Page 341 - Applied Probability
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15. Diffusion Processes
                              330
                                                             r i −1
                                                          1

                                                                                         (15.20)
                                                    =
                                                                 c ij→i+1,k p ij→i+1,k
                                             c i+1,k
                                                        p i+1,k
                                                              j=0
                                               is certain to belong to one of the intervals [a ij ,a i,j+1 ].
                              assuming that X t i
                                To carry out this updating scheme, we must approximate the integrals
                              p ij→i+1,k and c ij→i+1,k p ij→i+1,k . If the interval [a ij ,a i,j+1 ] is fairly narrow,
                              then the linear approximations
                                                    m(i, x)= µ ij0 + µ ij1 x
                                                     2
                                                    s (i, x)= σ  2                       (15.21)
                                                                 ij
                                                    f(t i ,x)= f ij0 + f ij1 x
                              should suffice for all x in the interval. The first two of these linear approx-
                              imations follow directly from the diffusion model. The constants involved
                              in the third approximation are determined by the equations

                                                a i,j+1
                                          =         (f ij0 + f ij1 x) dx
                                      p ij
                                               a ij
                                                               1
                                          = f ij0 (a i,j+1 − a ij )+ f ij1 (a i,j+1 + a ij )(a i,j+1 − a ij )
                                                               2
                                                a i,j+1

                                          =         x(f ij0 + f ij1 x) dx
                                   c ij p ij
                                               a ij
                                              1
                                          =    f ij0 (a i,j+1 + a ij )(a i,j+1 − a ij )
                                              2
                                                1     2                2
                                              + f ij1 (a i,j+1  + a i,j+1 a ij + a )(a i,j+1 − a ij )
                                                                       ij
                                                3
                              with inverses
                                                  2
                                            2p ij (2a +2a ij a i,j+1 +2a 2 i,j+1  − 3a ij c ij − 3a i,j+1 c ij )
                                                  ij
                                        =
                                   f ij0                                3
                                                             (a i,j+1 − a ij )
                                            6p ij (2c ij − a ij − a i,j+1 )
                                        =                        .                       (15.22)
                                   f ij1                   3
                                                (a i,j+1 − a ij )
                              Problem 13 asks the reader to check that the linear density f ij0 + f ij1 x is
                              nonnegative throughout the interval (a ij ,a i,j+1 ) if and only if its center of
                              mass c ij lies in the middle third of the interval.
                                Given the linear approximations (15.21), we now show that the double
                              integrals (15.18) and (15.19) reduce to expressions involving elementary
                              functions and the standard normal distribution function
                                                             1     x  −y /2
                                                                       2
                                                 Φ(x)  =   √        e     dy.
                                                             2π  −
                              The latter can be rapidly evaluated by either a power series or a continued
                              fraction expansion [10]. It also furnishes the key to evaluating the hierarchy
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