# 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.

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