B4090.HW4Solns

# B4090.HW4Solns - BTRY 4090 STSCI 4090 Spring 201 1 Homework...

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Unformatted text preview: BTRY 4090 / STSCI 4090 Spring 201 1 Homework 4 Solutions 1. Larsen & Marx p. 333, Question 5.7.2. Since ﬂ 2 O, for each 1'. E (1’?) z 02. By the weak law of large numbers demonstmted in Exempte 1 n 'J . . . 7 . . ’3 5.7.2, —— E K“ as a eonsmtent esnmntor of the mean of the 1? , 111 this case 0'. However, the 11 . (=1 2 proof given in the exempie requires that Var“:- )< m . This folioWS from an application of the moment generating function for the normal distribution since the fourth derivative of the moment generating function evaiueied 2110 is ﬁnite. 2. Larsen & Marx p. 333, Question 5 .74. (a) Late:- an). EM ~6>1]= EM “#5 +.u,. —&F] = E[(6",. — to; He. — e)? + 2(6". — met. — 6)] = Etté; — a. )3] + no. - 6f 1 + 2m. — one. ~ p. )3 = E[(§,. m ﬂ...)2]+(y,. * :9.)1 +0 or Egg; mail]: E[(é,, - 11,,)2]+(ﬂ..“ £932 The left hand side of the equation tends to 0 by the squared—error consistency hypothesis. Since the two smnmands on the right hand side are non-negative, each of them must tend to zero also. Thus, lim (11,, "— Ejz = 0. which implies ﬁrm (A, m 9} = 0, or lim ,1!“ z 6. (In-5w u-éuu n—iw {II} By Part (a) ﬂing" — 6: 0. i£-':=n For any E> O, ﬁm P( én“9125):£33ﬁ( <é.—u.>—<#.-e;yze) n—ém : 11111 P( {if — ,uﬂ 2 a) s VMSQ’) by Chebyshev’s Inequnﬁty. n—Lm: 5" V 6:3 E 63 — '1 ~ Bu; “E a) and by Part (a), Hm eke” “p,1)2] z e, a“ 5* 11—3“ at so the P —6‘l 2 a) = 0. Thus. 6 is consistent. I? Hmong 3. Larsen & Marx p. 344, Question 5.8.2 modiﬁed: r (a) r+S E[®} = 0.05 and P(0.02 < G) < 0.08) = 0.90]. That is, ﬁnd r satisfying: Diff= CDFbela(r,19r)(O.08) — CDicia(r,19,.)(0.O2) = 0.901. Try a few values of r, increasing or = E[®] = 0.05 so 5 =19r . Using computer software, ﬁnd 1* and S satisfying decreasing based on value of Diff; the search converges to r = 6.5, s = 19r = 123.5. In Minitab: Store the values 0.08 and 0.02 in column C1. For different pairs (r,s) with S = 191‘, say (r,s) = 6,19), (2,38), (3,57), ..., Cale > Probability distributions > Beta Cumulative probability; First shape parameter: r; Second shape parameter: s input column: C1 ; Optional storage: C2 If the difference between the values in C2 is less than 0.901 , increase (135); if the difference is greater than 0.901, decrease (r,s). Convergence to r z 6.5, s = 123.5 is quick. (b) As shown in Example 5.8.2, the posterior distribution is beta (k + r, n — k + s) , so Bayes estimate = mean ofbeta (k —i— r, n — k + s) = 4:1;— — .111 ._ k "l" 6'5 k+r+n¥k+s — n+r+s m n+130 k+6.5 n k 130 6.5 (C) = . — _ n+130 n+l30 n n+130 130 Notice what happens to the weights, n - and 130 n + 1 30 n + 130 (d) To produce the graphs using Minitab: Graph > Probability Distribution Plot > Beta ,as n—>oo. — prior "" posterior dbetatx. r. s) 0.00 0.05 0.1 D 0.15 0.20 4. Larsen & Marx p. 344, Question 5.8.4. (a) This is Exampie 5.8.2 (again), with the prior distribution m uniform = beta(1,1), so the squareduerror Bayes estimate is the posterior mean k+r _k+i_ n [£]+ 2 _1_ n+r+s n+2 n+2 n n+2 2 1 k r ‘0‘“) _+_ 06+O (b) As n—>°°, BaYes estimate= EL—ﬁ—"E—i—é ' 0=06 w(n+r+s) 1+:+i 1+0+ n n n (o) The Bayes estimate approaches the MLE, coinciding with it in the limit as n —> Do. S. Larsen & Marx p. 344, Question 5.8.6. The numerator of ggi 61 X z k) is frfﬁ @feigi r .r s r—I = 5 yragérgy ﬂ 93—ie—yﬁ‘ : ﬂ 3’ gr+xuigmty+mﬁ TU") ‘ PCS.) FUEL?) We recognize the part involving 635 the variable part of the gamma distribution with parameters r+sandy+pg so thatisge (61sz). 6. The posterior distribution of 9 is gamma (nr + s, w + y) , where w = yi . Therefore: (a) Larsen & Marx p. 345, Question 5.8.8. The squared-error loss Bayes estimate for 6 is the nr + 5 mean of the posterior distribution, which is n Zia] yl' + # (h) The absolute—error loss Bayes estimate for 6 is the posterior median, which is deﬁned as the value of m that solves Ige (6 | W = w)d6 : 0.5 , where 0 39 W = W) : Mauri‘s—Ie—(iwﬂw I“(nr + s) gL- (a) Pa,th {SE} W83) 2 W CL‘.) PGMLHBQPCBEES + (Paniaci‘ ’85)?(1:§’g)+ Pakéuémpﬂ 1« (‘4!) ,.(I,<,)+’f-(‘/z) + O (Var) ,_...—._ w C44] H L4 (:31?) 0 :O'F'FKPring fﬂi PM emu (BE) PWE’) (mac) "P 8810 e! a = ( QC 3 PG) Efackl381P\$¥5+Pm 6W1; [EQHBQ / 1.('/3) : «(m-r aft/3.) 73 :- V3 WEN!) 62m): F We (9mm = V; (C) 061*“): ’Fif‘sﬁ' '0 07E’F5P9in55 Ctr-e. é{0\CL'_ £7 (06M, 1:395)me PRES/Ohm) = N .96”. [313391ng PGQEHD (Eé)7>(13’6) I-la l0 1 ~('/3) :w .ito(9{?)+(fé)1¢(2/z) '/z “‘ % (WW) 4 Z zit—{‘19 (quof—E‘ =7) I Pﬁwoéw) : ’" WEE/“FM t 4+ 2." mwamww . dam; # SXJiCe) >/~N(e,lec203 W Wﬁﬁm! I I Lf-Q) r: 8‘35? MSG-61: W— “fa (c) L216 XNN(Q,U»1) We; .QMW (TL) “Tl/Han POS‘é’Efigr—h 0(1’5-éri6u‘ér‘0n 4:370 '6- L‘s qLSo mom/mo" wréLI mean TE Uni. mam“: ’t'L-i—el K + (It-+21% (1] and Variance. 0.1.2.?— x/m‘ “:2 (3) .L‘ (x It?) 4: (6'1 X ‘9 m f-Fxbde) 426(6de L356 (’15 COnC—Qhfrq‘éﬂ C3 n (X (C) (Con-£77m ad) Now “Elms. jib‘ﬂs'é—am 3‘5 Quay/- ij‘é‘ Fm’é‘D . C13 and (‘2') 3 Amanda 1 - I (e-“£( )7— Cé} s: : 2,1,7- 1030) [email protected] ( MFG) WIT,” U a Whara Z 1600- V00 0 z ——-a———-——--————-— :: 330 1600 + cmo QM! {é 8’60 00 —: m- _———r——- :00 5mm} (“OHM (as—o + [MGWOO I I: {O 90 Ben I L Sgéstlmziord =2 m0 9, L Gama-{600 u 2: —-—~ if 1220 69/00 +1500 M, 5 ) 6‘100 y I600 0 o = - W. £750+f§co ’0 0 + 6Y00+I§00 [000 (4) Ag? A, p¢ (6} ’505) :.PJ(91153 / foo { I Lima“ - {5‘91}; “ A 55-9 6 1’6 \3’3/0;O (a) { Z r ; v— a KGQI-RSOMOH (a! "—3 ' u 363 5) W0 6 -———-—--—-—* (g + 9’00 —" ‘38 3’00 '16 é(!o?’o3=M/0¥0+ '/ O 700+le léH’GO I 0 Z 10%,. {3' Ben T 1 [5- 6 #00 Q 1: M :- [5’ le+g900 ‘96 év ’6 5(mzo);: GO 10¥o+- ~M~M*Mﬂ0 61400 + («5 (6%?“ 8&9. (a) To produce the graphs using Minitab: Graph > Probability Distribution Plot > Beta dnormﬁc. mean =1100,sd = 20) 0.010 0.015 0.020 0.005 0.000 700 800 900 1000 1100 1200 1 300 X Cb) L(theta) 0.004 0.006 0.008 0.010 0.002 0.000 700 800 000 1000 "£100 1200 1300 iheta (C) 03.0 mwod 9.0.0 mood “ammutcm u 3 .89. u cmmE iEacu 000.0 900 1000 1100 1200 1300 800 700 (d) 9.0 00.0 00.0 00.0 Ems“: 00.0 00.0 900 1000 1100 1200 1300 800 W30 theta (e) dnurm(x. mean = 1071.15. sd = sqrt(15.38)j 0.04 0.06 0.08 0.10 0.02 L ll-ﬂ—n-uln‘l———nn-u uw-.....“.....—._—n ﬁr..___._ q... “ { C) Q C) we 800 930 1000 non 1200 1300 X 10. (a) ﬁrm/y” Mud eatafr‘pﬂ . ﬂ PCT-#3) 7—: I-r 1.075): 2—(rﬁcrwcs) 3" (“3:3 > I"l (fr-#5) Esl (Pfrxﬂs))n( 7%*3: )r-YTTU '10) H (at? 9m = 1X~I EQMW): : “I :9 I“ 611 a Z jaw = 5(8) ml) #4 iﬂa(ja~v“=§ﬂk—v+a(M: 5(e~s,n+:>+ 5w \$094) w) ——5‘(9-I, 14)] 2-; aagry.+z_ 5(6- 1, v0]: cal—giﬁéa—bm) 5mm) = if} [gm] 14+!)- “1 ’1 if (m a if E W—ﬂnaiL—o‘j E{9Ngj: 1 £4 Pam ,, Ga E4 Rig} EA“; = 5(6) n+3) *1“?(E'Dn‘éﬂéviﬁWW!7(~E—Dn=5(é—I,H+DD+15(9«'mj*0+ 5w [{[é‘EJzéﬁWMEQﬁQ—j)w} 5(9a13nﬂ = 6": \$5: [25(3'1)H+D+§{9*Lhﬂ Van (gm): —[E(§ML)]1 M §E(§ML)= V&n(§m:_) + Ewe-1&4 :rH 4 (C) : ‘- ’wnﬂﬂ. - d 8- Jan-tuLZ-ﬁlnﬂmi L n y] i'L'jv' try-CE‘DH ‘anfrgi “(1330] L) = V”? (gm) + [‘ 367“ Swim)? I wn-(w—p“ =: _. E- g. M} n—U n+1 _ n+1. M! wé?"1%me(32~l)‘+j='§ﬁ[eﬂ+t{3*0+[9‘0‘[9"3) m1 0 J - M : we 9*“9 ’ (ammmawwwwh ~ MW 6. Mmmmwwm%ﬁ%w %/ ...
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