[R] Output of arima

[Ashim Kapoor]

Dear Eric,

Many thanks for your reply.

Best Regards,
Ashim

On Wed, Nov 14, 2018 at 4:05 PM Eric Berger <ericjberger using gmail.com> wrote:

> Hi Ashim,
> Per the help page for arima(), it fits an ARIMA model to the specified
> time series - but the caller has to specify the order - i.e. (p,d,q) - of
> the model.
> The default order is (0,0,0) (per the help page). Hence your two calls are
> different. The first call is equivalent to order=c(0,0,0) and the second
> specifies order=c(1,0,0).
> In the first, since there is no auto-regression, all the variance is
> "assigned" to the innovations, hence sigma^2 = 5.233.
> The second case you understand.
> A clue that this was happening is that the first call only returns a
> single coefficient (where is the autoregressive coefficient? - not there
> because you didn't ask for it).
> The second call returns two coefficients, as requested/expected.
>
> HTH,
> Eric
>
>
> On Wed, Nov 14, 2018 at 12:08 PM Ashim Kapoor <ashimkapoor using gmail.com>
> wrote:
>
>> Dear Eric and William,
>>
>> Why do the 1st and 2nd incantation of arima return sigma^2 as 5.233 vs
>> .9999?
>> The help for arima says  --->  sigma2: the MLE of the innovations
>> variance.
>> By that account the 1st result is incorrect. I am a little confused.
>>
>> set.seed(123)
>> b <- arima.sim(list(order = c(1,0,0),ar= .9),n=1000000,sd=1)
>>
>> # Variance of the innovations, e_t = 1
>>
>> # Variance of b = Var(e_t)/(1-Phi^2) = 1 / (1-.81) = 5.263158
>>
>> arima(b)
>>
>> > arima(b)
>>
>> Call:
>> arima(x = b)
>>
>> Coefficients:
>>       intercept
>>         -0.0051
>> s.e.     0.0023
>>
>> sigma^2 estimated as 5.233:  log likelihood = -2246450,  aic = 4492903
>> >
>>
>>
>> arima(b,order= c(1,0,0))
>>
>> Call:
>> arima(x = b, order = c(1, 0, 0))
>>
>> Coefficients:
>>          ar1  intercept
>>       0.8994    -0.0051
>> s.e.  0.0004     0.0099
>>
>> sigma^2 estimated as 0.9999:  log likelihood = -1418870,  aic = 2837747
>> >
>>
>> On Tue, Nov 13, 2018 at 11:07 PM William Dunlap <wdunlap using tibco.com>
>> wrote:
>>
>> > Try supplying the order argument to arima.  It looks like the default is
>> > to estimate only the mean.
>> >
>> > > arima(b, order=c(1,0,0))
>> >
>> > Call:
>> > arima(x = b, order = c(1, 0, 0))
>> >
>> > Coefficients:
>> >          ar1  intercept
>> >       0.8871     0.2369
>> > s.e.  0.0145     0.2783
>> >
>> > sigma^2 estimated as 1.002:  log likelihood = -1420.82,  aic = 2847.63
>> >
>> >
>> > Bill Dunlap
>> > TIBCO Software
>> > wdunlap tibco.com
>> >
>> > On Tue, Nov 13, 2018 at 4:02 AM, Ashim Kapoor <ashimkapoor using gmail.com>
>> > wrote:
>> >
>> >> Dear All,
>> >>
>> >> Here is a reprex:
>> >>
>> >> set.seed(123)
>> >> b <- arima.sim(list(order = c(1,0,0),ar= .9),n=1000,sd=1)
>> >> arima(b)
>> >>
>> >> Call:
>> >> arima(x = b)
>> >>
>> >> Coefficients:
>> >>       intercept
>> >>          0.2250
>> >> s.e.     0.0688
>> >>
>> >> sigma^2 estimated as 4.735:  log likelihood = -2196.4,  aic = 4396.81
>> >> >
>> >>
>> >> Should sigma^2 not be equal to 1 ? Where do I misunderstand ?
>> >>
>> >> Many thanks,
>> >> Ashim
>> >>
>> >>         [[alternative HTML version deleted]]
>> >>
>> >> ______________________________________________
>> >> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> >> https://stat.ethz.ch/mailman/listinfo/r-help
>> >> PLEASE do read the posting guide
>> >> http://www.R-project.org/posting-guide.html
>> >> and provide commented, minimal, self-contained, reproducible code.
>> >>
>> >
>> >
>>
>>         [[alternative HTML version deleted]]
>>
>> ______________________________________________
>> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>

	[[alternative HTML version deleted]]