characteristic function

Let $X$ be a random variable. The characteristic function of $X$ is a function${\phi }_X:ℝ\to ℂ$ defined by

$${\phi }_X(t)=E{e}^{itX}=E\cos (tX)+iE\sin (tX),$$

that is, ${\phi }_X(t)$ is the expectation of the random variable $e^{itX}$.

Given a random vector $\overline{X}=(X_1,\mathrm{\dots },X_n)$, the characteristic function of$\overline{X}$, also called joint characteristic function of $X_1,\mathrm{\dots },X_n$,is a function ${\phi }_{\overline{X}}:{ℝ}^n\to ℂ$ defined by

$${\phi }_{\overline{X}}(t)=E{e}^{i\overline{t}\cdot \overline{X}},$$

where $\overline{t}=(t_1,\mathrm{\dots },t_n)$ and$\overline{t}\cdot \overline{X}=t_1{X}_1+\mathrm{\cdots }+t_n{X}_n$ (the dot product.)

Remark. If $F_X$ is the distribution function associated to $X$, by theproperties of expectation we have

$${\phi }_X(t)={\int }_{ℝ}{e}^{itx}𝑑F_X(x),$$

which is known as the Fourier-Stieltjes transform of $F_X$, and provides an alternatedefinition of the characteristic function. From this, it is clear that the characteristicfunction depends only on the distribution function of $X$, hence one can define the characteristicfunction associated to a distribution even when there is no random variable involved.This implies that two random variables with the same distribution must have the samecharacteristic function. It is also true that each characteristic function determinesa unique distribution; hence the , since it characterizes the distribution function (see property 6.)

Properties

1. The characteristic function is bounded by $1$, i.e. $|{\phi }_X(t)|\le 1$ for all $t$;

2. ${\phi }_X(0)=1$;

3. $\overline{{\phi }_X(t)}={\phi }_X(-t)$, where $\overline{z}$ denotes the complex conjugate of $z$;

4. ${\phi }_X$ is uniformly continuous in $ℝ$;

5. If $X$ and $Y$ are independent random variables, then ${\phi }_{X+Y}={\phi }_X{\phi }_Y$;

6. The characteristic function determines the distribution function; hence, ${\phi }_X={\phi }_Y$ if and only if $F_X=F_Y$.This is a consequence of the inversion : Given a random variable $X$with characteristic function $\phi$ and distribution function $F$, if $x$ and $y$are continuity points of $F$ such that , then

   $$F(x)-F(y)=\frac{1}{2\pi }\underset{s\to \mathrm{\infty }}{lim}{\int }_{-s}^s\frac{e^{-itx}-e^{-ity}}{it}\phi (t)𝑑t;$$

7. A random variable $X$ has a symmetrical distribution (i.e. one such that $F_X=F_{-X}$)if and only if ${\phi }_X(t)\in ℝ$ for all $t\in ℝ$;

8. For real numbers $a,b$, ${\phi }_{aX+b}(t)=e^{itb}{\phi }_X(at)$;

9. If , then ${\phi }_X$ has continuous $n$-th derivatives and

   $$\frac{d^k{\phi }_X}{d{t}^k}(t)={\phi }_X^{(k)}(t)={\int }_{ℝ}{(ix)}^k{e}^{itx}𝑑F_X(x),\mathrm{\hspace{0.33em}\hspace{0.33em}1}\le k\le n.$$

   Particularly, ${\phi }_X^{(k)}(0)=i^{k}E{X}^k$; characteristic functions are similar tomoment generating functions in this sense.

Similar properties hold for joint characteristic functions.Other important result related to characteristic functions is the Paul Lévycontinuity theorem.