Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

120 Multivariate Distributions

and only iff(x 1 ,x 2 )can be written as a product of a nonnegative function ofx 1

and a nonnegative function ofx 2. That is,

f(x 1 ,x 2 )≡g(x 1 )h(x 2 ),

whereg(x 1 )> 0 ,x 1 ∈S 1 , zero elsewhere, andh(x 2 )> 0 ,x 2 ∈S 2 , zero elsewhere.

Proof.IfX 1 andX 2 are independent, thenf(x 1 ,x 2 )≡f 1 (x 1 )f 2 (x 2 ), wheref 1 (x 1 )
andf 2 (x 2 ) are the marginal probability density functions ofX 1 andX 2 , respectively.
Thus the conditionf(x 1 ,x 2 )≡g(x 1 )h(x 2 ) is fulfilled.
Conversely, iff(x 1 ,x 2 )≡g(x 1 )h(x 2 ), then, for random variables of the contin-
uous type, we have

f 1 (x 1 )=

∫∞

−∞

g(x 1 )h(x 2 )dx 2 =g(x 1 )

∫∞

−∞

h(x 2 )dx 2 =c 1 g(x 1 )

and

f 2 (x 2 )=

∫∞

−∞

g(x 1 )h(x 2 )dx 1 =h(x 2 )

∫∞

−∞

g(x 1 )dx 1 =c 2 h(x 2 ),

wherec 1 andc 2 are constants, not functions ofx 1 orx 2 .Moreover,c 1 c 2 = 1 because

1=

∫∞

−∞

∫∞

−∞

g(x 1 )h(x 2 )dx 1 dx 2 =

[∫∞

−∞

g(x 1 )dx 1

][∫∞

−∞

h(x 2 )dx 2

]

=c 2 c 1.

These results imply that

f(x 1 ,x 2 )≡g(x 1 )h(x 2 )≡c 1 g(x 1 )c 2 h(x 2 )≡f 1 (x 1 )f 2 (x 2 ).

Accordingly,X 1 andX 2 are independent.

This theorem is true for the discrete case also. Simply replace the joint pdf by
the joint pmf. For instance, the discrete random variablesXandY of Example
2.4.1 are immediately seen to be dependent because the support of (X, Y)isnota
product space.
Next, consider the joint distribution of the continuous random vector (X, Y)
given in Example 2.1.3. The joint pdf is

f(x, y)=4xe−x

2

ye−y

2

,x> 0 ,y > 0.

which is a product of a nonnegative function ofxand a nonnegative function ofy.
Further, the joint support is a product space. Hence,X andY are independent
random variables.
Example 2.4.3.Let the pdf of the random variableX 1 andX 2 bef(x 1 ,x 2 )=
8 x 1 x 2 , 0 <x 1 <x 2 <1, zero elsewhere. The formula 8x 1 x 2 might suggest to some
thatX 1 andX 2 are independent. However, if we consider the spaceS={(x 1 ,x 2 ):
0 <x 1 <x 2 < 1 }, we see that it is not a product space. This should make it clear
that, in general,X 1 andX 2 must be dependent if the space of positive probability
density ofX 1 andX 2 is bounded by a curve that is neither a horizontal nor a
vertical line.