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136                    Fundamentals of Probability and Statistics for Engineers

           Upon evaluating   Y  (t), the moments of Y  are given by [Equation (4.52)]:
                                      
n …n†
                                 n
                             EfY gˆ j   …0†;     n ˆ 1; 2; ... :        …5:33†
                                         Y



             Example 5.9. Problem: a random variable X is discrete and its pmf is given in

           Example 5.1. Determine the mean and variance of Y where Y ˆ  2X ‡  1.
             Answer: using the first of Equations (5.31), we obtain
                                           X
                       EfYgˆ Ef2X ‡ 1gˆ       …2x i ‡ 1†p …x i †
                                                      X
                                            i

                                    1        1        1        1
                             ˆ…
1†     ‡…1†     ‡…3†     ‡…5†           …5:34†
                                    2        4        8        8
                               3
                             ˆ ;
                               4
                                         2
                                                        2
                           2
                       EfY gˆ Ef…2X ‡ 1† gˆ   X  …2x i ‡ 1† p …x i †
                                                          X
                                               i

                                   1        1        1         1        …5:35†
                             ˆ…1†     ‡…1†     ‡…9†     ‡…25†
                                   2        4        8         8
                             ˆ 5;
           and
                                                     3
                                                        2  71
                            2
                                         2
                                    2
                             ˆ EfY g
 E fYgˆ 5 
         ˆ   :          …5:36†
                            Y
                                                     4     16
             Following the second approach, let us use the method of characteristic func-
           tions described by Equations (5.32) and (5.33). The characteristic function of Y  is
                               X    jt…2x i ‡1†
                          Y …t†ˆ  e      p …x i †
                                          X
                                 i

                                    1     jt  1   3jt  1   5jt  1
                                
jt
                             ˆ e       ‡ e     ‡ e      ‡ e
                                    2       4        8        8
                               1   
jt    jt  3jt  5jt
                             ˆ …4e    ‡ 2e ‡ e  ‡ e †;
                               8
           and we have
                                                                3
                                             j
                               
1 …1†
                       EfYgˆ j   …0†ˆ j   
1   …
4 ‡ 2 ‡ 3 ‡ 5†ˆ ;
                                  Y
                                             8                  4
                                        1
                                 …2†
                          2
                      EfY gˆ
  …0†ˆ …4 ‡ 2 ‡ 9 ‡ 25†ˆ 5:
                                 Y
                                        8
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