2.7. Transformations for Several Random Variables 147f(0) = 0? Then our newSisS={−∞<x<∞butx =0}.Wethentake
A 1 ={x:−∞<x< 0 }andA 2 ={x:0<x<∞}.Thusy=x^2 ,withthe
inversex=−
√
y,mapsA 1 ontoT ={y:0<y<∞}and the transformation is
one-to-one. Moreover, the transformationy=x^2 ,withinversex=
√
y,mapsA 2
ontoT ={y:0<y<∞}and the transformation is one-to-one. Consider the
probabilityP(Y ∈B), whereB⊂T.LetA 3 ={x:x=−
√
y, y∈B}⊂A 1 and
letA 4 ={x:x=
√
y, y∈B}⊂A 2 .ThenY∈Bwhen and only whenX∈A 3 or
X∈A 4 .Thuswehave
P(Y∈B)=P(X∈A 3 )+P(X∈A 4 )=∫A 3f(x)dx+∫A 4f(x)dx.In the first of these integrals, letx=−
√
y. Thus the Jacobian, sayJ 1 ,is− 1 / 2
√
y;
furthermore, the setA 3 is mapped ontoB. In the second integral letx=
√
y.Thus
the Jacobian, sayJ 2 ,is1/ 2
√
y; furthermore, the setA 4 is also mapped ontoB.
Finally,
P(Y∈B)=∫Bf(−√
y)∣
∣
∣
∣−1
2√
y∣
∣
∣
∣dy+∫Bf(√
y)1
2√
ydy=∫B[f(−√
y)+f(√
y)]1
2
√
y
dy.Hence the pdf ofY is given by
g(y)=1
2
√
y[f(−√
y)+f(√
y)],y∈T.Withf(x) the Cauchy pdf we have
g(y)={ 1
π(1+y)√y^0 <y<∞
0elsewhere.In the preceding discussion of a random variable of the continuous type, we had
two inverse functions,x=−
√
yandx=
√
y. That is why we sought to partition
S(or a modification ofS) into two disjoint subsets such that the transformation
y=x^2 maps each onto the sameT. Had there been three inverse functions, we
would have sought to partitionS (or a modified form ofS) into three disjoint
subsets, and so on. It is hoped that this detailed discussion makes the following
paragraph easier to read.
Letf(x 1 ,x 2 ,...,xn)bethejointpdfofX 1 ,X 2 ,...,Xn, which are random vari-
ables of the continuous type. LetSdenote then-dimensional space where this joint
pdff(x 1 ,x 2 ,...,xn)>0, and consider the transformationy 1 =u 1 (x 1 ,x 2 ,...,xn),
...,yn=un(x 1 ,x 2 ,...,xn), which mapsSontoT in they 1 ,y 2 ,...,ynspace. To
each point ofSthere corresponds, of course, only one point inT; but to a point
inT there may correspond more than one point inS. That is, the transformation