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Unformatted text preview: Mathematics Department, UCLA T. Richthammer spring 09, sheet 4 Apr 17, 2009 Homework assignments: Math 170B Probability, Sec. 1 33. Let X 1 , . . . , X n a Bernoulli sequence with success rate p (i.e. the X i are independent, have values 0 or 1 and P ( X i = 1) = p ). Let Y denote the number of occurrences of 11s in the sequence. For example, for the sequence 1111101101 we have Y = 5. (a) What is the expected number of 11s and what is its variance? (b) What is the probability that there is no 11 in a sequence of length n = 5? Check your results in the extremal cases p = 0 and p = 1. Answer: (a) We have Y = n 1 ∑ i =1 1 A i , where A i is the event that there is are 1s at positions i and i + 1. We have E ( Y ) = ∑ i P ( A i ) = ∑ i p 2 = ( n 1) p 2 . To calculate P ( A i 1 ∩ A i 2 ) we have to distinguish whether i 2 = i 1 + 1, in which case we get P ( A i 1 ∩ A i 2 ) = p 3 , or whether i 2 > i 1 + 1, in which case we get P ( A i 1 ∩ A i 2 ) = p 4 . So E ( ( Y 2 ) ) = ∑ i 1 <i 2 P ( A i 1 ∩ A i 2 ) = ( n 2) p 3 + ( ( n 1 2 ) ( n 2)) p 4 = ( n 2) p 3 + ( n 2 2 ) p 4 , and we get V ( Y ) = 2( n 2) p 3 + 2 ( n 2 2 ) p 4 + ( n 1) p 2 ( n 1) 2 p 4 = ( n 1) p 2 + (2 n 4) p 3 (3 n 5) p 4 . If p = 0 we have E ( Y ) = 0 and V ( Y ) = 0, and if p = 1 we have E ( Y ) = n 1 and V ( Y ) = 0, which makes sense. (b) P ( Y = 0) = 4 ∑ l =0 ( 1) l ∑ i 1 <...<i l P ( A i 1 ∩ . . . ∩ A i l ) = 1 ∑ i P ( A i ) + ∑ i 1 <i 2 P ( A i 1 ∩ A i 2 ) ∑ i 1 <i 2 <i 3 P ( A i 1 ∩ A i 2 ∩ A i 3 )+ P ( A 1 ∩ . . . ∩ A 4 ). To calculate the probabilities of the intersection we have to distinguish how many of the indices are consecutive. We get P ( A i ) = p 2 , P ( A i 1 ∩ A i 2 ) = p 3 if the indices are consecutive and p 4 otherwise, P ( A i 1 ∩ A i 2 ∩ A i 3 ) = p 4 if all 3 indices are consecutive and p 2 · p 3 = p 5 otherwise, and P ( A 1 ∩ . . . ∩ A 4 ) = p 5 . Thus P ( Y = 0) = 1 4 p 2 + (3 p 3 + 3 p 4 ) (2 p 4 + 2 p 5 ) + p 5 = 1 4 p 2 + 3 p 3 + p 4 p 5 . For p = 0 we get 1 and for p = 1 we get 0, which makes sense. 34.* Let X 1 , . . . , X n be independent continuous RVs that all have the same distribution. We say that a record occurs at time k if X k > X 1 , . . . , X k 1 . Let Y denote the number of records. Show that E ( Y ) ∼ log n and V ( Y ) ∼ log n for n → ∞ . Answer: We have Y = ∑ i 1 A i , where A i is the event that a record occurs at time i . E ( Y ) = ∑ i P ( A i ). In order to calculate the probability of A i = { X i > X 1 , . . . , X i 1 } we observe that with probability 1 the values of X 1 , . . . , X i are all distinct (since their distribution is continuous), and because they have the same distribution each one is the largest one with the same probability, so P ( A i ) = 1 i , which implies E ( Y ) = n ∑ i =1 1 i ....
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 Winter '10
 RICHTHAMMER
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