# 611hw1sol - Homework 1 Solutions 7 Let M n = max X 1,X n...

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Unformatted text preview: Homework 1 Solutions 7. Let M n = max { X 1 ,...,X n } . For x ∈ R + , P( M n /n ≤ x ) = Y i P( X i /n ≤ x ) = Z nx 1 (1 + t ) 2 dt n = 1- 1 1 + nx n = ( 1- 1 1 + nx 1+ nx ) n 1+ nx n →∞-→ e- 1 x 11. Let Y n := ln ˜ X n = 1 n ∑ n i =1 ln X i . Then, Eln X i = Z 1 ln xdx = ( x ln x- x ) | 1 =- 1 E(ln X i ) 2 = [ln x ( x ln x- x )] 1- Z 1 (ln x- 1) dx =- [ x ln x- x- x ] 1 = 2 Var(ln X i ) = E(ln X i ) 2- (Eln X i ) 2 = 1 . By the CLT, √ n ( Y n + 1) ⇒ N (0 , 1). a) By the law of large numbers, Y n p → Eln X i =- 1. By the continuous mapping theorem, ˜ X n = exp { Y n } p → 1 /e . b) By the delta method, lim n →∞ √ n ( ˜ X n- 1 /e ) d = (1 /e ) lim n →∞ √ n ( Y n + 1) ∼ N (0 , 1 /e 2 ) . 1 12. Let Y i := X- 1 i . Then Y i s are i.i.d. and E Y i = Z 2 1 1 x dx = ln2 E Y 2 i = Z 2 1 1 x 2 dx = 1 / 2 Var( Y i ) = 1 / 2- (ln2) 2 . By CLT, √ n ( 1 n ∑ n i =1 Y i- ln2) ⇒ N (0 , 1 / 2- (ln2) 2 ). a) Since f ( x ) = 1 /x is continuous at ln2, by the law of large numbers and continuous mapping theorem, H n = 1 1 n ( Y 1 + ··· + Y n ) p → 1 / ln2 . b) By the delta method, lim n →∞ √ n ( H n- 1 / ln2) d = f (ln2) lim n →∞ √ n ( 1 n n X i =1 Y i- ln2) ∼ N , 1 / 2- (ln2) 2 (ln2) 4 ....
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611hw1sol - Homework 1 Solutions 7 Let M n = max X 1,X n...

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