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Unformatted text preview: Mathematics Department, UCLA T. Richthammer winter 09, sheet 7 Feb 13, 2009 Homework assignments: Math 170A Probability, Sec. 1 079. For a geometric RV X show that P ( X = n + k  X > n ) = P ( X = k ) by a calculation. Can you explain without any calculation why this should be true? Answer: P ( X > n ) = k = n +1 (1 p ) k 1 p = p (1 p ) n k =0 (1 p ) k = p (1 p ) n (1 (1 p )) = (1 p ) n , so P ( X = n + k  X > n ) = P ( X = n + k ) P ( X>n ) = p (1 p ) n + k 1 (1 p ) n = p (1 p ) k 1 = P ( X = k ). This makes sense because X describes the time of the first success in a Bernoulli sequence X 1 , X 2 , . . . : Let Y denote the waiting time for the first success after time n . Then { X > n } = { X 1 = 0 , . . . , X n = 0 } and { X = n + k } = { X 1 = 0 , . . . , X n = 0 , Y = k } , and in particular { X > n } describes the behavior of the first n experiments and { Y = k } describes the behavior after the first n experiments, so these events are independent. Thus P ( X = n + k  X > n ) = P ( Y = k  X > n ) = P ( Y = k ) = P ( X = k ), where in the last step we used that both X and Y are geometric RVs and thus have the same distribution. 080. Let X be a geometric RV with parameter p . Calculate E ( X ( X + 1)( X + 2)), and use this to determine the third moment. Answer: E ( X ( X + 1)( X + 2)) = k 1 k ( k + 1)( k + 2) pq k 1 = d 3 dq 3 k 1 pq k +2 = p d 3 dq 3 k q k = p d 3 dq 3 1 1 q = 6 p (1 q ) 4 = 6 p 3 . So E ( X 3 ) = E ( X ( X +1)( X +2) 3 X ( X +1)+ X ) = 6 p 3 6 p 2 + 1 p . 081. A fair coin is flipped several times. Calculate the probabilities for the following events: (a) In the first 10 coin flips we get at least 3 heads. (b) The first head appears at the 10 th flip or later. (c) The second head appears at the 10 th flip. Answer: (a) The number X of heads among the first 20 coin flips is a binomial RV with parameters n = 20 , p = 1 2 , so P ( X 3) = 1 P ( X < 3) = 1 2 k =0 ( 10 k ) 1 2 10 = 1 56 1024 = 94 . 5%. (b) The waiting time Y for the first success is a geometric RV with parameter 1 / 2, so P ( Y 10) = k =10 1 / 2 k = 1 / 512 . 2%. (c) It doesnt matter what happens after the 10 th flip, so we can choose S = { , 1 } 10 and P to be the equidistribution. The given event A has 9 elements (the first head can be the first, second,. . . ,9 th flip), so P ( A ) = 9 / 1024 = 0 . 9%. 082. Suppose X 1 , X 2 , . . . is a Bernoullisequence with parameter p . Determine the PMF and the expectation of (a) the waiting time for the second success. (b*) the waiting time for the rth success ( r 1). Answer: Let Y r denote the waiting time for the rth success....
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 Winter '10
 RICHTHAMMER

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