CS计算机代考程序代写 F70TS2: Time Series Exercise Sheet 5

F70TS2: Time Series Exercise Sheet 5
1. We consider three time series. In each case we assume that the unknown model is an AR(1) with unknown mean. We obtain the following statistics:
(i) 􏰢n t=1
(ii) 􏰢n t=1
(iii) 􏰢n t=1
(xt − x ̄)2 = 1224.79 and 􏰢n−1(xt − x ̄)(xt+1 − x ̄) = 995.11, t=1
(xt − x ̄)2 = 703.52 and 􏰢n−1(xt − x ̄)(xt+1 − x ̄) = 459.05, t=1
(xt − x ̄)2 = 662.33 and 􏰢n−1(xt − x ̄)(xt+1 − x ̄) = −446.17. t=1
In each case n = 400. Carry out the following:
(a) Estimate the unknown φ in each case and write down your estimated AR(1) model (by
means of a symbol μ for the unknown mean). Assume now that x ̄ = 10.25.
(b) Estimate γ(0), σε2, Var(x ̄) and the approximate 95%-CI of μ for the model in (i).
(c) Estimate γ(0), σε2, Var(x ̄) and the approximate 95%-CI of μ for the model in (iii). 2. From two time series x1, …, x500 and y1, …, y500 we obtain
nn
(a) 􏰢 xt = 9677.84, 􏰢(xt − x ̄)2 = 3630.92,
t=1 t=1
nn
(b) 􏰢 yt = 20050.48, 􏰢(yt − y ̄)2 = 8515.04,
t=1 t=1
Assume that the unknown models are AR(2) with unknown means. For both time series, estimate and write down the models. Then estimate σε2, Var(x ̄), Var(y ̄) and calculate the approximate 95%-CI of μX and μY .
t=1 t=1 n−1
n−2
􏰢 (xt − x ̄)(xt+1 − x ̄) = 2516.86 and 􏰢 (xt − x ̄)(xt+2 − x ̄) = 2183.88.
t=1 t=1 n−1
n−2
􏰢 (yt − y ̄)(yt+1 − y ̄) = 5591.87 and 􏰢 (yt − y ̄)(yt+2 − y ̄) = 2211.04.
1

3. Suppose we are given time series data x1, . . . , x100 which satisfies
100
􏱜(xt−x ̄)2 = 247.8,
t=1
􏱜(xt − x ̄)(xt+1 − x ̄) = −189.1 ,
t=1
where x ̄ is the mean of x1, . . . , x100. The last eight values of this time series data are shown in
the table below.
􏲢2
a) Calculate the value of the fitted parameters φ and σ􏲢ε when an AR(1) model is fitted to
the data x1,…,x100.
b) Using this fitted model, calculate 95% forecasting intervals for X101 and X102, assuming
that the white-noise terms εt have a Normal distribution.
c) Estimate the limit of the width of the 95% forecasting interval for X100+k as k → ∞.
4. Represent the following one-dimensional AR(2) process in a VAR(1) form: Xt = −0.2Xt−1 + 0.6Xt−2 + εt.
5. Let the ARCH(1) process (Xt) be defined by
X =ε􏰏α +αX2 ,
99
t
93
94
95
96
97
98
99
100
xt
1.993
-2.750
3.458
-2.850
3.743
-3.910
2.118
-1.844
t t 0 1 t−1
where α0 > 0, 0 < α1 < 1, and (εt) is a white noise process with variance σε2 = 1. Let ρ(k) be the autocorrelation at lag k of the process (Xt). Throughout this question you may assume that the process (Xt) is weakly stationary and that E(Xt4) < ∞. a) Calculate Var(Xt). b) Find ρ(k) for all k. c) Show that Cov(X2 , X2) = α Var(X2 ). t−1 t 1 t−1 2