Page 202 - Classification Parameter Estimation & State Estimation An Engg Approach Using MATLAB
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CRITERIA FOR SELECTION AND EXTRACTION                        191

              Example 6.1   Interclass and intraclass distance
              Numerical calculations of the example in Figure 6.2 show that before
              normalization J INTRA  ¼ trace(S w ) ¼ 0:016 and J INTER  ¼ trace(S b ) ¼
              0:54. Hence, the ratio between these two is J INTER =J INTRA  ¼ 33:8.
              After normalization, J INTRA  ¼ N ¼ 2, J INTER  ¼ J INTER=INTRA  ¼ 59:1,
              and J INTER =J INTRA  ¼ trace(S b )=trace(S w ) ¼ 29:6. In this example,
              before normalization, the J INTER =J INTRA  measure is too optimistic.
              The normalization accounts for this phenomenon.




            6.1.2  Chernoff–Bhattacharyya distance

            The interclass and intraclass distances are based on the Euclidean
            metric defined in the measurement space. Another possibility is to use
            a metric based on probability densities. Examples in this category are the
            Chernoff distance (Chernoff, 1952) and the Bhattacharyya distance
            (Bhattacharyya, 1943). These distances are especially useful in the two-
            class case.
              The merit of the Bhattacharyya and Chernoff distance is that an
            inequality exists with which the distances bound the minimum error
            rate E min . The inequality is based on the following relationship:

                                                p ffiffiffiffiffiffi
                                    minfa; bg     ab                   ð6:11Þ

            The inequality holds true for any positive quantities a and b. We will use
            it in the expression of the minimum error rate. Substitution of (2.15) in
            (2.16) yields:

                              Z
                        E min ¼  ½1   maxfPð! 1 jzÞ; Pð! 2 jzÞgŠpðzÞdz
                               z
                                                                       ð6:12Þ
                              Z
                            ¼    minfpðzj! 1 ÞPð! 1 Þ; pðzj! 2 ÞPð! 2 Þgdz
                               z

            Together with (6.11) we have the following inequality:


                                            Z
                               p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
                         E min    Pð! 1 ÞPð! 2 Þ  pðzj! 1 Þpðzj! 2 Þdz  ð6:13Þ
                                             z
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