ARIMA question

Do you have a particular version?  Mine does not have a method like that...

  

x.model
    

Call:
arima(x = x, order = c(0, 0, 2))

Coefficients:
         ma1      ma2  intercept
      0.3812  -0.1138     0.0936
s.e.  0.0956   0.0862     0.2631

sigma^2 estimated as 4.318:  log likelihood = -215.15,  aic = 438.31
  

class(x.model)
    

[1] "Arima"
  

methods(fitted)
    

[1] fitted.default*         fitted.isoreg*          fitted.nls*
[4] fitted.quantmod*        fitted.smooth.spline*   fitted.values.quantmod*

   Non-visible functions are asterisked
  

fitted(x.model)
    

NULL
  


On Nov 7, 2007 3:07 PM, Nathan Bryant <nbryant at optonline.net> wrote:
  

There is a method called "fitted()" that applies to most model classes
including Arima, which does the same thing.


Jeff Ryan wrote:
    

I think it is as simple as backing out from the residuals:

# an MA2 model
x <- arima.sim(list(ma=2),n=100)

#Fitted as such...
x.model <- arima(x,c(0,0,2))

# add the residuals to the original data
x.insample.fit <- x-residuals(x.model)

# and you can even see them:
plot(x)
par(new=TRUE)
lines(x.insample.fit,col=3,lty=2)

Jeff


On Nov 7, 2007 12:47 PM, Yalla, Swaroop (FID)
<Swaroop.Yalla at morganstanley.com> wrote:

      

Hi:

I have another ARIMA question for R. I was finally able to use ARIMA
modeling on my data. Now to forecast out of sample, we can use
predict(fit, n.ahead = 10) type of command and thats fine- but how can I
see the fit in-sample. I mean is there a easy way to just compare the
actual data with the fitted model in-sample?

thanks for all the help..
Swaroop
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