Proofs of the Inversion Formulas for Characteristic Functions

In order to study the properties of characteristic functions, we require the following basic facts concerning the conditions under which various limiting operations may be interchanged with the expectation operation. These facts are stated here without proof (for proof see any text on measure theory or modern integration theory).

We state first a theorem dealing with the conditions under which, given a convergent sequence of functions $g_n(\cdot )$ , the limit of expectations is equal to the expectation of the limit.

In particular, it may happen that (5.2) will hold with $G(x)$ equal for all $x$ to a finite constant $C$ . Since $E[C]=C$ is finite, it follows that (5.3) will hold. Since this is a case frequently encountered, we introduce a special terminology for it: the sequence of functions $g_n(\cdot )$ is said to converge boundedly to $g(\cdot )$ if (5.1) holds and if there exists a finite constant $C$ such that 

\left|g_{n}(x)\right| \leq C \quad \text { for all real } x \text { and integers } n. \tag{5.4} 

From theorem 5A it follows that (5.3) will hold for a sequence of functions converging boundedly. This assertion is known as the Lebesgue bounded convergence theorem. Theorem 5A is known as the Lebesgue dominated convergence theorem.

Theorem 5A may be extended to the case in which there is a function of two real variables $g(x,u)$ instead of a sequence of functions $g_n(x)$ .

Note that (5.5) says that $g(x,u)$ is continuous as a function of $u$ at each $x$ . If a Borel function $G(x)$ exists such that

|g(x, u)| \leq G(x) \quad \text { for all real } x \text { and } u \tag{5.6} 

and if $E[G(X)]$ is finite, then for any real number $u$ 

\lim _{u^{\prime} \rightarrow u} E\left[g\left(X, u^{\prime}\right)\right]=E[g(X, u)]. \tag{5.7} 

Note that (5.7) says that $E[g(X,u)]$ is continuous as a function of $u$ .

We next consider the problem of differentiating and integrating a function of the form of $E[g(X,u)]$ .

As one consequence of theorem 5C, we may deduce (2.10).

It should be noted that the integrals in (5.11) involving integration in the variable $u$ may be interpreted as Riemann integrals if we assume that (5.5) holds. However, the assertion (5.11) is valid even without assuming (5.5) if we interpret the integrals in $u$ as Lebesgue integrals.

Finally, we give a theorem, analogous to theorem 5A, for Lebesgue integrals over the real line.

Theorem 5E, like theorem 5A, is a special case of a general result of the theory of abstract Lebesgue integrals, called the Lebesgue dominated convergence theorem.

We next discuss the proofs of the inversion formulas for characteristic functions. In writing out the proofs, we omit the subscript $X$ on the distribution function $F_X(\cdot )$ and the characteristic function ${\varphi }_X(\cdot )$ .

We first prove (3.13). We note that \begin{align} \frac{1}{2 U} \int_{-U}^{U} e^{-i u x} \phi(u) d u & =\frac{1}{2 U} \int_{-U}^{U} d u\left[\int_{-\infty}^{\infty} e^{i u(y-x)} d F(y)\right] \\ & =\int_{-\infty}^{\infty} d F(y) \frac{1}{2 U} \int_{-U}^{U} \; du e^{i u(y-x)}, \end{align}in which the interchange of the order of integration is justified by theorem $5D$ . Now define the functions \begin{align} g(y, U) & = \begin{cases} \frac{1}{2 U} \displaystyle\int_{-U}^{U} e^{i u(y-x)} \, du = \frac{\sin U(y-x)}{U(y-x)}, & \text{if } y \neq x \\ 1, & \text{if } y = x. \end{cases} \\[2mm] g(y) & = \begin{cases} 0, & \text{if } y \neq x \\ 1, & \text{if } y = x. \end{cases} \end{align} 

Clearly, at each $y,g(y,U)$ converges boundedly to $g(y)$ as $U$ tends to $\mathrm{\infty }$ . Therefore, by theorem $5\mathrm{A}$ , \begin{align} \lim _{U \rightarrow \infty} \frac{1}{2 U} \int_{-U}^{U} e^{-i u x} \phi(u) d u & =\lim _{U \rightarrow \infty} \int_{-\infty}^{\infty} g(y, U) d F(y) \\ & =\int_{-\infty}^{\infty} g(y) d F(y)=F(y+0)-F(y-0). \end{align}

We next prove (3.12). It may be verified that $$\mathrm{Im}[e^{-iux}\varphi (u)]=E[\sin u(X-x)]$$ 

for any real numbers $u$ and $x$ . Consequently, for any  

\frac{2}{\pi} \int_{1 / U}^{U} \frac{\operatorname{Im}\left[e^{-i u x} \phi(u)\right]}{u} d u=\int_{-\infty}^{\infty} d F(y) \frac{2}{\pi} \int_{1 / U}^{U} \frac{\sin u(y-x)}{u} du, \tag{5.15} 

in which the interchange of integrals in (5.15) is justified by theorem 5D. Now it may be proved that \begin{align} \lim _{U \rightarrow \infty} \frac{2}{\pi} \int_{1 / U}^{U} \frac{\sin u t}{u} d u & = \begin{cases} 1, & \text{if } t > 0 \\ 0, & \text{if } t = 0 \\ -1, & \text{if } t < 0, \end{cases} \tag{5.16} \end{align}in which the convergence is bounded for all $U$ and $t$ .

A proof of (5.16) may be sketched as follows. Define

$$G(a)={\int }_0^{\mathrm{\infty }}{e}^{-au}\frac{\sin ut}{u}du.$$ 

Verify that the improper integral defining $G(a)$ converges uniformly for $a\ge 0$ and that this implies that

$${\int }_0^{\mathrm{\infty }}\frac{\sin ut}{u}du=\underset{a\to 0+}{lim}G(a).$$ 

Now

\int_{0}^{\infty} e^{-a u} \cos u t d u=\frac{a}{a^{2}+t^{2}}, \tag{5.17} 

in which, for each $a$ the integral in (5.17) converges uniformly for all $t$ . Verify that this implies that $G(a)={\tan }^{-1}(t/a)$ , which, as $a$ tends to 0, tends to $\pi /2$ or to $-\pi /2$ , depending on whether or . The proof of (5.16) is complete.

Now define \begin{align} g(y) & = \begin{cases} -1, & \text{if } y < x \\ 0, & \text{if } y = x \\ 1, & \text{if } y > x. \end{cases} \end{align}By (5.16), it follows that the integrand of the integral on the right-hand side of (5.15) tends to $g(y)$ boundedly as $U$ tends to $\mathrm{\infty }$ . Consequently, we have proved that

$$\frac{2}{\pi }{\int }_0^{\mathrm{\infty }}\frac{\mathrm{Im}(e^{-iux}\varphi (u))}{u}du={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}g(y)dF(y)=1-2F(x).$$ 

The proof of (3.12) is complete.

We next prove (3.4). We have

\begin{align} & \int_{-U}^{U}\left(1-\frac{|u|}{U}\right) \gamma(u) \phi(u) d u \tag{5.18}\\ &=\int_{-\infty}^{\infty} d F(x) \int_{-U}^{U} d u e^{i u x}\left(1-\frac{|u|}{U}\right) \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{-i u y} g(y) d y \\ &=\int_{-\infty}^{\infty} d F(x) \int_{-\infty}^{\infty} d y g(y) U K[U(x-y)] \end{align}

in which we define the function $K(\cdot )$ for any real number $z$ by

K(z)=\frac{1}{2 \pi}\left(\frac{\sin (z / 2)}{z / 2}\right)^{2}=\frac{1}{\pi} \int_{0}^{1} d v(1-v) \cos v z ; \tag{5.19} 

(5.18) follows from the fact that

\begin{align} \frac{1}{2 \pi} \int_{-U}^{U} d u e^{i u(x-y)}\left(1-\frac{|u|}{U}\right) & =\frac{U}{2 \pi} \int_{-1}^{1} d v e^{i v U(x-y)}(1-|v|) \\ & =\frac{U}{\pi} \int_{0}^{1}(1-v) \cos v U(x-y) dv. \end{align}

To conclude the proof of (3.4), it suffices to show that

g_{U}(x)=\int_{-\infty}^{\infty} d y g(y) U K[U(x-y)] \tag{5.20} 

converges boundedly to $g^{\ast }(x)$ as $U$ tends to $\mathrm{\infty }$ . We now show that this holds, using the facts that $K(\cdot )$ is even, nonnegative, and integrates to 1; in symbols, for any real number $u$ 

K(-u)=K(u), \quad K(u) \geq 0, \quad \int_{-\infty}^{\infty} K(u)du=1, \tag{5.21} 

In other words, $K(\cdot )$ is a probability density function symmetric about 0.

In (5.20) make the change of variable $t=y-x$ . Since $K(\cdot )$ is even, it follows that

g_{U}(x)=\int_{-\infty}^{\infty} g(x+t) U K(U t) dt. \tag{5.22} 

By making the change of variable t^{\prime}=-t in (5.22) and again using the fact that $K(\cdot )$ is even, we determine that

g_{U}(x)=\int_{-\infty}^{\infty} g(x-t) U K(U t) dt. \tag{5.23} 

Consequently, by adding (5.22) and (5.23) and then dividing by 2, we show that

g_{U}(x)=\int_{-\infty}^{\infty} d t U K(U t) \frac{g(x+t)+g(x-t)}{2}. \tag{5.24} 

Define $h(t)=\frac{g(x+t)+g(x-t)}{2-g^{\ast }(x)}$ . From (5.24) it follows that

g_{U}(x)-g^{*}(x)=\int_{-\infty}^{\infty} d t U K(U t) h(t). \tag{5.25} 

Now let $C$ be a constant such that $2|g(y)|\le C$ for any real number $y$ . Then, for any positive number $d$ and for all $U$ and $x$ 

\begin{align} & \left|g_{U}(x)-g^{*}(x)\right| \leq \sup _{|t| \leq d}|h(t)| \int_{|t| \leq d} U K(U t) d t \tag{5.26}\\ & \quad+\sup _{|t|>d}|h(t)| \int_{|t| \geq d} U K(U t) d t \leq \sup _{|l| \leq d}|h(t)|+C \int_{|s| \geq U d} K(s) d s. \end{align}

For $d$ fixed ${\displaystyle {\int }_{|s|\ge Ud}K(s)ds}$ tends to 0 as $U$ tends to $\mathrm{\infty }$ . Next, by the definition of $h(t)$ and $g^{\ast }(t)$ , $\underset{|t|\le d}{sup}|h(t)|$ tends to 0 as $d$ tends to 0. Consequently, by letting first $U$ tend to infinity and then $d$ tend to 0 in (5.26), it follows that $g_U(x)$ tends boundedly to $g^{\ast }(x)$ as $U$ tends to $\mathrm{\infty }$ . The proof of (3.4) is complete.