M562-HW9-6

# M562-HW9-6 - MTH/STA 562 Exercise 9.61 Calculate 1 =1 6...

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Unformatted text preview: MTH/STA 562 Exercise 9.61. Calculate 1 =1 6+1)y9+2 y 0+1 ’=EY : 1 9+1 = —( _ “1 M /(9+ )9 dy [ 0+2 (9+2 0 9:0 Equating the ﬁrst sample and population moments, we obtain the estimator of (9 as follows: 7223:; or (94974“ or (2r_1)=(1_v)aor 5:21;. 9+1 It follows from Example 9.2 that 7 is consistent for m. Then, using Theorem 9. 2, we can show that 5 is consistent for 0, since 5 converges in probability to ﬂl_2(0+1)—(6+2)_ {—3 ‘ (6+2)—(6+1) “ 0. TL Note that 6 is not a function of sufﬁcient statistic — ZlnY}. This is implies that i=1 gis not the MVUE of 0. Exercise 9.62. Note that ,u’l = E (Y) = A. Equating the ﬁrst sample and population moments, we obtain the estimator of /\ as follows: /\ = 7. Exercise 9.63.. Note that M2 2 EC”) 2 02 + ,uz = 02 (since ,u = 0). Equating the second sample and population moments, we obtain the estimator of TL 2 .A2_i 2 a as follows. a —n§ Yg. i=1 Exercise 9.64.. Note that ,u’l = E(Y) = ,u. and ,u’2 2 E(Y2) = 0'2 + #2- Equating the ﬁrst and second sample and population moments 1 7?: Y2}? and —ZY;2=32+/322 ”i=1 we obtain the estimators of ,u and a2 as follows: ﬂ = 7 and n A l n —-2 1 —2 02:5;1/3—1/ 2;;(12—37). Exercise 9.65. Let 1 if the ith ball is black Y; = 0 elsewhere. Since there are 19 black balls and N —-- 0 white balls, we note that for Y; to qual 1 we need to ﬁll position i with a black ball, which can be done in 9 ways, and the other 72 — 1 positions may be ﬁlled with any combination of the other N — 1 balls, which can be done in (N—1)(N—2)--~[(N—1)—(n—1)+1]=(N——1)(N—2)---(N—n+1) _ (N — 1)! — (N — n)! ways. All together, there are N! NN—l N—2---N~ :— < >< > < n+u (N—nﬂ ways to ﬁll the n positions. Thus, 6 N—1!! 0 —n ! P{Yi=1}: (121)2N- (N—n)! Thus, 0 m=Emr4umn=u+oPm=m=N. Equating the ﬁrst sample and population moments, we obtain the estimator of 0 as follows: sothst 6=N =N%ZY;=_, _ a Y—w? Where Y 2 2K. i=1 Exercise 9.66. (a) Calculate 0 =9 ,_ -3/ _ 2 __2_ gy_2_y_3y __2_ (13113 “Q “I“E(Y)“02O(gy ﬂaw—02 2 3y ‘02 2 3 “3' Equating the ﬁrst sample and population moments, we obtain the estimator of 0 as follows: A 72% so that 5:37. (b) The likelihood of the sample is n 2 2 n " L(yl,y2,“‘73/n;9) = 1155(9—92') = (—0—?) [lye—1%") 2 which cannot be factored into a function of Y and 9 and a function only of Y1, Y2, - - -, Kr Thus, Y is not a suﬁicient statistic for 0,. Since 0 is a function of Y, it cannot be a function of a sufﬁcient statistic. Exercise 9.67. Note that this a beta distribution with parameters a = ﬂ = 0. Then 0 1 I _ _ = _ which does not involve 6 and hence is of no use in ﬁnding a moment estimator. Since 6 0 1 2 2 V y = _____L___ : —, a H (9+9)2(9+0+1) 4(29+1) consider 1 1 0+ 1 ’=EY2= Y Ey2=__.__._ _:________, “2 ( ) W )+[ ( )] 4(20+1)+4 2(26+1) Equating the second sample and population moments 1 n __ 5 + 1 n i=1 2 (25 + 1) ’ and solving for 5, we obtain the estimator of 0 as follows: 1 — %ZY;2 g: i=1 4 ; 2Y3 — 1 i=1 Exercise 9.68. For a single observation Y, the ﬁrst sample moment is m’1 = Y. Since Y has a geometric distribution, ,u’l = E (Y) = i. Equating the ﬁrst sample and population moments, we obtain the estimator of p as follows: 1 so that 13 = —. Y: Y 1 1’5 Exercise 9.69. Since this is a uniform distribution on the interval (0, 36), ,u’l = E (Y) = %. Equating the ﬁrst sample and population moments, we obtain the estimator of p as follows: > | ...
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