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2010exam1soln - STAT 670B name February 2010 Exam 1 Show...

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Unformatted text preview: STAT 670B name ; February, 2010 Exam 1 Show all work. 5 Problems. 100 points total. (lOpts) 1. Let X1 ..... Xn be iid Gamma(a, ,6) random variables. The Gamma(a, fl) distribution has pdf xwle’m/‘B, O S x < oo, a,,8 > O. With E(X) :- afl and Va'r(X) = aflg. (a) Is this an exponential family? Explain. (b) Find a 2-dimensional sufficient statistic for (01,5). ' «-1 4/5 I \i .‘ 1 ..—“— PM “4 e < “\e—w {MW 9"?) ’lSM'MPva/vd‘ld PM"? Wm {:41 1 91(4)?) bowl/K b(x\:“:X‘»C(8\:'L"" I m m mg»- Suva 44x12 ~3de W hMW| M (3 H. I) 4mm: hm moi/AP (Zm (8ch (3M “1:3me _ (b3 f3; (a\ GW%(9<HE)\ LSOAAMPMU {mad I} wl 8: (N #53 r. "fig'wfitmo Tog: (ZH x4» ZJ: (ml T" (30pts) 2. Let X1, . . .',Xn be a random sample from a Gamma(3, 6) distribution with pdf 1 2 —a:/6 I : < . . f(:r[6) F(3)93£E6 ,0_a:<oo,6>0 (a) Find the maximum likelihood estimator (MLE) of 6, Call it 6. (b) Show the MLE 6 is an unbiased estimator of 6. ( . ( c) Find the variance of the MLE, 6. d) Find the Fisher information I (6) of a single random variable X from a Gamma(3, 6) distribution. (e) Find the Fisher information In(6) ofa random sample X1, . . . ,Xn from a Gamma(3, 6) distribution. ' (f) Is the MLE in part (a) 6 a uniform minimum variance unbiased estimator of 6? Explain. n Y] ' _ \ N l 4 — 8 _ I > n, A _ lo (a L 8]x\= ". Mane"?Y e : LEG» (éflfllfye 83:" y 0% F s1 ‘ 3-— + J: Zr. = o 3 B W 'I V‘ _ T ’ /\ _ V f = "‘ ,_ — .. — 1 a MLE- E 3“ Ext I 3 3 6 ‘5— 5 u... I/M‘J‘ic: Clo/ck. M‘ ’6': :X: is 01 MM 7 W 9““ dammiw euztéwold— 3% 3 . (“)o‘l Lion 2 §1_ 2:15 ‘ _ J?“ 1—- 0:?“ 5 _ -911— 2(an___) 40 OW“ 5v 2 (35' (5:1 “(Eng (snzlw (2'0 {ill ._, 1X3 3 . ‘ 1 blank page x A _ __ ' ' £4 (a) E(a\:.t(_§;\: gem =§39=® gtsmmbzwd 4 JOhMWfi—{Q u (c\ Vm(.§>\‘= VM (3;?) 1' 2:32ka(7\ = #- 3>_®_z : .52: . h 3Y3 ? ‘ a —7( ; L. (013 £(M8\= Wm 9‘ l9. cMTaere): .; 4. 3;, 0‘6 5 5* 0H3 ilhcm‘X : i - g); 0%” 3A 55 I(B\:*E CW =9hzm‘) :—E Egg—4X]: ‘9- +QE0Q @& 9’ 5‘ 83 : -2. +461? 8* gfi . , ~ :‘ 3 u (A jag): him: 3n (3W momma 8‘ \ 8" I /\ a. ‘ L‘_ (4“ SWU- (3 \5 unbtw.) -L-- = 2 © Mm CRLE \ n'fleX '35“. E1.) (BEL (c\ ) \JM(@\: “Q: : ‘ ‘ 33% NIQS‘ /\ j- @115 UVWME. 3 _ (20pts) 3. Let X1 ..... Xn be a random sample from a distribution with pdf f(x|6)=%(1+0m), —1gx<l, —1<9<1 and zero otherwise. (a) Find the method of moments estimator of 0. Call it 5. (b) Find the mean, variance, and mean squared error (MSE) of d. ID '5 (a3 E(X\ : SlX 7‘9:(I+S'x‘)0w ._ SEX *‘ §7<2>1M= fixz+ngt (D =(n +;--<—A-—~%x —- @3— _ I S §=§3=> 8=37 ”5:37 : fixh gm“ L. Wt“? '('i+%\=i;*t‘é MAVMLVA‘ $34.33: is? —— 3,33" 1; Vm(§\:vm(3‘§<\= 0mm :3; 1—}: = 1?, 2‘ Mae e\ Mm fifimrm WM “3?: (20pts) 4. Let X1, . . . ,Xn be a random sample from a distribution with pdf f(:r|6’)= %,0<6<x<oo and zero otherwise. (a) Find a minimal sufficient statistics for 6. Be sure to justify your answer. (b) Find the maximum likelihood estimator (MLE) of 0. Call it é. Be sure to justify your answer. h 9 __ ' n w NLW - LL! 3:5 item) : Q“ ( 1T 7‘5.) In «m 445%) W" , L ,«a n i ‘k “’1 —[8)v5) (Us k __ L’4 9L 0'\ V (1—1- 2.3:; “LEG“ (16-M\ Wm 7(0): Why/1x % 3L0}: m “(h ”M who- 5 (ml-awe an a “ham «if 8 um Irena») (m = 1 This. as Aims W xm ; \3““ Item Wad «5.. X“) \5‘ Q “112% HIM/Q owggiwmfir OWShe PM 9 NO)”: I . 1:19 a“ m 1 W03 .0 mm cam {Ta—mm ®n \5 am MNMWL? JRimc'Jhc/nwf EL. EA ®>7<UX L[§‘1\’O Fm Sf X0); LL®|X\ 13 am lVKNMIua/LWMLL wwwh Mm Mus M 69/;04 pmflole (20pts)‘ 5. Let X1, X2, . . . ,Xn be a random sample of size n from N(O, 0). Consider two estimators for 6, é=—1-ZXE (1) ”i=1 ~_ 1 n 2 €“n—1ZX5" (2) a) Are either of these estimators unbiased estimators of 6? Explain. b) Are either of these estimators asymptotically unbiased estimators of 6? Explain. b) What is the mean squared error (MSE) of each of these estimators? You do not have ( ( (c) Find the variance of each of these estimators. How do they compare? ( to simplify your expressions. 9. E<ZYL\= h® we": ng-WB' Luv “K: :7 vmtzxfl= 6) gm (o\E(’é\=fi(-lfi2xf\:~‘-E {EFZW in” 5 Hm MMZM figE(2i\= fi‘“8=£i& A ‘ ya. 6 l5 Wb‘md. (b\ ’Bol/h M methQ 4%de SW02 0 t3 mwhm E \QUM___ 8: 9 ““355 Y\-\ 2' (a var/ex: we; :x5\ 1vM<<fzfl=—:- ““5 ”4-? aheltrgh L V04 (§\ 3 VGA (lad ZXA Q4\VMKZ‘::X\ CHM-”‘3 (hvllz \o campau VW((_9\ s VM ((9\ )WM .5;— Mfll- [OE—3:1 0’\ —L" Ma m}... . Slit/U ”L < ”-L) 3 VCM {LEM LrlJm{® £5) , n5 rw n5 cm I ‘ l A a. mask vm©\+ @‘m®\ = We 2 n 9. :. gln Z 4,. _ . Mal. [(9\‘~ thc9\ +®Mo 93 5) . p‘»n“_¥9 8] {writ ...
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