Having problem with forecast.Arima() function

Dear Berend,

I am using R version 3.0.1 and I loaded code packages below

library(TSA)
library(lattice)
library(stats)
library(utils)
library(forecast)

I reisntalled R and loaded packages 'steries' and 'forecast' as you suggested. Everything works out fine. Maybe it was the 'TSA' package which gave me the warinig masseges  

On Mon, Aug 5, 2013 at 3:03 PM, Berend Hasselman <bhh at xs4all.nl> wrote:

On 05-08-2013, at 07:47, Salaam Batur <swordlight20 at gmail.com> wrote:

Dear R users,

I am having a problem using forecast.Arima fuction. Here is the whole code:

d=scan("D:/Data.txt")
d
D=ts(data=d, start=1981,end=2012, frequency=1)
D
Time Series:
Start = 1981
End = 2012
Frequency = 1
[1]  384  403  427  450  499  550  575  615  640  680  702  730  760  790
[15]  790  830  870  871  906  920  968 1010 1060 1111 1165 1191 1217 1221
[29] 1089 1089 1090 1103

Please use dput to show objects.

lnD=log(D)
lnD3=diff(lnD, differences=3)
adf.test(lnD3)
       Augmented Dickey-Fuller Test

data:  lnD3
Dickey-Fuller = -3.5315, Lag order = 3, p-value = 0.05795
alternative hypothesis: stationary

#########d parameter is determined by ADF test, which is 3#############

Now choosing p and q parameters

par(mfrow=c(2,1))
acf(lnD3, lag.max=20)
pacf(lnD3, lag.max=20)

######### from acf and pacf, p=2, q=1##############

Now fitting Arima

arima1=arima(lnD3, order=c(2,0,1))
arima1
Series: x
ARIMA(2,0,1) with non-zero mean

Coefficients:
         ar1      ar2      ma1  intercept
     -0.5189  -0.2033  -1.0000     -1e-04
s.e.   0.1806   0.1770   0.0993      5e-04

sigma^2 estimated as 0.00118:  log likelihood=54.24
AIC=-100.48   AICc=-97.87   BIC=-93.64

####### Which looks good#########

Using auto.arima() to see what R have in mind

autoarima=auto.arima(lnD, d=3)
autoarima
Series: lnD
ARIMA(2,3,0)

Coefficients:
         ar1      ar2
     -1.0282  -0.5851
s.e.   0.1524   0.1560

sigma^2 estimated as 0.001731:  log likelihood=50.37
AIC=-94.73   AICc=-93.77   BIC=-90.63

###### From AIC and BIC, I prefer arima1 instead of autoarima######
Now using forecast.Arima

forecastArima1=foreca.Arima(arima1, h=5)
*Warining message
Error in ts(x) : 'ts' object must have one or more observations*

I assume that this should be forecast.Arima

But forecasting autoarima is no problem
forecastAutoArima=forecast.Arima(autoarima, h=5, c=(0.95))
forecastAutoArima

    Point Forecast    Lo 80    Hi 80    Lo 95    Hi 95
2013       7.084688 7.031373 7.138004 7.003150 7.166227
2014       7.167079 7.049206 7.284951 6.986808 7.347349
2015       7.285478 7.069813 7.501142 6.955648 7.615308
2016       7.443636 7.085484 7.801789 6.895890 7.991383
2017       7.618670 7.081729 8.155612 6.797489 8.439852

Why???? Is there a bug probelm with arima() function itself???

This not reproducible code.
Which packages have you loaded?
adf.test, auto.arima and forecast.Arima are not standard.

So using package sos told me that I needed package tseries and forecast.

Using your data and your code gave me this for the code starting at auto.arima:

 autoarima=auto.arima(lnD, d=3)

autoarima

Series: lnD
ARIMA(2,3,0)

Coefficients:
          ar1      ar2
      -1.0282  -0.5851
s.e.   0.1524   0.1560

sigma^2 estimated as 0.001731:  log likelihood=50.37
AIC=-94.73   AICc=-93.77   BIC=-90.63

forecastArima1=forecast.Arima(arima1, h=5)
forecastArima1

     Point Forecast       Lo 80      Hi 80       Lo 95      Hi 95
2013  -0.0212777938 -0.06606177 0.02350618 -0.08976898 0.04721339
2014   0.0088824273 -0.07130043 0.08906529 -0.11374668 0.13151153
2015  -0.0004052589 -0.08451693 0.08370641 -0.12904295 0.12823243
2016  -0.0017177972 -0.08583008 0.08239448 -0.13035643 0.12692083
2017   0.0008517381 -0.08342997 0.08513345 -0.12804602 0.12974949

forecastAutoArima=forecast.Arima(autoarima, h=5, c=(0.95))
forecastAutoArima

     Point Forecast    Lo 80    Hi 80    Lo 95    Hi 95
2013       7.084688 7.031373 7.138004 7.003150 7.166227
2014       7.167079 7.049206 7.284951 6.986808 7.347349
2015       7.285478 7.069813 7.501142 6.955648 7.615308
2016       7.443636 7.085484 7.801789 6.895890 7.991383
2017       7.618670 7.081729 8.155612 6.797489 8.439852

So you must be doing something other than what you have shown.

Berend

If anyone knows the problem, or shows me a right direction, I would really
appreciate it!!!
Many many thanks!!!

Chintemur Batur

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