Consider the following time-series process given by

𝑦
𝑡
=
(
1
+
0.6
𝐿
+
0.1
𝐿
2
)
𝜀
𝑡

where
𝐿
denotes the lag operator and
𝜀
𝑡
∼
𝑊
𝑁
(
0
,
8
)
. What is the lag 1 autocorrelation of
𝑦
𝑡
,
𝜌
(
1
)
?

Question

Consider the following time-series process given by

𝑦
𝑡
=
(
1
+
0.6
𝐿
+
0.1
𝐿
2
)
𝜀
𝑡

where 
𝐿
 denotes the lag operator and 
𝜀
𝑡
∼
𝑊
𝑁
(
0
,
8
)
. What is the lag 1 autocorrelation of 
𝑦
𝑡
, 
𝜌
(
1
)
?

BlackTom AI · Accepted Answer

We’re given a time-series process y_t = (1 + 0.6 L + 0.1 L^2) ε_t, where L is the lag operator and ε_t ~ WN(0, 8). This expresses y_t as a finite moving-average representation of the white noise: y_t = θ0 ε_t + θ1 ε_{t-1} + θ2 ε_{t-2} with θ0 = 1, θ1 = 0.6, θ2 = 0.1.

Step 1: Compute the variance of y_t (γ0). For an MA(q) process, Var(y_t) = σ^2 ∑ θ_j^2, where σ^2 is Var(ε_t). Here σ^2 = 8 and the θ’s are {1, 0.6, 0.1}.
- θ0^2 = 1^2 = 1
- θ1^2 = 0.6^2 = 0.36
- θ2^2 = 0.1^2 = 0.01
Sum = 1 + 0.36 + 0.01 = 1.37
Thus γ0 = 8 × 1.37 = 10.96.

Step 2: Compute the lag-1 autocovariance γ1. For an MA(q), γ_k = σ^2 ∑ θ_j θ_{j-k}, summing over j where both indices are valid. For k = 1, the terms are θ0 θ1 and θ1 θ2:
- θ0 θ1 = 1 × 0.6 = 0.6
- θ1 θ2 = 0.6 × 0.1 = 0.06
Sum = 0.66
Thus γ1 = 8 × 0.66 = 5.28.

Step 3: Compute the lag-1 autocorrelation ρ1 = γ1 / γ0.
ρ1 = 5.28 / 10.96 ≈ 0.4816, which rounds to 0.482.

Summary: The lag-1 autocorrelation ρ(1) for y_t is approximately 0.482.

Consider the following time-series process given by 𝑦 𝑡 = ( 1 + 0.6 𝐿 + 0.1 𝐿 2 ) 𝜀 𝑡 where 𝐿 denotes the lag operator and 𝜀 𝑡 ∼ 𝑊 𝑁 ( 0 , 8 ) . What is the lag 1 autocorrelation of 𝑦 𝑡 , 𝜌 ( 1 ) ?数值题

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