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Unformatted text preview: STAT 333 Assignment 2 SOLUTIONS 1. Consider a sequence of independent tosses of a fair coin. Each toss results in a head H or a tail T. Let λ be the event that the total number of H equals exactly onethird the number of tosses, i.e. we say λ occurs on the n th toss if the total number of H up to and including the n th toss equals n/3. a. Give an argument that λ is a renewal event The first time we are waiting for this event to occur, if we want it to happen on trial 3n, we will need exactly n Heads and 2n Tails . Once the event occurs, say on trial 3k, we know there are exactly k Heads and 2k Tails so far. Now if we want it to occur 3n trials later (on trial 3k+3n), we need a total of k+n Heads and 2(k+n) Tails in the total 3(k+n) flips, which means we need an additional n Heads and 2n Tails in the next 3n trials. In other words, we need the exact same conditions for the event to occur on trial 3n the very first time as after an earlier occurrence. b. What is the period of λ ? It can only occur on a multiple of three trials, so the period is d=3 c. Prove that lambda is transient. Why does this make logical sense when you consider the law of large numbers? (that with a large number of trials, the percentage of successes will approach the probability of a success) There are a couple of ways to do this. We can see that the renewal sequence is r =1, r 3n+1 = 0, r 3n+2 = 0, and r 3n = (3n C n) (½) 3n for all integer n. Using Stirling’s approximation, 3nCn = (3n)!/(2n!)n! ≈ (3n) 3n e3n √(2π3n) / [ (2n) 2n e2n √(2π2n) (n) n en √(2πn) ] = c (27/4) n /√n So E[V λ ] = c Σ ((27/4)*(½) 3 ) n /√n = c Σ 0.84375 n / √n < ∞ since 0.84375 < 1. Since the expected number of occurrences is finite, λ is transient. This makes sense because as the number of trials approaches infinity, the proportion of heads will approach ½. It would be impossible for the proportion to hit 1/3 infinitely often since it is converging to ½. 2. Suppose a sequence of independent trials X 1 ,X 2 , . . . is generated by randomly picking digits (with replacement) from the set {0, 1, 2, . . ., 9}. Prove that the event “2 3 5” is recurrent by: a. Finding r n = P(“2 3 5” occurs on trial n) and showing that Σr n = ∞ r = 1, r 1 = r 2 = 0, and r n = (1/10) 3 = 0.001 for n ≥ 3 E[V 235 ] = Σ ∞ n=1 r n = 0 + 0 + Σ ∞ n=3 0.001 = ∞ Hence “235” will occur infinitely often and is recurrent. b. Obtaining the generating function R 235 (s) of the renewal sequence {r n }, finding the pgf F 235 (s) of the first waiting time T 235 and evaluating F 235 (s) at s = 1 to prove recurrence. R 235 (s) = Σ ∞ n=0 r n s n = 1 + 0 + 0 + Σ ∞ n=3 0.001s n = 1 + 0.001s 3 /(1s) F 235 (s) = 1 – 1/R 235 (s) = 1 – (1s)/(1 – s + 0.001s 3 ) = (0.001s 3 )/(1 – s + 0.001s 3 ) f 235 = F 235 (1) = 0.001/(11+0.001) = 1 so “235” is recurrent....
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This note was uploaded on 03/30/2012 for the course STAT 333 taught by Professor Chisholm during the Winter '08 term at Waterloo.
 Winter '08
 Chisholm

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