Page 269 - Introduction to Statistical Pattern Recognition
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5  Parameter Estimation                                      25 1












                     5.   Derive v, of  (5.92) for Data 1-1.
                     6.   A quadratic classifier for zero-mean distributions is given as

                                                             01
                                        h(X) =XT[C;I  - ZC,']X ><  t  .
                                                             02
                         In the design phase, C1 and C2  are estimated by
                                                                N
                                       I
                                          N
                                                             I
                                               T
                                                                     T
                                   *
                                  CI  = -  XiX;    and  C2 = -  Y,Y,  ,
                                       N  ;=I                N  ;=I
                         where XI and  Y, are samples from wl  and  w2  respectively.  For  testing
                                                        n
                         Xk, Xk is excluded from design to get Elk. Prove






                    7.   Modify the Procedure 111 of Section 3.2 to the leave-one-out method.
                     8.   Assuming  MI = 0,  M2 = M  = [m 0. . . O]',   and  XI = C2 = I,  compute
                         the  integral  of  (5.153)  along  the  Bayes  boundary  (xI = -/2)   to
                         obtain  PI.  Use  h(X)==x,     -MTM/2,  d':(X)=(C,:'=lx-f)2,  and
                                           1
                         p  (x) = (2~)-"'~exp[- -E  ,:'=  I x,' 1.
                                           2
                    9.   N  boxes  have  equal  probability of  getting  a  sample.  When  N  samples
                         are thrown, the  ith  box  receives k,  samples.  Defining w, = k,/N, prove
                         that
                         (1)   E{w,) = 1/N,
                         (2)  E((w, - l/N)(w, - l/N)} =6,/N2  - l/N3.
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