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Date Submitted: 04/09/2014 09:30 PM
Solution of Assignment 3
Problem 2.15 By the properties on the best linear prediction (Brockwell and Davis (2002), page 68), we have for n > p Pn Xn+1 = Pn (φ1 Xn + ... + φp Xn−p+1 ) + Pn (Zn+1 ) = Pn (φ1 Xn ) + ... + Pn (φp Xn−p+1 ) + Pn (Zn+1 ) = φ1 Pn (Xn ) + ... + φp Pn (Xn−p+1 ) + E[Zn+1 ] = φ1 Xn + ... + φp Xn−p+1 The mean squared erroris
2 E[(Xn+1 − Pn Xn+1 )2 ] = E[Zn+1 ] = σ 2 .
Problem 2.22 Let X1 , X2 , X3 , X4 , X5 be observations from AR(1) model. The results follow from equations (2.5.12)-(2.5.14) in BD pages 66-67. a. The best linear predictor of X3 given X1 , X2 is P (X3 /X1 , X2 ) = a1 X1 + a2 X2 where (a1 , a2 ) are such that 1 φ φ 1 a1 a2 = φ2 φ .
Then, a1 = 0 and a2 = φ. The best linear predictor of X3 given X1 , X2 is P (X3 /X1 , X2 ) = φX2 . b. The best linear predictor of X3 given X4 , X5 is P (X3 /X4 , X5 ) = a1 X4 + a2 X5 where (a1 , a2 ) are such that 1 φ φ 1 a1 a2 = φ φ2 .
Then, a1 = φ and a2 = 0. The best linear predictor of X3 given X4 , X5 is P (X3 /X4 , X5 ) = φX4 . 1
c. The best linear predictor of X3 given X1 , X2 , X4 , X5 is P (X3 /X1 , X2 , X4 , X5 ) = a1 X1 + a2 X2 + a3 X4 + a4 X5 where (a1 , a2 , a3 , a4 ) are such that
φ 1+φ2
1 φ φ3 φ4
φ 1 φ2 φ3
φ3 φ2 1 φ
φ4 φ3 φ 1
a1 a2 a3 a4
=
φ2 φ φ φ2
.
Then, a1 = 0, a2 = a3 = is
and a4 = 0. The best linear predictor of X3 given X1 , X2 , X4 , X5 φ (X2 + X4 ). 1 + φ2
P (X3 /X1 , X2 , X4 , X5 ) =
d. By equation (2.5.14) in BD page 67, the mean squared errors are a. 2 2 σ φ σ2 2 E[(X3 − P (X3 /X1 , X2 ))2 ] = − 0 φ 1−φ = σ 2 . σ2 φ 2 1−φ 1−φ2 b. σ2 − E[(X3 − P (X3 /X4 , X5 ))2 ] = 1 − φ2 c.
σ 2 φ2 1−φ2 σ2 φ 1−φ2 σ2 φ 1−φ2 σ 2 φ2 1−φ2
φ 0
σ2 φ 1−φ2 σ 2 φ2 1−φ2
= σ2.
σ2 . = 1 + φ2
σ2 − E[(X3 − P (X3 /X1 , X2 , X4 , X5 )) ] = 1 − φ2
2
0
φ 1+φ2
φ 1+φ2
0
Problem 3.1 a.We have φ(z) = 1 + 0.2z − 0.48z...