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Unformatted text preview: BSc and MSci EXAMINATIONS (MATHEMATlCS) May—June 2015 This paper is also taken for the relevant examination for the Associateship of the Royal College of Science. M351/M4Sl Statistical Theory 1 Date: Monday. 11th May 2015 Time: 2 pm — 4pm Solutions 5?) 2015 Imperial College Landon MESlfM-ﬁSl Page 1 of 5 seen it 1. (a) (i) The conditional distribution of X given T does not depend on 9. (ii) T is sufﬁcient and for any other sufﬁcient statistic S, T is a Function of 5'. (iii) The only function h which satisfies Elhffj] 2 0 V6 is MT) : D (aimost sureiy). 2 i (b) (i) 5(6) = niosﬁé’) - 621;; :31:- meth seen ..= '- U.(a) = M9) = g — Zl-Lle' = TL(p,[3) — X). By inspection 2? is the CRUE ; of “(9) = %. 3 1 3 — W(A) = Egg-g = P? l . (ii) Take gm) = / (mm-é dp = [ id,” =10”) (+0). E] (iii) Appiying Bayes theorem the posterior is 7t(6|x) oc 7r[6)L(9) DC go—le—ﬂegneeﬂ :31 z, : Swen—1 e-6(3+::‘:1 32‘) {X .fGamrna{or+n.B—.—mij (8): which is again a Gamma distribution. (iv) —d1°s;9(9"x3 : mung-1 — (.3 4. 23:1 21¢}. Solving W51 = 0 gives us the posterior mode ﬁll—1Y- 1E ._u=1 ‘ ‘ E9 M3451 Statistical Theory I (2015) Page 2 of S (iii) o; = 511135.EeD P.9(X E R). 3(6) : PHUC E R). A test is unbiased if 53(ng g 3091) for all 60 E 90, 81 E (5)1. Alternatively 5(61) 2 cu V916 81 Let Rwo) denote the critical region for a test of size o- of H0 : 6 = I99 against H1 :9 % I90. Let K]! = {60 :X E R(60)}. Then 11135 a 100(1 -— (1)5”? conﬁdence Interval since Pg{9 E ‘11): P9(X §E RGO): 1- cs. {(6} = —§10g(27r02) H 2—15 2111(3’ -19:L'1)2 . U. (6)— 721;13- (1;- —9It)— _ 23,31 (—3?“ 4;). 1— 'lxt Hence by inspection 2% is the CRUE of I9. The CRLB is +5- Zszl“ '1 r‘r . 2113‘ . . . 11]? E13] = -—1=T-;H—El. Hence an unbiased estimator 15 T = W. . 1:1 ‘ 3' )2 ”(72 var Sin—n ——‘—§. Hence ( 1: 1 \$5 23-) (\$213 2 . , . _ CRLB_ (Z? 1'} Efﬁmenq- (T) — W W. L(I9')— (l —6)l9'“' 18— 5“ =ﬁr1-meH96HyHl. T‘ t: x —l— y is minimal sufficient. Let 0 < 90 < (91 < 1. The likelihood ratio for H0 : I9 = I90 against H1 : 8 = 91 is A H (1— I91)e‘31 61 H _ (1 — 90384“ 90 a- >- 1 henCe this is an increasing function of the sufficient statistic If and the monotone likelihood ratio criterion is satisﬁed. Yes. The likelihood ratio test is uniformly most powerful because the monotone likelihood ratio criterion is satisfied. M3451 Statistical Theory | (2015} Page 3 of 5 seen i; I LO H l E 3 :. 1 meth seen i, T lel l lwl H [H 3. (a) S : 2?:1X5 is complete and sufficient because it is the natural statistic T l—parameter exponential family followed by X1 ._ . . . , X71. (b) Let its) be an unbiased estimate of 62. Then 00 Emma a me = 92 <23» :Mac )(1 — 6)I =6 <==5 5(0) —: t{1)(1 — a) = a 4:74» 3(0) 2 153(1): *1, &t(;t) = 0 otherwise. Hence an unbiased estimator 01’62 is given by T = 113320 — Il,v;=1. (c) {(6)— — n10g{6)-1— Slogl we )=£’(9 )='5‘e5—f—=5 1(X+1~»5)5 Lie) weave)? = E :5.»+ ﬁg] = 5 + e55 = m 1:; to 3a.! (cl) U. (6) cannot be written in the form w[T— 6) for any statistic T Ciel Alternativel . note that 6 i {Me} is not a statistic. 5’ Lie} (e) The Rao-Blackwell estimate is given by Elli-123:5 = s} = FUJI 2 ms -_— s) — PIS: 5-) (T1 —1+S—1)9n_1(1_9)5'9{1—6)0 S (n-I— s — 1)gn(1_ 9)5 s (1‘) Yes. The improved estimator in (f) is unbiased, and it is a function of the complete of the sufficient statistic 5. Hence by the Lehman—Scheffé theorem it is the MVUE. M3451 Statistical Theory 1 {2015) Page 4 of 5 I—““—l. I meth seen J; l_._._._._,.,_,_.,_l E lei '0 Lei iii 4- (a) sim. seen i) l , n 1 1 Ll6_\'16}‘)= H (lg—X 11X <51,“— 62}; 11} 331-) i=1 = Efﬂﬁli .Ai'1,,-<ﬂx RAMSEY what an MAYA)? 1111))! Hence (A'(ﬂ_).}"(n)) is sufficient by the Neyman factorisation theorem. l—2Hl (b) Both Hg and H1 are composite. E (C) (i) .. =L(4X'{nh )rn 3) “_T2” r— n r— 1’1 m _“ LKT. T')_ 5‘ tn} 3 (nI Hence A : 4nlog(T) —- 2n]og(XL-n}) — Bologﬂ’tnj). E Assuming Hg is true. then I9 : 64x = 6?}- is a scale parameter. and AreX1,.. . .9141) = 4nlog(9T) — 2n 103(I9J\'(n}) — 2n, IOWA/1,11) = A(X1. . . . 1K1) —— 4n-log(€} —— 2n logOEi) — 2r: log[6) : A(X1:...1}’,1). Hence A is ancillary for 6. 2 i (ii) T is a complete sufficient statistic and A is ancillary. Hence by Basu's theorem T is independent or A. [3 Using the same approach as in (c)(i), we may write A = 4n. log(3'1T) — 2ri.log(I9—1X{ﬂ)) — 2n10g(9"13"(n}), which can be rearranged as —2r1 log(I9"1X[n]) — 2n.log[6‘_1}’(n)) = A — 4n log(6_1T). (Alternatively note that we may assume I9 : 1 without affecting the the distribution of A because it is ancillary.) From facts 4 and 3, —2nlog(6‘1X[n}) ~ 31-13 independently of —2nlog{6_ 11-1,.) 1» x5“;- Hence by fact 1 the LHS is I13, From facts 5 and 3 ~4n10g{6"1T} ~ X3. and we showed that it is independent of A. Hence 31-3 = A+ x5. Now A ~ )3 follows from fact 2. E (iii) The critical region is A > z, where r is chosen such that PIA > leg): 1 — FXE (z I: 0.. Using the hint this simplifies to e ‘1':— ~ :1 and the solution :5 c = —210g(o-_}. Equivalently the critical region can be written as R={[)Ir ...... X ,1 i1 ..... I-;.I A_(X1 ...... X“ ,,. Y1 ..... ms —21og(o.i}. l? (cl) Under H1 the parameter space is 2-dimensional and under Hg it is l-dimensional. Based on Wilks' Theorem one would expect A i x? where the degrees of freedom is the difference between the two dimensionalities. However this is contraddicted by (c)(ii) which states that A w x3. Wilks‘ Theorem does not apply here because one of the regularity conditions requires that the range of the samples does not depend on the par.ameter(s). '- 4 M3481 Statisocal Theory 1 (2015) Page 5 of 5 ...
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