670bexam1soln

# 670bexam1soln - STAT 670B name February 2009 Exam 1 Show...

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Unformatted text preview: STAT 670B name February, 2009 Exam 1 Show all work. 4 Problems. 100 points total. (30pts) 1. Let X1, . . . ,Xn be a random sample from a'EmponentialW) distribution with pdf - f(m) : grim/9,0 S a: < oo, 6 > 0 and E(X) = 0 and 'Var(X) : 02. (a)_Find the maximum likelihood estimator (MLE) of 9, call it 5 (b) Find the Fisher information 1(0) of a single random variable X from an Emponential(0) distribution. _ (c) Find the Fisher information'Inw) of a random sample ’from an Eprnentz'aKQ) dis- tribution. (d) Is the MLE in part (a), a uniform minimum variance unbiased estimator (UMVUE) of 6’? Explain. - (e) Find a complete suﬂiCient statistic for Is the MLE in part (a) a function of the complete sufﬁcient statistic? Explain. ' '\ \Vl 9 m Llele 83 l .P blank page > 0 85-2)? ‘ v _ .L -n‘ +nﬁ? =0 8:” Mom/W47“ 612/ d lav/H5} A ~ ®< g— 3: A 0% 4 O (9 > y <3 =Y NE a m. max: ~ Ewing ms} i%wmh:‘§e‘”®- n = —l‘ 5 - 1 IJQ£HH9\ 0? Q G\%L\$Z. :- -'—\—“— "T l 42%“ 5 8"“ dxzmuﬁw ,: 1.2 — Q'k m” 9 53 : dzlﬂﬂﬂg) : L’AEQYE : +—-— —9\8' : \__. L< cm} > 5L @3 <91 @3 6* - "‘“‘ _ __ " 2 {ﬁx _, h __ J. : J—q 1 ‘ t ‘05 v < ®l\ S2 ow I, 21, Cc} In (9\ 2 H1085 = D. (SM/ti 0» vamch ﬂamﬁe. Cam») ‘ €52 /\ lo T r amok I“ (9 1'5 mmnbnweJaoJ'lth-oawﬂé) . \ 2.. \ a L a ; CQL6 s —‘—— = «- z @— wmk- .8; ———~—» nTLQ ILL ID. ﬂ “32(9) # 8 | A _ h 5:“ xx 9 : X MyaﬂAﬂl‘Vﬁ‘ [a C63. :H'MQ\= Jé Cfle rs amumiJM/Q fem/m5” lC=L U0, k\(\/\\:‘ \ Q(9\:J§) (N,L©\= —@>—t1(3{\;7<. SWU -J§ Mmbww Om Wumaqu try/“2‘ O Baggy/g) 36.- - IA \‘\ 3 ZY; K 0L COM/Fuji oqjtmﬂ'jbphshg “gm 69’ biz: swag. 4 is LAAS, CDI" (X ’- L-rl—a I 3 ‘U. MLE :' Y )5 Q xyﬂ-WMJNM an (25pts) 2. Let X1, . . . , Xn be a random sample from a distribution With pdf f(:z:|6)=(9+1)x9, 0<m<1, 6>0 and zero otherwise. ( ( ( ( S a) Find the methOd of moments estimator of 6, call it 5. b) Find the maximum likelihood estimator (MLE) of 6, call it é. c) Find a sufﬁcient statistic for 6. d) Which of your estimators, if any, from part (a) and (b) are functions of the sufﬁcient tatistics? Explain. . l I, I a <00 Em : Smuwg‘m : 8.0“ t&+»a\‘m<9*‘att=£§ﬂ W“? . ‘ 92 < b\ : (8—95 6 at; at: §DO~QW rpm ®+A <Q+01\\:)Z = @‘r‘ @(«T—D = 1‘47 6:135 3.1.1.; 3 @sztu’i‘ «74 "7‘ 7 —t L(811\:TKT (sow? = (omv‘i‘mf mWMTD : h ‘(%(&+\\h ‘+ mint; d-im’\Li®iY\)=: n t it . “mg—W » V - T—.‘ "KL deg @+! (F’ C? J}. + Z iothL= @144 H Chuckq r3 54mm: 4‘ 3 ‘ ‘ dab Mom _. _ n i, x I ¥ ~ —- ————- /_ ’. ﬂ WW mat (WV ez‘é‘ : — «f ,. CSIWQ.@+1>ZI>O :ir\>0\:, Zion“; . wt“. . 4_, '_.D_ -1 ‘§= “I” -: ‘~ 8%: z”: x M ' ZWKJ we? ‘ 5 blank page CC\ £[YI@\: QBWa’X lSom wKPaer ‘waut ,t=l. w! Mud =l ,ch = (9m ,M(e\ = 8‘ :1 1g (7%: Max Li. WM (9,9..‘(3 —‘ L2; X Is a Wlkwé—ahhlsdut ‘80? ' \ ((A W W 94% Fa/C/LUWEQ’AJWG‘A‘ ’W W‘ U1 J L (5 /\ _ 6mm " l“ ‘2 X )5 hb+ 0\ cM/Aoi‘wvt (“f Wﬁkmtwaé' owsi‘nc x 4 A. h Syn/ca :4 IS 0. \D/wvxC/EXCVLW‘ :X; m+ H Xk. l (E: ' /\ SMLE - " Kn -- 3 ‘5 0k 4M/VLCJWM ave “OULﬁwlo/[CLWJEaQél/gla? EaaaxL ‘qx L IL 4 \ SW'- 2 L’ 15 [‘1 7 n < i M (/GwagZyM’v (20pts) 3. Consider a random variable X With pdf f(m|6)=—%,O<0§x<oo, cc and zero otherwise. (a) Let X1, . . . ,Xﬁ be a random sample of size n. Find a minimal suﬂicient statistics for 6. .5 (b) Let {(1, . . . ,Xn be a random sample of size n. Find the maximum likelihood estimator ( LE) 6’ of 6. h“ g --.—" // L" \ \ J i g, - n~’\ w ‘ MUS” T? wan Wﬁ’aiww , 'l - . wm‘qtl3~W\:/WYL7 \Q‘Ulywu L‘ _ WWW \6 W’OEEUAJLS' (1/) “MC/{hang Q N MA W : ﬁts l9 Jjg‘g ’Xm : kéLHl \ ‘ >< ( m l 5 CL ML/A \ WWJL 05% t gmmgjr 5357J\s<l7\g OY‘ . we. m «a magi: 2121th he — Elwmzmxcg f, . \\ r :"—'7 ‘ B 2 8n 'X L- i) u” 555‘; {v t)\ L7 I v. K ®h is M Woman/VAC? Q’WViJRoL/Léhé (6‘ l3 \(LL >6 Fm B .> XL.) )Ll8l14\=0 l F05, 9 é er } HQ} 15 We l‘aomw’? 3. lwt Uaaxxi- <57; two/{1, 4 a t , /\ l m teem}; 4 igmial‘gh Nut 9 X“) )5 (25pts) 4. Let X1,X2, .. estimator for 02, . ,Xn be a random sample of size n from N01,, 02). Consider an (a) Find the E(&2). (b) Find the Var(&2). (c) Find the mean squared error (MSE) of 62. More. in m N yin-n. E( §(X»';73L\ = m => E (5% = (ra- (T V04 (:(X3—7Y\ =9\(n—I\ ==> VOW<SI\ :22rﬁl—l Q—L ‘0 (ﬂ Ldr S1 = i Elk—7y m2: h_l aL‘—\r\_\7~_»-'—'L_.,_Z am:— e. (ms 3,13: _(s)—b._"<r &Cb\\/ Gz\=\/ mi\=n-_Ll m-_-I’*~” O M( “(N—x S (“+3 VW‘ (S\ ' WH 7% ': (MA) 910}! (h—H)L z 9 (c) may): vmcaﬂ + [Ema—ﬂ = 4031+ 0‘1- 3'1]; MM)?" {n+9} “‘"t; :‘ £31) 10““ + [On-sz “01(12):;‘1’1‘1 ‘ (n+1) hm 1° (m?)qu + H03 (In-le (“+E)L ‘: O'ZQ‘L} n+3 ...
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## This note was uploaded on 10/04/2011 for the course STAT STAT670B taught by Professor Drbailey during the Spring '11 term at San Diego State.

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670bexam1soln - STAT 670B name February 2009 Exam 1 Show...

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