3
$\begingroup$

I studied at BA level, that ARIMA(0,1,0) on $y_t$ and ARIMA(0,0,0) on diff($y_t$) are the same models. I am doing the Box–Jenkins model estimation on the historic data of US unemployment rate.

My results (in R):

> arima010a

Call:
arima(x = munrate, order = c(0, 1, 0))


sigma^2 estimated as 19.23:  log likelihood = -2126.56,  aic = 4255.11
> arima010b

Call:
arima(x = diff(munrate), order = c(0, 0, 0))

Coefficients:
      intercept
         0.0708
s.e.     0.1618

sigma^2 estimated as 19.23:  log likelihood = -2126.46,  aic = 4256.92

As you can see, the information criterion (AIC) differs and the $log(L)$, too. My question is, why do they differ in the two model?

I'm looking forward to any answer.


I tried the same later in gretl, and there wasn't any difference between the two models.

$\endgroup$
  • $\begingroup$ Another possibility: Estimating these models requires numerical optimization, wich does not neccesarily converge to the true solution. $\endgroup$ – InfiniteVariance Apr 13 '16 at 22:17
2
$\begingroup$

I don't know R code but are you estimating an intercept in the ARIMA(0,1,0) model? Because if not, this could be why there is a difference, since you are estimating an intercept in the ARIMA(0,0,0) model.

$\endgroup$
1
$\begingroup$

I believe that @Andrew_M is right, this is caused by differences in default options in the implementations of ARIMA across statistical applications.

library(forecast)
births <- scan("http://robjhyndman.com/tsdldata/data/nybirths.dat")
birthstimeseries <- ts(births, frequency=12, start=c(1946,1))
level010 <- arima(births, order = c(0,1,0), include.mean=TRUE)
diff000 <- arima(diff(births), order = c(0,0,0), include.mean=FALSE)
print(level010)
print(diff000)

Results:

> print(level010)
Call:
    arima(x = births, order = c(0, 1, 0), include.mean = TRUE)


sigma^2 estimated as 2.266:  log likelihood = -305.25,  aic = 612.51
> print(diff000)

Call:
    arima(x = diff(births), order = c(0, 0, 0), include.mean = FALSE)


sigma^2 estimated as 2.266:  log likelihood = -305.25,  aic = 612.51
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.