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Mth562-HW9-2

# Mth562-HW9-2 - MTH/STA 562 Exercise 9.1 Since this is an...

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Unformatted text preview: MTH/STA 562 Exercise 9.1. Since this is an exponential distribution with mean 6, we see that u = 6 and 02 == 62. Now Var (A1) = V617”(Yi=eg Va?" Q2) 2 Var (Yli' 2 Y2) =%1[ Va?" (Y1) + Var (3/2)] = i (292) :9; Var (53) = Var (Y; E2: 22:9) ~[Var(Y1) + 4Va7" (1,2)] g (592) = 5792 Va?“ (55) = Var<7 ” = 0-93—29 Hence, A A Var g 93 1 eff (“5) Z wig? 2 ii A A V é a WM) ___ gig—e: AA _ Var<§5)_ﬂ_3 Eff (937 65) — “Va?" (g3) — i — 5 Exercise 9.2. This is a random sample from a population With mean ,u and variance 0'2. (a)Since A Y+Y 1 1 EM) = E 12 2)=§[E(Y1)+E(Y2)]=-2—(2M)=n A 1 Y2+~-+Yn—1 1 E = — —Yn (“2) E 4y“ 2(n—2) +4 1 1 1 = —~ Y — EY Yn_ — Yn 4E(1)+ (n—~2)[ (2)+ +E( 1)]+4E( ) _ 1 1 1 _M 11 2.. E023) = E(7)=M, each of the three estimators is unbiased. (b)Now Var (M) = Var (Yl 3Y2) = i [Var (Y1) + Va?" (1/5)] = 2-11— (202) = __ [\Dq A 1 +Y+ Y”- 1 1Var<Y>+ 1 [v (Y) +v (Y >1+1v (Y) = — a?" — 16 1 4(6—2)2 2 a1" 1 16111 " ._ 1 2 1 2 1 2 __ 1—6—( )+4(n_2)2[(n—2)a]+i€(a) _a_2+_a_2_+a_2 :11 a2 _ _ 16 4(n 2) 16 8 4(n—2)_8(n——2) 2 VGT(ﬁ3) = Var (7>=% Hence, ,1 A _Var(ﬂ1)_U—22_n and 2 V_a7“(ﬁ 3): 88:: 2) n2 eff (#371112): Va —,l/1:7"( 3) 02 m Exercise 9.3. Since this is a uniform distribution over (0, 0 + 1), we see that 11: E(Y)= 6—1-1211—226+% 02=Var(Yi)=———~——~—1(0+i;_61 =1—12. (1)FOI'§1, A __ 1 — 1 1 1 E(91)=E(Y—§>=E(Y)—§=<0+§)——=6 Va?" (51) =Va7“ (7—5) =Va7" (7) = a = (2) F016<y<0+1, F<y>=P1Y§y}=/dt=y—6. Thus, the probability density function of Yb) is given by M (y) = n [F<y>1“‘1 f (y) = n (y — a)“ forg<y<9+11Now 0+1 0+1 E(Y<n>—9) = /(y-9)N(y~9)"“1dy= n(y --9)”dy 0 0 1 n n y=9+1 TL = 1 (y - 9) 11 — TL + 1 yzg TL + 1 and y=0+1 TL || TL i (y ) 31:0 77: + 2 Thus, 71 E [no] — 9 = so that E [3%)] = a + 5:? and so that E [1/5”] = 26E [YW] — 62 + —— ll 26(9+ n+1 n _ TL = 62 20 . + n+1 n+2 Hence, n EYn— i” n+1 n_ n n+1 n+1 =E[Y(n)]"n:1 E (E2) =6 6+ and Var (02) = Va?" [Yon _ n : 1] 2 VW" [YW] : E [3132)] _ (E [Emir 2 92+29—L+ 77’ —<6+ ” ) n+1 n+2 n+1 n n n n 2 = 02 29 ~92 ——29-—— —— (—m) + n+1+n+2 n+1 n+1 H n:2I—(n:1)2: (n+2)T(Ln+1)2' Consequently, the efﬁciency of él relative to \$2 is TL A A _ V” (52) _ <n+2><n+T>2 _ 12712 eff (91,92) - Va?" (51) — \$7; — (n+2)(n+1)2« Exercise 9.4. Since this is a uniform distribution over (0, 6), we see that ~E(Y)—Q and 2—VT(Y-)——Qi “_ 1—2 0"“’_12‘ With the distribution function 0 forySO F(y)= % for0<y<6 l foryZli, (1) The probability density function of Y0) is given by 9<1><y> = n [1 — F(y)]"‘1f(y) = ”(1%)“? = ii (1%)“ for O < y < 6. Thus, 0 71—1 1 E[Y(1)] = /% <1 — %) dy (letting u = g and du = adry) nH/u (1 —- 1071—1 du = n63 (2, n) 0 P(n)___ 6 (n+1)nI‘(n)_n+1 : ”a? (2) r (n) F (n + 2) l l | 1 n6 so that 51 = (n + 1) Y0) is an unbiased estimator of (9. Now ‘9 2 2 _. ”AL A.“ - _£ .1) E[Y(1)] — / 6 (1 0) dy (lettlngu— 9 and duw edy 1 _ F(3)F(7z) _ 2 2 _ n 1 = 2 = 2 __ 716’ [u (1 u) du n6B(3,n) m9 F(n+3) _ 702 2F(n) _ 262 ‘ l (n+2)(n+1)nr(n)‘(n+1)(n+2)' so that __ 2 2 _ 292 _ _ 92 Va7"[i/'(1)] ’ E iymi ‘" {E in)” _ (n+ 1) (n +2) (71+ 1)? 2 62 72 n02 n+1.(n+1)(n+2) (n+1)2(n+2) Thus , n02 n62 Va?" (51) =01 +1>2Variy<1>i = (n + ”2 (—71—:1)2(n + ) n + 2' (b) The probability density function of Y2”) is given by gm) (y) = n [F on“ f (y) = n (if-f; ”ya:— 4 for 0 < y < «9. Thus, a 0 =9 ny” n n yn+1 y n 6"“ n6 EYn =/—d =_/nd=— =_. = lﬂlo y 099 WL+J y=0 so that 52 : "THE/(n) is an unbiased estimator of 6. Now 9 n+1 9 n+2 y=9 9n+2 62 EY2=/W d=£/”%=33’— =3- =” [(71)] 9” y 9“ y y 9” n+2 -0 6" n+2 n+2 0 0 41“ so that W] = E [W l — {E [W = — 43-9—- 2 i W n+2 n+1 2 : n62 [—1— __ n 2] : “—7126 N n+2 (n+1) (-11) (n+2) Thus, A n + 1 2 (n + 1)2 n62 62 Var (62) :( n ) Var[l/(1)] : —- n2 In +1)2(7—1—:?) = n—(n + 2). (3) Consequently, the efﬁciency of 51 relative to ég is 2 eff (51,52) 2 M _ ”(3H, 2 1 A -—_ 92 2 v Var (61) #2 n Exercise 9.5. It has been shown that Va (32) 2 Va (32) = 204 1 n — 1" Since Y1 —— Yg N N (0, 202), it follows that Y1 -— Y2 N N 0, 1 . x—202 ( ) Thus, (Y1 — Y2)? 202 has a chi-square distribution with 1 degree of freedom. Hence, (Y1 —— 1/2)? ________ = 2 Va?“ [ 202 which implies that (Y1 ‘ Y2)2 Var (8;) = Var [ 2 =204 Consequently, the efﬁciency of 6% relative to 6% is Var (62) 204 A2 A2 2 71—1 Exercise 9.6. Since this is a Poisson distribution with mean /\, we see that ,u = /\ and 02 = A. Then 1 A Y1 + Y2 1 Eel) = E(- 2 —) = 51E(Y1)+E(Y2)l = §<2A>=A E(/\2) = 19(7) =M=AH Thus, both X1 and X1 are unbiased estimators of A. N 0w A _ Y1 +Y2 _ 1 _ 1 _ /\ Va?" (A1) — Var< 2 —> —— Z [Var (Y1) + Var (1/2)] — 1(2A) — —2— A _ 02 A Va?" (A2) = Var (Y) = E = 5. Hence, A A A Va?" (A ) A 2 e A ,A = A = 3 = —‘, ff< 1 2) Var (A1) ’3 71 Exercise 9.7. Since this is an exponential distribution with mean 6, we see that ,u = 6 and a2 = 62 with the distribution function 1—6—9/9 foryZO F(y)—{0 fory<0. The probability density function of Y0) is given by 9(1) (y) = n i1 — F(y)]n_1f(y) = n(e—y/9)n_1_l_e~y/9 : EeﬂLy/a 6 6 for y 2 0, which is an exponential distribution with mean 6/71. Thus, 6 62 E [1/0)] = E and VG?" [Yin] = E so that 61 = nYm is an unbiased estimator of 6 with MSE (01) = Var (61) = ngVar[1/(1)] = 712- 2—: = 92. 6 Now <1 9 ‘3 L‘D H <1 Q ‘3 Q Hence, Exercise 9.8. (a) This is a normal distribution with mean u and variance 2 a . Then lnf(y)=ln{—1 expl~(y—M)2]}=C—(y2;5)2- whereC=ln< 1 > a 27r 202 a 27T' Thus, 01nf(y)_y—u_ d 821nf(y)__i 3/1 — 02 an 8M2 T 02' Using the Cramer-Rao lower bound, we see that 1 1 a2 I(“)=_an— 00:“ 1:7; “E (~ ) “E (a) — 2 which is equal to Va?" (Y) 2 £5. However, Var (ﬂ) > I ([1,) for any unbiased estimator ,H of ,u. Hence, 7 is an eﬂicient estimator of u. (b)This is a Poisson distribution with mean A. Then Aye—A lnp(y)=ln< ' )zylnA~A~ln(y!), y. Thus, 2 8111M?» y a 1nf(y) y = —— — 1 d ——~————~ = ———-‘, 8A A an 8)? A2 Using the Cramer—Rao lower bound, we see that W)__1__ 1 _ 1_t_;_a TL- which is equal to Var (7) = ”72 = A However, Var X) Z I (A) for any unbiased f A. estimator X of A. Hence, 7 is an efﬁcient estimator o ...
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