HW3 - Mathematics Department, UCLA T. Richthammer spring...

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Mathematics Department, UCLA T. Richthammer spring 09, sheet 3 Apr 10, 2009 Homework assignments: Math 170B Probability, Sec. 1 19. Calculate the expectation and the variance for the sum of 12 rolls of a fair die. Answer: X = X 1 + . . . + X 12 , where the X i are the values of the 6 rolls. The X i are independent and we can calculate E ( X i ) = 1 6 (1 + . . . + 6) = 7 2 and E ( X 2 i ) = 1 6 (1 2 + . . . + 6 2 ) = 91 6 , so V ( X i ) = 35 12 . So E ( X ) = 12 E ( X i ) = 42 and V ( X ) = 12 V ( X i ) = 35. 20. Let X n be the time of the n -th success in independent trials with success rate p . (a) Calculate E ( X n ) and V ( X n ). (Hint: Use 08(c)) (b) Explain briefly why you would expect ρ ( X n , X n + k ) 0, then calculate it. Answer: (a) By 08(c) we have that X n = Y 1 + . . . + Y n , where the Y i are independent geometric RVs. Since E ( Y i ) = 1 p and V ( Y i ) = 1 - p p 2 for all i , we get E ( X n ) = E ( Y 1 ) + . . . + E ( Y n ) = n p and V ( X n ) = V ( Y 1 ) + . . . + V ( Y n ) = n (1 - p ) p 2 . (b) If X n is known to be large, then X n + k is more likely to be large than if X n is known to be small. Thus X n , X n + k should be positively correlated. Using 08(c), Cov ( X n , X n + k ) = Cov ( Y 1 + . . . + Y n , Y 1 + . . . + Y n + k ) = n i =1 n + k j =1 Cov ( Y i , Y j ) = n i =1 Cov ( Y i , Y i )+ i ± = j Cov ( Y i , Y j ) = n i =1 V ( Y i ) + 0 = n (1 - p ) p 2 , where we have used that independent RVs are uncorrelated. So ρ ( X n , X n + k ) = Cov ( X n ,X n + k ) V ( X n ) V ( X n + k ) = q n n + k . 21. Suppose there are two methods (A, B) for measuring the distance from the earth to the moon. A scientist using method A gets the values 402, 405, 399, 398, 399 (in 1000 km). A second scientist using method A gets the values 407, 399, 400, 398, 399, 400, 397, 402, 401, 397. A third scientist using method B gets the values 399, 400, 402, 399, 402, 399. (a) What estimate does each scientist get for the true value of the distance? (b) Explain briefly why you would expect the estimate of the second scientist to be better than the estimate of the first scientist. (c) Which method seems to be better? Confirm your answer by a calculation. Answer: (a) The estimate for the true value is the sample mean ¯ X = X 1 + ... + X n n . Here we get 400 . 6 , 400 and 400 . 2 respectively. (b) The average deviation of ¯ X from the true value is given by V ( ¯ X ). By the lecture V ( ¯ X ) = 1 n V ( X i ). So a larger n gives a smaller deviation. (c) The variance of the measurement can be estimated by the sample variance ¯ V = ( X 1 - ¯ X ) 2 + ... +( X n - ¯ X ) 2 n - 1 . Here we get 8 . 3 , 8 . 7 , 2 . 2 respectively. So the second method is better.
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22. Show that for a sample of size n as in 1.3 of the lecture X i - ¯ X and ¯ X
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This note was uploaded on 07/07/2010 for the course MATH 170B 170B taught by Professor Richthammer during the Winter '10 term at UCLA.

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HW3 - Mathematics Department, UCLA T. Richthammer spring...

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