Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

2.1. Distributions of Two Random Variables 91

x 2

x 1 12 p 1 (x 1 )

1 101 101 102

2 101 102 103

3 102 103 105

p 2 (x 2 ) 104 106

The joint probabilities have been summed in each row and each column and these
sums recorded in the margins to give the marginal probability mass functions ofX 1
andX 2 , respectively. Note that it is not necessary to have a formula forp(x 1 ,x 2 )
to do this.

We next consider the continuous case. LetDX 1 be the support ofX 1 .For

x 1 ∈DX 1 , Equation (2.1.7) is equivalent to

FX 1 (x 1 )=

∫x 1

−∞

∫∞

−∞

fX 1 ,X 2 (w 1 ,x 2 )dx 2 dw 1 =

∫x 1

−∞

{∫∞

−∞

fX 1 ,X 2 (w 1 ,x 2 )dx 2

}

dw 1.

By the uniqueness of cdfs, the quantity in braces must be the pdf ofX 1 ,evaluated

atw 1 ;thatis,

fX 1 (x 1 )=

∫∞

−∞

fX 1 ,X 2 (x 1 ,x 2 )dx 2 (2.1.9)

for allx 1 ∈DX 1. Hence, in the continuous case the marginal pdf ofX 1 is found by
integrating outx 2. Similarly, the marginal pdf ofX 2 is found by integrating out
x 1.

Example 2.1.5(Example 2.1.2, continued). Consider the vector of continuous
random variables (X, Y) discussed in Example 2.1.2. The space of the random
vector is the unit circle with center at (0,0) as shown in Figure 2.1.2. To find the
marginal distribution ofX,fixxbetween−1 and 1 and then integrate outyfrom
−

√

1 −x^2 to

√
1 −x^2 as the arrow shows on Figure 2.1.2. Hence, the marginal pdf
ofXis

fX(x)=

∫√ 1 −x 2

−

√

1 −x^2

1

π

dy=

2

π

√

1 −x^2 , − 1 <x< 1.

Although (X, Y) has a joint uniform distribution, the distribution ofXis unimodal

with peak at 0. This is not surprising. Since the joint distribution is uniform, from

Figure 2.1.2Xis more likely to be near 0 than at either extreme−1 or 1. Because

the joint pdf is symmetric inxandy,themarginalpdfofYis the same as that of

X.

Example 2.1.6.LetX 1 andX 2 have the joint pdf

f(x 1 ,x 2 )=

{

x 1 +x 2 0 <x 1 < 1 , 0 <x 2 < 1

0elsewhere.