M562-HW9-5

M562-HW9-5 - MTH/STA 562 Exercise 9.48 In Exercise 9 30(1...

This preview shows pages 1–9. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 chi-square 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 771-705) = 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(1---p)- (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=1|W}+(0)P{T7é1}W}=P{T=1|W} : —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(T|W)]=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 Lehmann-Scheffé 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(yz-M)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. ...
View Full Document

This note was uploaded on 04/20/2008 for the course MTH 562 taught by Professor Cheng during the Spring '08 term at Creighton.

Page1 / 9

M562-HW9-5 - MTH/STA 562 Exercise 9.48 In Exercise 9 30(1...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online