Dear Eric,
Many thanks for your reply.
Best Regards,
Ashim
On Wed, Nov 14, 2018 at 4:05 PM Eric Berger 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 defa
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)
Dear Eric and William,
Why do the 1st and 2nd incantation of arima return sigma^2 as 5.233 vs
.?
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),
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
Try google'ing for 'variance of an AR(1) process'.
With the same seed, if you set n=100, you will get something that will
compare well with what you discover from your search.
On Tue, Nov 13, 2018 at 2:04 PM Ashim Kapoor wrote:
> Dear All,
>
> Here is a reprex:
>
> set.seed(123)
> b <- arima
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
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