STAT 514 - HW 6 Solution Key

STAT 514 - HW 6 Solution Key - A 2 7{qr b log L(5,0'2Iyi =...

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Unformatted text preview: A . 2 7% {qr b) log L(5,0'2Iyi) = 421— log(21r)—% iogaz—WLZY? +%Efiyi—§é—2xi2. I For a fiat—ed value of a2, 61031, A . . W”%Exiyi"%2x{2 avg-t 0 => 13: 2‘32“. Also a a in aziogL_ 1 2 Wm; in <9, so it is a maximum. Since 3 does not. depend on 02, it is the MLE. And 3 is unbiased because ' ur‘ Efi zmizmz . 2x? 2x? 5 c) f? = )2ng when: ai = xi] are constants. By Corollary 4.6.2, 3 is normally distributed wiih mean ,8, and n 2 E ‘ QED?) (2:3) Ex? 3 _ _1-u D W 1 n 7.2a ) Eng-x; w giEIEYI —- I“; z ,6 EY- n 2 b v 1 z 1 z )3” m M2 “33‘ ) “(i - (Exp? a?“ (taxi)g n22? " as? {In fact, 5 is BLUE (Best Linea: Unbiased Estimator of 5), as discussed in Section 12.2.2.) £4 Egg-£5 logf(X|9)] 2' E9[§§(§§ logfixlg))] 2 _ E a 335mm) “ figqxze) 5535mm) “‘ 9 "5'5 We: a) " E0 Wfixw) ‘ Toma) Now consider the first. term: flifixw) 2 2 a . 34% m I [é-iflxiflfldx = 3d? 53 f(x{9)dx (assumption) $.— adfi E3 [39—9 maxim] -.-.- o - (7.3.7) and the identity is proved. _-.- gri’gffi ' since the crowterm is zero. Hence, Ea? is minimized w __ fit a o --n- 1., _ a . .. ‘ flies W911) —~ 5510:; i5} px‘flmp) x‘ ~— Eiglleosp + (1-1;) 105(1 p) - .. n . —(l- ._ w .. __ - E 1—: :‘z 3%“ 111—13 2 punwp) hula} Mumativeiy, we could calculate -nEs(-§§§}ogf(xlfl)) _ ._ 33., x _ 1-K _. _,n a2 _ - nE(apzlog[p (1 p) ])w [KEEP-*0. XBOSG'PH) =_ns(%[%_£§;§])=_nn(:§ma%§)=-nfiTgfi): may Then using 119): p and +19) = 1, 1’09) 1 filer») — a m 2 V X. 32 n p 1-1)) 5 w FREE mlogflxlfl) We know that £5: = 9. Thus, it attains the Creamer-B30 bound. :1 ' n n , - n - a.) EaiElgmzeip=pZai=p Eencethcestimatorisunbiased. i=1 i=1 i=1 ifil b) V _)n:aixi)=§e12vmg=f;a§a2=a3ia?. ' 1:1 i==1 1:1 i:1 Therefore, we need to 211?, subject to the constraint 231 = I. Add and subtract the mean of the 33, 1/11, to get by choosing a»: z lln for all '1. Thus, 2 (1/11)}{il :2 X has the minimum variance among all linear unbiased estimators. in ...
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