# Hw 7 - Stat 116 Homework 7 Solution 1(a)Show that if Xi...

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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 0 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 0 E v X dv v = e t = Z 0 -∞ e t E e tX dt = Z 0 -∞ 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 < 0 . 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. 1

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solution: E ( X k ) = k α 1 2 - k α 1 2 = 0 V ar ( X k ) = E ( X 2 k ) - ( E ( X k )) 2 = 1 2 ( k 2 α + k 2 α ) = k 2 α When α 0, V ar ( X k ) = k 2 α 1,by Chebyshev’s inequality we know P ( | n i =1 X i | ≥ n ) V ar ( X i ) n 2 2 n n 2 2 = 1 n 2 0 as n → ∞ .
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