2003 ASSIGNMENT

# 2003 ASSIGNMENT - (C) N=200 As the increase in sample size,...

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5 10 15 20 0.0 0.2 0.4 0.6 0.8 1.0 Theoretical ACF Index n 2003 ASSIGNMENT Q1 (a) P(Xn=2|Xn-1=0)=1/4 Xn-1=e n-1+e n-2 (en-1=-1 and e n-2=1or en-1=1 and e n-2=-1 ) If Xn=2 so en=1 and en-1=1 as well and en-1=1 and e n-2=-1 in Xn-1 The probability of en=1 is 1/2, en-1=1/2 as well,so the answer is 1/4. (b) P(Xn=2|Xn-1=0,Xn-2=2)=0 For Xn-2=2 en-2=1 and e n-3=1 Xn-1=0, e n-2=1and en-1=-1 Xn=2 en=1 and en-1=1 but en-1 =-1 according to Xn-1 (c) The process is not Markov chain, because the probability is not only depend on one past state. It is also depend on the several past states. Q3 (a)For AR(1)model

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0 10 20 30 40 50 60 70 0.0 0.2 0.4 0.6 Lag Partial ACF Theoretical PACF 5 10 15 -0.2 Lag 50 realization of PACF (b) N=50
0 5 10 15 -0.4 0.0 0.4 0.8 Lag ACF 50 realizations of ACF 0 5 10 15 20 -0.2 0.2 0.6 1.0 Lag 200 realization of ACF

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Unformatted text preview: (C) N=200 As the increase in sample size, the plot become more likely to the Theoretical one. 5 10 15 20 0.0 0.2 0.4 0.6 Lag Partial ACF 200 realization of PACF R-code Q3 ###(a) n<-ARMAacf(ar = 0.75,ma = 0, lag.max = 20,pacf = FALSE) plot(n,main=’Theoretical ACF’) p<-arima.sim(model=list(ar=0.75),10000000) o<-pacf(p) plot(o,main="Theoretical PACF") ###(b) x<-arima.sim(model=list(ar=0.75),50) y<-acf(x) plot(y,main="50 realizations of ACF") v<-pacf(x) plot(v,main="50 realizations of PACF") ###(c) g<-arima.sim(model=list(ar=0.75),200) l<-acf(g) plot(l,main="200 realizations of ACF") t<-pacf(g) plot(t,main="200 realizations of PACF")...
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## This note was uploaded on 06/12/2011 for the course ASB 2003 taught by Professor Kim during the Three '11 term at University of New South Wales.

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2003 ASSIGNMENT - (C) N=200 As the increase in sample size,...

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