A Regression with ARIMA errors is given by the following formula (saw on Hyndman et al, 1998):

$Y_t = b_0 + b_1 X_{1,t} + \dots + b_k X_{k,t} + N_t$

where $N_t$ is modeled as an ARIMA process.

If we have that the model for $N_t$ is ARIMA$(0,0,0)$, then $N_t = e_t$, and $Y_t$ is modeled by an ordinary regression.

Suppose the following data:

a <- structure(c(29305, 9900, 9802, 17743, 49300, 17700, 24100, 11000, 
10625, 23644, 38011, 16404, 14900, 16300, 18700, 11814, 13934, 
12124, 18097, 30026, 3600, 15700, 12300, 14600), .Tsp = c(2010.25, 
2012.16666666667, 12), class = "ts")
b <- structure(c(1.108528016, 1.136920872, 1.100239002, 1.057191265, 
1.044200511, 1.102063834, 1.083847756, 1.068585841, 1.084879628, 
1.232979511, 1.168894672, 1.257302058, 1.264967051, 1.234793782, 
1.306452369, 1.252644047, 1.178593218, 1.124432965, 1.132878661, 
1.189926986, 1.17249669, 1.176285957, 1.176552, 1.179178082), .Tsp = 
c(2010.25, 2012.16666666667, 12), class = "ts")

If I model it using auto.arima function, I have:

auto.arima(a, xreg=b)
Series: a 
ARIMA(0,0,0) with zero mean     

s.e.   1773.186

sigma^2 estimated as 101878176:  log likelihood=-255.33
AIC=514.65   AICc=515.22   BIC=517.01


lm(formula = a ~ b)

(Intercept)            b  
      48638       -26143  

Coefficients from the models differ. Shouldn't they be the same? What am I missing?

  • 1
    $\begingroup$ your first (auto.arima) model doesn't have an intercept... try using a-mean(a) and b-mean(b) to eliminate the need for one. You'll get the same coefficient estimate that lm gives if you do. $\endgroup$ – jbowman May 14 '12 at 18:04
  • $\begingroup$ Yes, if I remove the mean, the auto.arima model generates the same coefficients that lm. But why isn't it estimating an intercept when I use the raw data? Shouldn't the procedures be the same inside auto.arima? $\endgroup$ – João Daniel May 14 '12 at 18:07
  • $\begingroup$ If you just use arima, you can specify include.mean=TRUE in which case you'll get the lm coefficient estimate; I too find it a little odd that auto.arima doesn't allow this option (I just tested it) but I expect there's a good reason for it. $\endgroup$ – jbowman May 14 '12 at 18:20
  • $\begingroup$ I believe it's because the purpose of auto.arima is automatically estimates all coefficients, including the choice of include or not the intercept. So, it seems the results are different because the intercept is not being calculated in the "regression" part of the model, but in the ARIMA. And in this specific case, they disagree: (a) lm chooses to include an intercept and (b) ARIMA chooses not to include. Is it correct? EDIT: Actually it doesn't seem to be correct, because if the intercept is on the $N_t$ part, it's the same thing if it's on the regression when we substitute. $\endgroup$ – João Daniel May 14 '12 at 19:23

As pointed out in the comments, the difference between the models is that auto.arima() has not included an intercept. It selects a model, possibly including the constant, using the AICc. With one covariate, the model is $$y_t = \beta_0 x_t + n_t$$ where $n_t$ is an ARIMA process. Note that the intercept is shifted to the ARIMA process. In this example, the selected model for $n_t$ does not include a constant.

If you know what model you want, why use auto.arima()? Instead, you could use


which gives

Series: a 
ARIMA(0,0,0) with non-zero mean 

      intercept          b
       48638.40  -26143.23
s.e.   32410.27   27893.41

sigma^2 estimated as 93138232:  log likelihood=-254.25
AIC=514.5   AICc=515.7   BIC=518.03

This is the same as the model obtained using lm(a~b). The estimates are identical, but the standard errors are different because they are estimated in a different way (numerically from the hessian matrix rather than using the inverse of $(X'X)$.)


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