solutions3 - X i pyg up' ip (1) } z s { z i b~|Uwiyuvkyg }...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: X i pyg up' ip (1) } z s { z i b~|Uwiyuvkyg } z s p { z ab~|thy P 1 i 2vkyg i s i s p ip uvTxUwvuvrt'srpq' s and are independent then 12PCD9 8 C o D$i H C V9hI CDe e i `9lHk e ie c 8 $i i j m m nC g R e 57 m P'U d q i g e 555 d` "u C i vph# j ic hef 5 6 g 8 4555 H@C i c@ P$`PC C i H@ i DebEC eGC c H C yebH yGC D9e 8 i@C vFPrd 3rwvD BPx3rwvutU3rph# q i @ C q i s q i g 57 555 5 4555 (b) , , and , . So 1c d fe9 C c H H W @ H R A @ C 9"ba`Y3XVFVUTSQFPI H 9"GC A@ FEC CD9" 8 A@ B 8 9 8 6 3# 1 ) # % # 20$('&$"! Chapter 1, Problem 1. Homework 3: Solutions for , On the other hand, suppose that (1) holds. Then (b) The next one is true by definition. Chapter 3, Problem 5. If Chapter 3, Problem 4. (a) 1 E E , correspond to the following zones for : . The three "zones" Y X 8 % 1 '` i i Vvk'D tPQ pT and let denote p'`Vxp' i D i ip 8 $ V U W V P s the range of . Then i ip Qxpq' R P 2Sp Q . U # s khD 8 12b}yzw U'wi{yzy~ G uBABykI I{ !s s D 8 } z s 6 z A } z s p 6 z EB~|tPI{ !s @ z s D p { z 8 !} bU'Hhyy G . Then, F } z i D 7 yphECi { z d 8 7 9 6 { } { C 3 1 # V5 xv4$V 2v$V ' C C V0 xrh$V)# ' (`% I C V& x V%nx C C qV$ sks IXq`$ I Vd kwwV C C s d V#d s !" !s `dtpy d s d VD # !V(nC . Thus, d d sp $t9 s and hence, Chapter 3, Problem 7. Note that and hence and are independent. . Therefore, C u and Thus, Chapter 3, Problem 12. Let Chapter 3, Problem 10. Fix . is independent of and denote the range of 2 Y X # uvkyg i A B } and let if and only if and and for 8 d s C Ep 8 3@ USs 7 T R i i g Tph# 1 i j p ) # g rqvk ) !uves krvekrvky# i # i # i # i s i g C D 8 s Chapter 3, Problem 15. Note that and hence . . Now, 8 2 H 3 2 u`3r0 1 ' "! "! C 2 Db 2 2 Q P ' "! E 7 @ BA @ 87 C AI 1H )G '9 )' A 7@ @ 7 2 FE D' CBA '9 )8' 2 6 s R j "! ' ' t' g u53q q 2 4 2 ' s h' 3q R 2 0 a`3r1# 3 3 1 ' " ) 1 " '` ' # ) 3 ( ) ! C i i up i u' i % % & " $ ) ! C ' i upg ) ! C i i yupe srpe etp 8 has range s i g ksrpy# s Y 8 X V U % ip p' (2) i rvk'D 8 0 Vx V U Qxp'D i i Y 8 X phD i E Y X 8 0 P i uvkyg So 1 i i &Vvk'D `x `x C Also, Thus, So Differentiate to get by (2). Chapter 3, Problem 14. Chapter 3, Problem 13. for . 3 . And, 3 { z sp uSyst9seui . Then . For ui # Now let { C TD PCXp E2Pp 8 Pp 8 p 8 p C 8 s@ v$ W 1 C v&nrV Cp C p Cp p q` iX`xPp 8 `IP2 8 vQ 1 R C v yrV 3 { C a 8 @ ' R C { z s p 2(VWI 29btyqyV$DeyV# { l$hC C p C C p C p P`R QPp P`IP2VIp 8 C DD0hC s . Then $# x y C lT 8 T T #uR # # 3uq$xn ) kyy # C Pp 8 7 T 3@ UR " i ) Tv # For and since Then ' 3 C { sp C V&nI 2y@ z y V$uVQ)# Thus, 8 1C D3 8 8 hC C R C V triangle with corners and 1 C3 ui and that , let Chapter 3, Problem 20. Let Let be the triangle with corners at . , let for every where and note that be the triangle with corners at 4 . So, . Then, . Then . For , let which means be the . 8 1 c 18 1 Y A iR Y A C H H A A d HY D YR Y D H u!YS s R d a d (e) (d) .37 ui # C d D For , let 3 C { sp nxSy@ z bt9yui C P`IP2 8 `xE2 8 p 8 @Cp Cp p Cp { and 1 C ' 3 ui So, 1C o `DD ' C D 8 3 @ vPC 3 PC @ ui This is equal to Chapter 3, Problem 18. (c) 9.58 (b) .89 (a) .84 be the triangle with corners when . 5 . Then ...
View Full Document

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.

Ask a homework question - tutors are online