Page 325 - Applied Probability
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14. Poisson Approximation
                              314
                                10. Consider an infinite sequence W 1 ,W 2 ,... of independent, Bernoulli
                                   random variables with common success probability p. Let X α be the
                                   indicator of the event that a success run of length t or longer begins
                                                                t
                                                                    W k and
                                   at position α. Note that X 1 =

                                                                k=1
                                                                      j+t−1

                                                    X j  =(1 − W j−1 )     W k
                                                                       k=j
                                   for j> 1. The number of such success runs starting in the first n po-

                                   sitions is given by S =   X α , where the index set I = {1,...,n}.
                                                         α∈I
                                   The Poisson heuristic suggests the S is approximately Poisson with
                                              t
                                   mean λ = p [(n−1)(1−p)+1]. Let B α = {β ∈ I : |β −α|≤ t}. Show
                                   that X α is independent of those X β with β outside B α . In the Chen-
                                   Stein bound (14.3), prove that the constant b 2 = 0. Finally, show
                                                                                  t
                                                                   2
                                   that the Chen-Stein constant b 1 ≤ λ (2t +1)/n +2λp for 1 <t ≤ n.
                                   (Hint:
                                                             2t
                                                       2t
                                              b 1  = p +2tp (1 − p)
                                                                            2t
                                                              2
                                                      +[2nt − t + n − 3t − 1]p (1 − p) 2
                                   exactly. Note that the pairs α and β entering into the double sum for
                                   b 1 are drawn from the integer lattice points {(i, j): 1 ≤ i, j ≤ n}.An
                                   upper left triangle and a lower right triangle of lattice points from
                                   this square do not qualify for the double sum defining b 1. The term
                                   p 2t  in b 1 corresponds to the lattice point (1, 1).)
                                11. Let X 1 ,...,X n be n independent, exponentially distributed waiting
                                   times with common intensity λ, and define M n = max 1≤i≤n X i . Show
                                   that λM n − ln n converges in distribution to the extreme value sta-
                                   tistic having density e −e −u e −u . (Hints: This assertion can be most
                                   easily demonstrated by considering the moment generating function
                                                                            )
                                   of λM n −ln n. Since M n has density n(1−e −λx n−1 λe −λx , prove that

                                                            ∞
                                                                                )
                                       E[e s(λM n −ln n) ]=   e s(λx−ln n) n(1 − e −λx n−1 λe −λx dx
                                                           0

                                                            ∞          1
                                                                su
                                                                           )
                                                                               e
                                                      =        e (1 −   e −u n−1 −u du.
                                                           − ln n     n
                                   Argue that in the limit
                                                                      ∞       −u

                                                                         su −e
                                              lim E[e s(λM n −ln n) ]=  e e     e −u du
                                             n→∞
                                                                     −

                                                                      ∞
                                                                =       w −s −w dw        (14.9)
                                                                           e
                                                                     0
                                                                =   Γ(1 − s),
                                   where Γ(x) is the gamma function. )
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