144 Multivariate Distributionstogether with the inverse functionsx 1 =w 1 (y 1 ,y 2 ,...,yn),x 2 =w 2 (y 1 ,y 2 ,...,yn),...,xn=wn(y 1 ,y 2 ,...,yn)define a one-to-one transformation that mapsSontoT in they 1 ,y 2 ,...,ynspace
and, hence, maps the subsetAofSonto a subsetBofT. Let the first partial
derivatives of the inverse functions be continuous and let thenbyndeterminant
(called the Jacobian)J=∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
∂x 1
∂y 1∂x 1
∂y 2 ···∂x 1
∂yn
∂x 2
∂y 1∂x 2
∂y 2 ···∂x 2
∂yn
..
...
...
.
∂xn
∂y 1∂xn
∂y 2 ···∂xn
∂yn∣∣
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣not be identically zero inT.Then
∫
···∫Af(x 1 ,x 2 ,...,xn)dx 1 dx 2 ···dxn=∫
···∫Bf[w 1 (y 1 ,...,yn),w 2 (y 1 ,...,yn),...,wn(y 1 ,...,yn)]|J|dy 1 dy 2 ···dyn.Whenever the conditions of this theorem are satisfied, we can determine the joint pdf
ofnfunctions ofnrandom variables. Appropriate changes of notation in Section
2.2 (to indicaten-space as opposed to 2-space) are all that are needed to show
that the joint pdf of the random variablesY 1 =u 1 (X 1 ,X 2 ,...,Xn), ..., Yn=
un(X 1 ,X 2 ,...,Xn), where the joint pdf ofX 1 ,...,Xnisf(x 1 ,...,xn), is given by
g(y 1 ,y 2 ,...,yn)=f[w 1 (y 1 ,...,yn),...,wn(y 1 ,...,yn)]|J|,where (y 1 ,y 2 ,...,yn)∈T, and is zero elsewhere.
Example 2.7.1.LetX 1 ,X 2 ,X 3 have the joint pdf
f(x 1 ,x 2 ,x 3 )={
48 x 1 x 2 x 3 0 <x 1 <x 2 <x 3 < 1
0elsewhere.
(2.7.1)IfY 1 =X 1 /X 2 ,Y 2 =X 2 /X 3 ,andY 3 =X 3 , then the inverse transformation is given
by
x 1 =y 1 y 2 y 3 ,x 2 =y 2 y 3 ,andx 3 =y 3.
The Jacobian is given by
J=∣ ∣ ∣ ∣ ∣ ∣
y 2 y 3 y 1 y 3 y 1 y 2
0 y 3 y 2
001∣ ∣ ∣ ∣ ∣ ∣=y 2 y^23.Moreover, inequalities defining the support are equivalent to0 <y 1 y 2 y 3 ,y 1 y 2 y 3 <y 2 y 3 ,y 2 y 3 <y 3 ,andy 3 < 1 ,