# HW7 - Mathematics Department UCLA T Richthammer winter 09...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 doesn’t 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 Bernoulli-sequence with parameter p . Determine the PMF and the expectation of (a) the waiting time for the second success. (b*) the waiting time for the r-th success ( r ≥ 1). Answer: Let Y r denote the waiting time for the r-th success....
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

HW7 - Mathematics Department UCLA T Richthammer winter 09...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online