ANSWER: I have attached the R code for the given problem. First we simulated the AR(2) process using arima.sim command. In part A we used pacf command to plot the pacf of the series. The plot clearly shows it as AR(2) since only two values are outside the confidence bounds. In part C, we estimated the series using arima function.
################ R CODE BEGINS #########################
rm(list=ls(all=TRUE))
set.seed(12345)
p1=1;p2=-0.6;n=48
## Simulate AR(2) with t distribution innovations
X=arima.sim(48, model=list(ar = c(1,-0.6)), innov = rt(48, df = 5))
## Part A
pacf(X,lag.max = 48)
## Part B
arima(X,order=c(2,0,0),method="CSS",include.mean = FALSE)
## Part C
X1=arima.sim(48, model=list(ar = c(1,-0.6)), innov = rt(48, df =
5))
pacf(X1,lag.max = 48)
arima(X1,order=c(2,0,0),method="CSS",include.mean =
FALSE)
################# R CODE ENDS ##################################
need code from R. Time series with applications in R class 7. Simulate an AR(2) series with фі--1.0, P2 ;--0.6, n-48, but error terms from a t-distribution with five degrees of freedom. (a) Displ...