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M562-HW9-4

# M562-HW9-4 - MTH/STA 562 Exercise 9.29 This distribution...

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Unformatted text preview: MTH/STA 562 Exercise 9.29. This distribution can be written as p(xi;p) = P{X¢ = x1} = 10””i (l —p)1-\$i for 56¢ = 0,1” The likelihood of the sample is n n . ——-x' Em; n—Zwi L(w1,w2,---,mn;p)=Hp(xt;p)=Hp””‘(1-—p)1 1=pi=l (1—19) izl - 1'21 i=1 Consider n i351; ’I’LWiCL‘i g<zmhp> :pizl (1 up) i=1 and h(\$1,\$2,"',\$n)=1t 2'21 71 Then it follows from Theorem 9.4 that Z Xi is a sufﬁcient statistics for p. ' 1 1: Exercise 9.30. The likelihood of the sample is = <2:02>"”exp[—2:2;W] Notice that 22:041— “)2 = gm ~ 3) + (-g— to]? = i<yz—y)[email protected]—M)2 because 1:1 2;:(211 —@)(@—M) = 2(ﬁ-M);(yt —§) = 2(y—M) (:13, _ W) = 2(ﬁ—u)(ny—ny) 2 Thus, H A p-i \_/ § \ 10 CD >4 ’U l lr—A 3 ET I I: V [O g.._i £5 'U | I.— M: A “S | Q] ‘13 (a) When a2 is known7 consider 9 (y, M) = exp [—ﬁn (@ - m2] and h(y1,y2,-~,yn)=( 1 )n/2exp[-%Zn:(yi—y)2] 27ro2 Then it follows from Theorem 9.4 that 7 is a sufﬁcient statistics for ,u. (b) When ,u is known, consider 9 (for —?)2702> =( 1 )n/Zexp [—ZLZHXM —§)2] 2 i=1 27m and 1 _ 2 h(y1,y2,~-,yn) =exp[—272n(y—M) ]- Then it follows from Theorem 9.4 that E (K — ‘17.)2 is a sufficient statistics for i=1 2 o . (c) When both it and o2 are unknown, consider n 1 “/2 Z<yi—o)2+n(‘y‘—u) ‘ _ — 2 2 = __ _i=1— _.____ 9 (i=1 (3/; y) 70 > (02) exp 202 and 1 72/2 h(y1,y2,---,yn) = (E) v Then it follows from T heorem 9.4 that 7 and 2 (K- —- T7)2 are jointly sufﬁcient i=1 for ,u and 02. Exercise 9.31. Each Yt- has a Poisson distribution with mean /\ and Y1, Y2, - - -, Yn are independent. It is easy to verify by the method of moment—generating 71 function that :16- has a Poisson distribution with mean n/\. The conditional i=1 n distribution of Y1, Y2, - - -, Yn given ZYi is i=1 Inyi %x, then P{Y1 =y1,Y2 =y2,---,Yn zyn,ZK =13} =0 and thus i=1 i=1 P{Y1=y1>Y2:y27”',Yn=yn If Zyz- = x, then i=1 P{Y1=y1,Y2=y2,---,Yn=yn,ZK=x} i=1 P{Y1=y17Y2:y27'”,Yn=yn} H 'r—1 b—l and thus This shows that P {YI = y1, Y2 = 3/2, - - - ,Yn = yn ZK = as} does not depend on i=1 A. Hence, 2K is sufﬁcient for A. i=1 Exercise 9.32. The likelihood of the sample is L(y1,y2,---,yn;0) = f(y1;9)f(y2;9)-~f(yn;9) 2y ~“y2/0 2y w 2 2y __ 2 : — 1 . — y2/0 ..... _ yn/H 6 e 0 e 6 6 TH% 2% 9n exp 0 Consider n 1 i: ' n n 9<Zy§,9>=6—ne><p — 16 and h(yuy2w~,yn)=2 Ill/22- i=1 i=1 Then it follows from Theorem 9.4 that 2 Y1? is a sufﬁcient statistics for 6. i=1 Exercise 9.33. The likelihood of the sample is n L(y1ay2>"'7ynia): Hf< yiaa z: m—l n my _ mn n 1 n m . . no...) way» 0—1 Consider “ 1 1 n "“1 9 (29171179) = JGXP (—3 :91”) and h(3/1,3/27" 'lyyn — WHO—[91> ., i=1 i=1 7?, Then it follows from Theorem 9.4 that ZYz-m is a sufﬁcient statistics for a. i=1 Exercise 9.34. The likelihood of the sample is [41112927 ' ' - 731nm) = {INK/up) = Hp(1—p)yi—1 H ’E 1: I 3 II H 'U 3 T: 1 E 3 ‘T 3 Consider _ g(y,p) : p" (1 —p)ny‘n and h (931,332, - - - ,xn) = 1. Then it follows from Theorem 9.4 that Y is a sufﬁcient statistics for p. Exercise 9.35.. The likelihood of the sample is TL n n n a—l L(y1,y2,---7yn;049) = Hf(y¢;a,0) = Hgyfy—l = \$70; (HM) . i=1 i=1 Consider n n n “—1 OZ 9 (H Elba) = 6710c (H 92‘) and h(y17y27"'7yn)=1 i=1 i=1 Then it follows from Theorem 9. 4 that I:[Y is a suﬂicient statistics for a. Exercise 9.36. The likelihood of the sample is TL TL ~(a+1) L(y1,y2,---,yn;oz,ﬁ) : Hf (yi;a ﬁ)=Ha18ay—(a+l) :anﬂna (HM) . 7121 1'21 4 Consider n n —(a+1) .9 (H9176) =an/Bna (HM) and h(y17y27”'7yn) = 1“ i=1 i=1 TL Then it follows from Theorem 9.4 that H Y; is a sufﬁcient statistics for (1. i=1 Exercise 9.37. The likelihood of the sample is ll Kr: A “S Cb V H S A % V c~ A E v ml )1 3 55‘: ‘S L(y17y27”'7yn;6) 2:1 Consider n n —-c(6)Zd(y;-) n 9(Zd<yi>,e) =[a(6>1 e and h<y1,y2,-~,yn>=Hb(yi>w i=1 i=1 Then it follows from Theorem 9.4 that Z d (Y;) is a sufﬁcient statistics for 6. i=1 Exercise 9.38. The exponential distribution is given by f (y; 6) = %e_y/9 for y > 0. Letting (1(6): > b(y):17 C(mz—‘év andd(y)=y, we see that f (y; 6) is in the exponential family. It was shown in Exercise 9.37 i=1 statistics for 6. This implies that 7 is a sufﬁcient statistics for 6. TL TL that Z d (Y;) is a sufﬁcient statistics for 6, which implies that Z Y; is a sufﬁcient i=1 Exercise 9.39. Write f (y; a, 6) = ﬂew—mug for y > 0. Letting a(a)=6—a, b(y)=1, c(a)=a—1, andd(y)=lny, we see that f (y; a, 6) is in the exponential family. It follows from Exercise 9.37 Tl that ZlnY; is a sufﬁcient statistics for a. It was shown in Exercise 9. 34 that .21 77; z n 'n H Y; is a sufﬁcient statistics for a. Since In H Y; = Zlniﬂ, we have no contra— iz'l . i=1 i=1 dlction. Exercise 9.40. Write f (y; 045) = aﬁae_(a+1)lny for y > 0. Letting (1(0) : 0460', b(y)=1, C(01) = 04+ 1, and d(y) = lny, we see that f (y; (1,5) is in the exponential family. It follows from Erercise 9.37 that Ean is a sufficient statistics for a. It was shown in Exercise 9. 36 that 221 n fll Y is a sufﬁcient statistics for a Since ln H Y— Zan, we have no contra— 2: 2': 1 2:1 diction Exercise 9.41. The likelihood of the sample is n —1— . ‘ , L(y1,y2,---,yn;6)= Hf( 3/299 ):{ 9n 1f0<yz<6Vz i 110 elsewhere. = lf0<y(n): max{ylay27"'7yTL} <6 elsewhere. an ihoe [yen] , where _ 1 if0<y(n)<‘9 [(0,9) [M70] ._ i 0 elsewhere. ' Consider g(y(n),6)= g—lnfmeiwml and h(y1,y2,-~,yn)=1ﬂ Then it follows from Theorem 9.4 that Y(n) is a sufﬁcient statistics for 0. Exercise 9.42. The likelihood of the sample is L(y17y27"')yn;61792) : 1—1f(1 92761702) 2 1 ifl91 <yz<62Vi elsewhere. if 91 < 9(1)_ < y(n) < 92 elsewhere 002— —1)'191 | o 8 5|? H (62 _ 17291) WW co) [31(1)] I(_oo 02) [gm] , where ya) = min {3/17 3J2: ' ' ' 73/71} 7 y(n) : max{y17 3/27 ' l ' 73/71} 7 _ 1 lf 01 < y(1) < 00 [(91700) [2/0)] _ { 0 elsewhere. 7 ﬂ 1 if —oo<y(n)<92 [(079) [Mel ‘— { 0 elsewhere. I Consider 9 (9(1),y(n),61,62) = Wham) [ml I<~oo,92> [you] and h(y1,y2,---,yn)=1v Then it follows from Theorem 9.4 that YO) and KM are jointly sufﬁcient for 61 and 02. Exercise 9.43. The likelihood of the sample is 1‘ ex — i+~n¢9 if i>9Vi L(y1,y2,--nyn;9) = Hf(yz;9)={ p< g3! ) y i=1 0 elsewhere. __ { exp (—_Zyz+n6> if 9(1) =minf917927”'ayn} > 0 ‘ i=1 0 elsewhere. exp < i 312‘ + "9) Ram) [31(1)] , i=1 where [(0700) [gm] — { 0 elsewhere. 0 Consider 9(y(1),9) = eXp (729) 109,00) [21(1)] and h (311,312, - - ~ ,yn) = exp <- 2%) n i=1 Then it follows from Theorem 9.4 that Y0) is a sufﬁcient statistics for (9. Exercise 9.44. The likelihood of the sample is L(y17y27"'7yni6) : Hf(yii9): i=1 2) elsewhere. 9337112912 ifym) =min{y1,y2,---,yn} S 9 || 0 1:1 elsewhere. 37L TL = “3;; (1.11%?) [[091 [3/00] 7 where . ﬂ 1 ifOSy(n)39 110,0] 13/01)] _ { O elsewhere. ., Consider 9 (90079): 9:” I(600)[y(1)1 and h(y1,y2,- "7%) = 3” 121%2 i=1 Then it follows from Theorem 9.4 that Y(n) is a sufﬁcient statistics for 6. Exercise 9.45. The likelihood of the sample is ” {H2533 ify1>6Vi Hf<yi;9): =1 ’ L(y17y27”'7yni6) 1 i=1 0 elsewhere. __ 27102" H is ify(1) = min{y1,y27 - ~ - vyn} > 9 ~ i=1 1 0 elsewhere. ’I'L Tl n 1 = 2 02 H— [(eoo)[y(1)1 i=1 y? where __ 1 if ya) > (9 [(9:00) 1y(1)1 _ { 0 elsewhere. 0 Consider 7‘ 1 9 (34(1)? 9) : 0271.1(9m) [34(1)] and h (y1, y2, ' ' ' 7yn) = Til—I E' Then it follows from Theorem 9.4 that YO) is a sufﬁcient statistics for 6. Exercise 9.46. The likelihood of the sample is L(yl:y2a'”7ynia50) ” {ﬁﬂyﬁ‘l ifOSyigﬁw I—l ll 11f(%;a 9): 7,— 1:1 0 elsewhere. elsewhere, >01 n 0‘ —1 '99“ (Hly)<1_1%> ifOSym)=maX{yi,y2,-~,yn}£6 0 r 01” II n n —1 6— (131%) (11%) [[0,0] [yew] 7 where _ 1 if 0 S y(n) S 9 [[0,9] 1y(n)1 " { 0 elsewhere. . Consider n n n a n —1 g (H yi>y(n)aa79) = (56:7(1191') and h (3/1792, ‘ ' '7yn) = (H M) - i=1 i=1 i=1 Then it follows from Theorem 94 that H yz- and YM) are jointly sufficient for a i=1 and 0 Exercise 9.47. The likelihood of the sample is L(y1,y2,---,yn;a,ﬂ) : ﬁfWﬁCLﬂ) 2{ anﬂm Elli—(M1) if 311 Z ﬂVi i=1 0 elsewhere. :3 l | _a n —1 £21 £21 0 elsewhere. n _a n —1 0/13"“ (H yr) (H91) Inca) [9(1)]: i=1 2:1 where __ 1 if ya) 2 ,3 Ilﬁw) [9(1)] _ { 0 elsewhere. " Consider n n —0‘ n ‘1 9 (Hyi.y(1),0z,ﬁ> = anﬁm (H 31¢) and h (211,112, - - - ,yn) = (H m) ‘. i=1 i=1 i=1 Then it follows from Theorem 9.4 that H Y; and Y0) are jointly suﬂicient for oz 2'21 and ﬂ. ...
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