CS计算机代考程序代写 1.D 2.C

1.D 2.C
3.A 4.A
5.B
6.D 7. B 8.A 9.A
10. D 11. C 12.A 13. A 14.D 15.C 16. D
17. B
F70TS2: Time Series Multiple Choice Revision Questions: Solutions
Under H0, number of TPs P ∼ N(132, 3171/90) approx.
P-value = P (P ≥ 145|H0) = P [Z > (144.5 − 132)/5.936)] = P (Z > 2.106) = 0.018
p=3,m=2 Theequationsare: 5a0+10a2 =􏰢yt and 10a0+34a2 =􏰢t2yt ⇒7a ̃0=3.4􏰢yt−􏰢t2yt ⇒a ̃0=1[−3,12,17]
35
γ4 =γ0 ×ρ4 =3×0.62 =1.08
Yt =(1−0.9B+0.2B2)Zt =(1−0.5B)(1−0.4B)Zt ⇒ Zt = (1 − 0.5B)−1(1 − 0.4B)−1Yt
= (1+0.5B+0.25B2 +…)(1+0.4B+0.16B2 +…)Yt = (1+0.9B+0.61B2 +…)Yt ⇒Yt =−0.9Yt−1 −0.61Yt−1 +…+Zt
Yt =A(B)Zt where A(B)= 1(−B2 +2B+3+2B−1 −B−2) 5
γ(Y) = 1 (1+4+9+4+1)σ2 = 19σ2 0 25 Z 25Z
CY(z)= 1 (−z2 +2z+3+2z−1 −z−2)2σ2 25 Z
⇒GY(z)= 1 {19+8(z+z−1)−2(z2 +z−2)−4(z3 +z−3)+(z4 +z−4)} 19
⇒f∗(ω)=1+16cosω−4 cos2ω−8 cos3ω+2 cos4ω 19 19 19 19
area is πσY2 = π(1 + 0.82 + 0.12)σZ2
werequireα±0.6<1, α>−1
modelis(1−B)(1+0.4B)Yt =(1−B)Zt,i.e.(1+0.4B)Yt =Zt
ρ2 = γ2/γ0 = 5/22.5
modelisYt ={1+(α+β)B(1−αB)−1}Zt i.e.(1−αB)Yt =(1+βB)Zt
MA(2) part: Zt + β1Zt−1 + β2Zt−2 we require β2 ± β1 > −1, β2 < 1 {Yt}isMA(2) modelisYt =(1−B)−2(1+βB)Zt =[1+(2+β)B+(3+2β)B2 +...]Zt y50(2) = 50.4 = 0.1y50(1) + 0.8y50 ⇒ y50 = 56.7875 ⇒ y51(1) = 0.1y51 + 0.8y50 = 50.44 Yt = (1 − 0.6B − 0.2B2)−1(1 − B)−1(1 − 0.8B)Zt ⇒Yt =(1+0.6B+···)(1+B+···)(1−0.8B)Zt =Zt +(0.6+1−0.8)Zt−1 +··· ⇒ψ1 =0.6+1−0.8=0.8 The prediction limits are 57.9 ± 1.6449􏰜(1 + 0.82)1.65 = (55.19, 60.61) 1