# [R] AR(2) coefficient interpretation

Prof Brian Ripley ripley at stats.ox.ac.uk
Tue Dec 23 09:06:31 CET 2008

```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:

>
> 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
>   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
>
>> predict
>
> it came out the value of 9.397515 below and I have no idea about how to
>
> \$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:
>>
>> 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:
>>
>> attach(r)
>> fit<-arima(a, c(2,0,0))
>>
>> Thank you for your help first.
>>
>>
>
> --
> View this message in context: http://www.nabble.com/AR%282%29-coefficient-interpretation-tp21129322p21138255.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> and provide commented, minimal, self-contained, reproducible code.
>

--
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

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