2.6. Extension to Several Random Variables 135
and (b) its integral over all real values of its argument(s) is 1. Likewise, a point
functionpessentially satisfies the conditions of being a joint pmf if (a)pis defined
and is nonnegative for all real values of its argument(s) and (b) its sum over all real
values of its argument(s) is 1. As in previous sections, it is sometimes convenient
to speak of the support set of a random vector. For the discrete case, this would be
all points inDthat have positive mass, while for the continuous case these would
be all points inDthat can be embedded in an open set of positive probability. We
useSto denote support sets.
Example 2.6.1.Let
f(x, y, z)=
{
e−(x+y+z) 0 < x,y,z <∞
0elsewhere
be the pdf of the random variablesX,Y,andZ. Then the distribution function of
X,Y,andZis given by
F(x, y, z)=P(X≤x, Y≤y, Z≤z)
=
∫z
0
∫y
0
∫x
0
e−u−v−wdudvdw
=(1−e−x)(1−e−y)(1−e−z), 0 ≤x, y, z <∞,
and is equal to zero elsewhere. The relationship (2.6.2) can easily be verified.
Let (X 1 ,X 2 ,...,Xn) be a random vector and letY =u(X 1 ,X 2 ,...,Xn)for
some functionu. As in the bivariate case, the expected value of the random variable
exists if then-fold integral
∫∞
−∞
···
∫∞
−∞
|u(x 1 ,x 2 ,...,xn)|f(x 1 ,x 2 ,...,xn)dx 1 dx 2 ···dxn
exists when the random variables are of the continuous type, or if then-fold sum
∑
xn
···
∑
x 1
|u(x 1 ,x 2 ,...,xn)|p(x 1 ,x 2 ,...,xn)
exists when the random variables are of the discrete type. If the expected value of
Yexists, then its expectation is given by
E(Y)=
∫∞
−∞
···
∫∞
−∞
u(x 1 ,x 2 ,...,xn)f(x 1 ,x 2 ,...,xn)dx 1 dx 2 ···dxn (2.6.3)
for the continuous case, and by
E(Y)=
∑
xn
···
∑
x 1
u(x 1 ,x 2 ,...,xn)p(x 1 ,x 2 ,...,xn) (2.6.4)
for the discrete case. The properties of expectation discussed in Section 2.1 hold
for then-dimensional case also. In particular,Eis a linear operator. That is, if