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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 avgt 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 _ _1u D W 1 n
7.2a ) Engx; 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(X9)] 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 [éiﬂxiﬂﬂdx = 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 + (11;) 105(1 p)  ..
n . —(l ._ w .. __
 E 1—: :‘z 3%“ 111—13 2 punwp) hula} Mumativeiy, we could calculate nEs(§§§}ogf(xlﬂ))
_ ._ 33., x _ 1K _. _,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 11)) 5 w
FREE mlogﬂxlﬂ) We know that £5: = 9. Thus, it attains the CreamerB30 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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 Spring '07
 SENTURK,DAMLA
 Statistics

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