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 <wdun...@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 <ashimkap...@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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>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
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