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# M3S1 2016 A.pdf - BSC and MSci EXAMINATIONS(MATHEMATICS...

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Unformatted text preview: BSC and MSci EXAMINATIONS (MATHEMATICS) May-June 203.6 This paper is also taken for the relevant examination fo? the Associateship of the Royal College of Science. M381/M4Sl Statistical Theory l Date: Tuesday. 10th May 2016 Time: 9:30 — 12:00 Solutions © 2016 Imperial College London M351/M451 Page 1 of 8 1. (a) Neyrnan Factorization Criterion: Suppose that X = {X1,...,Xn) has a joint distribution fg(x]. Then T = T(X) is a sufficient statistic for 0 if and oniy if no) = s (To), 6) Mm). Proof: First suppose that T is sufficient. Then few) 2 P5(X = 3:) = 2 P30: = s,T z t) = P9(X = cs“ = Tm) = PM“ = T{\$))1’;(X = sIT = Tit-n = amt), out) Now, suppose that fats) = g (T(x),l9) Mm). Then. if T(:c) = P5(X :35] * P9(X = 2:) Pe(T=i) # Z Palx = 3') ll P3(X : :L’lT = t) TWP!- = g Totem) = film) 2 9(T(y)i6)h(y) Z My) T {wet Tui1=t which does not depend on 8. If T(:i:) % t. then Png = mlT = t) = 0. In both cases. P5(X = \$111” = t) is independent of 6 and so T is sufficient. (b) (i) According to Neyrnan Factorization Criterion. since 11 n . ‘_ n _ f9(:1;) : H93i(1_ 3)1—\$t = 6E1 1(1 c g)“ (2:132, £21 31'. T = 2 Xi is a sufficient statistic for 3. i=1 Yes. T is complete because Bernoulli distribution is a member of "full rank" exponential family of distributions since fs(\$) = exp (1n'(«1—_)E\$t+nln(l—9]). Also (2" Is minimal sufficifent J{no—(:35 se it I81“: cmo Oil§te an? sufficient statistic. 1 = 2 == . (ii) An unbiased estimator o By the Lehmann— 20138 Scheffe Theorem, the UMVUE of 6(1 — 6) is P(X1= (LXQ = 1’ 2X5 = i) i=1 n P(X1 =0,X2 = 1,2 Xi =3) 5:]. E(I(X1 : 0,X2 = 1m“ = t) 41%; X;- = z) I: PUY} =0,Xg=l12Xi=t"-1) i=3 13(2 Xi = 2) {:1 6(1 -— 3) ( ::f ) ei—lu _ 6)n—2——t+1 Therefore. kl—Hn—(HTIT— is the UMVUE of 6(3. -+ 6). (iii) The Cramer—Rao lower bound here is (3% (6(1 — 6mg 2 (1 — 29)? .. (1 — 29? * 9(1 —9)(1 — 26')2 1(6) nIx1(6') ‘“ 31—1—65 ‘* n. ‘ 7}. Only estimators of the form {a E X,- +b} achieve the Cramer-Rao lower i=1 bound. So the variance of the UMVUE of 6(1 u 9) does not attain the lower bound. The Iog—Iikelihood function is 2 _ 'n. 2 1 _ ‘ 2 n ‘_ _ 2 Km. meme )— “Elogﬂw )~ 2-55 {21% - m) + 12309» #1) } and the MLEs of the parameters are “Jew; a2_1 " X£+Ye 2 " Jar-+14 MH 2 and a —;{Z Xg— 2 +2 3’;- 2 (b) (i) Since Zz'Lg N[ D 202) the iog—iikelihood function based on Z1... Zn is n 1 “9-2) = w~2-log;(21n:r2) _ 472 Z :53 and then the NILE of 02 is :72 = ~27! E Z? MLEs are consistent under the regluiarity conditions. Because normai distribution satisfies the reguiarity conditions. the MLE 32 of 02 is consistent. (ii) Since 13(32): 202. method of moments (MM) estimator of 0'2 202=—223 => 01%!!M=§1E:ZE' a=l ni=1 (iii) The family {N(O,202): or2 > 0} has monotone likelihood ratio in z of. :1 Using the Karim-Rubin Theorem. the UMP test at ievel o: is 1 1 if 22:? > is out) = “:1 0 if z; z? < 3: i=1 where k can be chosen so that n k at = 1303(2252 2 k) x 130304») > *3). i=1 We then get A: *- Zaﬁxﬂn ) (iv) The power of the UMP test obtained' In (iii) Is non- decreasing In 6 because of the monotone iikeiihood ratio property. 50 the UMP test here' Is an unbiased test since its power is not less than 0:. 4of8 ﬂ 3. (3) Applying the Bayes theorem, we can write the posterior distribution as follows AeﬁAagne"31§(¢i—2} ﬁne—9C}; (mi—-2)+A) ‘9 mgr-‘2 ~3 i— f? AGHMEHB i§=31{ )dﬂ face ﬁne (:2 (I ”‘11) d6 C where c is a constant which does not depend on 9. Because the posterior Tl . distribution is proportional to 6”e(“K9}, where K is 2 (mi — 2) +A. the posterior 5:1 is Gomme(n + l, K). In fact Six m Gamma (n + l, i: (m,- — 2) + A). i=1 Alternatively. one can obtain the above posterior distribution straightfomarclly by calculation of the integral in the denominator (i.e.. the constant e) as Follows: Ae-Aﬁane_8s§z(x‘_2) Que—6(rg(xi_2)+’\) 7r(€|:2:) = n = ﬂ —3 35—2 —E {— fﬂ‘” Ara-Mane Eli )d3 [5” ans (5:31“ 2H) d6 Que—I9 (ﬁg (:3,- ‘2)+A) = 1 n n n E (Exponential m3: — 2 + A ) ) i Z {mi-2)+A) ( (i; ( ) ) 1:1 n —a " rap-2 A (2 (scre2)+/\) ﬁne (E5 1+ ) = 1:1 in! i E1 (mi-23+A) 71 ”+1 —9 5 5—2 A (2 (2:1: - 2) + A) We (1%“ H) # i=1 _— n! T}. which is again Gamma. (n + 11 2 (so; — 2) + A). i=1 (b) Yes. because both the prior and the posterior are Gamma distributions. Note that exponential distribution is a special case of Gamma distribution. (c) Under the squared error loss. the Bayes estimator is the posterior mean. Because the posterior is Gamma distribution. we can easily obtain n n. + 1 eBayes : n E(\$g—2)+/\ i=1 50f8 (d) The above Bayes estimator can alternatively be obtained as follows: n+1 ‘— 1: "—- (f: (131: - 2) + A) fine SKEW 2}+A) LN n! ll ﬁBayes d9 Ewe) = fame (1; (m1: — 2) + A)” ll n n+1 E (ExponentiaKZ (xi - 2) + M) i=1 n! _ n+1 Z(\$i—2)+A i=1 Because the Bayes estimator obtained in (c) is unique. therefore it is admissible. Under the whole parameter space, the MLEs of 61 and 32 are §1MLE = -1— and X 32;?ng = ~1— respectively. And under H0, the MLEs of 91 and 82 are yl a A m + n = 6 : w‘ (61x: ( 2)0 mX+nY Hence. the likelihood ratio statistic is as follows 353% W) _ L ((93)n , (900) Egg 25(9) L (guns, g2MLE) 1| A(any) 1|. A m m3:- .._ n 3m ((900) e 3 E ((600) e ”3 A m ... n a. m “9131”: 2 m A n -92MLE 2: y; (91MLE) 8 ‘21 (32MLE) 8 ”1 m + n E _m n + m X "n _m.+n m+nX m-i-n m+n§7 ‘ ll We know the likelihood ratio test rejects Hg for small values of M33324). Now, because M33, 3;) depends only on T 2 £5» and we can make A(m,y) small by making T small or T large. so a test based on T = é would reject Hg for small or large values of T. in fact, a level 0: test based on T = 9}; rejects Ho : 91 = 62 if T 5 c1 orT 2 Cg. where cl and (:2 can be chosen so that PHD {T 5 clj+PHn(T __>_ Cg) = 0:. where the distribution of T = ”:33. under Hg. is F(2m,2n). By considering equal tails of the F distribution. we can reject Hg if T g Fina/29771311) or T Z Faf2(2m, 2n). Gofs (c) Under Ho and under regularity conditions. the asymptotic distribution of ~2l0g(/\(&=,y)) is 98(1)- The likelihood ratio level or test based on the asymptotic distribution rejects Ho if —2ios(k(\$.y)) 5 X141} [d) From (in) we have 9 X PH0:91=32 (Flea/2(2m:2n) S 5:7 S F aizigm 2n)) = 1 " a: and hence using the connection betWeen confidence intervals and hypothesis tests. a confidence interval For 9—3; with confidence coefficient 1 — o: is 9 (3; F1 0,2(2m 270,11 X Fa/2( (mm, 211)) M3451 Statistical Theory 1 (2016) Page ‘17 of 8 TONS 5. (a) The likelihood equation is n 20131—6) _ 8(6):;1+(\$é—6)2 —0 Because 5(6) is not monotone in 6. the equation 51%“) = 0 may have more than one solution for given sample 31.....In (b) Using the Newton-Raphson method a new estimate is given by goal) I an +. mew) Hwﬁkl) where 3(6) is given in (a) and = n l—(xi—€)2 22(1+{m,,—_9)) ‘ (c) The Fisher scoring algorithm gives a new estimate as follows SW) we): 3500+ A Hecate?) where H*(6j = E(H(9)) Considering the hint, we get ll H*(3) E 22% =2n/m .In—(ml—ﬂi_dmi (1 + (\$1 a e) -m 'n' (1+ (mi — 6)2)3 _ 4:1 °° 1Wei-9) MEL _ -2 # 'lT El (1+[:in—3)2)3dﬂ%m'.:'l:b(:\$')_21 and hence (d) Because E(Xi) is not well-defined, the sample mean may not be a good initial estimate of 19. Since the density of the Xis is symmetric around 6, one may use the sample median as an initial estimate. (e) The convergence of the Newton-Raphson algorithm is often faster (when both algorithms converge) because it uses the observed Fisher information rather than the expected Fisher information which needs integral calculation. M3451 Statistical TheOry | {2016) Page 8 of 8 Sof8 BEE ﬂ ...
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• Fall '17
• Statistical theory, exponential family, Sufficient statistic, likelihood ratio, Bayes estimator

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