Hw 7 - Stat 116 Homework 7 Solution 1. (a)Show that if X i...

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Unformatted text preview: Stat 116 Homework 7 Solution 1. (a)Show that if X i follows a binomial distribution with n i trials and probability of success p i = p , where i = 1 ··· n and the X i are in- dependent, then ∑ n i =1 X i follows a binomial distribuion. Also show that if the p i are unequal, the sum does not follow a binomial dis- tribuion.(Hint: apply moment generating function.) (b)Let X ∼ Bin ( n,p ) show E 1 1+ X = 1- (1- p ) n +1 p ( n +1) ( hint : write 1 1+ x = R 1 t x dt and use the generating function) Solution: (a) E ( e t ∑ X i ) = Y Ee tX i independence of binomial (1) = Y (1- p + pe t ) n i (2) = (1- p + pe t ) ∑ n i (3) We recognize this as a binomial ( ∑ n i ,p ). If p i ’s are unequal, then E ( e t ∑ X i ) = Q (1- p i + p i e t ) n i . When we multiply the expression, it is not of the form (1- p i + p i e t ) n . Hence it is not binomial. (b) E 1 1 + X = Z 1 E v X dv v = e t = Z-∞ e t E e tX dt = Z-∞ e t ( q + pe t ) n dt = 1- q n +1 p ( n + 1) . 2. Ross Page 457, problem 9 solution: We use the Chebyshev’s inequality, P ( | X n- 1 | > . 01) < E ( X n- 1) 2 (0 . 01) 2 = 1 n (0 . 01) 2 . Hence in order to get 1 n (0 . 01) 2 < . 01, we need n > 100. 3. X k are independent and identically distributed as P ( X k = k α ) = P ( X k =- k α ) = 1 2 ,k = 1 , 2 , ··· . Use Chebyshev’s inequality to show that (a) for α ≤ 0 , ∑ n i =1 X i n → 0 in probability as n → ∞ (b) show that for α = 1 4 , the result of (a) still holds....
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This note was uploaded on 01/13/2010 for the course STATS 116 taught by Professor Staff during the Spring '07 term at Stanford.

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Hw 7 - Stat 116 Homework 7 Solution 1. (a)Show that if X i...

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