# Density Func tions

An important concept is that of a density function. Density functions are usually associated to differentiable distribution functions:

Definition 1.13: The distribution of a random variable X is called absolutely continuous if there exists a nonnegative integrable function f, called the density function ofX, such that the distribution function F of X can be written as the (Lebesgue) integral F (x) = f (u)du. Similarly, the distribution of a random  vector X є Rk is called absolutely continuous if there exists a nonnegative integrable function f on Rk, called the joint density, such that the distribution function F of X can be written as the integral

where x = (x1,…, xk)T.

Thus, in the case F(x) = /TO f (u)du, the density function f (x) is the derivative of F (x) : f (x) = F'(x), and in the multivariate case F (x1 ,…,xk) = /—TO… /—TO f (u 1,…, uk )du1 …duk the joint density is f (xb •••, xk) = (9/9×0…(d/dxk)F(x1, …,xk).

The reason for calling the distribution functions in Definition 1.13 abso­lutely continuous is that in this case the distributions involved are absolutely continuous with respect to Lebesgue measure. See Definition 1.12. To see this, consider the case F(x) = /-TOf (u)du, and verify (Exercise) that the corre­sponding probability measure г is

li(B) = I f(x)dx, (1.27)

where the integral is now the Lebesgue integral over a Borel set B. Because the Lebesgue integral over a Borel set with zero Lebesgue measure is zero (Exercise), it follows that г(B) = 0 if the Lebesgue measure of B is zero.

For example, the uniform distribution (1.25) is absolutely continuous be­cause we can write (1.25) as F(x) = /-TO f (u)du with density f (u) = 1 for 0 < u < 1 and zero elsewhere. Note that in this case F(x) is not differen­tiable in 0 and 1 but that does not matter as long as the set of points for which the distribution function is not differentiable has zero Lebesgue mea­sure. Moreover, a density of a random variable always integrates to 1 be­cause 1 = limx^TOF(x) = /TO f (u)du. Similarly, for random vectors X є

Rk: TOTO o—TO ••• TOTO f (u1>… uk )du1 …duk =1

Note that continuity and differentiability of a distribution function are not sufficient conditions for absolute continuity It is possible to construct a contin­uous distribution function F(x) that is differentiable on a subset D c R, with RD a seo with Lebesgue measure zero, such that F'(x) = 0 on D, and thus in this case /TO F'(x)dx = 0. Such distributions functions are called singular. See Chung (1974,12-13) for an example of how to construct a singular distribution function on R and Chapter 5 in this volume for singular multivariate normal distributions.