STAT_333_Test_2_Solutions

STAT_333_Test_2_Solutions - STAT 333 Spring 2010 Test 2...

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STAT 333 Spring 2010 Test 2 SOLUTIONS Tues, July 6 2:30 – 4:00 pm First (given) name:__________________ Last (family) name:____________________ Student ID #:__________________ UW userid:____________________ Instructions: 1. Please fill in the above information 2. This test has 7 pages, including this cover page 3. Answer all questions in the space provided 4. You have 90 minutes for the test 5. Show all your work and justify your steps 6. Good luck! Question Marks available Marks obtained 1 8 2 11 3 10 4 9 5 14 6 8 Total 60

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1. Suppose an infinite sequence of letters is selected randomly from the 26-letter alphabet. [2] a) Find the expected # of trials until the first occurrence of the pattern “E F G E E F G E” Delayed renewal event with maximum overlap “E F G E” So T EFGEEFGE = T EFGE + T EFGEEFGE-bar But “E F G E” is delayed renewal too with maximum overlap “E” So T EFGEEFGE = T E + T EFGE-bar + T EFGEEFGE-bar = 26 + 26 4 + 26 8 using the Renewal Theorem = 208,827,521,578 (2.088275216 x 10 11 is OK) [4] b) Derive the pgf F EFGE (s) of the first waiting time until we observe “E F G E” Delayed renewal sequence: d 0 = 0, d 1 = d 2 = d 3 = 0, d n = (1/26) 4 , for n ≥ 4. Associated renewal sequence: 0 r = 1, 1 r = 2 r = 0, 3 r = (1/26) 3 , n r = (1/26) 4 for n ≥ 4. So D EFGE (s) = [s 4 /26 4 ]/(1 – s) R EFGE-bar (s) = 1 + s 3 /26 3 + [s 4 /26 4 ]/(1 – s) = [1 – s + (1 – s)s 3 /26 3 + s 4 /26 4 ]/(1 – s) So F EFGE (s) = [s 4 /26 4 ]/[1 – s + (1 – s)s 3 /26 3 + s 4 /26 4 ] [2] c) Find the probability that “E F G E” occurs for the first time on trial 8. (Hint: you can do this without expanding F EFGE (s)) We want f 8 = P(“E F G E” occurs for the first time on trial 8) We need _ _ _ _ E F G E but not E F G E E F G E since in that case it has occurred first on trial 4, and not _ E F G E F G E, since in that case it has occurred first on trial 5. F 8 = P(“EGFE” occurs on trial 8)–P(“EFGE” occurs on 4 & 8)–P(“EFGE” occurs on 5 & 8) = (1/26) 4 – (1/26) 8 – (1/26) 7 = 0.0000021882
2. Andrew and Betty are playing separate independent and identical games. On each game, they each have probability 0.3 of winning. We say λ occurs on game n if Andrew and Betty have exactly the same ordered sequence of wins and losses up to and including game n . For example, λ would occur on trial 2 if they both had (W, W) on the first two games, or both had (L, L), or both had (L, W), or both had (W, L). [2]

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STAT_333_Test_2_Solutions - STAT 333 Spring 2010 Test 2...

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