminus \set {X \le a} }\) Measure of Set Difference with Subset, as $\set {X \le a} \subseteq \set {X \le b}$
\(\ds \) \(=\) \(\ds \map \Pr {a < X \le b}\)


In the following, we shall show the second equality.

Let $\mu$ be the probability distribution of $X$.

Let $m$ be the Lebesgue measure on $\R$.

For $T > 0$ let $m_T$ be the restriction of $m$ to $\closedint {-T} T$, i.e.:

$\forall A \in \map \BB \R : \map {m_T} A := \map m { A \cap \closedint {-T} T}$

Let $\mu \times m_T$ be the product measure.


Then:

\(\ds \int^T _{-T} \dfrac {e^{-ita} - e^{-itb} }{it} \map \phi t \rd t\) \(=\) \(\ds \int^T _{-T} \int \dfrac {e^{-ita} - e^{-itb} }{it} e^{i t x} \rd \map \mu x \rd t\)
\(\ds \) \(=\) \(\ds \iint \dfrac {e^{it \paren {x - a} } - e^{it \paren {x - b} } }{it} \rd \map \mu x \rd \map {m_T} t\)
\(\ds \) \(=\) \(\ds \iint \map f {x,t} \rd \map \mu x \rd \map {m_T} t\) $(1)$

where:

$\ds \map f {x,t} := \dfrac {e^{it \paren {x - a} } - e^{it \paren {x - b} } }{it}$

The $f$ is essentially bounded with respect to $\mu \times m_T$ , since:

$\forall \struct {x, t} \in \R \times \R_{\ne 0} : \cmod {\dfrac {e^{it \paren {x - a} } - e^{it \paren {x - b} } }{it} } \le \dfrac 3 2$

by Bounds for Complex Exponential.

In particular, $f$ is $\mu \times m_T$-integrable, as $\mu \times m_T$ is finite.


Thus by Fubini's Theorem:

\(\ds (1)\) \(=\) \(\ds \int f \; \rd \paren {\mu \times m_T }\)
\(\ds \) \(=\) \(\ds \iint \map f {x,t} \rd \map {m_T} t \rd \map \mu x\)
\(\ds \) \(=\) \(\ds \int F_T \rd \mu\)

where:

\(\ds \map {F_T} x\) \(:=\) \(\ds \int \map f {x,t} \rd \map {m_T} t\)
\(\ds \) \(=\) \(\ds \int _{-T}^T \dfrac {e^{it \paren {x - a} } - e^{it \paren {x - b} } }{it} \rd t\)
\(\ds \) \(=\) \(\ds \int _{-T}^T \dfrac {e^{it \paren {x - a} } }{it} \rd t - \int _{-T}^T \dfrac { e^{it \paren {x - b} } }{it} \rd t\)

Observe:

\(\ds \int _{-T}^T \dfrac {e^{it \paren {x - a} } }{it} \rd t\) \(=\) \(\ds \int _{-T}^T \dfrac {\map \cos {t \paren {x-a} } + i \map \sin {t \paren {x-a} } }{it} \rd t\) Euler's Formula
\(\ds \) \(=\) \(\ds \int _{-T}^T \dfrac {\map \sin {t \paren {x-a} } }{t} \rd t\) Definite Integral of Odd Function
\(\ds \) \(=\) \(\ds 2 \int _0^T \dfrac {\map \sin {t \paren {x - a} } } t \rd t\) Definite Integral of Even Function
\(\ds \) \(=\) \(\ds 2 \; \map \sgn {x - a} \int _0^T \dfrac {\map \sin {t \size {x - a} } } t \rd t\) Sine Function is Odd
\(\ds \) \(=\) \(\ds 2 \; \map \sgn {x - a} \int _0^{T \size {x-a} } \dfrac {\sin s } s \rd s\) Integration by Substitution with $s = t \size {x-a}$
\(\ds \) \(=\) \(\ds 2 \; \map \sgn {x - a} \; \map \Si {T \size {x - a} }\)

where:

$\sgn : \R \to \set {-1, 0, 1}$ is the signum function
$\Si : \R \to \R$ is the sine integral function

Similarly:

$\ds \int _{-T}^T \dfrac {e^{it \paren {x - b} } }{it} \rd t = 2 \; \map \sgn {x - b} \; \map \Si {T \size {x - b} }$

Thus we have:

$\ds \map {F_T} x = 2 \; \map \sgn {x - a} \; \map \Si {T \size {x - a} } - 2 \; \map \sgn {x - b} \; \map \Si {T \size {x - b} }$

By Limit at Infinity of Sine Integral Function, for all $x \in \R \setminus \set {a,b}$:

\(\ds \lim _{T \mathop \to +\infty} \map {F_T} x\) \(=\) \(\ds \pi \paren {\map \sgn {x - a} - \map \sgn {x - b} }\)
\(\ds \) \(=\) \(\ds 2 \pi \; \map {\chi _{\hointl a b} } x\)

As $\map \mu {\set {a, b} } = 0$ by hypothesis, this means in particular:

$\ds \lim _{T \mathop \to +\infty} F_T = \chi _{\hointl a b}\quad$ $\mu$-almost surely

On the other hand, Sine Integral Function is Bounded:

$\ds \sup_{T > 0, \; x \in \R} \size {\map {F_T} x} \le 4 \norm \Si_\infty < +\infty$

Note that the constant function $4 \norm \Si_\infty$ is $\mu$-integrable.

Therefore:

\(\ds \lim _{T \mathop \to +\infty} \int F_T \rd \mu\) \(=\) \(\ds \int \lim _{T \mathop \to +\infty} F_T \rd \mu\) Lebesgue's Dominated Convergence Theorem
\(\ds \) \(=\) \(\ds 2 \pi \int \chi _{\hointl a b} \rd\mu\)
\(\ds \) \(=\) \(\ds 2 \pi \; \map \mu {\hointl a b}\)
\(\ds \) \(=\) \(\ds 2 \pi \; \map \Pr {X \in \hointl a b}\) Definition of Probability Distribution
\(\ds \) \(=\) \(\ds 2 \pi \; \map \Pr {a < X \le b}\)



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Source of Name

This entry was named for Paul Pierre Lévy.


Sources

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