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15. Diffusion Processes
                              322
                              lation geneticists substitute p(x)= x in formula (15.7) defining the infin-
                                               2
                              itesimal variance σ (t, x). This action is justified for neutral and recessive
                              inheritance, but less so for dominant inheritance where the allele frequency
                              x is typically on the order of magnitude of the mutation rate η.It isalso
                              fair to point out that in the presence of inbreeding or incomplete mixing of
                              a population, the effective population size is less than the actual pop-
                              ulation size [3]. For the sake of simplicity, we will ignore this evolutionary
                              fact.
                              15.4 First Passage Time Problems
                              Let c<d be two points in the interior of the range I of a diffusion process
                              X t . Define T c to be the first time t that X t = c and similarly for T d. The
                              process X t exits (c, d) at the time T =min{T c,T d}. We consider two related
                              problems involving these first passage times. One problem is to calculate
                              the probability u(x) = Pr(T d <T c | X 0 = x) that the process exits via d
                              starting from x ∈ [c, d]. It is straightforward to derive a differential equation
                              determining u(x) given the boundary conditions u(c) = 0 and u(d)= 1.
                              With this end in mind, we assume that X t is time homogeneous.
                                For s> 0 small and x ∈ (c, d), the probability that X t reaches either c
                              or d during the time interval [0,s]is o(s). Thus,
                                               u(x)=E[u(X s ) | X 0 = x]+ o(s).

                              If we let ∆X s = X s −X 0 and expand u(X s ) in a second-order Taylor series,
                              then we find that

                                      u(X s )= u(x +∆X s )
                                                                  1                   2
                                             = u(x)+ u (x)∆X s +    u (x)+ r(∆X s ) ∆X ,  (15.8)


                                                                                      s
                                                                  2
                              where the relative error r(∆X s ) tends to 0 as ∆X s tends to 0. Invoking
                              equations (15.1), (15.2), and (15.8) therefore yields
                                        u(x)  = E[u(X s )] + o(s)
                                                                   1  2


                                              = u(x)+ µ(x)u (x)s + σ (x)u (x)s + o(s),
                                                                   2
                              which upon rearrangement and sending s to 0 gives the differential equation
                                                                  1  2
                                                0= µ(x)u (x)+ σ (x)u (x).                 (15.9)


                                                                  2
                                It is a simple matter to check that equation (15.9) can be solved explicitly
                              by defining
                                                             x
                                                                 "  y 2µ(z)
                                                  v(x)  =      e −  l  σ 2 (z)  dz dy
                                                            l
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