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### m507b-hw3soln-s08

Course: MATH 507A, Spring 2008
School: USC
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507b MATH ASSIGNMENT 3 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 5: (1.2) 1 2 in two steps means 1 3 2, so p2 (1, 2) = p(1, 3)p(3, 2) = (.9)(.4) = .36. 2 3 in 3 steps means 2 1 3 3 or 2 2 1 3 or 2 1 1 3, so p3 (2, 3) = (.7)(.9)(.6) + (.3)(.7)(.9) + (.7)(.1)(.9) = .63. (1.3) We claim that for all and all n 0, P (Xn = 0) = Equivalently, subtracting from 1, P (Xn = 1) = - (1 - - )n (0) - + + ....

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507b MATH ASSIGNMENT 3 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 5: (1.2) 1 2 in two steps means 1 3 2, so p2 (1, 2) = p(1, 3)p(3, 2) = (.9)(.4) = .36. 2 3 in 3 steps means 2 1 3 3 or 2 2 1 3 or 2 1 1 3, so p3 (2, 3) = (.7)(.9)(.6) + (.3)(.7)(.9) + (.7)(.1)(.9) = .63. (1.3) We claim that for all and all n 0, P (Xn = 0) = Equivalently, subtracting from 1, P (Xn = 1) = - (1 - - )n (0) - + + . + (1 - - )n (0) - + + . For n = 0 the first equality says P (Xn = 0) = (0) which is true by definition. Suppose the claim is true for some n. Then P (Xn+1 = 0) = P (Xn+1 = 0 | Xn = 0)P (Xn = 0) + P (XN =1 = 1 | Xn = 1)P (Xn = 1) + (1 - - )n (0) - = (1 - ) + + + - (1 - - )n (0) - + + = + (1 - - )n+1 (0) - , + + so the claim is true for n + 1. (1.5) Since just one son and daughter from each generation are mated, and any other siblings are irrelevant, we may assume that each generation contains exactly one male and one female. The son effectively chooses a gene at random from each parent, independently. The daughter does the same, independently of the son. If, for example, the parents are Aa, Aa then the son and daughter are (without regard to order): 1 1 AA, AA with probability 1 1 = 16 , 4 4 AA, Aa with probability 2 1 1 = 1 , 4 2 4 1 AA, aa with probability 2 1 1 = 8 , 4 4 Aa, Aa with probability 1 1 = 1 , 2 2 4 1 Aa, aa with probability 2 4 1 = 1 , 2 4 1 1 aa, aa with probability 1 4 = 16 . 4 By similar calculations we get the rest of the transition matrix (for the states in the order above), with the 4th row given by the probabilities above: 1 0 0 0 0 0 1/4 1/2 0 1/4 0 0 0 0 0 1 0 0 1/16 1/4 1/8 1/4 1/4 1/16. 0 0 0 1/4 1/2 1/4 0 0 0 0 0 1 (1.6) Let Fn = (1 , .., n ). Now Xn+1 = so Xn + 1 Xn if n+1 {1 , .., n }, / if n+1 {1 , .., n } Xn Xn , P (Xn+1 = Xn | Fn ) = , n n k Since this depends only on Xn , {Xn } is a Markov chain, and p(k, k +1) = 1- N , p(k, k) = with all other p(k, l) = 0. P (Xn+1 = Xn + 1 | Fn ) = 1 - k , N (1.7) Suppose that for some n, k, we have Xn-1 = k - 1, and Xn = k. Then k is a new maximum value achieved by {Sj } at time n, meaning Sn = k. Conditionally on all this, a new maximum equal to k + 1 occurs at time n + 1 provided n+1 = 1. This shows that P (Xn+1 = k + 1 | Xn-1 = k - 1, Xn = k) = P (n+1 = 1 | Xn-1 = k - 1, Xn = k) = P (n+1 = 1) 1 = . 2 (1) In contrast, suppose Xn-3 = k - 1, Xn-2 = k, Xn-1 = k, Xn = k. Then k was a new maximum value achieved at time n - 2, meaning Sn-2 = k and then Sn-1 = k - 1 and Sn = either k - 2 or k, with probability 1/2 each. If Sn = k - 2 then there is 0 probability of a 2 new maximum at time n + 1, while if Sn = k then as in (1) there is probability 1/2 of a new maximum. Thus we have P (Xn+1 = k + 1 | Xn-3 = k - 1, Xn-2 = Xn-1 = Xn = k) = P (Xn+1 = k + 1 | Xn-3 = k - 1, Xn-2 = Xn-1 = Xn = k, Sn = k) P (Sn = k | Xn-3 = k - 1, Xn-2 = Xn-1 = Xn = k) + P (Xn+1 = k + 1 | Xn-3 = k - 1, Xn-2 = Xn-1 = Xn = k, Sn = k - 2) P (Sn = k - 2 | Xn-3 = k - 1, Xn-2 = Xn-1 = Xn = k) 1 1 1 = +0 2 2 2 1 = . 4 If the process {Xj } were Markov, then this and (1) would both be equal to P (Xn+1 = k + 1 | Xn = k). The fact they are not equal thus shows that {Xj } is not Markov. (1.8) Suppose i1 , .., in are equal each to 1, including k 1's and n - k -1's. Then (using the Beta distribution formula to calculate the integral), P (X1 = i1 , .., Xn = in , Xn+1 = 1) = E [P (X1 = i1 , .., Xn = in , Xn+1 = 1 | )] = E k+1 (1 - )n-k 1 = 0 k+1 (1 - )n-k d (k + 1)!(n - k)! , (n + 2)! k!(n - k)! , (n + 1)! = and similarly P (X1 = i1 , .., Xn = in ) = so P (Xn+1 = 1 | X1 = i1 , .., Xn = in ) = P (X1 = i1 , .., Xn = in , Xn+1 = 1) k+1 = . P (X1 = i1 , .., Xn = in ) n+2 (2) +n Since Sn = k - (n - k) = 2k - n, we have k = Sn2 , meaning the left side of (2) depends only on Sn ; therefore (2) is equal to P (Xn+1 = 1 | Sn = 2k - n), and {Sn } is Markov. Thus (2) says P (Sn+1 = 2k - n + 1 | Sn = 2k - n) = P (Xn+1 = 1 | Sn = 2k - n) = 3 k+1 , n+2 or letting j = 2k - n, P (Sn+1 = j + 1 | Sn = j) = n+j+2 . 2(n + 2) Since this transition probability depends on n, the Markov chain {Sn } is time-inhomogeneous. (A)(i) Clearly (0) = 1 and limt (t) = 0. Consider t 0; then (t) = -q (t) (1 + q(t))2 Calculating by differentiating this gets messy, so instead we will consider the increasing/decreasing and positive/negative properties of the numerator and denominator. Letting (t) = log c + 1 , t 1 q (t) = t log c + t -1 - t1 1 2 1 + log c + t c+ t = (t) - (t)-1 . ct + 1 1 Note (t) > 0 for all t > 0, since c > 1. Using (t) = - t(ct+1) we get q (t) = (t)-2 ct + 1 -1 + (t) t(ct + 1) c 1 - ct + 1 t , which is negative since both terms in the parentheses are negative. Thus q is decreasing, and since limt q (t) = (log c) > 0, q (t) is also positive for all t > 0, i.e. q is increasing. Hence in the above formula for , the numerator is increasing and negative, while the denominator is increasing and positive, so is increasing and negative. Therefore is convex. By the Polya Criterion (3.10 in Chapter 2, p. 102), is a characteristic function. (ii) The problem should say show E|i | = , not Ei = . This follows from the fact that a finite mean would imply (by Dominated Convergence) that is differentiable, but here is not differentiable at 0 since is an even function satisfying (1 - (t))/t as t 0. To prove recurrence we use Theorem 2.9 of Chapter 3 (p. 189). Note is real-valued. 4 We have (using the change of variable u = log 2 ): t - 1 dt = 1 - (t) 1+ - 1 q(t) dt = 2 + 2 0 1 t log c + 1 t dt 1 2 0 t log t 1 =2 du 2/ u 2 = . dt Here in the inequality we assumed that < 1/c, so c + 1/t 2/t for all t (0, ]. It follows from Theorem 2.9 of Chapter 3 that the random walk is recurrent for all 1 and c > 1. (B) Let Tk be the kth index n 1 for which Sn 0, Sn+1 > 0, with Tk = if there is no such index. Then Tk is a stopping time, so by the Strong Markov property, on the event {Tk < }, P0 (Tk+1 < | FTk ) = P0 (T1 Tk < | STk ) = PSTk (T1 < ). Since STk > 0, we have PSTk (T1 = ) PSTk (Sn > STk for all n > Tk ) = P0 (Sn > 0 for all n 1) = > 0. (This defines .) Hence P0 (Tk+1 < | FTk ) 1 - , which implies P0 (Tk+1 < | Tk < ) 1 - , and then, inductively, P0 (Tk+1 < ) (1 - )k P0 (T1 < ) (1 - )k+1 . Thus P n 1{Sn 0,Sn+1 >0} k = P (Tk < ) (1 - )k , so P (Sn 0, Sn+1 > 0) = E n n 1{Sn 0,Sn+1 >0} < . 5
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