STAT520 Hwk2 Solution
Courtesy of Wei Li
Q2.1 Solution:
Since {Xn } is a stationary,
E(Xn ) = µ
E(Xn Xn+h ) = γ(h) + µ2 = ρ(h)γ(0) + µ2 .
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To minimize E (Xn+h − aXn − b) , we need to set
∂
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E (Xn+h − aXn − b) ⇒ a(γ(0) + µ2 ) + bµ = ρ(h)γ(0) + µ2
∂a
∂
2
0=
E (Xn+h − aXn − b) ⇒ b = µ − aµ.
∂b
0=
Solve the equations, we have
a = ρ(h), b = µ(1 − ρ(h)).
Therefore, a = ρ(h) and b = µ(1 − ρ(h)) minimize the MSE. The estimator
X̂n+h = aXn + b is thence the best estimator.
Q2.2 Solution:
E(Xt ) = E(A cos(ωt) + B sin(ωt)) = E(A) cos(ωt) + E(B) sin(ωt) = 0
γ(h) = Cov(Xt , Xt+h )
= E(Xt Xt+h )
= E(A2 cos(ωt) cos(ωt + ωh) + AB cos(ωt) sin(ωt + ωh)
+ AB sin(ωt) cos(ωt + ωh) + B 2 sin(ωt) sin(ωt + ωh))
= cos(ωt) cos(ωt + ωh) + sin(ωt) sin(ωt + ωh)
= cos(ωh)
Therefore, {Xt } is a stationary series with mean zero and ACVF cos(ωh). Since
κ(h) = cos(ωh) is the ACVF of a stationary time series, it follows from theorem
2.1.1 that cos(ωh) is non-negative definite.
Remark: To show the non-negativity, it’s easier to construct a process with acvf
being the given function.
Q2.4 Solution:
Use the result as above, we may construct the stationary series
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(a) Xt = Z cos(πt), where Z ∼ W N (0, 1), t = 0, ±1, ±2...
(b) Xt = A+B cos( π2 t)+C sin( π2 t)+D cos( π4 t)+E sin( π4 t), t = 0, ±1, ±2...
where A, B, C, D, E are iid N (0, 1)
(c) Xt = √25 Zt + √15 Zt−1 , t = 0, ±1, ±2... Zt ∼ W N (0, 1)
Q2.8 Answer:
Only need to show {Xt } non-stationary. Suppose otherwise, then
γ = V ar(Xt ) = φ2 V ar(Xt−1 ) + σ 2 = φ2 γ + σ 2 = γ + σ 2
Note σ 2 > 0, then we must have γ = V ar(Xt ) = ∞. This fact contradicts its
stationarity.
Remark: This is a convenient way to show there is no stationary solution. If you
follow the hint in the textbook, you still need to show V ar(Xt ) = ∞ eventually.
QA 1. Solution:
R̂101 = E(Z101 + 0.2Z100 |Z100 ) = E(Z101 + 0.2 × 0.01) = 0.002
R̂102 = E(Z102 + 0.2Z101 |Z100 ) = E(Z102 ) + 0.2E(Z101 ) = 0.
Clearly, by definition the forecast error ê101 = Z101 and the forecast error ê102 =
Z102 + 0.2Z101 . Thus, the standard deviation of the forecast errors are
sd(ê101 ) = 0.025
sd(ê102 ) = 0.025 ×
√
1.04 = 0.0255
the autocorrelations are
cov(Rt , Rt+1 )
= 0.1923
V ar(Rt )
cov(Rt , Rt+2 )
ρ(2) =
= 0.
V ar(Rt )
ρ(1) =
QA 2. Solution:
Similarly, plug in, we have
R̂101 = 0.014
R̂102 = 0.008
sd(ê101 ) = 0.1414
sd(ê102 ) = 0.1414
ρ(1) = 0
ρ(2) = 0.2
E(Rt ) = 0.0125
V ar(Rt ) = 0.0208
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