Output of 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 1 ? Where do I misunderstand ?

Many thanks,
Ashim

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arima(b, order=c(1,0,0))

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]]

arima(b)

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 at 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]]

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 at 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 at 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]]

        [[alternative HTML version deleted]]

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 at 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 at 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 at 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]]

        [[alternative HTML version deleted]]