p3soln09

# p3soln09 - STAT 330 SOLUTIONS PART III 45(a Note that n P...

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STAT 330 SOLUTIONS PART III 45 . ( a ) Note that n P i =1 ¡ X i ¯ X ¢ = μ n P i =1 X i n ¯ X =0 and n P i =1 ( s i ¯ s )=0 . Therefore n P i =1 t i X i = n P i =1 ³ s i ¯ s + s n ´ ¡ X i ¯ X + ¯ X ¢ (1) = n P i =1 h s i ¡ X i ¯ X ¢ + ³ s n ¯ s ´ ¡ X i ¯ X ¢ +( s i ¯ s ) ¯ X + s n ¯ X i = n P i =1 s i U i + ³ s n ¯ s ´ n P i =1 ¡ X i ¯ X ¢ + ¯ X n P i =1 ( s i ¯ s )+ s ¯ X = n P 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 μ n P 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 μ μ n P i =1 t i + 1 2 σ 2 n P i =1 t 2 i . Therefore by (1) and (2) E exp μ n P i =1 s i U i + s ¯ X ¶¸ = E exp μ n P i =1 t i X i ¶¸ (3) x p μ μ n P i =1 t i + 1 2 σ 2 n P i =1 t 2 i . 45 . ( b ) n P i =1 t i = n P i =1 ³ s i ¯ s + s n ´ = n P i =1 ( s i ¯ s n s n =0+ s = s (4) n P i =1 t 2 i = n P i =1 ³ s i ¯ s + s n ´ 2 (5) = n P i =1 ( s i ¯ s ) 2 +2 s n n P i =1 ( s i ¯ s n n P i =1 ³ s n ´ 2 = n P i =1 ( s i ¯ s ) 2 +0+ s 2 n = n P i =1 ( s i ¯ s ) 2 + s 2 n 1

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45 . ( c ) M ( s 1 ,...,s n ,s )= E exp μ n P i =1 s i U i + s ¯ X ¶¸ = E exp μ n P i =1 t i X i ¶¸ =e x p μ μ n P i =1 t i + σ 2 2 n P i =1 t 2 i by (3) x p ½ μ s + 1 2 σ 2 n P i =1 ( s i ¯ s ) 2 + s 2 n ¸¾ by (4) and (5) x p μ s + 1 2 σ 2 μ s 2 n ¶¸ exp 1 2 σ 2 n P i =1 ( s i ¯ s ) 2 ¸ 45 . ( 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 P 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 . 3 , ¯ X and U are independent random variables which implies ¯ X and n P i =1 U 2 i = n P i =1 ¡ X i ¯ X ¢ 2 are independent random variables. 2
46 . 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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47 . 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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p3soln09 - STAT 330 SOLUTIONS PART III 45(a Note that n P...

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