2.4. Independent Random Variables 119
SinceX+Y≤4, it would seem thatXandYare dependent. To see that this is
true by definition, we first find the marginal pmf’s which are:
pX(x)=
( 10
x
)( 15
4 −x
)
( 25
4
) , 0 ≤x≤4;
pY(y)=
( 8
y
)( 17
4 −y
)
( 25
4
) , 0 ≤y≤ 4.
To show dependence, we need to find only one point in the support of (X 1 ,X 2 )where
the joint pmf does not factor into the product of the marginal pmf’s. Suppose we
select the pointx=1andy= 1. Then, using R for calculation, we compute (to 4
places):
p(1,1) = 10· 8 ·
(
7
2
)
/
(
25
4
)
=0. 1328
pX(1) = 10
(
15
3
)
/
(
25
4
)
=0. 3597
pY(1) = 8
(
17
3
)
/
(
25
4
)
=0. 4300.
Since 0. 1328
=0.1547 = 0. 3597 · 0 .4300,XandYare dependent random variables.
Example 2.4.2.Let the joint pdf ofX 1 andX 2 be
f(x 1 ,x 2 )=
{
x 1 +x 2 0 <x 1 < 1 , 0 <x 2 < 1
0elsewhere.
We show that X 1 andX 2 are dependent. Here the marginal probability density
functions are
f 1 (x 1 )=
{∫∞
−∞f(x^1 ,x^2 )dx^2 =
∫ 1
0 (x^1 +x^2 )dx^2 =x^1 +
1
2 0 <x^1 <^1
0elsewhere,
and
f 2 (x 2 )=
{∫∞
−∞f(x^1 ,x^2 )dx^1 =
∫ 1
0 (x^1 +x^2 )dx^1 =
1
2 +x^20 <x^2 <^1
0elsewhere.
Sincef(x 1 ,x 2 )
≡f 1 (x 1 )f 2 (x 2 ), the random variablesX 1 andX 2 are dependent.
The following theorem makes it possible to assert that the random variablesX 1
andX 2 of Example 2.4.2 are dependent, without computing the marginal probability
density functions.
Theorem 2.4.1. Let the random variablesX 1 andX 2 have supportsS 1 andS 2 ,
respectively, and have the joint pdff(x 1 ,x 2 ).ThenX 1 andX 2 are independent if