p3soln08

# p3soln08 - STAT 330 SOLUTIONS PART III 42.(a) Note that n X...

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STAT 330 SOLUTIONS PART III 42 . ( a ) Note that n X i =1 ¡ X i ¯ X ¢ = Ã n X i =1 X i ! n ¯ X =0 and n X i =1 ( s i ¯ s )=0 . Therefore n X i =1 t i X i = n X i =1 ³ s i ¯ s + s n ´ ¡ X i ¯ X + ¯ X ¢ (1) = n X i =1 h s i ¡ X i ¯ X ¢ + ³ s n ¯ s ´ ¡ X i ¯ X ¢ +( s i ¯ s ) ¯ X + s n ¯ X i = n X i =1 s i U i + ³ s n ¯ s ´ n X i =1 ¡ X i ¯ X ¢ + ¯ X n X i =1 ( s i ¯ s )+ s ¯ X = n X i =1 s i U i + s ¯ X. Also since X 1 ,X 2 ,...,X n are independent N ¡ μ , σ 2 ¢ random variables E " exp Ã n X i =1 t i X i !# = n Y i =1 E [exp ( t i X i )] = n Y i =1 exp μ μ t i + 1 2 σ 2 t 2 i (2) =e x p Ã μ n X i =1 t i + 1 2 σ 2 n X i =1 t 2 i ! . Therefore by (1) and (2) E " exp Ã n X i =1 s i U i + s ¯ X !# = E " exp Ã n X i =1 t i X i !# (3) x p Ã μ n X i =1 t i + 1 2 σ 2 n X i =1 t 2 i ! . 42 . ( b ) n X i =1 t i = n X i =1 ³ s i ¯ s + s n ´ = n X i =1 ( s i ¯ s n s n =0+ s = s (4) 1

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n X i =1 t 2 i = n X i =1 ³ s i ¯ s + s n ´ 2 (5) = n X i =1 ( s i ¯ s ) 2 +2 s n n X i =1 ( s i ¯ s )+ n n X i =1 ³ s n ´ 2 = n X i =1 ( s i ¯ s ) 2 +0+ s 2 n = n X i =1 ( s i ¯ s ) 2 + s 2 n 42 . ( c ) M ( s 1 ,...,s n ,s )= E " exp Ã n X i =1 s i U i + s ¯ X !# = E " exp Ã n X i =1 t i X i !# =e x p Ã μ n X i =1 t i + σ 2 2 n X i =1 t 2 i ! by (3) x p ( μ s + 1 2 σ 2 " n X i =1 ( s i ¯ s ) 2 + s 2 n #) by (4) and (5) x p μ s + 1 2 σ 2 μ s 2 n ¶¸ exp " 1 2 σ 2 n X i =1 ( s i ¯ s ) 2 # 42 . ( d ) Since M ¯ X ( s M (0 ,..., 0 )=exp μ s + 1 2 σ 2 μ s 2 n ¶¸ and M U ( s 1 n M ( s 1 n , 0) = exp " 1 2 σ 2 n X i =1 ( s i ¯ s ) 2 # we have M ( s 1 n M ¯ X ( s ) M U ( s 1 n ) and therefore by Theorem 3 . 8 . 2 , ¯ X and U are independent random variables which implies ¯ X and P n i =1 U 2 i = P n i =1 ¡ X i ¯ X ¢ 2 are independent random variables. 2
44 . G n ( y )= P ( Y n y P (min ( X 1 ,...,X n ) y ) =1 P ( X 1 > y,. ..,X n >y ) n Y i =1 P ( X i ) since X 0 i s are independent random variables Z y e ( x θ ) dx ¸ n e n ( y θ ) ,y > θ (6) Since lim n →∞ G n ( y ( 1 if y> θ 0 if y< θ therefore Y n p θ and by the Limit Theorems U n = Y n θ p 1 . Now P ( V n v P ( n ( Y n θ ) <v P ³ Y n v n + θ ´ e n ( v/n + θ θ ) using (6) e v ,v 0 which is the c.d.f. of an EXP (1) random variable. Therefore V n v EXP (1) for n , 2 ,... which implies V n D V v EXP (1) . Since P ( W n w P ¡ n 2 ( Y n θ ) <w ¢ = P ³ Y n v n 2 + θ ´ e n ( w/n 2 + θ θ ) e w/n ,w 0 therefore lim n →∞ P ( W n w )=0 for all w < which is not a c.d.f. Therefore W n has no limiting distribution. 3

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45 . We f rst note that P ( Y n y )= P (max ( X 1 ,...,X n ) y ) = P ( X 1 y,. ..,X n y ) = n Y i =1 P ( X i y ) since X 0 i s are independent random variables = n Y i =1 F ( y ) =[ F ( y )] n ,y support of X i . (7) Therefore G n ( z P ( Z n z ) = P ( n [1 F ( Y n )] z ) , 0 <z<n = P ³ F ( Y n ) 1 z n ´ = P ³ Y n F 1 ³ 1 z n ´´ since F is increasing function imples F 1 is also an increasing function. Thus G n ( z )=1 P ³ Y n <F 1 ³ 1 z n ´´ =1 h F ³ F 1 ³ 1 z n ´´i n by (7) ³ 1 z n ´ n , 0 <z<n.
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## This note was uploaded on 04/27/2010 for the course STAT 330 taught by Professor Paulasmith during the Fall '08 term at Waterloo.

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p3soln08 - STAT 330 SOLUTIONS PART III 42.(a) Note that n X...

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