1.8. Expectation of a Random Variable 63

Because

∑

x∈SX|g(x)|pX(x) converges, it follows by a theorem in calculus

(^6) that
any rearrangement of the terms of the series converges to the same limit. Thus we
have,
∑
x∈SX
|g(x)|pX(x)=
∑
y∈SY
∑
{x∈SX:g(x)=y}
|g(x)|pX(x) (1.8.4)

∑
y∈SY
|y|
∑
{x∈SX:g(x)=y}
pX(x) (1.8.5)

∑
y∈SY
|y|pY(y), (1.8.6)
whereSY denotes the support ofY.SoE(Y) exists; i.e.,
∑
x∈SXg(x)pX(x)con-
verges. Because
∑
x∈SXg(x)pX(x) converges and also converges absolutely, the
same theorem from calculus can be used to show that the above equations (1.8.4)–
(1.8.6) hold without the absolute values. Hence,E(Y)=
∑
x∈SXg(x)pX(x),which
is the desired result.
The following two examples illustrate this theorem.
Example 1.8.4.LetYbe the discrete random variable discussed in Example 1.6.3
and letZ=e−Y.Since(2e)−^1 <1, we have by Theorem 1.8.1 that
E[Z]=E
[
e−Y
]

∑∞
y=0
e−y
(
1
2
)y+1
= e
∑∞
y=0
(
1
2
e−^1
)y+1

e
1 −(1/(2e))

2 e^2
2 e− 1
.
Example 1.8.5.LetXbe a continuous random variable with the pdff(x)=2x
which has support on the interval (0,1). SupposeY=1/(1+X). Then by Theorem
1.8.1, we have
E(Y)=
∫ 1
0
2 x
1+x
dx=
∫ 2
1
2 u− 2
u
du=2(1−log 2),
where we have used the change in variableu=1+xin the second integral.
Theorem 1.8.2 shows that the expectation operatorEis a linear operator.
Theorem 1.8.2. Letg 1 (X)andg 2 (X)be functions of a random variableX.Sup-
pose the expectations ofg 1 (X)andg 2 (X)exist. Then for any constantsk 1 andk 2 ,
the expectation ofk 1 g 1 (X)+k 2 g 2 (X)exists and it is given by
E[k 1 g 1 (X)+k 2 g 2 (X)] =k 1 E[g 1 (X)] +k 2 E[g 2 (X)]. (1.8.7)
(^6) For example, see Chapter 2 on infinite series inMathematical Comments, referenced in the
Preface.