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Unformatted text preview: @ The population pmf is f(x9) = 9(1 — 9)x " 13061030 ol", an exponential family with
g .1? t(x) = x. So ZXi is a complete, sufﬁcient suﬁicient statistic by Theorem 6.1.3 and Thurem 6.1.6. 2 xi — 11 '~ negative binomial(n,€). 6.7.2.
w a) The sample density 18 Hf(:s0)= [1025’] = GnuI 11991 , so HXiissuﬁicient for9,
not 2%. 0—1 to . . .
b) Since f(xi  9) = 9e( ) gx)’ an exponential family with t(x) = log}, 2 (03% = logﬂig is
‘ complete and'sumcient by Theorem 6.1.6. Since HXi is a onetoone function of logniﬁ,
HE is also a complete, suﬁicient statistic. ' a I To ﬁnd a best unbiased mtimator of 9, ﬁrst ﬁnd a complete suﬁcient statistic. The joint
pdf is I f(xl0)= (71,; —")1IIIH.)(x,)= —(217 )nl 1[09)(muﬁl) By the Factorization Theorem, max I Xi' is a sufﬁcient statistic. To check that it is a
complete sufﬁcient statistic, let Y = max [Xi I. Note that the pdf of Y is fY(y) = nyn—l/On,
0 < y < 0. Suppose g(y) is a function such' that mm =/: “n —n—z(y)dy— — o for an 0
Taking derivatives shows that all—15(3) = o, for all 9. So 3(a) = o, for all a, and Y =
max Xi  is a complete sufﬁcient statistic. Now 1 BY: 12ij Tifdy=m9=>E(E—:1Y)=9. Thereforen —n—max Xi  is a best unbiased estimator for 0 since it is a function of a
complete sufﬁcient statistic. (Note that (X (1), X00) 18 not a minimal suﬁicient statistic
(recall Exercise 5.22). It is for 9 < Xi < 20, —20 < Xi < 9, 40 < Xi < 60, etc., but not when the rangeis symmetric about zero. Then max  Xil is minimal suﬁicient. =7~ s8
#13) The CramérRao Lower Bound for unbiased estimates of p is , 2
[3%.]  _ 1
_ __T—
“llBiff log L(pIX) nE { it; log [px(1—p)1X]} ___p(1p)
_ x1(l—Xi ‘1 ’
=nE{ 3 (11)) _7—} since EX: p. The MLE ofp is 3 = 1132311, with EB: p and Varfi = p(l—p)/n. Thus 6
attains the CRLB and is the best unbiased estimator of p. 'Qus b) By independence, E(X1X2X3X4) = 11;; Exi = p4, so the estimator is unbiased. Since
ZXi is a complete sufﬁcient statistic, Theorems 7.3.2 and 7.3.5 imply that E(X1X2X3X4 
in) is the best unbiased estimator of p4. Evaluating this yields E(x x x ix“ x n) < 2mm)
12X34 i=1 1,— — HEP: Xi=t)
if 4 11.4 t—4 n—t‘
=” (H) “ ‘” =<H>/< >
. (I: PlUp)1H "4
for t Z 4. For t < 4 one of the Xia must be zero, so the estimator is E(X1X2X3X4 l + tn 
m a. From Theorem 5.3.2, Y : X0) has pdf n—1
fY(Y)— (n lIAWY/A[1_(1_eY/A)] =XC’“Y/"_ Thus Y ~ exponential (A/n) so EY = A/n and nY is an unbiased estimator of A.  b. Since fx(x) is in the exponential family, 2% is a complete sufﬁcient statistic and
E(nX(1) I EXi) 18 the best unbiased estimator of A. Since E(): X): nA, We must have E(nX(1)I 2K)=n1i}:X by completeness. Of course, any function of EXi that 13 an
imbiased estimator of A' is the best unbiased estimator of A. Thus, we know directly that since EEK) = nA, XXI/n' 1s the best unbiased estimator of A. c. From part a); A = 601.2 and from part b) A = 128.8. Maybe the exponential model is not a
good assumption. ﬁt) Since the Poisson family is an exponential family with t(x) = x, 2X1 is a complete
suﬁcient statistic. Any function of 2% that is an unbiased estimator of A is the unique
but unbiased estimator of A. Since i is a function of 21g and EX = A, X is the but
unbiased estimator of A. b) 52 is an unbiased estimator of the population variance, that 18, E82: A. X 1s a oneto—one function of EXi. So X 18 also a complete suﬁicient statistic. Thus, E(S2X)' 1s an
unbiased estimator of A and, by Theorem 7.3.5, it is also the unique best unbiased estimator of A. Therefore 15(82 X) = X. Then we have
VarS2 = Var(E(S2IX)) + Evm(521'() = VarX + EVar(S2X),
so Var S2 > VarX. c) We formulate a general theorem. Let T(X) be a complete sufﬁcient statistic, and let T(X)
be any statistic other than T(X) such that ET(X) = ET’(X). Then E[T’(X) [T(X)] = T(X)
and Var T'(X) > Var T(X). ...
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 Spring '07
 SENTURK,DAMLA
 Statistics

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