[R] Output of arima

[Ashim Kapoor]

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]]
>>
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>
>

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