2.3

a

x_{t} = 0.8x_{t−1} − 0.15x_{t−2} + w_{t} − 0.3w_{t−1} x_{t} − 0.8x_{t−1} + 0.15x_{t−2} = w_{t} − 0.3w_{t−1} (1 − 0.8B + 0.15B^{2})x_{t} = (1 − 0.3B)w_{t} (1 − 0.3B)(1 − 0.5B)x_{t} = (1 − 0.3B)w_{t} (1 − 0.3B)x_{t} = w_{t} x_{t} − 0.5x_{t−1} = w_{t} x_{t} = 0.5x_{t−1} + w_{t} ARMA(1, 0) it’s causal: (1 − 0.5B) = 0 or B = 2 > 1. invertible: Yes. w_{t} = x_{t} − 0.5x_{t−1}.

b

x_{t} = x_{t−1} − 0.5x_{t−2} + w_{t} − w_{t−1} x_{t} − x_{t−1} + 0.5x_{t−2} = w_{t} − w_{t−1} (1 − B + 0.5B^{2})x_{t} = (1 − B)w_{t} (1 − 0.5B)(1 − 0.5B)x_{t} = (1 − 0.3B)w_{t} ARMA(2, 1) it’s causal: (1 − B − 0.5B^{2}) = 0 roots: 1 ± i with module greater than 1 not invertible:(1 − B) = 0, root lies on unit circle

2.6

a

x_{t} + 1.6x_{t−1} + 0.64x_{t−2} = w_{t}, ϕ_{1} = −1.6, ϕ_{2} = −0.64

(1 + 1.6B + 0.64B^{2})x_{t} = w_{t}

ϕ(B)x_{t} = w_{t}
ψ(B)ϕ(B)x_{t} = ψ(B)w_{t}

(ψ_{0} + ψ_{1}B + ψ_{2}B^{2} + …)(1 + 1.6B + 0.64B^{2}) = 1

B^{0} : ϕ_{0}  = 1
B^{1} : ϕ_{1} + 1.6ϕ_{0}  = 0
B^{j} : ϕ_{j} + 1.6ϕ_{j−1} + 0.64ϕ_{j−2}  = 0, j ≥ 2

ACF: ρ(h) − ϕ_{1}ρ(h − 1) − ϕ_{2}ρ(h − 2) = 0, h = 1, 2, … ρ(0) = 1 see example 2, page 103

ρ(−1) = \frac{ϕ_{1}}{1 − ϕ_{2}} , ρ(1) = ρ(−1)

plot:

ρ(0) = 1
ρ(1) = −\frac{1.6}{1 + 0.64} = −0.9756
ρ(2) = −1.6 × (−0.9756) − 0.64 = −0.9210
ρ(3) = −1.6 × 0.9210 − 0.64 × (−0.9756) = −0.8492
ρ(4) = −1.6 × (−0.8492) − 0.64 × (0.9210) = −0.7095

b

x_{t} − 0.40x_{t−1} − 0.45x_{t−2} = w_{t}, ϕ_{1} = 0.40, ϕ_{2} = 0.45
(1 − 0.4B − 0.45B^{2})x_{t} = w_{t}
ϕ(B)x_{t} = w_{t}
ψ(B)ϕ(B)x_{t} = ψ(B)w_{t}
B^{j} : ϕ_{j} − 0.4ϕ_{j−1} − 0.45ϕ_{j−2} = 0, j = 0, 1, 2, …

ACF:

ρ(h) − ϕ_{1}ρ(h − 1) − ϕ_{2}ρ(h − 2) = 0, h = 1, 2, …

ρ(0) = 1 see example 2, page 103

ρ(−1) = \frac{ϕ_{1}}{1 − ϕ_{2}} , ρ(1) = ρ(−1)

see plots. c same idea as a&b. 2.7 ARMA(1,1)

x_{t} = ϕx_{t−1} + θw_{t−1} + w_{t}, |ϕ| < 1

from (2.43)

r(h) − ϕr(h − 1) = 0, h = 2, 3, …

so the general solutions is r(h) = cϕ^{n}, h = 1, 2, … solve for c using (2.44)

r(0) = ϕr(1) + σ_{w}^{2}[θ_{0}ψ_{0} + θ_{1}ψ_{1}] θ_{0} = ψ_{0} = 1
r(0) = ϕr(1) + σ_{w}^{2}[1 + θ(ϕ + θ)] see example 2.10, ψ_{1} = ϕ + θ
r(0) = ϕr(1) + σ_{w}^{2}[1 + θϕ + θ^{2}]
r(0) = ϕr(1) + σ_{w}^{2}[1 + θϕ + θ^{2}]
r(1) = ϕr(0) + σ_{w}^{2}θ_{1}ϕ_{0}
r(1) = ϕr(0) + σ_{w}^{2}θ

solving for r(0)&r(1),

r(0) = σ_{w}^{2} \frac{1 + 2θϕ + θ^{2}}{1 − ϕ^{2}}
r(1) = σ_{w}^{2} \frac{(1 + θϕ)(ϕ + θ)}{1 − ϕ^{2}}

because r(1) = cr(0) ⟹ c = \frac{r(1)}{ϕ} so:

r(h) = r(1)ϕ^{h−1}

or

ρ(h) = \frac{(1 + θϕ)(ϕ + θ)}{1 + 2θ + θ^{2}}ϕ^{h−1}, h ≥ 1

ARMA(1,1), AR(1) page 92, ρ(h) = ϕ^{h}, h ≥ 0,

ARMA(0,1), MR(1) page 96,

ρ(h) = \frac{θ}{1 + θ^{2}}  h = 1
= 0  h > 1

plots ϕ = 0.6, θ = 0.9 see 2.8.

2.12 Cardiovascular mortality:

M_{t} = 11.45 + 0.43M_{t−1} + 0.44M_{t−2} + w_{t}
V ar(w_{t}) = 32.5
M_{n+m}^{n} = 11.45 + 0.45M_{n+m−1}^{n} + 0.44M_{n+m−2}^{n}

using 2.78

P_{n+m}^{n} = σ_{w}^{2} ∑_{j=0}^{m−1} ψ_{j}^{2}.

2.13 a

E[(y − g(x))^{2}] = E[E[(y − g(x))^{2}|x]]
E[(y − g(x))^{2}|x] = E[(y − E[y|x] + E[y|x] − g(x))^{2}|x]
= E[(y − E[y|x])^{2}|x]E[(E[y|x] − g(x))^{2}|x] + 2E[(y − E[y|x])(E[y|x] − g(x))|x]

E[y|x] − g(x) is a constant, so it’s easy to show:

E[(y − E[y|x])(E[y|x] − g(x))|x] = 0.

and as a result:

E[(y − g(x))^{2}|x] ≥ E[(y − E[y|x])^{2}|x]

or

E[(y − g(x))^{2}] ≥ E[(y − E[y|x])^{2}]

and E[(y − g(x))^{2}] is minimized when g(x) = E[y|x].

b

y = x^{2} + z

To minimize MSE, set:

g(x) = E[x^{2} + z|x]

= E[x^{2}] + E[z|x] = 1

E(y) = E(x^{2}) + E(y) = var(x) + 0 = 1

MSE = E[(y − g(x))^{2}]

= E[(y − 1)^{2}] = var(y) = var(x^{2} + y) = var(x^{2}) + var(y) = 1

c

MSE = E[(y − a − bx)^{2}]
= E[y^{2}] − 2aE[y] − 2bE[xy] + a^{2} + 2abE[x] + b^{2}E[x^{2}]

differentiate by a and b, and set both partial derivatives to zeros, we get:

a = E[y] − bE[x] = 1, b = \frac{E[xy]}{E[x^{2}]} = 0

so MSE = E[(y − 1)^{2}].