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AR(2) coefficient interpretation

3 messages · Stephen Oman, Brian Ripley

Stephen Oman
# 
I am a beginner in using R and I need help in the interpretation of AR result
by R.  I used 12 observations for my AR(2) model and it turned out the
intercept showed 5.23 while first and second AR coefficients showed 0.40 and
0.46. It is because my raw data are in million so it seems the intercept is
too small and it doesn't make sense. Did i make any mistake in my code? My
code is as follows:

r<-read.table("data.txt", dec=",", header=T)
attach(r)
fit<-arima(a, c(2,0,0))

Thank you for your help first.
Stephen Oman
# 
As I need your urgent help so let me modify my question. I imported the
following data set to R and run the statements i mentioned in my previous
reply 
   Year Month   Period     a    b      c
1  2008   Jan 2008-Jan 105,536,785  9,322,074  9,212,111
2  2008   Feb 2008-Feb 137,239,037 10,986,047 11,718,202
3  2008   Mar 2008-Mar 130,237,985 10,653,977 11,296,096
4  2008   Apr 2008-Apr 133,634,288 10,582,305 11,729,520
5  2008   May 2008-May 161,312,530 13,486,695 13,966,435
6  2008   Jun 2008-Jun 153,091,141 12,635,693 13,360,372
7  2008   Jul 2008-Jul 176,063,906 13,882,619 15,202,934
8  2008   Aug 2008-Aug 193,584,660 14,756,116 16,083,263
9  2008   Sep 2008-Sep 180,894,120 13,874,154 14,524,268
10 2008   Oct 2008-Oct 196,691,055 14,998,119 15,802,627
11 2008   Nov 2008-Nov 184,977,893 13,748,124 14,328,875

and the AR result is
Call:
arima(x = a, order = c(2, 0, 0))

Coefficients:
         ar1     ar2  intercept
      0.4683  0.4020     5.8654
s.e.  0.2889  0.3132     2.8366

sigma^2 estimated as 4.115:  log likelihood = -24.04,  aic = 56.08

The minimum mount of a is more than 100 million and the intercept is 5.86
based on the result above. 
If I placed all values into the formula then Xt=5.8654+0.4683*(184,977,893
)+0.4020*(196,691,055 )= 165,694,957.27. Do you think that makes sense? Did
i interpret the result incorrectly?

Also, i submit the following statement for the prediction of next period
it came out the value of 9.397515 below and I have no idea about how to
interpret this value. Please help. 

$pred
Time Series:
Start = 12 
End = 12 
Frequency = 1 
[1] 9.397515

$se
Time Series:
Start = 12 
End = 12 
Frequency = 1 
[1] 2.028483
Stephen Oman wrote:

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Brian Ripley
    

    

      
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You forgot to RTFM.  From ?arima

      Different definitions of ARMA models have different signs for the
      AR and/or MA coefficients.  The definition used here has


  'X[t] = a[1]X[t-1] + ... + a[p]X[t-p] + e[t] + b[1]e[t-1] + ... + 
b[q]e[t-q]'


      and so the MA coefficients differ in sign from those of S-PLUS.
      Further, if 'include.mean' is true (the default for an ARMA
      model), this formula applies to X - m rather than X.

Since you have not yet produced a reproducible example (at least in a 
single email), we don't have enough information to reproduce your reults.
But I hope we are not fitting AR(2) models to (potentialy seasonal) time 
series of length 11.

        
On Mon, 22 Dec 2008, Stephen Oman wrote: