第二版部分习题解答；.pdf

+064-3.* 75 1 ,12/ λ = 1.0; λ = 1.5 . 4. o!u0Xq(X`u0aqXq:0K:w$. 2. π(λ1 |3) = 0.2457; π(λ2 |3) = 0.7543, 6 (1) π(θ|12) =  1, 2 11.5 < θ < 12.5, K; 0,  1 x = (11.1, · · · , 12) 8. θ 0XqCX π(θ|x) = 420 θ(1 − θ) (0 < θ < 1), x = (x , · · · , x ), x X P 0-1 Æ , x = 1. 10. M;} (2) π(θ|x) = 10, 11.5 < θ < 11.6, K. 0, 19 1 20 i 20 i i=1 1= Z fθ (x)dx = C(θ) X 1 = C(θ) o2: Z Z exp X exp X k nX j=1 k nX j=1 o θj Tj (x) h(x)dx ⇐⇒ o θj Tj (x) h(x)dx. ;Æ θ C$uqT=q-Aw 12. %7y!>PlGA θ \N. |z!Ct7Vj+sx h(x) = . 1 n Q xi . i=1 14. M;Æ T (X ) = (X , X ), x. (1) f (x, θ) = (n) ( θ − 12 < x(1) 6 x(n) < θ + 12 1 K 0 |z!Ct7w. 16. |z!Ct7w T < CS 1.4.2 w T XT(S . 18. Æ 1 Z= m+n Z m X Xi + i=1 n X j=1 Yj ! "m # n X X 1 2 2 , S = (Xi − X) + (Yj − Y ) m + n − 2 i=1 j=1 T S CX a T σ 0 UMVUE. 2 2 o f (x, θ) ;A$0"+k:|7 2 9<?:>;8 2 n 20. θ̂ = X = 1X Xi n i=1 X θ 0 UMVUE. Æ T (X ) = P (X − X) , J$[N0iqBX α 0iqRAX n 22. 2 i i=1 ( 1, , T (x) < σ χ 2 0 n−1 (1 − α/2), ℄ T (x) > σ χ 2 0 K, χ (α) X"}: n − 1 0}AC02 α CYA. P X , J$[N0iqBX α 0iqRAX 24. Æ T (X ) = ϕ(x) = 0, n−1 (α/2), n−1 n i i=1 8 < 1 1 1 , T (x) < χ (1 − α/2), ℄ T (x) > χ (α/2), 2λ 2λ ϕ(x) = : 0 K, χ (α) X"}: 2n 0}AC02 α CYA. 2n 2n 0 0 2n 75 2 ,12/ E 2.2.2 r0 “d;E+?$aq:. 4. E 2.2.3 |aqC0CYA*7A0A?$/sP. 6. aqCX Γ(20, 2). 2. 8. m(x) = n » Y i=1 – λr Γ(xi + r) . Γ(r) xi ! (λ + 1)xi +r E 2.3.3 | :o aqC0xA?/sP. 12 (2) :A$π(β) = 1/β; (4) x X :Aπ(x ) = 1/x , 0 < x < x. 10. 0 0 1 2 13 (2) π(θ) ∝ 1/[θ(1 − θ) ] 0 0 (0 < θ < 1). (4) π(λ) ∝ 1/λ, λ > 0. M;$! Fisher i^Jeffreys aq< Fisher i^0AJ. 16. M;|E 4 2.4.2 T 2.4.3 r0A?. 18. Æ p 0aqCX Be(a, b), XqCX Be(x + a, x − k + b), K θ = max{θ , x }, b+Ka qC>< r C X |θ ∼ N (θ , 30 ) (i = 1, · · · , n) π π 1 i i i 2 θi |λ ∼ N (µ, τ 2 ) (i = 1, · · · , n), λ = (µ, τ 2 ) > > : µ ∼ N (100, 202 ), π(τ 2 ) ≡ 1, µ, τ 2 0aqC9 . 75 3 ,12/ 2. θ 0XqCX N (174.05, 1.265). 4 (2). n > 36. 6. π(θ1 , θ2 , |x) = π1 (θ1 |θ2 , x)π2 (θ2 |x), “ 1 1 2 2 X 8. θ 0XqCX Be(5, 297). 10 (1) 'u$O7 θ 0XqCX Be(2, 3); (2) '1$O7 θ 0XqCX Be(4, 8). 12. M;E7 3.3.1 (2) 0. 14. θ 0XqCX Γ(13/3, 37/3). 16 (1). λ 0aqCX Γ(4, 20000); (2). λ 0XqCX Γ(n + 4, nx̄ + 20000). 18. M;E7 3.5.2 (2) 0. 20. θ 0XqCX Γ (α + 1, β + x ). 22. M;E7 3.2.1 (3) 0. 24. M;Xq:0 GK:$/ θ 0Xq:X π2 (θ2 |x) n P n Γ(n/2 + α, D), D = λ + (xi − x̄)2 + x̄2 . n+1 i=1 −1 r o 1 (θ − t)T Σ−1 ∗ (θ − t) , 2 ` −1 ` ´ ` ´ −1 Σ∗ = Σ + A−1 )−1 , t = Σ−1 + A−1 Σ−1 x + A−1 µ . t p 1 π(θ)|x) = (2π)− 2 |Σ∗ |− 2 exp /JssP. n − ; 0':ev 75 4 ,12/ 1. _N 3 ` INXqUL d{0XqA/` 1+α (1 + α)(x + β − 1) , V π (x) = . α+β+x (α + β + x)2 (α + β + x + 1) 15. µπ (x) = x̄, V π (x) = x̄/n (nx̄ > 0). 9 (2). µπ (x) = 17. µπ (x) = ” π (θ |θ , x) X N n +n 1 x̄, 2(n +1 1)θ , nx̄ + β (nx̄ + β)2 , V π (x) = , nr + α > 2. nr + α − 1 (nr + α − 1)2 (nr + α − 2) 9<?:>;8 4 19. µπ (x) = x+β 2(x + β)2 n , V π (x) = , + α > 2. n + 2α − 2 (n + 2α − 2)(n + 2α − 4) 2 2 Bm 2Bm , V π (x) = , m > 4, 2 m−2 (m − 2) (m − 4) m X mk(x̄ − µ)2 Bm = (xi − x̄)2 + . m+k i=1 21 (2). µπ (x) = 2 (m + r − 1)τm V π (x) = , m + r − 3 > 0, m+r−3 »X – m 1 2 τm = (xi − x̄)2 + λ . m(m + r − 1) i=1 22 (3). µπ (x) = x̄, µ + nτ x̄ 2τ β̃ , V π (x) = , 1 + nτ (n + 2α − 2)(1 + nτ ) n 1X n(x̄ − µ)2 β̃ = β + (xi − x̄)2 + . 2 i=1 2(1 + nτ ) 23 (3). µπ (x) = 2. _N 3 ` INXqAL 12. θ̂M D = Xq|A℄/` nr + α − 1 , α + β + nx̄ − 2 ˆ ˜2 β − α + n(x̄ − 2r) (nr + α)(nx̄ − nr + β) P M SE(θ̂M D ) = + (α + β + nx̄)2 (α + β + nx̄ + 1) (α + β + nx̄)2 (α + β + nx̄ − 2)2 n+3 n+5 16. θ̂M D = , P M SE(θ̂M D ) = . nx̄ + 20000 (nx̄ + 20000)2 ˘ ¯ 2θ12 . 18. θ̂M D = θ̂1 = max θ0 , X(n) /2 , P M SE(θ̂M D ) = (α + n − 1)2 (α + n − 2) mx̄ + kµ 21 (3). θ̂M D = , m+k »X – m 1 mk(x̄ − µ)2 (xi − x̄)2 + . P M SE(θ̂M D ) = (m − 2)(m + k) i=1 m+k 23 (3). θ̂M D = µπ (x) = µ + nτ x̄ , 1 + nτ P M SE(θ̂M D ) = V π (x) = n β̃ = β + 24. } 1X n(x̄ − µ)2 (xi − x̄)2 + . 2 i=1 2(1 + nτ ) 2τ β̃ , (n + 2α − 2)(1 + nτ ) 0XqCX p LC<;CMXqAL XqUL dRz# θ ` ´−1 ` −1 ´ ` ´−1 θ̂M D = Σ−1 + A−1 Σ X + A−1 µ = X − Σ Σ + A (X − µ), XqhAX `Σ + A ´ . 3. _N 3 ` INXqYAL Xq|A℄ (hA) /` −1 −1 −1 √ 18. θ̂M e = θ1 α+n 2, θ1 = max{θ0 , X(n) /2}, ? 4 9<=; 5 P M SE(θ̂M e ) = » – √ 2 (α + n)θ12 (α + n)θ1 α+n + − θ 2 . 1 (α + n − 1)2 (α + n − 2) α+n−1 2uN 21 (3) 0sPdR. 23 (3). 2uN 23 (3) 0sPdR. 21 (3). 4 (1). µπ (x) = 2/5, V π (x) = 1/25; (2). µπ (x) = 1/3, V π (x) = 2/117. 6 (1) (a) µπ (x) = 2.67, , V π (x) = 0.44; (b) (2) (a) m 1 = 2.53, P M SE(m 1 ) = 0.4596 2 2 µπ (x) = 2.7, V π (x) = 0.54 . (b) m 1 = 2.63, P M SE(m 1 ) = 0.545 . 2 2 8 (1) Be(x+2, n−x+1); Be(x + 4, n − x + 1); n = 1000, x = 710. (2) µπ1 (x) = 0.71, µπ2 (x) = 0.7104. XqUL (2) XqAL 10 (1) XqA V (x) = 9.187; Xq|A℄ P M SE(θ̂ ) = 11.056. µπ (x) = 10.93, π θ̂M D = 9.563, xi + αi − 1 xi + αi 12. θ̂iM D = , θ̂iE = , i = 1, · · · , r. N +α−2 N +α n 2(λ + A) 1X 2 14. , A = xi . χ2n+2α (0.90) 2 i=1 16. MD 6|>0 1 − η 0i%hX " 2(t + β) 2(t + β) , χ2f (η/2) χ2f (1 − η/2) # = [13806, 161263]. α = 2.9914, β = 59741; t = 40000, r = 0; f = 2(r + α) ≈ 6, η = 0.05 . 18 (1) [3.34, 5.75]; (2) [3.32, 5.78]. 20. [1.24, 1.52]. 22 (1) 24. [2.75, 5.49]; p? H . (2) [3.95, 7.35]. 0 26. α0 < 0.005, α1 > 0.995, α0 /α1 Æ VeMy{ H . 0 σ µ+τ z σ2τ 2 2 28. Z = X̄ − Ȳ, µ(z) = , η = , σ2 + τ 2 σ2 + τ 2 µ = µ1 − µ2 , −µ(z)/η > 0 H0 , H0 . 30. 2 w$ , 7p? 2 E y{ σ = m1 + n1 , τ = τ + τ , 2 2 Z 1 `n´ x P (X = x|θ) · g1 (θ)dθ = θ (1 − θ)n−x dθ = 1/6, x 0 0 ` ´ `5´ 1 5 P X = x|θ = 1/2 ( ) 15 π B (x = 3) = = 3 2 = ≈ 2, m1 (x) 1/6 8 m1 (x = 3) = 2 1 2 2 Z 1 } aq^[ X 1M tDz!w 1, Mp? π H0 . 0 1 9<?:>;8 6 p? H . 34. |4=gO#~%6 32. 2 b = µ1 = Z |4=gO#~%6 h 2. µ1 − uα/2 q 0X Z nτ 2 σ12 + nτ 2 Z x̄ + σ12 . µ == 10.12 2 σ1 + nτ 2 0 95% %hX η12 + σ22 , µ1 + uα/2 q i η12 + σ22 = [ 9.0933, 11.1467 ] 75 5 ,12/ &l8 a , l8 a ;Q`f . 1 1 n+α 4. θ̂B = . nx̄ + α + β n+r 1 λ + nx̄ 6. θ̂B = , η= η̂B = . λ + nx̄ θ n+r−1 (x + 1)(N + 2) 1 8. p̂B = − . N (n + 2) N n P x2i + 2β i=1 10. τ̂B = . n + 2α + 2 x+β 12. θ̂B = . n + 2α + 2 x . 14. θ̂B = α+1 √ 16. θ̂B = θ1 · n+α 2, θ1 = max{θ0 , 2x(n) }. Æ 18. θ̂B = 115.996 ≈ 116 . M;a$! p(x|θ)/p(x|a) 0':+XC*0 L T L , vF G. 22. M;| 20 NI5RA0sP GXqD 9XqD '.&e, / tD L 0':. 20. e H k X xj + αj (xj + αj )[(n + α) − (xj + αj )] , Vjπ = , j = 1, · · · , k, α = αi . n+α (n + α)2 (n + α + 1) i=1 24. θ̂jB = &l8 a , b,XpjX AB j. 28. M;&aqCX π(θ) ≡ 1, -! tDL 0': K0D RA< 26. 1 A 1/n. 0 tDL X θ̂ = x/r. K e'AI5`0D RA<A 1/r. 32. M;s# minimax L TaqC2#7 5.6.1 0 3 HPlb. P c . |$Pl`0A?w, c = c = · · · = 34. w$ θ̂ 0D RA R(θ̂ , θ) = X P c = 1/n 7R(θ̂ , θ) = c Pl P c = 1 '.`eb θ̂ = n1 X = X 79 /D RA R(θ̂ , θ) ;u" θ ∈ R '.&e, M7yK<.n0. 30. θ B n n n 2 i 1 n n i=1 n n n i=1 n 2 i i i=1 n 2 i i=1 ? 6 9<=; 7 Xq:TXqD bXqUI507yw. 38. C 0XqGV 0.89 T 0.96 hÆ= (k C ; ε- *ZX X = (X , · · · , X )( X)5Az). dOAz X Y = (y , · · · , y , E , · · · , E ), T(Az`;AE+RAX l (θ|Y, X ), E l (θ) = E [l (θ; X , Y]. z# >X|C0f3`, /P4=H=qH1> Q t, BdX X , X Pl X > t `0CX U [t, θ ]. %<;J θ < θ , I(0 < X < θ) #PlC`[v0G& 0, - l (θ) ;J θ < θ & , & 9/ EM G?5g. 6. |H^~8 Metropolis G?4\/ θ 0XqUX 0.62, Xq X 0.05 10. (1) | π(θ, λ) = π(θ|λ)π(λ) )0 G_Cb. (2) x π(θ|λ, x) ∝ f (x|θ)π(θ|λ) Gb/. E3, | π(λ|θ, x) ∝ π(θ|λ)π(λ) Gb. i 1 1 X|Y,θ (j−1) n 1 i m m (j) c c m m (j) m (j) (j) m (j) 75 7 ,12/ } 2.5.3 λ 0XqC-X)C Γ(nx̄ + b, n + a), z# E(λ|X , · · · , X ) = Jz(A, x̄ → 1 vG 1 , ;8 8 } BIC 7y BIC = −2ln[` ´p̂ (1 − p̂ ) 6. (1) zX µ 0XqCX N (µ̂ , σ ), µ = n/σ ). } G = N (µ , σ ), n t t M LE M LE n 2 n n 4. (1) 2 ] + lnn b. (2) . P 2 µ0 /τ02 + n i=1 xi /σ , σn2 = 1/(1/τ02 + 2 1/τ0 +n/σ 2 2 t n−t Z Z 1 σ 2 + (µt − µ̂n )2 + σn2 lnf (z|µ)π(µ|xn )dθdG(z) = − ln(2πσ 2 ) − , 2 2σ 2 n Z n 1X 1 1 X (xi − µ̂n )2 + σn2 η̂(G) = lnf (xi |µ)π(µ|xn )dµ = − ln(2πσ 2 ) − . n i=1 2 n i=1 2σ 2 η(G) = %<80X b(G) = Exn [η̂(G) − η(G)] = Exn h1 2 + n (µt − µ̂n )2 1 X (xi − µ̂n )2 i − . 2σ 2 n i=1 2σ 2 p G/. lnf (x |µ̂ ) − E x [lnf (x |µ)] = nσ /(2σ ), lnπ(µ̂ ) − E x [lnπ(µ)] = σ /(2τ ), va n 2 n n µ n 2 0 n X Qn (µ̂n ) = i=1 2 n n 2 n µ| n [(xi − µ̂n )/σ 2 + (µ0 − µ̂n )/(nτ02 )]2 /n T S (µ̂ ) = 1/(nσ ). }7 7.1 b/0nuL X n 2 n nb̂(Ĝ) = − (2) « n (3) pD = 2lnf (xn |µ̂n ) − 2 + Sn−1 (µ̂n )Qn (µ̂n ) + 1 . 2 R t 2 n 2 n 2 n n 2 lnf (x|µ)π(µ|xn )dµ = nσn2 /σ 2 → 1. 75 9 ,12/ |N 1 0sP}`:bk 4. λ ‹ 0 6 P (|fn (x) − f (x)| > ε) ≤ E|fn (x) − f (x)|2 ε2 → 0, , n → ∞. 0 tDL T PEB L /` λ̂B = 6. nσn2 σ2 + n2 2 2σ 2τ0 Jz(A, , n → ∞ 7W, µ̂ → µ , σ → 0, nσ → σ , Q (µ̂ ) → σ , z#, b̂(Ĝ) → 1. 2. „ x+τ , r+α+1 λ̂P EB = x + τ̂n , r+α+1 τ̂n = α−1 X. r A θ 0 NPEB L X θ̂N P EB = #" X = X 0L . x − r + 1 p̂n (x + 1) · , x p̂n (x) n+1 , #(A) #{i : 1 6 i 6 n + 1, Xi = x} , n+1 Z 1 . p̂n (x) X pG (x) = p(x|θ)dG(θ) ; A F0HA p̂n (x) = < 0 : 0 ? 9 9<=; 8. 9 A θ 0 NPEB L X (1) θ̂N P EB = − fn (x) (2) fn (x) ; fn(j) (x) = « „ n x − Xi 1 X K , j = 1, 2, j hn nh1+j n i+1 Z ∞