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Assignment 1- solution

# Assignment 1- solution - THE UNIVERSITY OF HONG KONG...

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT 1302 PROBABILITY AND STATISTICS II (2009-10) Assignment 1 solutions 1. (a) P M n > x n · = P min { X 1 , . . . , X n } > x n · = h P X 1 > x n ·i n = 1 - x n · n . (b) Consider the c.d.f. of nM n = F n ( x ), F n ( x ) = P ( nM n x ) = 1 - P ( nM n > x ) = 1 - P M n > x n · = 1 - 1 - x n · n . As n → ∞ , 1 - ( 1 - x n ) n 1 - e - x , i.e. F n ( x ) 1 - e - x (which is the exponential distribution function of unit rate). Therefore, nM n converges in distribution to a exp(1) random variable. 2. (a) By Weak Law of Large Numbers, ¯ X D -→ μ = E [ X 1 ] = Z 1 0 ( x · 1) dx = x 2 2 fl fl fl fl 1 0 = 1 2 , n → ∞ . (b) By Weak Law of Large Numbers, n i =1 X 2 i n D -→ μ X 2 i = E [ X 2 1 ] = Z 1 0 ( x 2 · 1) dx = x 3 3 fl fl fl fl 1 0 = 1 3 , n → ∞ . (c) Note that E [ X 1 ] = 1 / 2, Var( X 1 ) = E [ X 2 1 ] - ( E [ X 1 ]) 2 = 1 / 12. By Central Limit Theorem, n ( ¯ X - 1 / 2) p 1 / 12 D -→ N (0 , 1) n ( ¯ X - 1 / 2) D -→ N (0 , 1 / 12) , n → ∞ . 1

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(d) Note that n - 1 / 2 n i =1 ( X 2 i - 1 / 3) = n ( ¯ X 2 i - 1 / 3), E [ X 2 1 ] = 1 / 3, Var( X 2 1 ) = E [ X 4 1 ] - ( E [ X 2 1 ]) 2 = 1 / 5 - (1 / 3) 2 = 4 / 45.
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Assignment 1- solution - THE UNIVERSITY OF HONG KONG...

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