时间序列|Time Series and Risk Name Models MA-347/MA-M47

这是一个英国的时间序列与风险模型作业作业
Marking Criteria: Full marks will be awarded for work that 1) is mathemat
ically correct, 2) shows an understanding of material presented in lectures, 3)
gives details of all calculations and reasoning, and 4) is presented in a logical
and clear manner.
Please attach your working, with this sheet at the front.
1. (i) Given an expression for 2Xt!5Xt!1+4Xt!2!Xt!3 in terms of second
order dierences. [10 Marks]
(ii) Write the following time series process as a vector-valued process that
possesses the Markov property
Xt = 5Xt!1 ! 4Xt!2 + Xt!3 + Zt, t ” 4.
[10 Marks]
(iii) A time series is defined by the following relationship
Xt = Xt!1 + Zt, t 2 N
with initial X0 being a given square integrable random variable, where {Zt, t 2
N} is a sequence of iidrv’s with E(Z1) = 0 and V ar(Z1) = !2. Assume that
X0 and the sequence {Zt, t 2 N} are mutually independent. Show that
V ar(Xt) = V ar(X0) + t!2, 8t 2 N.
[10 Marks]
2. Calculate the values o 1 and 2, the autocorrelation function k at lag
k = 1 and lag k = 2 respectively, for the stationary AR(2) time series process
Xt = !0.8Xt!1 + 0.1Xt!2 + Zt, t ” 2
with initials X0 = X1 = 0, where {Zt, t 2 N} is a sequence of square integrable
iidrv’s. [40 Marks]
3. Let {Xt, t 2 N0} be an ARMA(1,2) time series process defined as follows
Xt = Xt!1 + Zt + $Zt!2, t 2 N
with X0 = 0, where {Zt, t = !1, 0, 1, 2, ……} is a sequence of iidrv’s with
E(Z!1) = 0 and V ar(Z!1) = !2. Show the following for t ” 1
(i) Cov(Xt, Zt) = !2
(ii) Cov(Xt, Zt!1) = !2
(iii) Cov(Xt, Zt!2) = (2 + $)!2.

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