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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—2N~ :—
< >< > < 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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 Spring '08
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