Multivariate Transformations Not One-to-One

With the definitions of $X_1,X_2,\ldots,X_n, U_1,U_2,\ldots,U_n$ as in section 3.3, suppose now that to each point of ${\cal A}$ there corresponds exactly one point of ${\cal B}$, but that to each point of ${\cal B}$ there may correspond more than one point of ${\cal A}$.

Assume that we can represent ${\cal A}$ as the union of a finite number, $k$, of disjoint sets ${\cal A}_1,\,{\cal A}_2,\, \dots ,\, {\cal A}_k$, such that (2.4) does represent a one-to-one mapping of each ${\cal A}_j$ onto ${\cal B}, \ j=1,\,\dots ,\,k$. That is, for each transformation of ${\cal A}_j$ onto ${\cal B}$ there is a unique inverse transformation

$$x_i =G_{ij}(u_1,\,u_2,\,\dots ,\,u_n), \ \ i=1,\,2,\, \dots ,\,n; \ \ j=1,\,2,\,\dots ,\,k,$$

each having a non-vanishing Jacobian, $\vert J_j\vert, \ j=1,\,2,\,\dots ,\,k$. The joint pdf of $U_1,\,U_2,\,\dots ,\,U_n$ is then given by

$$\sum^k_{j=1}\mbox{abs}\vert J_j\vert f\left[G_{1j}(u_1,\,\dots ,\,u_n),\,\dots ,\,G_{nj}(u_1,\,\dots ,\,u_n)\right]$$

for $(u_1,\,u_2,\,\dots ,\,u_n)\in {\cal B}.$
The marginal pdf's may be found in the usual way if required.

Example 3..5
Given $X_1$ and $X_2$ are independent random variables each distributed $N(0,1)$, so that

$$f(x_1,\,x_2)=(2\pi )^{-1}\,e^{-(x^2_1+x^2_2)/2}, \ \ -\infty < x_1 < \infty ; \ -\infty < x_2 <\infty ,$$

define $U_1=(X_1 + X_2)/2$, $U_2=(X_1 - X_2)^2 /2$ and find their joint distribution. The transformation is not one to one since to each point in
${\cal B}= \{ (u_1,u_2): \ -\infty < u_1 < \infty , \ \ 0 \leq u_2 <\infty\}$ there corresponds two points in
${\cal A} = \{ (x_1,x_2): \ -\infty < x_1 < \infty, \ \ -\infty < x_2 < \infty \}$. There are two sets of inverse functions.

(i)

$x_1=u_1-(u_2 /2)^{\frac 12}; \ \ x_2 =u_1+(u_2 /2)^{\frac 12}$.

(ii)

$x_1=u_1+(u_2 /2)^{\frac 12}; \ \ x_2=u_1-(u_2 /2)^{\frac 12}$.

From the definition of $U_2$, there is one type of mapping when $x_1 > x_2$ and another when $x_2 > x_1$. Consequently we define

$${\cal A}_1 = \{ (x_1\,x_2); \ \ x_2 > x_1\}$$

and

$${\cal A}_2 = \{ (x_1,\,x_2); \ \ x_2 < x_1\} \ .$$

Note that the line $x_1=x_2$ has been omitted since when $x_1=x_2$ we have $u_2=0$. However, since $P(X_1 =X_2)=0$, excluding this line does not alter the distribution and we therefore consider only ${\cal A} = {\cal A}_1 \cup {\cal A}_2$.

Then (i) defines a one-to-one transformation of ${\cal A}_2$ onto ${\cal B}$ and (ii) defines a one-to-one transformation of ${\cal A}_1$ onto ${\cal B}$. Thus the joint pdf of $(U_1,\,U_2)$ is given by

$$\begin{eqnarray*} f_{U_1,U_2}(u_1,u_2) & = & \frac{1}{2\pi} \exp \left\{ -\frac... ...-1} \ e^{-u_2/2}\ , \ \ \mbox{for} \ \ (u_1,u_2)\in {\cal B}. \end{eqnarray*}$$

Comment: This also shows that $U_1$ and $U_2$ are stochastically independent.