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Unformatted text preview: Sufﬁciency 7.1.2 is X2(n). Hence E = n, var = 2n. Accordingly, E [2X3] :02 andvar<ZXi2) = (ﬁ>2(2n) = E14 n n n n 7.1.3 This is a rather lengthy exercise. One observation that might help is illustrated
with the second part. The pdf of Y2 is 91W 13:1! (1 ‘ ii z 6y“) ” mm?” 0 < y < 9' f9 634209  Ell/93 dy
0 ll E(Y2) Let y = 6w (this is the observation and this substitution can be used in each part); so
1 60r(3)r(2) 9
— 2 — :: ————————— = —,
E(Y2) —— 60/0 11) (1 w)dw m5) 2
Likewise
F(4)I‘(2) 302 02
2 __ 2 z . = .
E(Y2 ) _ 66 N6) 10 , var(Y2) 20
So E(2Y2) = 0 and Var(2Y2) = (4)(92/20) = elf/5. The marginal p.d.f. of Yl is given by gl(y1) = %(1 — 3%)2, 0 < yl < 0. We have E(Y1) = 3/00% (1 — %)2dy = 30/01 w(1 — w)2dw = 30118232) 2 3—. Hence E(4Y1) = 9, i.e., 4Y1 is an unbiased estimator of 9. For the variance, we ﬁrst
calculate E(Y12)= 30 foe (gr (1 —, 3‘92 dy = 302/01 w?(1 5 w)2dw = 302m2g§3> = g. Then Var(Y1) 2: ii; — ($2 = $02, and so Var(4Y1) = 16Var(Y1) : $02. The rest is
analoguous calculations and the results are Var(2Y2) = 95—2 and Var(%Y3) 2. 61/15 , STAT Luo [MATH qeq
I i Exﬁfcisi 7.2..l, The joint p.d.f. of the X, is 1 (27r0)"”/2 exp [—— Z ,
.21 29‘ so it follows from the factorization theorem that ELI X is a sufﬁcient statistic for 0. MM 71,0912) )xn = xf+xj+...+x,,2
WC Cam) fmb aka/MM) atequ
1 /
[(102136) = (2W6) "2 exp[ 21/26] ) g>O
[320(4) 2) ’A/h) = 1 )_00<X,<oo ,J—oozx‘wo Exercise 7.2. 2.. The joint p.d.f. of the X. is e—n902c; Hm! . n _1
: [e—nogﬂzi] wig] ,
i=1 , so, by the factorization theorem, 2;] X. is a. sufﬁcient statistic for 0. New/ﬂaming“ 416.63 kz(7‘1J~) X“) z [TlX3 9]. a
a e 3:011J2131“' )
1 Excimse 7. 2. 3. It is enough to do the second part. If f(z;0) = Q(0)M(z)I(o,9)(m), then the joint
p.d.f. is ‘ ‘ HQ(0)M($B;)I(0’9)($5) = {[Q(0)]“I(0,9)(maxmi)} . 4’qu u1(x.,xz),,,)xn) = MmeXUXL X} r“) n k1(3}el = [Q(6)J" ) 0<g<6l
kz(?<«,..)Xu> = TlHOQ ’ x,>o It follows from the factorization theorem that max X, is a sufﬁcient statistic for 0. )...I9(“>O, 7.2.8The joint p.d.f. is given by n _1 _1 _ . . 0—1
1‘} (1 — 2309 _ [F(6)]2n (1),;(1 — ,
5° H?=1lX£(1 — X0] is sufﬁcient for 0, Problem 7 .3.3.
X17X27 : %e~$/6, 0 < CU < 00, f($1,$2,9) : Elie—(1131+:22V0
Y1=X1+X2, Y2=X2
(ma/1,33): X1 2 Y1 '“Y27 2 X2 2 Y2, D 2 1
9(y1,y2,9) = f(¢1(1€,Y2),¢2(Y1,Y2))D = age—(Yl—wmo * 1 z
= :358‘3’1/0, 0 < 3/2 < 211 < oo
Y2 = X2 N E$p(6), = 6, V(Y2) : 92
9(91) 2 gm 9(y17y2)d31/2 =/f0y1 alga—“NW2 = ﬁe‘yl/Oﬁ < yl < oo _y1 9
9(yzly1)=%%l=§§m=yil, 0<y2<y1 2 45(91) = E(y2lyl) = 0m ygg(y21y1)dy2 = 01” y2yi1dy2 _—_ Z 1121.
Y1 = X1 + X2 ~ I‘(2,6), Var(Y1) : 202
V(E(Y2IY1)) = V(¢(Y1)) = %V(Y1) = g < 02 = V(Y2) Exercise. 7. 3. 2. ' We know from Chapter5 that the p.d.f. on; is 93(y3) = 3§(yoi)2(1—%°—)2, 0 < y3 < 0.
So 9 1 = 30/; (yoj)3(1 — yogfdya = 300/0 w3(1 — w)2dw = 300I‘—(:—%%§3:2 = Thus E(2Y3) = 0. The joint p.d.f. of Y3 and Y5 is g—2y§(y5 — ya), 0 < 3/3 < 315 < 0, and
the p.d.f. of Y5 is ggyg. From these, we see that the conditional p.d.f. of Y3 given Ys is 1292;;(Vs—ya), 0 < ya < gs < 0, and we haVe 9;
i 24 2/5 3 1 3 6
E(2Y3ly5) = E 313(195  y3)dys = 243/5 w (1 — wld'w = 5‘95
5 0 0 Here we used the change of variable y; : 3,510. So (p(y5) : gy5_ For the variances, using
9 y y 1 2 2 300/ (7540 ”‘ jaydys = 3092/ 104(1 — w)2dw = $62.
0 o and hence Var(Y3) = §92 — (%’)2 = g;, we get Var(2Y3) = 97:. Moreover, after easy integrations, we ﬁnd 2 £0, = $62, and so Var(<p(Y5)) = g, which is smaller
than Var(2Y},), as it should be. 0/0624 MVPZ algmd wt ...
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This note was uploaded on 06/11/2010 for the course STAT 410 taught by Professor Alexeistepanov during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 AlexeiStepanov
 Statistics, Probability

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