This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full
Document
This note was uploaded on 01/13/2010 for the course STATS 116 taught by Professor Staff during the Spring '07 term at Stanford.
 Spring '07
 Staff
 Binomial, Probability

Click to edit the document details