M562-HW9-5

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

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

View Full Document Right Arrow Icon
Background image of page 1

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

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

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

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

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

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

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

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9
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

Page1 / 9

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

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online