Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

122 Multivariate Distributions

Example 2.4.4(Example 2.4.2, Continued).Independence is necessary for condi-

tion (2.4.2). For example, consider the dependent variablesX 1 andX 2 of Example

2.4.2. For these random variables, we have

P(0<X 1 <^12 , 0 <X 2 <^12 )=

∫ 1 / 2

0

∫ 1 / 2

0 (x^1 +x^2 )dx^1 dx^2 =

1

8 ,

whereas
P(0<X 1 <^12 )=

∫ 1 / 2

0 (x^1 +

1

2 )dx^1 =

3

8

and

P(0<X 2 <^12 )=

∫ 1 / 2

0 (

1

2 +x^1 )dx^2 =

3
8.
Hence, condition (2.4.2) does not hold.

Not merely are calculations of some probabilities usually simpler when we have
independent random variables, but many expectations, including certain moment
generating functions, have comparably simpler computations. The following result
proves so useful that we state it in the form of a theorem.

Theorem 2.4.4. SupposeX 1 andX 2 are independent and thatE(u(X 1 ))and
E(v(X 2 ))exist. Then

E[u(X 1 )v(X 2 )] =E[u(X 1 )]E[v(X 2 )].

Proof. We give the proof in the continuous case. The independence ofX 1 andX 2
implies that the joint pdf ofX 1 andX 2 isf 1 (x 1 )f 2 (x 2 ). Thus we have, by definition
of expectation,

E[u(X 1 )v(X 2 )] =

∫∞

−∞

∫∞

−∞

u(x 1 )v(x 2 )f 1 (x 1 )f 2 (x 2 )dx 1 dx 2

=

[∫∞

−∞

u(x 1 )f 1 (x 1 )dx 1

][∫∞

−∞

v(x 2 )f 2 (x 2 )dx 2

]

= E[u(X 1 )]E[v(X 2 )].

Hence, the result is true.

Upon taking the functionsu(·)andv(·) to be the identity functions in Theorem
2.4.4, we have that for independent random variablesX 1 andX 2 ,

E(X 1 X 2 )=E(X 1 )E(X 2 ). (2.4.3)

We next prove a very useful theorem about independent random variables. The
proof of the theorem relies heavily upon our assertion that an mgf, when it exists,
is unique and that it uniquely determines the distribution of probability.

Theorem 2.4.5.Suppose the joint mgf,M(t 1 ,t 2 ), exists for the random variables
X 1 andX 2 .ThenX 1 andX 2 are independent if and only if

M(t 1 ,t 2 )=M(t 1 ,0)M(0,t 2 );

that is, the joint mgf is identically equal to the product of the marginal mgfs.