Stationary_Models

Moving Average Models

See the theory here

Simulation of MA Process

The code below simulates the MA process

set.seed(1)
b <- c(0.8,0.6,0.4)
x <- w <- rnorm(1000)
for (t in 4:1000){
  for (j in 1:3) x[t] <- x[t]+b[j]*w[t-j]
}

Lets plot x

plot(x,type="l")

lets take the ACF of (x)

acf(x)

As expected, the first three autocorrelations are significantly different from 0.

Fitted MA Model

c An MA(q) model can be fitted to data in R using arima function with order function parameters set to c(0,0,q)

x.ma <- arima(x,order=c(0,0,3))
x.ma

## 
## Call:
## arima(x = x, order = c(0, 0, 3))
## 
## Coefficients:
##          ma1     ma2     ma3  intercept
##       0.7898  0.5665  0.3959    -0.0322
## s.e.  0.0307  0.0351  0.0320     0.0898
## 
## sigma^2 estimated as 1.068:  log likelihood = -1452.41,  aic = 2914.83

We have taken coefficients as 0.8, 0.6 and 0.4, so the function works ok.

x <- read.table("ts/pounds_nz.dat",header=T)
x.ts <- ts(x,st=1991,fr=4)
x.ma <- arima(x.ts,order=c(0,0,1))
x.ma

## 
## Call:
## arima(x = x.ts, order = c(0, 0, 1))
## 
## Coefficients:
##         ma1  intercept
##       1.000     2.8329
## s.e.  0.072     0.0646
## 
## sigma^2 estimated as 0.04172:  log likelihood = 4.76,  aic = -3.53

acf(x.ma$res[-1])

Mixed Model: The ARMA Process

Simulation and Fitting- ARMA model

The ARMA process can be simulated using R function arima.sim, with the order function paramameter set to c(p,0,q). Let alpha = -0.6 and beta = 0.5

set.seed(1)
x <- arima.sim(n=10000,list(ar=-0.6,ma=0.5))
coef(arima(x,order=c(1,0,1)))

##          ar1          ma1    intercept 
## -0.596966371  0.502703368 -0.006571345

As we can see the sample estimates of alpha and beta are close to the model parameters

Lets try to fit the Exchange rate series using MA(1), AR(1) and ARMA(1,1)and compare it using AIC.

x.ma <- arima(x.ts,order=c(0,0,1))
x.ar <- arima(x.ts,order=c(1,0,0))
x.arma <- arima(x.ts,order=c(1,0,1))
AIC(x.ma);AIC(x.ar);AIC(x.arma)

## [1] -3.526895

## [1] -37.40417

## [1] -42.27357

AIC is lowest for x.arma

x.arma

## 
## Call:
## arima(x = x.ts, order = c(1, 0, 1))
## 
## Coefficients:
##          ar1     ma1  intercept
##       0.8925  0.5319     2.9597
## s.e.  0.0759  0.2021     0.2435
## 
## sigma^2 estimated as 0.01505:  log likelihood = 25.14,  aic = -42.27

Lets find the acf

acf(resid(x.arma))

This is fitting well.

You can choose the best fitting ARMA (p,q) model is chosen using the smallest AIC either by using a for loop with upper bound for p and q- let us say 2. In each step of the for loop, the AIC of the fitted model is compared with the currently stored smallest value. if the model has a smaller AIC, then the new value and model is stored. To start with best.aic is initialized to infinity (INF), after the loop is completed, the best model can be found in best order.

best.order <- c(0,0,0)
best.aic <- Inf
for (i in 0:2) for (j in 0:2) {
fit.aic <- AIC(arima(x.ts, order = c(i, 0,
j)))
if (fit.aic < best.aic) {
best.order <- c(i, 0, j)
best.arma <- arima(x.ts, order = best.order)
best.aic <- fit.aic
}
}
best.order

## [1] 2 0 1

Here we found that 2,0,1 is the best order, lets see its AIC

x.arma1 <- arima(x.ts,order=c(2,0,1))
AIC(x.arma1)

## [1] -42.80284

Here AIC is the least

save.image("workspace.Rdata")