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430     8 Bayesian statistics and parameter estimation



                   where H   α   > 0 is the Hessian of h   α  (θ  α  )at θ  α  . Substituting this quadratic expan-
                                                            M
                                                                              1
                   sion into (8.226) and assuming I θ (θ  α  ) = 1 everywhere where exp{− Nh   α  (θ  α  )} is
                                               α
                                                                              2
                   significantly nonzero yields
                                                  9              :
                                                     1
                                         Q   α   = exp − Nh  α   θ  α   T   α        (8.228)
                                                     2       M
                   where
                               '                                       1

                         T   α   =  exp −  1   θ  α   − θ  α  T  NH   α     θ  α   − θ  α     dθ   α   (8.229)
                                                 M
                                                                    M
                                         2
                               P  α

                   Using the Gaussian integral

                                    1  T     T       (2π)        1  T  −1
                            '                            P/2
                              exp − z Az + b z dz =         exp   b A b              (8.230)
                                    2                 |A| 1/2    2
                            P

                   we obtain
                                           (2π) P  α  /2    2π    P  α  /2   α  −1/2
                                     T   α   =       =          |H   |               (8.231)
                                           |NH  α  1/2   N
                                                 |
                   Thus, from (8.228) and (8.223), we have
                                                                    P   α  /2
                                                             1
                                                   N            2π
                              p  α  (y|M α ) = C   α   exp −  h  α   θ  α            (8.232)
                                                          M
                                                   2            N
                   where
                                                           c
                                                       σ  α   α
                                                           0
                                        C   α   =                                    (8.233)
                                                                    1/2
                                                         )
                                              (2π) N/2 (σ   α  (N+1)   H   α
                   We have then an approximation for the Bayes factor (8.214):
                                                                (
                                                              α        P  α  /2
                                                       S   α   θ M  2π
                                             C  α
                                                 exp −
                                                           α     2  N
                                   p(y|M α )           2 σ MLE
                          B αβ (y) =       ≈                    (                    (8.234)
                                   p(y|M β )           S   β     θ   β     2π    P   β  /2
                                                            M
                                             C   β   exp −      2
                                                       2 σ   β     N
                                                          MLE
                   where we have used (8.225). Taking the natural logarithm of (8.234), we have
                                                         α
                                                  S   α   θ M  P   α   2π
                              ln[B αβ (y)] ≈ ln C   α     +        ln
                                                      α   2    2       N
                                               −
                                                  2 σ
                                                     MLE

                                                            β
                                                     S   β   θ M  P   β   2π
                                          − ln C   β   −      +       ln             (8.235)
                                                            2
                                                     2 σ   β      2       N
                                                        MLE
                   If we assume that C   α   and C   β   are of the same order of magnitude, and define Schwartz’s
                   Bayesian Information Criterion (BIC),

                                                       α
                                                S   α   θ M  P   α    N
                                       α
                                  BIC (y) =−            +         ln                 (8.236)
                                                     α   2    2      2π
                                                2 σ MLE
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