Solved – auto.arima Not Minimizing AIC

Question

I simulated a MA(3) process using:

set.seed(66)
w <- rnorm(100,0,3.6)
p1 <- 0.4; p2 <- -0.2; p3 <- 0.3;
ma3 <- w[1]
ma3[2] <- w[2] + p1*w[1] 
ma3[3] <- w[3] + p1*w[2] + p2*w[1]
for (t in 4:100) ma3[t] <- w[t] + p1*w[t-1] + p2*w[t-2] + p3*w[t-3]

Running auto.arima on the time series gives:

> auto.arima(ma3)                                   
Series: ma3 
ARIMA(0,0,1) with zero mean     

Coefficients:
         ma1
      0.3854
s.e.  0.1152

sigma^2 estimated as 14.41:  log likelihood=-275.39
AIC=554.77   AICc=554.89   BIC=559.98

However, fitting the series to a MA(3) model gives a lower AIC:

> arima(ma3, order=c(0,0,3))

Call:
arima(x = ma3, order = c(0, 0, 3))

Coefficients:
         ma1      ma2     ma3  intercept
      0.4039  -0.0836  0.5125     0.2752
s.e.  0.1158   0.0905  0.1039     0.6078

sigma^2 estimated as 11.2:  log likelihood = -264.67,  aic = 539.34

I'm not sure what's going on. I thought that auto.arima selected the best model based on the AIC.

Answer 1

By default, auto.arima uses a stepwise search and there is no guarantee that it will find the best model. You can do a more complete search by setting stepwise=FALSE in the call. Like this:

> library(forecast)
> auto.arima(ma3,stepwise=FALSE)

Series: ma3 
ARIMA(3,0,0) with zero mean     

Coefficients:
         ar1      ar2     ar3
      0.3203  -0.0556  0.2816
s.e.  0.0976   0.1025  0.0984

sigma^2 estimated as 13.74:  log likelihood=-271.57
AIC=551.15   AICc=551.57   BIC=561.57

However, this still hasn't found an MA(3) model. That is because the fitted MA(3) model is almost non-invertible.

> fit <- Arima(ma3, order=c(0,0,3))
> plot(fit)

Note the root lies almost on the unit circle. auto.arima will not consider models that are close to the stationarity or invertibility boundary, as they are typically numerically unstable.