Page 289 - Applied Probability
P. 289

277
                                                                12. Models of Recombination
                                 9. In the count-location model, suppose that the count distribution is
                                   concentrated on the set {0, 1} or on the set {1, 2, 3, 4}. Prove that
                                   the model exhibits positive or no interference [23]. (Hint: Check the
                                                                                       2
                                   log-concavity property by showing that Q (s)Q(s) − Q (s) reduces


                                   to a sum of negative terms.)
                                10. In the count-location model, suppose that the count distribution has
                                   generating function
                                                               8    1  4  4  5
                                                     Q(s)=       s + s + s .
                                                               35   5     7
                                   Prove that the opposite of inequality (12.16) holds for x 1 = .4 and
                                   x 2 = .1 and, therefore, that negative interference occurs [23].
                                11. The Poisson-skip model with skip distribution s 1 = p and s 3 =1 − p
                                   has interarrival density
                                                                         x 2  −x
                                                              −x
                                                   f(x)  = pe    +(1 − p)   e  .
                                                                          2
                                   Show that f(x) has decreasing hazard rate for x small and positive
                                   [18].
                                12. The collection of count-location models that are also stationary re-
                                   newal models is very small [3]. Haldane’s homogeneous Poisson model
                                   is one example. Another is the trivial model with no chiasmata. The
                                   admixture of two such independent processes furnishes a third exam-
                                   ple subsuming the first two. To prove that this exhausts the possi-
                                   bilities, consider a count-location process that uniformly distributes
                                   its pointson[0, 1]. Let q n denote the nth count probability. If the
                                   process is also a stationary renewal process, let F ∞ (x) be the distri-
                                   bution of X 1 and F(x) be the distribution of the subsequent X i ,in
                                   the notation of Section 12.4. Show that
                                                                   ∞

                                                    F ∞ (x)  = 1 −    q n (1 − x) n
                                                                  n=0
                                   and that
                                               F(x)   =  Pr(X 2 ≤ x | X 1 = y)
                                                               ∞  nq n (1 − x − y)
                                                                               n−1
                                                               n=0
                                                      =1 −                         .
                                                                 ∞  nq n (1 − y) n−1
                                                                 n=0
                                   Use these identities and the identity F (x)= [1−F(x)]/µ to demon-

                                                                    ∞

                                   strate that φ(x)=  ∞  nq n (1 − x) n−1  satisfies the functional equa-
                                                      n=0
                                   tion φ(x)φ(y)= µ −1 φ(x + y), where µ is the mean of F(x). Setting
                                   ψ(x)= µÐ (x), it follows that ψ(x)ψ(y)= ψ(x + y) and ψ(0)=1.
   284   285   286   287   288   289   290   291   292   293   294