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Appendix B: The Normal
                              Distribution


                              B.1 Univariate Normal Random Variables

                              A random variable X is said to be standard normal if it possesses the
                              density function

                                                                 1    x 2
                                                     ψ(x)  =   √   e −  2 .
                                                                 2π
                                                              ˆ
                              To find the characteristic function ψ(s)= E(e isX )of X, we derive and
                              solve a differential equation. Differentiation under the integral sign and
                              integration by parts together imply that

                                       d            1     ∞  isx   x 2
                                         ˆ
                                         ψ(s)  =   √       e   ixe −  2 dx
                                       ds           2π  −
                                                      i     ∞  isx  d  x 2
                                               = −√          e     e −  2 dx
                                                      2π  −    dx
                                                   −i        2 ∞      s     ∞      x 2
                                                              4
                                                                                 e
                                               =   √   e isx −  x 4 4  − √   e isx −  2 dx
                                                          e
                                                             2
                                                    2π         −     2π  −
                                                     ˆ
                                               = −sψ(s).
                                                                                        ˆ
                              The unique solution to this differential equation with initial value ψ(0) = 1
                                 ˆ
                                          2
                              is ψ(s)= e −s /2 . The differential equation also yields the moments
                                                          1 d
                                                              ˆ
                                               E(X)=          ψ(0)
                                                          i ds
                                                      =0
                                                          1 d 2
                                                                ˆ
                                                   2
                                               E(X )=          ψ(0)
                                                           2
                                                          i ds 2
                                                          1           2 ˆ
                                                                ˆ
                                                      =      − ψ(s)+ s ψ(s)
                                                          i 2               s=0
                                                      =1.
                                An affine transformation Y = σX + µ of X is normally distributed with
                              density
                                                1    y − µ        1     (y−µ) 2
                                                  ψ          =  √     e −  2σ 2  .
                                                σ     σ           2πσ
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