104 Multivariate Distributions
x 2
x 1 = 0x 1 = 1
x 2 = 1
(0, 0) x 2 = 0
x 1
S
Figure 2.2.2:The support of (X 1 ,X 2 ) of Example 2.2.3.
ofT:
x 1 =0 into 0=^12 (y 1 +y 2 )
x 1 =1 into 1=^12 (y 1 +y 2 )
x 2 =0 into 0=^12 (y 1 −y 2 )
x 2 =1 into 1=^12 (y 1 −y 2 ).
Accordingly,T is shown in Figure 2.2.3. Next, the Jacobian is given by
J=
∣ ∣ ∣ ∣ ∣ ∣ ∣
∂x 1
∂y 1
∂x 1
∂y 2
∂x 2
∂y 1
∂x 2
∂y 2
∣ ∣ ∣ ∣ ∣ ∣ ∣
=
∣ ∣ ∣ ∣ ∣ ∣
1
2
1
2
1
2 −
1
2
∣ ∣ ∣ ∣ ∣ ∣
=−
1
2
.
Although we suggest transforming the boundaries ofS, others might want to
use the inequalities
0 <x 1 <1and0<x 2 < 1
directly. These four inequalities become
0 <^12 (y 1 +y 2 )<1and0<^12 (y 1 −y 2 )< 1.
It is easy to see that these are equivalent to
−y 1 <y 2 ,y 2 < 2 −y 1 ,y 2 <y 1 y 1 − 2 <y 2 ;
and they define the setT.
Hence, the joint pdf of (Y 1 ,Y 2 )isgivenby
fY 1 ,Y 2 (y 1 ,y 2 )=
{
fX 1 ,X 2 [^12 (y 1 +y 2 ),^12 (y 1 −y 2 )]|J|=^12 (y 1 ,y 2 )∈T
0elsewhere.