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Unformatted text preview: MTH/STA 562 Exercise 9.72. (a) The likelihood function is n yi
” Awe—A /\i=1 6—71"
L(y17y27'”7yni)‘)=H : TL ~= 112‘! z 1 H 3%!
i=1 Then the log of likelihood function is TL lnL (y1,y2,   ~,yn; /\) = (2%) ln)\ — n/\ — Zlnyil.
i=1 i=1 Thus,
ilnL( 'A)—li i—n
d/\ 3/1792, 73/717 “‘ Ai=1y1 ~
Equating the derivative to 0, we obtain the M LE as follows:
A 1 z _
/\ = — Y; = Y.
n i=1 ((9) Since ,u = a2 = A, it follows that E<X>=E<7>=M=A and VGT<X>=VCLT<7>=0§:%, (0) Since X is unbiased for A and lim Va?" = = 0, it follows from n—>oo Sly Theorem 9.1 that X is a consistent statistic for /\. _
(d) Since the MLE of /\ is A = Y, the MLE for e_’\ is e‘“Y. Exercise 9.73. The likelihood function is
n l _ 4 1 1 "
Lf917y27 ' ' ' ,yn;9) = H 56 yz/e : 57; exp (—5 ~
i=1 i=1
Then the log of likelihood function is
1 n
1nL(y17y27 ' ' = —nln6 _ g i=1 Thus, d n 1n @lnLWLWV'Wynig) : Equating the derivative to O, we obtain the M LE as follows:
n 1 n
—7 + 7
9 = 2 YQ=O or —n+%ZK=0 or d: ZYz=?,
i=1 ' z 1 which is the MVUE of 6 as shown in Example 94,9" Since 6 > 0, 15(0) 2 02 is a one—to—one function of 6. Therefore, 72 is the M LE of 62, which is not the M VU E of 62 as shown in Example 9. 9. Exercise 9.74. (a) The likelihood function is n 7. TH W 1‘ 7.7L TL Th— 1 n 7
L(y17y27"'7yn;6):H—éyi 16 yi/6:%<H% 1) exp i=1 i=1 i=1 Consider TL 9 (Xi/2,9) = %; exp (—5 23/1) and h(y1,y27 . . . 7y”) : Hyg—lw
i=1 '— i=1 It follows from Theorem 9.4 that Z Y] is a sufﬁcient statistic for «9.
'=1 (b) The log of likelihood function is n 1 n ‘
lnL(y1,y2,~,yn;9) =nlnr—nln(9+(r—1)::lnyi— i=1 i=1 Then d 1 n
n
_ . . . n; 6 = —— — delnL(y17y27 7y ) 6 +62
Equating the derivative to 0, we obtain the M LE as follows:
77, 1 n 1 n A 1 n ‘
~7+~ 17:0 or ~n+—A K": or 6: 6 92 9 n (0) Consider E (Y’) = 7
t ,1 °° 1 ‘
gym—16‘?! Mdy = /gwe”w/edw (with w = y’ and dw = Ty’_1)
0
(W) = 97
since the second integral is equal to the expected value of an exponential random
variable with mean (9. This implies that that is, g is an unbiased estimator of 9. Since 5 is a function of the sufﬁcient statistic ZY], it is the M VUE for 6.,
i=1 Exercise 9.75. (a) As deﬁned in Exercise 9.41, the likelihood function is
written as “ { 1 if0<y¢<26+1V73 L ... n. g 2 i. : (20+1)"
(24173127 ’y l ) f (y ’6) 0 elsewhere.
1 z { (20+1)” if 0 < Mn) = max {y17y2, '  ' ,yn} < 26 +1 0 elsewhere. 1
mﬁogeﬂ) [you] 7 where
1 ifO<y(n)<20+1 Jame“) iy(“)1 : { 0 elsewhere.
To maximize L(y1,y2,    ,yn;0), we minimize (26 + 1)” subject to 26 + 1 > gm)
or 0 > % [yw — 1]. Hence, the MLE of 9 is 1 g: E [Mm — 1] ‘, (b) The variance of the distribution is 2
02 = Va?" (Y) = Since 6’ > 1, this is a one~to—one function. Thus, the M LE for the variance of the
underlying distribution is (2§+r1>2 {2% [Mm — 1] +1}2 _ E 12 12 12 " Exercise 9.76. (a) The likelihood function is n 1 _ ‘ 1 n 1 n
[IQ/1,927 ' ' ' 7yn; 9) = H @5in We = 55;( gm) exp (—5 2%) .,
i=1 ':1 i=1 Z The log of likelihood function is ’n 1 n
lnL(y1,y2,~,yn;6) = —2nln6+Zlny¢ —— i=1 i=1 Then
d 2n 1 " EEIHL(y1:y2a"'7yn;6) = *7+ﬁ;yi' Equating the derivative to O, we obtain the M LE as follows: 2n 1 n 1 n A n OI —2n+5;lﬁ=0 OI‘ Q—ZnZK— 3 Calculate 7 g (120 +130 + 128) = 126. Hence, the MLE of 9 is 7 126
—— —— = 63. 2 2 (b) This is a gamma distribution with parameters a = 2 and ﬂ = 9. We have a: :
,u = E(Y) = a5 = 20 and 02 = Var (Y) = 0152 = 262‘. Hence, and Va?" = iVaT (7) = —— = (c) The bound on the error of estimation is T M2 (6:935
205:2 Var(6)=2 Ezz T~=51i44i (d) The MLE for the variance of Y, 202, is 252 = 2 (63)2 = 7, 938. Exercise 9.77. (a) The likelihood function is _ __ n 1 a—l —y¢/0 __ 1 n a——1 “l .
L (917 :92: 7 y'n, a) — F (a) 9a e —" 67m H yi exp 6 i=1 The log of likelihood function is
lnL (y1,y2,    ,yn;oz,0) = —vnln1‘(oz) — nalnd + (a — 1) Zlnyi — g Zyi.
i=1 ' If a is known, then d “E 1 n Equating the derivative to 0, we obtain the M LE as follows: —%g+A1—ZY;=0 or ~na+§ZY¢=O or 52:216152 :1 “0‘ i=1 ([9) This is a gamma distribution With parameters a and B = 6‘. We have ,u = E(Y) = a5 = a6 and 02 2 Va?" (Y) = a52 = d6? Hence,
E(§) = imngtfeezeand var(e)=§v(7)
_102_1 a92_92
— Jar—5571; 4 (c) By Example 9.2, 17 is consistent for M = a0, Then, by Theorem 9. 2, E = g is consistent for 0‘39 = (9.
(d) Using Lehmann and Scheffé’s method, we have L(x17m27"'7$n;a79) L(y17y27"'ayn;a76) 1 n a_1 n
W I exp H
l (WM—m)" i=1 In order for this likelihood ratio to be independent of 9, we need 2 3:, = E y, so
i=1 i=1
n
that Z Y, is the minimal sufﬁcient statistic for 0..
2:1 n
(e) The moment—generating function of U = Z Y, is
i=1 my (t) : ﬁmy, (t) = H (1 — 075)” = (1 — 6t)ﬁm = (1 — 9t)_10 with n = 5 and a = 2,. i=1 i=1
Now consider W = %1 with
‘ t 2t 2t “10 _
mW (t) = E (etW) = E [(342%)] = E (ego—W] = my = <1~ 65) = (1— 2t) 10
which implies that W has a chi—square distribution with 2 (10) = 20 degrees of freedom. Using W as a pivoting statistic, write 95g20 < V < 05%)} I OI‘ 2 < (9 < ._2 }= X00520 X0 95;20 and the 90% conﬁdence interval for 0 is
2:32 2:32 2:12 22%
i=1 i=1 i=1 i=1 , or r,
X?) 05;20 X39520 31,41 10,85 Exercise 9.78. The likelihood function is L ($1,332? ' ' ' 75cm) 91:92, ' ' Wyni/J’lnu’ZaO—Z)
m 1 The log of likelihood function is lnL ($171.27 ' ' ' 7$m7y17y27 ' ' ' JyniM17M2702) _ (m;n>1n(27r)—. (m;”)1n02_2%2 liwfmeriwrmfy i=1 i=1 Then
ilnL($1 3:2 m yl y2 y Iu,1 #2 0'2) 2—].. Em 8M1 7 7 7 my: 7 7 7 7’17 7 7 U2 i=1 1 7
—8 lnL(x1x2x 9192"‘y'/~‘1/~‘202)=i EanMz)
6/112 7 7 7 m7 7 7 7 n7 7 7 02 i=1 1, 7
a 2
i8021nL($171327.i'a'rm7y17y27"'7yni/“L17lu’270— ) + “is [m — m2 _ 772] w. i=1 i=1 Equating the derivative to 0, we obtain the M LE3 as follows: 2(Xi—ﬂ1)=0 or ZXi—mﬂ1=0 or ﬁ1=*ZXi=X
i=1 i=1 m i=1
n n 1 n w
2(K—ﬁ2)=0 or ZK—Nﬂg: or ﬂ2=EZK=Y
7:1 i=1 i=1
m + n 1 m A 2 n A 2
i Z _ X1 _ Y; —
26.2 2 (02)2 ( lu’l) + ( 1112) :l
or
1 m A n A
771+”: § [2% — 771)? +201 we)?
7:1 i=1
or Exercise 9.79. Let p1, pg, 173 be the proportions of voters in the popula—
tion favoring candidates A, B, and C, respectively. Further, deﬁne the random
variables n1, n2, 713 as the numbers of voters in a random sample of size n who
favor candidates A, B, and C’, respectively. Note that p1 + p2 + 193 = 1 and
n1 +n2 +713 2 n so that we may write p3 = 1 —p1 — p2 and 713 = n — n1 — 712. The
random variables n1, n2, n3 follow a multinomial probability distribution and the
likelihood function is
n! n3 ,1)??? (1 — p1 — p2) L(n1,n2,n3;p1,p2,p3)= , ,
711172.713. 6 The log of the likelihood is lnL(n1,n2,n3;P1,P2,P3)
= lnn! — lnnll — lnngl —lnn3! +~n11np1 + 7121an + mm (1 —— p1 —p2), Differentiating with respect to p1 and p2, we have a n1 713
—lnL n n, ; , , =  — ——
apl ( 17 2 H3 101 P2 193) pl 1 _p1 _p2
and a
n n
—lnL(n1,n2,n3;p1,p2,ps) = —2  —§———
3191 P2 1 —P1—P2 Set these two partial derivatives equal to 0 and solve simultaneously for Z31 and
132, that is, $———————:13 A20 and Zia—#113 A=0w
P1 1—p1:p2 p2 1—291292
Then
“1(1“‘J31 — 132) = 713151 and n2 (1 —231 — 152) = 713232 ((*)) Adding these two equations, we have (711 + 7190—231 — 132) = 713(171 +152) 01" n1+n2"(711 + 712) (131 +252) = “3 (231 +132) A A A A A A 711 + “2
0f n1+n2 = (711 + 712 + 713) (P1 +102) 0T n1+n2 = n (101 +192) 01' p1+p2 = '  Plugging this result into (*), we have n A n +77, A
n1<1—n1+ 2)=n3p1 and n2 <1——1 2>=n3p2. n n
Since
1— n1+ng W n—(m +712) _ E
n n n
Then
713 A W3 A
711 <——>  713171 and n2 = 713192
Thus,
A 711 A TL2
P1  “— and p2 — “'2
n 71
Hence, n1 ’I’Lg n3 133:1‘151—172=1———=—u
n n n If 1’51 = 030,132 = 038, and 133 = 0.32, then 151 — 132 = 0.30 — 038 = —0.08‘, Note that 7 A A n n 1
Var (p1 _ P2) : V0171 "‘ = (Tl/1 — 712)
1
= E [Var (m) + Var (n2) — 2001) (m, 712)]
1 i
Z [71171 (1 ‘ P1) + “102 (1 “ 102) * 2 (—npip2ll
1_ _
2 p1( 2Q+p2(1 p2)+2p1p2
n n n which may be estimated as (030) (0.70) (0.38) (0,62) 2 (0.30) (0.38) ._.__ + ___..___._ + _—_— = 0.006736 100 100 100 and the approximate bound on the error of estimation is 2 Var (131 3523 % 2x/0.()0673 = 0,1641. Exercise 9.80. The likelihood of the sample is given by L(0) =
= (9+1)”(yiy2yn)9 The logarithm of the likelihood is lnL(9) = nln(9+ 1) +61n(y1y2~yn) Then
dlnL(9) n 72
d6 ~——6—+1+ln(y1y2~yn)—O or +1
or n
6 + 1 = —
1n (3/192 ' ' Hence, the M LE of 6 is
5: n —1. Exercise 9.810. The probability distribution is P {Y = y} = <:>py (1 * p)” ‘ 8 (9+1)yi'(9+1)ygm~(HIM/Z (a) 13 = % maximizes P {Y = 0} = (1 — p)2 so that it is the MLE of p.
(b) Either 13 = i or 13 = % maximizes P{Y = 1} = p (1 —p) so that both are
the MLES of p. (c) 13: % maximizes P {Y = 2} = p2 so that it is the MLE of p. Exercise 9.82" The likelihood of the sample is given by L==p§(1—lemp$(l*pwdmw If pM = pw 2 plan + 75ln(1—pM), then L = p55 (1 ——p)145 so that lnL =
551np+145ln(1—p). Now d 55 145 ~bLz—nu——=0 dp p 1 —p
which implies that 55 (1 —p) = 145p or 55 = 20011, Hence7 the MLE of p is
A_ 55 _
p e 555 _ 0.275. Exercise 9.83. The likelihood function is written as { 1 if0<yi<26Vi (20)" L (3/17 3/2: ' ‘ ' 7 9n; 9) 2 U f @196) Z 0 elsewhere. 1' 1
(25)” if 0 < ym) = max {91; 3/2,    , yn} < 29
0 elsewhere. 1(029) [mm] , where _ 1 if0<y(n)<29
1(029) [MM] _ { 0 elsewhere. ' To maximize L (y1,y2,   . ,yn; 6’), we minimize (26)” subject to 29 > gm) or 6’ > $2111“ Hence, the M LE of 9 is
A Ye)
9 = t,
2 Exercise 9.84. (a) In Exercise 9.44, we showed that 3701) = max {Y1, Y2,    , Yn}
is a sufﬁcient statistic for 9 and the likelihood function is given by L(yi,y2, ' ' ' 7%;9) 2 6E [[0,9] [you],
i=1 where __ 1 ifosy(n)39
[[0,0] [CL/(“ll — { 0 elsewhere. " To maximize L(y1,y2, ~   ,yn; 6), we minimize 03” subject to (9 2 gm). Hence, the MLE of0is A
621%)" (b) In Exercise 9. 55, we showed that i/(n) = max {Yb Y2,    , Yn} has probabil
ity density function given by 9c) (M) = { @%? ﬂoswmse
0 elsewhere" ' Let W = i/(n) / 9. Then the probability density function of W is given by dﬁw 3n ((97.0)371“1 3n_1
fw (w; 9) = you (911%) lau = 63” ~6 = 3nw for 0 3 w 3 1. Hence, W = i/(n)/ 6 is a pivotal quantity.
((2) Consider P{a S W s b} = 1 — a. Without loss of generality, we may
assume that P{W<a}=g— and P{W>b}=%.
NOW a
9 : P{W < a} = /3nw3”_1dw = [103"]wza = a3“
2 O i w=O
so that a = (gr/3”. Similarly, 1
g : P{W > b} = /3nw3n‘1dw = [w3”]:: 21— b3"
12 )1/3“. Thus, sothatb=<1——3— or
a 1/3n YE“) V a 1/3n
P —— < < 1—— =1—
{<2> * 6 —( 2) a
or
1 0 l
P{ 1/3n}l—a
(1"?) n (5)
or Exercise 9.85. (a) The likelihood function is given by 1—12—59; if6<yi<ooforalli
1:1 I 0 elsewhere. L(ylay27”'7yn;6) : { — { if6<y(1):min{ylay2>"'7yn}
— 7/ 0 elsewhere.
n 1
= 292 (ll *3) 109,00) [9(1)] »
i=1 M
where
1(0’00) [Mm] _ { 0 elsewhere. "
To maximize L (y1, y2,    ,yn; 6), we maximize 262 subject to 6 < ya). Hence, the
M LE of 6 is A
6 : 3/21)”
(1)) The distribution function of Y is given by
y  t:
202 92 y 92
F(y;6)=P{YSy}=/—3dt=[——2 =1—~—2for9<y<oow
0 t t tzg y Then Ya) = min {Y1, Y2,    , Yn} has probability density function given by 92 “‘1 292 _ 277,62” 9(1) (M9) = nl1— W359)?"1 f (y;9) = n yg — y2n+1 for «9 < y < 00. Let W = Yin) / 6. Then the probability density function of W is
given by " (6w)2n+1 — w2n+1 dﬂw 2nd?" 2n
fW (“’19) = 9(1) (91M) BTU—l for 1 < w < 00. Hence, W = l/(n)/ 6 is a pivotal quantity.
(0) Consider P{a 3 W S b} = 1 —— a. Without loss of generality, we may assume that a a
P{W<a}=—2 and P{W>b}=§..
Now a
a 2n 1 “’Z“ 1
E = < CL} =/w2n+1dw 1': :'W:w:l = 1 — 1
so that a = (1 — gylﬂn” Similarly,
a °° 2n 1 w=°° 1
5 : Piw > b} 2/w2n+1dw : l—w2nlwzb : '62}:
b 11 >—1/2n so that b = . Thus, or
a —1/2n 3/0) a —1/2n
P 1~~ <—_< — =1—
{( 2) — 0 —(2> a
01'
1 9 1
P — <————< —1~a
a —1/2n '— Y — a 1/272}
{5) (1) (1—5)
01' Exercise 9.86. Let E = t(0) so that 9 = F1 If the maximum of the
likelihood is attained when «9 = 5, then L (9) g L (9) for all 9. Deﬁne [3 : t ((9)
and denote the likelihood as a function of B with L1 (5) E L [t—1 Then, for
any B, A L1 (5) = 4171(5)] = L (a) g L (5) = L [Vt—1 = L1 (5) . Hence, the MLE of e is B, that is, the MLE of t (9) is t Exercise 9.87. It was shown in Emample 9.14 that gt? 2 % is the M LE of p.
Then it follows from Exercise 9.76 that E 2 IE; is the M LE of R = 3%. Exercise 9.88. It was shown in Example 9.15 that S2 is the M LE of 02.
Then it follows from Exercise 986 that S is the M LE of a. 12 ...
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 Spring '08
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 probability density function, Likelihood function

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