Time-Series – ARIMA (0,1,1) or (0,1,0): Which Model to Use and Why?

arimamodelingtime series

I've just started learning time series so please excuse me if it's painfully obvious; I haven't managed to find the answer elsewhere.

I have a data series showing a pretty obvious trend although it's quite noisy. I can take pretty much any division of the data and run classical tests to show a highly significant difference in means.

I decided to have a look at time series analysis to see if it could help describe the trend. An ARIMA(0,1,1) model comes out with AIC,BIC=34.3,37.3 (Stata), whilst an ARIMA(0,1,0) model comes out with AIC,BIC=55.1,58.1 – so I understand I'm supposed to prefer the (0,1,1) model.

However, the coefficient for the MA(1) is displaying as -0.9999997 (and not showing any p-values). If I try the same in SPSS I get an MA(1) coefficient of 1.000 (I assume SPSS uses opposite signs) with a p-value of 0.990 – does this mean it suggests I drop the term?

My understanding is that the effect of a MA(1) coefficient of -1 is basically to remove the old error term and convert the whole series to a linear trend. Does this mean ARIMA is totally unsuitable for my needs? On the plus side it gives me a sensible value for the trend. If I use the (0,1,0) model then I still get a reasonable value for the trend but it's not significant any more.

Thanks for your help!

EDIT:
Thanks for looking in. The trend looks like a fairly linear decrease; the data points seen to fairly noisily rattle around above and below a trend line. The ARIMA (0,1,1) model produces something that's not far off a straight line decrease which seems sensible – the (0,1,1) produces what is essentially a lagged version of the data, translated down by one month of trend.
The data aren't stationary (due to the trend) – though the first differences seem to be. I don't think the (0,1,1) is a bad model – I'm just a little confused by the p-value seeming to suggest I should drop the MA term – or wondering if it means I should bin ARIMA entirely!

EDIT2
@vinux – thanks for the suggestion; that makes a lot of sense (and seems to be what the -1 MA term is trying to create?).
I've uploaded as many graphs as I could think of as people had requested.

tsline y - graph of raw values
tsline D.y - graph of differences
ac y - autocorrelations of y
pac y - partial autocorrelations of y
ac D.y - autocorrelations of first differences
pac D.y - partial autocorrelations of first differences

I've also put the monthly data up in CSV format at pastebin

Best Answer

It is difficult to give the right answer without looking at the data. Here are some points that may help you in your modelling.

The coefficient of MA(1) very close to 1 indicates the sign of overdifferencing. This means unit root in Moving averages.

My suggestion would be: Check the original series is stationary (visually) or check the presence of unit root. If you observe deterministic trend (eg: linear), add the trend part with time series model. If the original series is stationary build the time series without differencing.

Related Solutions

Solved – Extract BIC and AICc from arima() object

For the BIC and AIC, you can simply use AIC function as follow:

> model <- arima(x=sunspots, order=c(2,0,2), method="ML")
> AIC(model)
[1] 23563.39
> bic=AIC(model,k = log(length(sunspots)))
> bic
[1] 23599.05

The function AIC can provide both AIC and BIC. Look at ?AIC.

Seasonal ARIMA – Constraints on the Coefficients and Possible Software Bug in ITSM

I don't have the software ITSM but I could do some inspection upon the data and output that you posted. I fitted the model with different possible mistakes to see which one gives results closer to your output. I used a simplified version of the conditional sum of squares to fit this particular model.

My conclusion is that the code of ITSM that makes the computations is probably correct. The error arises most likely when reporting the results and printing the output. Below I give some details based on some examples that I run in the R software.

I fitted the ARIMA(1,1,0)(1,1,0) model written in expanded form $1 - \phi_{1}B - \Phi B^{12} + \phi_{1} \Phi B^{13}$ by minimizing the squared sum of residuals of the following equation with two parameters ($\phi_1$ and $\Phi$):

$$ y_t = \phi_1 y_{t-1} + \Phi y_{t-12} - \phi_1\Phi y_{t-13} + \hbox{error}_t \,. $$

y <- log(window(AirPassengers, end = c(1959, 12)))
# compute residual sum of squares for ARIMA(1,1,0)(1,1,0) with monthly data
# and minimize using BFGS
# this approach and code is intended just for testing purposes
ssq <- function(coefs, y)
{
  n <- length(y) - 13
  z <- diff(diff(y), 12)
  e <- rep(NA, n-13)
  for (i in seq.int(14, length(z)))
    e[i-13] <- z[i] - coefs[1]*z[i-1] - coefs[2]*z[i-12] + prod(coefs)*z[i-13]
  sum(e^2)
}
# minimize the sum of the squared residuals
res <- optim(ssq, par=c(0,0), method="BFGS", y=y)
# estimated AR coefficients (phi1 and Phi, respectively)
res$par
# [1] -0.3942546 -0.4516274
res$par[1] * res$par[2]
# [1] 0.1780562

The fitted model is therefore (in the expanded form and moving the terms to the right-hand-side):

$$ y_t = 0.3942546 y_{t-1} + 0.4516274 y_{t-12} - 0.1780562 y_{t-13} + \hbox{error}_t \,. $$

As the OP said, the sign of the AR coeffcient related to $B^{13}$ should be the opposite of those for the other coefficients and this is what we get here.

Now, I make intentionally a mistake in the sign of the $B^{13}$ coefficient by typing - prod(coefs)*z[i-13] instead of + prod(coefs)*z[i-13]. Then the following is obtained:

ssq.bug <- function(coefs, y)
{
  n <- length(y) - 13
  z <- diff(diff(y), 12)
  e <- rep(NA, n-13)
  for (i in seq.int(14, length(z)))
    e[i-13] <- z[i] - coefs[1]*z[i-1] - coefs[2]*z[i-12] - prod(coefs)*z[i-13]
  sum(e^2)
}
res.bug <- optim(ssq.bug, par=c(0,0), method="BFGS", y=y)
res.bug$par
# -0.2459668 -0.3511647
res.bug$par[1] * res.bug$par[2]
# [1] 0.08637484

The results are far from the values reported by the OP. The results obtained before are much closer to the output returned by ITSM (except for the sign related to $B^{13}$). Thus, I would say there is no bug in the code that makes the computations.

We could also suspect that the restriction was actually not enforced, but that cannot be the case because .1591 is exactly equal to the product .3485*.4565, so I don't think it is the case either.

It's interesting to see that if the restriction for $B^{13}$ is not enforced and the AR(13) is fitted (with zeros for the AR coefficients of order 2 to 11), then the estimated parameters are very close those obtained above for the restricted model (this may suggest that, given a model of order 13, the multiplicative seasonal ARIMA model is appropriate for the data).

The unrestricted model contains three coefficients, $\phi_1$, $\Phi$ and $\phi_{13}$. The optimization algorithm searches for values of theses coefficients that minimize the sum of squared residuals:

$$ y_t = \phi_1 y_{t-1} + \Phi y_{t-12} - \phi_{13} y_{t-13} + \hbox{error}_t \,. $$

ssq.ur <- function(coefs, y)
{
  n <- length(y) - 13
  z <- diff(diff(y), 12)
  e <- rep(NA, n-13)
  for (i in seq.int(14, length(z)))
    e[i-13] <- z[i] - coefs[1]*z[i-1] - coefs[2]*z[i-12] + coefs[3]*z[i-13]
  sum(e^2)
}
res.ur <- optim(ssq.ur, par=c(0,0,0), method="BFGS", y = y)
res.ur$par
# [1] -0.3951237 -0.4539019  0.1493490
res.ur$par[1] * res.ur$par[2]
# [1] 0.1793474

As we would expect, the parameters estimated in the unrestricted model do not satisfy the restriction exactly, but the last coefficient turns out to be close to the product of the two others. If we ran a test, the restriction would probably not be rejected.

Best Answer

Related Solutions

Solved – Extract BIC and AICc from arima() object

Seasonal ARIMA – Constraints on the Coefficients and Possible Software Bug in ITSM

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