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            052184472Xc09  CUNY148/Severini  June 2, 2005  12:8





                            262                       Approximation of Integrals

                            Since
                                                 1
                                            ∞                       ∞
                                                       x−1    j         j
                                                  (1 − u)  (yu) /j! ≤  y /j! = exp(y),
                                            j=0  0                  j=0
                            we may interchange summation and integration. Hence, by Theorem 9.2,

                                                               ∞   j     1
                                                                 y          x−1  j
                                                     x
                                            γ (x, y) = y exp(−y)       (1 − u)  u du
                                                                  j!
                                                              j=0    0
                                                               ∞  y  ( j + 1) (x)
                                                                   j
                                                     x

                                                  = y exp(−y)                  ,
                                                                  j!  (x + j + 1)
                                                              j=0
                            which, after simplification, is identical to (9.2).
                            Standard normal distribution function
                            Let
                                                     1          1  2
                                             φ(z) = √    exp − z    , −∞ < z < ∞
                                                     (2π)       2
                            denote the standard normal density function and let
                                                          z
                                                 (z) =     φ(t) dt,  −∞ < z < ∞
                                                        −∞
                            denote the standard normal distribution function.
                              Although  (z)is defined by an integral over the region (−∞, z), calculation of  (z),
                            for any value of z, only requires integration over a bounded region. Define

                                                              z
                                                      0 (z) =  φ(t) dt, z ≥ 0.
                                                            0
                            The following result shows that  (z) can be written in terms of   0 (z); also, by a change-
                            of-variable in the integral defining   0 it may be shown that   0 is a special case of the
                            incomplete gamma function γ (·, ·). The proof is left as an exercise.

                            Theorem 9.8.
                                                           1
                                                          −   0 (−z) if z < 0
                                                           2
                                                   (z) =
                                                           1
                                                           +   0 (z)  if z ≥ 0
                                                           2
                            and, for z > 0,
                                                              1     1 z 2
                                                       0 (z) = √ γ   ,    .
                                                            2 π     2 2

                              Hence, using Theorem 9.8, together with the series expansions for γ (·, ·)given in The-
                            orem 9.7, we obtain the following series expansions for   0 (·); these, in turn, may be used
                            to obtain a series expansion for  (·). The result is given in the following theorem, whose
                            proof is left as an exercise.
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