# CS计算机代考程序代写 F70TS2: Time Series Exercises 4 – Solutions

F70TS2: Time Series Exercises 4 – Solutions

1. The asymptotic normality of x ̄ is a very important property. By means of this property we can construct an approximate confidence interval for μ with a given confidence level α (the so-called interval estimation). For instance, the approximate 95% confidence interval for μ is

μ ∈ [x ̄ − 2SDx ̄, x ̄ + 2SDx ̄].

For our question we have x ̄ = 1.65 and SDx ̄ = Var(x ̄) = 0.1. Inserting these values into

the above formula, we have

μ ∈ [1.56 − 0.2, 1.56 + 0.2] = [1.36, 1.76].

2. The general formula for the asymptotic variance of x ̄ is

1 (qi=0ψi)2 Var(x ̄) ≈ n (1 − pi=1 φi)2 ,

because σε2 = 1 by assumption.

For a) we have Var(x ̄) ≈ (1 + 0.4)2/(1 − 0.5 − 0.2)2/900 = 0.0242. For b) we have Var(x ̄) ≈ (1 + 0.3 + 0.6)2/(1 − 0.2)2/900 = 0.00627. For c) we have Var(x ̄) ≈ (1 + 0.2)2/(1 + 0.6)2/900 = 0.000625.

For the above examples, n and σε2 are the same. But the asymptotic variances change from case to case. We can see that, for both MA and AR parts, the larger the sums qi=1 ψi and/or pi=1 φi (without ψ0 and φ0), the larger the asymptotic variance of x ̄, and vice versa.

Remark: Again, it is easy to see that as qi=1 ψi tends to −1 then Var(x ̄) vanishes. On the other hand, as pi=1 φi tends to 1, then Var(x ̄) diverges.

3. This is similar to the solution to the last question.

For a) we have Var(x ̄) ≈ (1 − 0.7)−2/400 = 0.02778.

For b) we have Var(x ̄) ≈ (1 + 0.3)−2/400 = 0.00148.

For c) we have Var(x ̄) ≈ (1 − 0.45 − 0.3)−2/400 = 0.040.

Furthermore, we have to calculate Var(Xt) = γ(0) in each case to obtain Var(y ̄).

For a) (an AR(1)), γ(0) = 1/(1 − φ2) = 1.961 and Var(y ̄) = γ(0)/n = 0.0049 << Var(x ̄). For b) (an AR(1)), γ(0) = 1/(1 − φ2) = 0.1.099 and Var(y ̄) = γ(0)/n = 0.002754 > Var(x ̄).

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For c) (an AR(2))

γ(0) = 1 − φ2 = 1 − 0.3 = 1.873 (1 + φ2)[(1 − φ2)2 − φ21] (1 + 0.3)[(1 − 0.3)2 − 0.452]

and Var(y ̄) = γ(0)/n = 0.00468 << Var(x ̄).
The above models show that the sample mean of a time series may have much larger variance compared with that obtained from independent data with the same variance. However, some- times the variance of the sample mean of a time series can also be much smaller than that for corresponding i.i.d. data.
4. For n = 400, the ± 2 bounds are ±0.1. And 5% or more out of 20 are one or more estimates. n
Hence, in case a) we can say that Xt is probably not an i.i.d. white noise. Indeed, Xt should also not be uncorrelated.
In case b), we can only say that Xt are probably uncorrelated. However, whether or not they are i.i.d. is not clear. Further diagnosis is required to answer this question. For example, to display the ACF of x2t . If Xt are i.i.d., then Xt2 are also i.i.d.. But, if Xt are only uncorrelated, Xt2 may be correlated to each other.
For n = 1600 the above bounds are ± 2 = ±0.05. And 5% out of 40 are 2 estimates. Further n
decisions are similar to those given above. Note that, if the process is stationary, the larger n, the better the estimation quality.
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