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                            266                       Approximation of Integrals

                                                Table 9.1. Approximations in Example 9.1.

                                                x       f (x)     ˆ f (x)   ˆ f (x)
                                                                             3
                                                                   2
                                                 1     0.596     0          2.000
                                                 2     0.722     0.500      1.000
                                                 5     0.852     0.800      0.880
                                                10     0.916     0.900      0.920
                                                20     0.954     0.950      0.955
                                                30     0.981     0.980      0.981


                            Integration-by-parts
                            One useful technique for obtaining asymptotic approximations to integrals is to repeatedly
                            use integration-by-parts. This approach is illustrated in the following examples.

                            Example 9.2 (Incomplete beta function). Consider approximation of the integral
                                                         x

                                                          t α−1 (1 − t) β−1  dt,
                                                        0
                            where α> 0 and β> 0, for small values of x > 0. This is the integral appearing in the
                            distribution function corresponding to the beta distribution with density
                                                 (α + β)  α−1     β−1
                                                        x   (1 − x)  , 0 < x < 1
                                                 (α) (β)
                            and it is known as the incomplete beta function.
                              Using integration-by-parts,
                                       x                1            x  β − 1     x
                                                                                α
                                                          α
                                       t α−1 (1 − t) β−1  dt =  t (1 − t) β−1    +  t (1 − t) β−2  dt
                                     0                  α           0    α    0
                                                        1  α     β−1  β − 1     x  α  β−2
                                                     =   x (1 − x)  +          t (1 − t)  dt.
                                                        α               α    0
                              For β ≥ 2,
                                                     x                1
                                                      α
                                                     t (1 − t) β−2  dt ≤  x α+1 ,
                                                   0                α + 1
                            while for 0 <β < 2,
                                                 x                        1
                                                  α
                                                 t (1 − t) β−2  dt ≤ (1 − x) β−2  x  α+1 ;
                                               0                        α + 1
                            hence, we may write
                                          x                1
                                                             α
                                          t  α−1 (1 − t) β−1  dt =  x (1 − x) β−1 [1 + o(x)] as x → 0.
                                        0                  α
                              Alternatively, integration-by-parts may be used on the remainder term, leading to the
                            expansion
                                  x                1                β − 1   x
                                                                                    2
                                                      α
                                  t α−1 (1 − t) β−1  dt =  x (1 − x) β−1  1 +  + o(x )  as x → 0.
                                0                  α                α + 1 1 − x
                            Further terms in the expansion may be generated in the same manner.
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