Page 231 - Introduction to Statistical Pattern Recognition
P. 231

5  Parameter Estimation                                       213



                   values of n and MTM, we may generally conclude that vg is larger than v, for n
                   >>  1.  This  implies  that  many  more  samples  are needed  to properly  design a
                   quadratic  classifier  than  a  linear  classifier.  It  is  believed  in  general  that  the
                   linear  classifier  is  more  robust  (less  sensitive to parameter estimation  errors)
                   than  the  quadratic  classifier,  particularly  in  high-dimensional  spaces.  The
                   above results support this belief both theoretically  and experimentally.
                        Also  note  that  for  large  n, v lY!  is  proportional  to  Ilk.  This  indicates
                   that,  as  far  as the  design  of  a  linear  classifier  is  concerned, a  fixed  multiple
                   could be used to determine the sample size from the dimensionality.  However,
                   (5.92) indicates that  the  value  of  the multiple depends on MTM, which  meas-
                   ures  the  separability  between  two  distributions  with  a  common  covariance
                   matrix  1.  In particular,  the  less the separability  between the  two distributions,
                   the greater k must be for a fixed bias.

                                              A
                        Variance: The variance of  E may be computed from (5.59) and (5.64) as




                                         1
                                  - (E - -)2  .                                (5.93)
                                        2

                   Applying the same approximation  as (5.60) and keeping up to the second order
                   terms of Ah,










                   where

                                                      jm,
                                                       2
                                      AC,(X) = Ah(X) + -Ah2(X)   .             (5.95)
                   Thus, (5.93) can be expanded to
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