solutions5

# solutions5 - w|q6"xc}|8~c|c{\$bwu si A qE z y x v tr f...

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Unformatted text preview: w|q6"xc}|8~c|c{\$bwu si A qE } z y x v tr f f k i u"R i pH S Q n m R f f i ubR i lH R k n m i h jB g f I W R `"e W `d H H A ( C R ( E W ( 8`1) ( ( g0) I A Y A a A Y A a a Y W R T cb`X H 7 r q Y W ( fs`rpih Q Y W gfX`U Y W R b`"d I cp6 y x5v y x5v Ccpw% t u2 H H ( ) Q ( S Q R H A E 2 a Y W R T edcb`XV UH ( 10) G A ( 1CB) ' ( 10) ' I PA @ A A E [email protected] 7 &\$" 6 C4 %# ! 5 A 7 %# ! 5 @9 8&\$" 6 432 %# ! &\$" Chebyshev's inequality gives, Chapter 5, Problem 1. Since Chapter 5, Problem 4. If Chapter 5, Problem 2. Homework 5: Solutions then 1 f i h 1 o g Q . So, , . Hence, ( C Q R ( S Q R t i 5 t p6 i v i x t 43 ' & 5 'A I ( g)A T H q A ) ( 5 " A T ! q A 5 )( ' i ! A 5 " () ( ! q A )( ' i ! A 5 " ( ( ! q A )( ' i ! A " ( ( A 5 ! q B( i ! \$' A %" ( 5 ! \$' A A i "# 5 ( ! ' Ai ( A i H A i PPI 5 I I A 5 ( A ( ( H q A A H Ai ( ' H H A ( A i I i A ( H )( Ai i n completeness, here it is. Let us write lowing fact. If ' Thus, I a PA @ A T H A ) ' ( H 5 1 2 v ' i H A ( ' and the second sum has Now, where the last line follows from plugging in the definition of Chapter 6, Problem 1. (a) We showed this in a previous homework. For are real numbers then . To compute 2 5 e A i ' 0 terms. So, we will make use of the fol. A H i T q i G ( H i ) ' ( ! i ) G ( i T I A A H i T cuf E q A H i T @q A i H i ) ' T ( ' H i T @fs i H i ) ' T q A ( T i H i ) ' ( H i T s i H i cuq A T ( E A H i T q A i H i T ( H i T i ) ( 5 I a PA ` A UH A ) ' T ( H Ai ( 5 H H A ( i ! H A T A T H i ( 5 5 I A ) ' ( ! H A ( H i 5 5 I A ) ' ( ! A ( i I Ai ( A 5 Ai H H o( i ! H e From this last expression we see that if A H i ' ( H i T q i ) G ( T q i H i ) ' ( i T H Ai ( Also, by the law of large numbers, H Since By the law of large numbers, function, Chapter 6, Problem 2. Let (b) We can write . Conversely, if " A H i T and A s . So, , we also have that . 3 i ) G ( A then . Then, . Since and is a continuous then . Hence, H i ) ' ( I E g m " # Q m I I !I H i ( Q A a T i ( I f 5 ! H A ( i f I I f f A f H ' ( H H su A q A s3 A ) ' ( i ( S q H i ) ' ( n s H A H i ( m A r A q S A H r A A A i ) ' ( g A n q n m C3R i ( S Q R i ) Q ( m fR i ( C Q R m n so Chapter 6, Problem 4. Fix Chapter 6, Problem 5. f A A ) ' ( H f ( ) ' g A ) G ( does not converge in quadratic mean. . Thus, . So f f f f f f f H C s q So, 5 ! A ( i ' and H 5 AH ! ( i 5 ! f A ( i G So, Since convergence in quadratic mean implies convergence in probability, By problem 2, Chapter 6, Problem 6. By the central limit theorem, . Then, for all 4 !I H m I H i ( Q m i ( Q and so , . But . . #! "4 i i ( " #! " H H Q Q H H H H H 5 ) H 8 !x ~ v ) H 8~ v x H 5 i H i m " Q 5 i m 5 " Q m i " 5 m i " PPI " }| Q H I I I I 5 " }| i " PPI II i " sPI 5 " }| i ( " s A H i piW A ( H i ) ( ( ( H i ( W u( R H i R( ( @ i ( gCn m "R by L'Hopital's rule. The density of ( H i ( H i ( S Q R and hence Chapter 6, Problem 13. Let Chapter 6, Problem 9. . Hence, . Now be the CDF of . Then, the CDF of is thus 5 when . . . Otherwise is i ) Q ( H i ...
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## This note was uploaded on 05/25/2008 for the course STAT 36-625 taught by Professor Larrywasserman during the Fall '02 term at Carnegie Mellon.

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