Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

146 Multivariate Distributions

Hence the joint pdf ofY 1 ,Y 2 ,Y 3 is

g(y 1 ,y 2 ,y 3 )=y 32 e−y^3 , (y 1 ,y 2 ,y 3 )∈T.

The marginal pdf ofY 1 is

g 1 (y 1 )=

∫ 1 −y 1

0

∫∞

0

y^23 e−y^3 dy 3 dy 2 =2(1−y 1 ), 0 <y 1 < 1 ,

zero elsewhere. Likewise the marginal pdf ofY 2 is

g 2 (y 2 )=2(1−y 2 ), 0 <y 2 < 1 ,

zero elsewhere, while the pdf ofY 3 is

g 3 (y 3 )=

∫ 1

0

∫ 1 −y 1

0

y^23 e−y^3 dy 2 dy 1 =

1

2

y^23 e−y^3 , 0 <y 3 <∞,

zero elsewhere. Becauseg(y 1 ,y 2 ,y 3 ) =g 1 (y 1 )g 2 (y 2 )g 3 (y 3 ),Y 1 ,Y 2 ,Y 3 are dependent
random variables.
Note, however, that the joint pdf ofY 1 andY 3 is

g 13 (y 1 ,y 3 )=

∫ 1 −y 1

0

y^23 e−y^3 dy 2 =(1−y 1 )y^23 e−y^3 , 0 <y 1 < 1 , 0 <y 3 <∞,

zero elsewhere. HenceY 1 andY 3 are independent. In a similar manner,Y 2 andY 3
are also independent. Because the joint pdf ofY 1 andY 2 is

g 12 (y 1 ,y 2 )=

∫∞

0

y^23 e−y^3 dy 3 =2, 0 <y 1 , 0 <y 2 ,y 1 +y 2 < 1 ,

zero elsewhere,Y 1 andY 2 are seen to be dependent.

We now consider some other problems that are encountered when transforming
variables. LetXhave the Cauchy pdf

f(x)=

1

π(1 +x^2 )

, −∞<x<∞,

and letY=X^2. We seek the pdfg(y)ofY. Consider the transformationy=x^2.
This transformation maps the space ofX,namelyS={x:−∞<x<∞},onto
T ={y:0≤y<∞}. However, the transformation is not one-to-one. To each
y ∈T, with the exception ofy = 0, there correspond two pointsx∈S.For
example, ify= 4, we may have eitherx=2orx=−2. In such an instance, we
representSas the union of two disjoint setsA 1 andA 2 such thaty=x^2 defines
a one-to-one transformation that maps each ofA 1 andA 2 ontoT.IfwetakeA 1
to be{x:−∞<x< 0 }andA 2 to be{x:0≤x<∞},weseethatA 1 is
mapped onto{y:0<y<∞},whereasA 2 is mapped onto{y:0≤y<∞},
and these sets are not the same. Our difficulty is caused by the fact thatx=0
is an element ofS. Why, then, do we not return to the Cauchy pdf and take