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150                    Fundamentals of Probability and Statistics for Engineers


           where  r  is  the  number  of  solutions  for  x of  equation  y ˆ  g( x),  and  J j is
           defined by
                                    qg    qg        qg
                                      
1    
1        
1
                                      j1    j1

                                                      j1

                                     qy 1  qy 2
                                                     qy n
                                      .               .
                                      .               .                 …5:69†

                                      .               .
                               J j ˆ
                                      
1    
1
                                    qg    qg        qg  
1
                                      jn    jn

                                                      jn

                                     qy 1  qy 2      qy n
           In the above, g j1 , g j2 ,... , and  g jn  are components of g .
                                                         j
             As we mentioned earlier, the results presented above can also be applied to
           the case in which the dimension of Y is smaller than that of X. Consider the
           transformation represented in Equation (5.60) in which m <  n. In order to use
           the formulae developed  above, we first  augment  the m-dimensional random
           vector Y by another (n 
  m) – dimensional random vector Z. The vector Z can


           be constructed as a simple function of X in the form
                                        Z ˆ h…X†;                       …5:70†
           where h satisfies conditions of continuity and continuity in partial derivatives.
           On combining Equations (5.60) and (5.70), we have now an n-random-variable
           to n-random-variable transformation, and the jpdf of Y and Z can be obtained


           by means of Equation (5.67) or Equation (5.68). The jpdf of Y alone is then
           found through integration with respect to the components of Z.
             Example 5.18. Problem: let random variables X 1  and X 2  be independent and




           identically and normally distributed according to
                                   1        
x 2 1
                        f  …x 1 †ˆ     exp      ;  
1 < x 1 < 1;
                         X 1        1=2
                                 …2 †       2
           and similarly for X 2 . Determine the jpdf of Y 1 ˆ  X 1 ‡  X 2 , and Y 2 ˆ  X 1 
  X 2 .
             Answer: Equation (5.67) applies in this case. The solutions of x 1  and x 2  in
           terms of y 1  and y 2  are
                             
1
                                                   
1
                       x 1 ˆ g …y†ˆ  y 1 ‡ y 2  ;  x 2 ˆ g …y†ˆ  y 1 
 y 2  :  …5:71†
                             1         2           2        2
           The Jacobian in this case takes the form
                                   
1
                                  qg 1  qg 
1       1
                                         1         1

                                                          1
                                 qy 1            2  2
                                       qy 2     ˆ     ˆ
 :
                            J ˆ    
1
                                 qg    qg  
1       1  1     2
                                   2
                                         2
                                               2    2

                                 qy 1  qy 2

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