2
$\begingroup$

I have the following output from the auto.arima function with specified xreg:

Series: Individuals 
Regression with ARIMA(2,1,1) errors 

Coefficients:
          ar1      ar2     ma1  inflation  GDP growth  EURIBOR
      -0.8366  -0.3928  0.4008    -0.0013      -8e-04   0.0139
s.e.   0.2155   0.0863  0.2333     0.0023       1e-03   0.0036

sigma^2 estimated as 0.0004495:  log likelihood=453.32
AIC=-892.63   AICc=-892   BIC=-870.09

Does it hold that

$y_t = -0.0013*inflation_t -8*10^{-4}*GDP_t + 0.0139*EURIBOR_t + n_t$

$n_t = -0.8366*n_{t-1} - 0.3928*n_{t-2} + e_t + 0.4008*e_{t-1}$

or does the differencing change something?

I get the residuals by the following code

nt <- residuals(fitInd, type = "regression")
et <- residuals(fitInd, type = "innovation")

When I check with the residuals "nt" it does match, for example 

nt[4]=0.2621466

and per the first equation:

$y_4 - (-0.0013*inflation_4 -8*10^{-4}*GDP_4 + 0.0139*EURIBOR_4)= n_4 = 0.2621466 $

However, it does not match for the innovation errors "et" per the second equation which should be equal to nt[4]:

-0.8366*nt[3] - 0.3928*nt[2] + et[4] + 0.4008*et[3] = -0.3318558

Could someone clarify what I am doing wrong. Thanks in advance.

Improve this question
$\endgroup$
1
  • $\begingroup$ Did you test for and possibly consider lags of any of the three predictors ? Were there any untreated anomalies in the residuals ? Was the error variance of the residuals constant over time ? These are items that need to be considered in determining a useful model. $\endgroup$
    – IrishStat
    Oct 7, 2019 at 14:32

1 Answer 1

Reset to default
1
$\begingroup$

Your equation would have been right if you used $\Delta n_t$ instead of $n_t$. It's the differenced model, like ARMA of $\Delta n_t$:

$$\Delta n_t = -0.8366*\Delta n_{t-1} - 0.3928*\Delta n_{t-2} + e_t + 0.4008*e_{t-1}$$

Improve this answer
$\endgroup$
3
  • $\begingroup$ good eyes ! as usual $\endgroup$
    – IrishStat
    Oct 7, 2019 at 14:34
  • $\begingroup$ Thank you, this gives more reasonable results. Interestingly, the figures do not match exactly: nt[4]-nt[3] = -0.002577296 and coefs[1]*(nt[3]-nt[2])+coefs[2]*(nt[2]-nt[1]) + et[4] + coefs[3]*et[3] = -0.002600029 What could be the reason for this slight mis-match? $\endgroup$
    – anonqqquest
    Oct 8, 2019 at 6:50
  • $\begingroup$ The difference could be in how you and R seed the initial values of errors given the ma and ar terms $\endgroup$
    – Aksakal
    Oct 8, 2019 at 12:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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