Monrad.410.30

Monrad.410.30 - STAT qlo / MATH HM EXgrgige 2‘1 2. 772124...

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Unformatted text preview: STAT qlo / MATH HM EXgrgige 2‘1 2. 772124 uéc (a) E9[u(Xl] = E600 (6) EeluOOJ = E900 K = X‘+X,_+ +Xn. The joint p.d.f. of X1,. . . ,Xn is 02"(1 — 0)""E"',.so Yl is sufficient for 19. Clearly the distribution of Y1 is b(n, 9). To show that Y1 is complete, let us suppose E[u(Y1 = 0. ‘ Then ' n n gunfigaku — 0W = (1 ~ 6)" gunman, where p = 0/(1 — 6). Using the fact that if co + clz + C22:2 + - o - + cnz“ : 0 for more than n values of z, then co = c1 = v -- = cn = 0, we conclude that u(0) : u(1) = «-- = u(n) 0. This Show: Hul- Y‘ is complete. (Since WC are loot:th al- a Legal” upoMnRI milét at Fmbatilgb Mass fimch‘ons inie) : o<e< E) Tl’W 7.5.2 also tells us Hm Y. is a (ample: Suteclent statistic tar 5;) , Y, is binomial = “G. Hana _; X? is an unbiased “Ema/m- out 6. LehMaun—Schette'z theorem (on pa, 387) Hts us that this finchon. at Y, is gm (and. %_.i.,,-,.-"0F. of 5mg.llalr..osslble, Mariana is. minimum Vadange unbiasgd mlimlzfir m.‘ 6. ' Exercise 7.53.00 We arc (coking a4 a Lead” QMMMHM {ami‘a a? FAQS: r 6>OI £636} = 67(6)“l = Upwovxbd - 64X) + 049)] )oom. 1% mm m TM 7.5.2 HM— Y,= Ewe) is a comple’ce smr-Rciwé 9&5ch Qré. "“ Sina Hm MCHOA gfi 6w" is One-h-OHE, We shfiskc (XX wXY/n = aw“ is odso compkk $4490;er «far 6. (E) In EXeM/ise 61.30:) M {wt/1d %a* W NLE (’maxfmum “Edd/toad MIX/maer a? 8 Is A _‘ __ n _ __ 1 9 " 2m x1) ‘ 6n (xtxzwxnw ’ ch is a ‘chkon 04 Hm geomd'n'c mean. 3on 7. 45.9 H; W \r' ' W MM‘me/M’) @4041 (mm pd? 4((x; a} j W {vol/(4 Y = WPQQ Meanm 8< X = w W. 1’ I? Y ha? (male {76 Compézk SKQQOW‘L a; e Jinx as M UMVUEOee, (X is JOEWJMA‘g u/néia/Aed wvwk a Wc/{M o§ Hoe/HZ) b! LCAMann-Sckcflé‘s H‘eonem. Sinu V 13 MEYMEAJ 5mm =- UMVUE = 56' _ [ is Q $535.5} (becausc X15 sugficim'f) Walt is a ‘F‘uMCWDH m5 X and has Hm 5am: mean as Y ) 44103- ?s 9. $0 EONsz is ' a‘so M UMVUE of 9 LekMam-Sc UCFC’J ; £ The, PM 3(336) o€ Hm Median saksfies g(e+'339\ = 3(6'336) )firaflg. V (See pug/1 2'40.) Problem 7.5.10. a) : 62$e—6x : 6—6x+log:v+21096 Z 810(9)K(:z:)+S(:c)+q((9)7 0 < a: < 00. proves f (:6, 6) is an exponential family and, by theorem 7.5.2. the statistic Y : 22:1 K : 221:1 3:1- is complete sufficient. b) X ~ H2, 5) a Y = :2; Xi ~ mm, 5-), my, 6) = min) 1 _ 1 * 00 16271 271—16—1/9 w 00 02n—13(2n—1)~ e y9 _ El?) _ 0 5 F(2n) dy ‘ F(2n) 0 F(2n—l) dy — F(2n——1)6 *1 : 6 E<2n—1) : (2n—1)F(2n~1) 2n~1’ Y So gngl is an unbiased statistics for 6 and it is function of the complete sufficient statistics Y . By Theorem 7.4.! it is the unique unbiased minimum variance estimator for (9. Problem 7.6.l.Xi N N(9,1) From Example 7.5.2 we know X is a complete sufficient statistic for <9. X7 N N(6,%) :;> EOE”) : V0?) + [E(X)]2 : %+ 62 :> EOE2 — : 62. So X2 — 3; is an unbiased statistics for 62 and it is function of the complete sufficient statistics X . By Theorem 7.4.! it is the unique unbiased minimum variance estimator for 62. ...
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Monrad.410.30 - STAT qlo / MATH HM EXgrgige 2‘1 2. 772124...

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