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15. Diffusion Processes
                              328
                                                             2N(1−f)
                                                                         z
                                                                                  2
                                                    k 4
                                             =
                                                 4N(1 − f)
                                                                     ∞
                                                                        m
                                                                         +2Nη−1 −z
                                                                                e
                                                                                   dz
                                                                       z 2
                                             ≈
                                                 2[2N(1 − f)] 2
                                                         +2Nη)
                                                   k 4 Γ( m k 4    0 m +2Nη    
 0 2N(1 − f)   m+4Nη−2  e −z dz
                                                        2
                                             =              m +2Nη  .
                                                 2[2N(1 − f)] 2
                              Taking m = 0 identifies the normalizing constant
                                                           2[2N(1 − f)] 2Nη
                                                   k 4  =                 .
                                                               Γ(2Nη)
                              With this value of k 4 in hand, the mean of f(x)is
                                                                          1
                                                                 Γ(2Nη + )
                                                                          2
                                                xf(x) dx  ≈                    .
                                               I               2N(1 − f)Γ(2Nη)
                              When Nη is large, application of Stirling’s formula implies that the mean

                              is close to the deterministic equilibrium value  η/(1 − f). In practice, one
                              should be wary of applying the equilibrium theory because the approach
                              to equilibrium is extremely slow.
                              15.7 Numerical Methods for Diffusion Processes
                              It is straightforward to simulate a diffusion process X t . The definition tells
                              us to extend X t to X t+s by setting the increment X t+s −X t equal to a nor-
                                                                       2
                              mal deviate with mean µ(t, x)s and variance σ (t, x)s. The time increment
                              s should be small, and each sampled normal variate should be independent.
                              Techniques for generating random normal deviates are covered in standard
                              texts on computational statistics and will not be discussed here [9, 10].
                              Of more concern is how to cope with a diffusion process with finite range
                              I. Because a normally distributed random variable has infinite range, it
                              is possible in principle to generate an increment that takes the simulated
                              process outside I. One remedy for this problem is to take s extremely small.
                                                                   2
                              It also helps if the infinitesimal variance σ (t, x) tends to 0 as x approaches
                              the boundary of I. This is the case with the neutral Wright-Fisher process.
                                Simulation offers a crude method of finding the distribution of X t .Sim-
                              ply conduct multiple independent simulations and compute a histogram
                              of the recorded values of X t . Although this method is neither particu-
                              larly accurate nor efficient, it has the virtue of yielding simultaneously the
                              distributions of all of the X t involved in the simulation process. Thus, if
                              1000 times are sampled per simulation, then the method yields all 1000
                              distributions, assuming that enough computer memory is available. Much
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