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STATLECT > Additional topics in probability theory

Joint characteristic function

In the lecture entitled Characteristic function we have introduced the concept of characteristic function (cf) of a random variable. This lecture is about the joint cf, a concept which is analogous, but applies to random vectors.

Definition_ Let X be a Kx1 random vector. The joint characteristic function of X is a function [eq1] defined by:[eq2]where $i=sqrt{-1}$ is the imaginary unit.

Observe that [eq3] exists for any $tin U{211d} ^{K}$, because[eq4]and the expected values appearing in the last line are well-defined, because both the sine and the cosine are bounded (they take values in the interval [eq5]).

Deriving cross-moments

Like the joint moment generating function of a random vector, the joint cf can be used to derive the cross-moments of X, as stated below:

Proposition_ Let X be a random vector and [eq6] its joint characteristic function. Let $nin U{2115} $. Define a cross-moment of order n as follows:[eq7]where [eq8] and [eq9]. If all cross-moments of order n exist and are finite, then all the n-th order partial derivatives of [eq10] exist and [eq11]where the partial derivative on the right-hand side of the equation is evaluated at the point $t_{1}=0$, $t_{2}=0$, ..., $t_{K}=0$.

nav_button Proof

See Ushakov, N. G. (1999) Selected topics in characteristic functions, VSP.

When we need to derive a cross-moment of a random vector, the practical usefulness of this proposition is somewhat limited, because it is seldom known, a priori, whether cross-moments of a given order exist or not. The following proposition, instead, does not require such a priori knowledge:

Proposition_ Let X be a random vector and [eq12] its joint cf. If all the n-th order partial derivatives of [eq13] exist, then:

  1. if n is even, for any [eq14] all $m$-th cross-moments of X exist and are finite;

  2. if n is odd, for any [eq15] all $m$-th cross-moments of X exist and are finite.

In both cases:[eq16]where the partial derivatives on the right-hand sides of the equations above are evaluated at the point $t_{1}=0$, $t_{2}=0$, ..., $t_{K}=0$.

nav_button Proof

Again, see Ushakov, N. G. (1999) Selected topics in characteristic functions, VSP.

Characterizing joint distributions

The joint cf can also be used to check whether two random vectors have the same distribution.

Proposition_ Let X and Y be two Kx1 random vectors. Denote by [eq17] and [eq18] their joint distribution functions and by [eq19] and [eq20] their joint cfs. Then:[eq21]

nav_button Proof

See e.g. Ushakov, N. G. (1999) Selected topics in characteristic functions, VSP.

Stated differently, two random vectors have the same distribution if and only if they have the same joint cf. This result is frequently used in applications, because demonstrating equality of two joint cfs is often much easier than demonstrating equality of two joint distribution functions.

More details

Joint cf of a linear transformation

Let X be a Kx1 random vector with characteristic function [eq22]. Define:[eq23]where A is a $L imes 1$ constant vector and $B$ is a $L imes K$ constant matrix. Then, the joint cf of Y is:[eq24]

nav_button Proof

This is proved as follows:[eq25]

Joint cf of a random vector with independent entries

Let X be a Kx1 random vector. Let its entries X_1, ..., $X_{K}$ be K mutually independent random variables. Denote the cf of the $j$-th entry of X by [eq26].

Then, the joint cf of X is:[eq27]

nav_button Proof

This is demonstrated as follows:[eq28]

Joint cf of a sum of mutually independent random vectors

Let X_1, ..., X_n be n mutually independent random vectors. Let Z be their sum:[eq29]Then, the joint cf of Z is the product of the joint cfs of X_1, ..., X_n:[eq30]

nav_button Proof

Similar to the previous proof:[eq31]

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