{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info iconThis preview shows pages 1–7. 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 Document Right 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 Document Right 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 Document Right Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}