[R] ARMAX model fitting with arima

[Rafael Laboissiere]

This is a follow-up to my request of yesterday.  Someone sent a private
reply indicating a chapter in the Shumway & Stoffer's book "Time Series
Analysis and Its Applications" [1], where some pitfalls of the arima
function are discussed.

In the meanwhile, I realized that the estVARXar function of the dse
package does estimate correctly the ARMAX parameters from my data.  Here
is the code, if someone is interested:

################################################################
### First, the data as I posted previously:
x <- u <- c (rep (0, 50), rep (1, 50))
x [1] <- 0
set.seed (0)
for (i in 2 : length (x)) {
    x [i] <- 0.3 * u [i] + 0.8 * x [i - 1] + 0.01 * rnorm (1)
}
### Second, the trivial DSE fitting:
library (dse1)
estVARXar (TSdata (input = u, output = x))
################################################################

[1] http://www.stat.pitt.edu/stoffer/tsa2/Rissues.htm

Cheers,

Rafael

* Rafael Laboissiere <rafael.laboissiere at inserm.fr> [2009-11-16 08:44]:

> I am trying to understand how to fit an ARMAX model with the arima
> function from the stats package.  I tried the simple data below, where
> the time series (vector x) is generated by filtering a step function
> (vector u, the exogenous signal) through a lowpass filter with AR
> coefficient equal to 0.8.  The input gain is 0.3 and there is a 0.01
> normal white noise added to the output:
> 
>     x <- u <- c (rep (0, 50), rep (1, 50))
>     x [1] <- 0
>     set.seed (0)
>     for (i in 2 : length (x)) {
>         x [i] <- 0.3 * u [i] + 0.8 * x [i - 1] + 0.01 * rnorm (1)
>     }
>     
> Then, I fit the model:
> 
>     arima (x, c (1, 0, 0), xreg = u, include.mean = FALSE, method = "ML")
>     Coefficients:
>              ar1       u
>           0.9988  0.2995
> 	       
> Why don't I get ar1 close to 0.8?  If I use lm to regress the data, it works:
> 
>     lm (x [2 : length (x)] ~ x [1 : (length (x) - 1)] + u [2 : length (u)] - 1)
>     Coefficients:
>     x[1:(length(x) - 1)]        u[2:length(u)]  
>                   0.7989                0.3015  
> 	      
> Any help will be appreciated.
> 
> Best,
> 
 
-- 
Rafael