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Unformatted text preview: MTH/STA 562 Exercise 9.48. In Exercise 9. 30(1)), it was shown that :01 — ,u)2 is sufﬁ— i=1
cient for 02 when n is known. Since ,u is known, let us consider the statistic W=l§2(Yi—M)2. n i=1
It is easy to verify that
1 n 2 1 2 2
E W = — E K — = — = ,
< > n [< u) ] nm a that is, W is an unbiased estimator of 02. Since W is a function of sufﬁcient statistic Z (Yi — m2, W is the MVUE for 02.
1'21 xxx—a2 Exercise 9.49. From Example 98, it was shown that 512 = {T ' t 2 . . 2 — 1 n _ 2
is the MVUE for a from the sample X1, X2,   , Xn. Similarly, 82 = E Z (Y; — Y)
i—l is the M VU E for 0'2 from the sample Y1, Y2,   , Yn. Now the sample pool—ed vari—
ance, ZEQQ —7) +£(Y; —7)2
512 2 i=1 i=1 __
p 2 (n — 1)
1 1 n _ 1 n _
: Angel—manage”)l
= as; + 53),
is also an unbiased estimator of 02, since
E(S§) = g [19(512) + E (53)] = g . 202 2 02” Using the result in Exercise 8. 2, we see that S; has minimum variance and is thus
an M VU E i. Exercise 9.50. In Exercise 9.32, it was Shown that Z Y;2 is suﬁicient for 6‘,
i=1
Now 00 2 00
E (Y2) = / 5y36”y2/9dy = / “gum/9cm (with u = 92 and du = 2W)
0 0 = E(U)=6, since U is an exponential random variable with mean 6. Thus, 1 n 1 n 1
E — Y2 =— E Y2 =— 6:6“ 2) M w
Since W = i Y? is a function of sufﬁcient statistic 2Y3, W is the M VU E
i=1 1:1
for 6. Exercise 9.51. With Y a Poisson random variable with parameter A, it is
necessary to ﬁnd the M VU E for E (C) = 3E (Y2) = 3 {Var (Y) + [E (1012} = 3 (A + A2) n. In Exercise 9‘, 31, it was shown that 216 is sufﬁcient for /\ and thus for /\2 and
i—l 3 (/\ + A2). Now note that E (7) = A, and E(?2)=Var(Y)+[E(7)J ——2+/\2
E<172—§> =E(172) =/\2
E{3<72—§+7>] = [E<72—%>—E(7)1=3(/\+A2)=E(O) Since W = 3 (—2 — +7) is a function of sufficient statistic :K, W is the
i=1
MVUE for E Exercise 9.52. (a) The given probability density function can be written in
the form 9—1
€1n<0y ) I eln0€(6—1)lny = dew—mm} for 0 < y < 1 and 0 > 0, f (y; 9) which is the exponential family with (1(6) = (9, My) =
d(y) = —1ny. In Exercise 9.37, it was shown that Zd(K) = ~ZlnY; is sufﬁcient for 6. (2)) Let w = — lny. Then 3/ = 6—") and 32% = —»e_“’. Thus, w d _
meme) Hawaii =9e(9’1)1“e l  = (9e_(9_1)we“w = ﬁe’ew for w > O, which is an exponential distribution with mean 1/6.
(0) Consider the transformation T : 20W“ Then to = 15/26 and $71: = 1/26.
Thus, . _ t. dw _ —9(t/29)1 _1 ~t/2
ﬁ@ﬁ%:M(%ﬂ)bt_6e 26—26 bm>0, which is the form of a chisquare distribution with 2 degrees of freedom. Therefore, 2 in = 26 2 W1
i=1 i=1 which is a chi—square distribution with 2n degrees of freedom.
(d) In Exercise 4.90, it was shown that if U has a chi—square distribution with 1/ degrees of freedom, then E = Based on this result and the result of
part (c), (e) In parts (a) and (b), it was shown that Z m = — Zan is sufﬁcient for
i=1 i=1 6. It follows from part (d) that E 2—1 =a
EN“
i=1
Thus,
3—1 _ 71—1
M Dun i=1 i=1 being a function of sufﬁcient statistic  Zan, is the M VU E for 9.
i=1 Exercise 9.534. In Exercise 9.41, it was shown that IQ") is sufﬁcient for 6?.
Also, it was shown in Example 9.1 that n
n+1 Ehnﬂ= a 3 Thus, 1
71+ ]=n+1 72 9:6” E [——Y(n> '
n n n + 1
Since W = "THE/(n) is a function of sufﬁcient statistic YW), W is the M VU E for
0. Exercise 9.54. Calculate in Exercise 9.43, it was shown that Y0) is sufﬁcient
for {9. Now nye_”(y'6)dy II E lle n (u + 9) 6—”“du (with u = y — 0 and du = dy) H H
°\8 °\8 O\8 nae—“udu + 9 / newudui
0 Note that the ﬁrst integral, being the expected value of an exponential random
variable with mean 1/71, is equal to 1/n while the second integral, being the
integral of an exponential random variable with mean 1 / n, is equal to 1. Thus, 1 1 1
Hence, the Statistic W = Y(1)——1T; is an unbiased estimator of 6. Since W = Y(1)—% is a function of sufﬁcient statistic Y“), W is the M VU E for 6. Exercise 9.55. Calculate in Exercise 9.44, it was shown that 1/0,) is sufﬁcient for 6.
(a) The distribution function of Y is H y 3t2 t3 by 3,3
F(y)=P{YSy}=/:p—3—dt=l5§] 5g forOSng,
0 i=0 that is, 0 y < 0
3
F (.9) = 75—3 0 S y S 6
1 y > 0.
Thus, Y(,,) has probability density function 9(n)(y;9) = n[F(y)l”‘1f(y) n(%§)n_1%%2— 033/30
0 elsewhere 937:.
0 elsewhere. {3ng3n'i (b) Now 6’ =0
3ny3n 3n 3n+1 9 3n
E [You] = / dy [9 = 9. (93” :55; 3n+1 3n+1 0 11:0 Hence, the statistic W = will/(H) is an unbiased estimator of 6. Since W = BEIYW is a function of sufﬁcient statistic Y0), W is the M VU E for 9, Exercise 9.56. Let Y1, Y2, , Yn be a random sample from as normal
’— = 1 distribution with mean ,u and variance 1. Then E (7) = a and Var (Y Also, it follows from Example 9.8 that 7 is sufﬁcient for ,u.
_ __ _  2
(a) Since Var (Y) = E (Y2) — [E , it follows that E (72) 2 mm + [13(7)]2: 1+ #2. Let ,1? = 72 — Then E/3=E72 ——‘=~— ———= . (M ) ( ) + M n M Hence, the statistic I? = 72 1 is an unbiased estimator of #2. Since ,L/La = 72 —%
A ——2 is a function of sufﬁcient statistic 7, u2 = Y — i is the M VU E for #2.
(b) Note that Y, being a normal random variable With mean u and variance
1/n, has moment—generating function given by t2
771705) = exp (at + n, Then 2
771/705) 2 (,u—l —> exp <ut+ 5—) =< + m7(t)
711%(75) :1;le (t) + (M + £>2m7(t)
771$“) — %(/~b + mﬂt) + %(M+ mix(t) + (Mr gym—370)
= 2(M+%)m—y—(t)+(u+%)3mv<t>
777%) = gnaw; (wﬂzmwn 3 (u+%)2m7<t>
+ (u + ~94 my (2:)
= %m7(t>+ g (mgfmw) + (u+;:)4mv<t> Thus, Now
A2 _ 1 2 — 2— 1
Em = El<y2——>l=E<Y4——Y2+—2>
n ‘ n n
_ 3 6M2 4 2 1 2 1
“ n2+ n +M —n<n+ )+n2
3 6,112 4 2 2M2 1
— 722+ 72 +'u —n2 n +n2
2 4M2 4
— n2+ n +M
Hence,
A A2 A 2 2 4112
2 _ 2 __ 2 __._ _ 4_ 4
Var(,u) — E<,u> —n2+ n +M M
2 42 2 1
n n n n Exercise 9.57. (a) Since T_ 1 ifY1=1andY2=O
— 0 elsewhere, E(T) = (1)P{Y1=1,Y2=0}+(0)P{Y1yélorléyéO}
P{11=1,Y2=0}=P{3€=1}P{Y2=0}
= 191(1~10)“+p0(1—p)1_°=p(1p) (b) Since W = Z X, W is a binomial random variable with 12 trials and success
i—l probability p. Now~
P{T=1,W=ui}:_ P{Y1 =1,Y2=0,W=w} New} For the numerator, since a success must occur on trial 1, but not on trial 2, the
remaining w — 1 successes must be distributed in n — 2 trials, which can be done n — 2
' . H
in (w _ 1) ways ence, <Z>pw(1—p)””w — 6 P{T=1[W=w}= (w—1)'(n—w— 1)' n'
_ 10' (71—10)! (71— 2)I
— (w—1)'(n—w—l)' n'
I w.(n_w) 1 :w(n—w) (0) Since = (1)P{T=1W}+(0)P{T7é1}W}=P{T=1W}
: —W(”"W)= ” (14:) (1—K): ” 37(1—17), n(n—1) n—l n 72 71—1 wee see that n—1 E[ ” 7(1—7)] =E[E(TW)]=E(T)=p(1—p). Note that W = EK = n? is sufficient for p [and hence for p(1  Since
i=1 #1” (1 — Y) is a function of sufﬁcient statistic W, n—fl—Y <1 — Y) is the M VU E
for p (1 —— p). Exercise 9.58. (a) Since n, n—ix,
L(x1,x2,~,xn;p) = Hp“ (1‘p)1_”” =pi=1 (1 —p) i=1
i=1
= —~—p (1 —p)”
1 —p 7
we have
L(3717$2»"'7$n;19) : (ﬁliM—ﬂ (1 j p n :< p )gwt—gw
L(y1)y27"'7ynip) i2” (are <1 w If :33, = 2%, then the likelihood ratio in part equals 1 and is,
i=1 ":1 ’L 71 Tl
thus, independent of p. If 2 36¢ 7E Zyi, then the ratio is not independent of p. i=1 i=1 (iii) Let g(y1,y2,    ,yn) : Zyi. By the Lehmann—Scheffé method, the
i=1 TL
result in part (ii) shows that Z Y, is a minimal sufﬁcient statistic for p. i=1 (b) From Example 9.7, n 1 n n
L(ylay2a"'ayni0) : exp i=1 i=1 Then A
tblw
V 3
CD
>4
’0
A

QIH
s». t M 3
H
e.“
\_/ a £4: 3
H
s L(':C17‘T27' ' L(y1,y27,yn;9) < Gallo
V
3
(D
><
’73
l
chi»;
iM:
Cd
w
V
in:
is The likelihood ratio in part is independent of 6 if and only if
E = Z .. Then, by the LehmannScheffé method, 2 Y; is a minimal suffi—
‘—1 i—l i=1 1,— _
cient statistic for 9. Exercise 9.59. Since the likelihood is  L(yl,y2,,yn;u,02) — ameXP[—ﬁ(yzM)2l
i=1
1 (2 W2 0” 2‘7 i=1
we have
7} ex —i n x, — 2
L (2:17 3:27 ' ' ' axn; luv 02) __ (2F) ﬂan p 202 ( L(y17y27”'7yni/J’702) _ n 2
(27f)7i/2an exp _—2%'2_ Z .— i=1
_ 1 n 2 n 2
— exp 33—5 Zoe—M) ~2<yi—M)
i=1 1:1 H
(D
>4
’5
A
“l
p...
KO
fa
M:

M:
$0

[\3
7;
A
M:
§
1
M:
g
V
—1
HH 0
Let g(y1,y2,   Wyn) = ghzg/f). Then the above likelihood ratio is in—
i=1 i=1 dependent of (M,02) if and only if 2 23/, and = 2313, that is,
i=1 i=1 i=1 i=1
9 (331, 5132,    ,ain) = g (y1, y2,    , yn). Hence, by the Lehmann—Scheffé method dis TL cussed in Exercise 9.58, K, E Y?) jointly form a minimal sufﬁcient statistic
i=1 i=1 for u and 02. Exercise 9.60. Since g1 (U) and 92 (U) are unbiased estimators of 0, we
have E[g1(U)]=6 and E[gg(U)]=9 sothat E[91(U)—92(U)]:0. Since the family of density functions of U is complete, it follows from deﬁnition
that 91 (U) E 92 (U That is, 91 (U) must equal 92 (U Thus, there is a unique
function of U that is unbiased estimator of «9. ...
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 Spring '08
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