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





                                                  9.2 Some Useful Functions                  263

                        Theorem 9.9. For z ≥ 0,
                                                                           ∞
                                            ∞
                                       1       (−1) j  z 2 j+1  1      −z 2        z 2 j+1
                                0 (z) = √                 = √ exp                        .
                                       (2π)    2 j! 2 j + 1  2 2       2      2  ( j + 3/2)
                                                j
                                                                               j
                                            j=0                            j=0
                          Since φ(z) > 0 for all z,  (·)isa strictly increasing function and, since it is a distribution
                        function,
                                             lim  (z) = 1 and  lim  (z) = 0.
                                             z→∞              z→−∞
                          The following result gives some information regarding the rate at which  (z) approaches
                        these limiting values; the proof follows from L’Hospital’s rule and is left as an exercise.

                        Theorem 9.10.
                                             1 −  (z)                (z)
                                         lim         = 1 and   lim        =−1.
                                         z→∞ φ(z)/z           z→−∞ φ(z)/z

                        The following result gives precise bounds on
                                                  1 −  (z)        (z)
                                                           and
                                                   φ(z)/z       φ(z)/z
                        that are sometimes useful.

                        Theorem 9.11. For z > 0,
                                                 z                  1
                                                    φ(z) ≤ 1 −  (z) ≤  φ(z).
                                               1 + z 2              z
                        For z < 0,

                                                 |z|              1
                                                     φ(z) < (z) <   φ(z).
                                                1 + z 2           |z|
                        Proof. Fix z > 0. Note that

                                ∞                 ∞  1                1   ∞
                                        2                    2                     2
                                  exp(−x /2) dx =     x exp(−x /2) dx ≤     x exp(−x /2) dx.
                               z                 z   x                z  z
                        Since
                                              d        2              2
                                                 exp(−x /2) =−x exp(−x /2),
                                              dx

                                               ∞
                                                                       2
                                                         2
                                                 x exp(−x /2) dx = exp(−z /2)
                                              z
                        so that

                                              ∞                  1
                                                      2                 2
                                                exp(−x /2) dx =≤  exp(−z /2)
                                             z                   z
                        or, equivalently,
                                                              1
                                                    1 −  (z) ≤  φ(z).
                                                              z
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