Question

5. For the processes X 0.4X,-1 Zt -0.7Zi-1, (i) Simulate and plot 100 values of the processes; (ii) Compute and graph their theoretical ACF and PACF using R. (iii) Compute and graph their sample ACF and PACF using R. How do sample functions compare to their theoretical counterparts? (iv) Analyze smoothness of the simulated processes using their ACF's. Please include the code with clear comments explaining the meaning of the code. Make sure to label the graphs.

Answer 1

SOLUTION

rm(list=ls())

# (a) For First process

# (i) Simulate and plot 100 values of the given process
# Initialoize values
x0<-0;z0<-0
x<-0;z<-0
for(t in 1:100)
{
z[t]<-rnorm(1,0,1) # Draw SN(0,1) random variable
x[t]=-0.6*x0+z[t]+1.2*z0 # Simulate one value of process
x0<-x[t]
z0<-z[t]
}
plot(1:100,x,xlab = "Number",ylab = "Simulated Values",main = "Graph of process 1","l")

# (a) For Second process
# Initialoize values
y1<-0;y2<-0
y<-0;z<-0
for(t in 2:100)
{
z[t]<-rnorm(1,0,1) # Draw SN(0,1) random variable
y[t]=-0.2*y1+0.48*y2+z[t] # Simulate one value of process
y1<-y[t]
y2<-y[t-1]
}
plot(1:100,y,xlab = "Number",ylab = "Simulated Values",main = "Graph of process 2","l")

########### (ii) theorectical ACF and PCAF using R

### (a) For Process 1

acf_1<-ARMAacf(ar=0.6,ma=1.2,lag.max=5)
# Theoretical ACF<-
# 0 1 2 3 4 5
# 1.0000000 0.7979381 0.4787629 0.2872577 0.1723546 0.1034128
pacf_1<-ARMAacf(ar=0.6,ma=1.2,lag.max=5,pacf=T)

# Theoretical PACF<-
# 1 2 3 4 5
# 0.7979381 -0.4347500 0.2920742 -0.2145219 0.1651622

### (a) For Process 2
acf_2<-ARMAacf(ar=c(0.2,-0.48),lag.max=5)
# Theoretical ACF<-
# 0 1 2 3 4 5
# 1.0000000 0.1351351 -0.4529730 -0.1554595 0.1863351 0.1118876

pacf_2<-ARMAacf(ar=c(0.2,-0.48),lag.max=5,pacf=T)
# Theoretical PACF<-
# 1 2 3 4 5
# 1.351351e-01 -4.800000e-01 3.673579e-17 -6.570475e-18 2.398725e-17

########## (iii) Sample ACF and PACF graph
# For Process 1
samp_acf_1<-acf(x,lag.max=20,main="ACF of process 1")
samp_pacf_1<-pacf(x,lag.max=20,main="PACF of process 1")

# For Process 2
samp_acf_2<-acf(y,lag.max=20,main="ACF of process 2")
samp_pacf_2<-pacf(y,lag.max=20,main="PACF of process 2")

Output: By using sample and theoretcial values of ACF and PACF, we can say that there is no any significance spike of ACF and PACF values for both process.

Simulated values plot and ACF, PACF graph are given below with clearly mentioning its axis lables,

Graph of process 1 (0D 0 20 40 60 80 Number

ACF of process 1 C) CN CN 0 5 10 15 20 Lag

PACF of process 1 CN C) f1 CN 5 10 15 20 Lag

ACF of process 2 C) CN CN 0 5 10 15 20 Lag

PACF of process 2 CN f1 CN 5 10 15 20 Lag