2
$\begingroup$

I am wondering about the exact definition of ARIMA model in function arima in R when exogenous regressors are included.

I understand that arima(y, order=c(p,0,q), xreg=x) is equivalent to estimating the following equation (where $\mu_y$ and $\mu_x$ stand for the means of $y$ and $x$, respectively):

(1) $(y_t-\mu_y)=\varphi_0+\phi_1(y_{t-1}-\mu_y)+...+\varphi_p(y_{t-p}-\mu_y)+\varepsilon_t+\theta_1\varepsilon_{t-1}+...+\theta_q\varepsilon_{t-q}+\beta_1x_t$

Or is it

(2) $(y_t-\mu_y)=\varphi_0+\phi_1(y_{t-1}-\mu_y)+...+\varphi_p(y_{t-p}-\mu_y)+\varepsilon_t+\theta_1\varepsilon_{t-1}+...+\theta_q\varepsilon_{t-q}+\beta_1(x_t-\mu_x)$

(only the last term differs between (1) and (2))?

Or perhaps I got both of them wrong?

Edit: I now realize that including both {$\mu_x$ and $\mu_y$} and $\varphi_0$ in (2) was superfluous.

$\endgroup$
3
$\begingroup$

Looking at the documentation and the code of arima, I conclude that the following linear model with ARIMA errors is fitted when exogenous regressors are included:

\begin{eqnarray} \begin{array}{l} (y_t-\mu-X_t^{'}\vec{\beta})&=&\phi_1(y_{t-1}-\mu-X_{t-1}^{'}\vec{\beta})+...+\phi_p(y_{t-p}-\mu-X_{t-p}^{'}\vec{\beta}) \\ &+&\varepsilon_t+\theta_1\varepsilon_{t-1}+...+ \theta_q\varepsilon_{t-q} \,. \end{array} \end{eqnarray}

$X_t^{'}$ is a row vector containing the values of the external regressors at time $t$ and $\vec{\beta}$ is a column vector containing the coefficients related to those regressors.

Thus, there is no $\varphi_0$ term and the mean $\mu_x$ is not removed from the exogenous regressors.

$\endgroup$
  • 1
    $\begingroup$ Thanks for your answer! Are you sure you got all the time indices for $X$ right? They are all $t$, but maybe they should match those of $y$? Also, shouldn't $\mu_y$ be replaced by a constant like $\beta_0$ (which need not equal the mean of $y$)? $\endgroup$ – Richard Hardy Oct 28 '14 at 20:51
  • $\begingroup$ I think both of my own guesses provided together with the question are wrong. A good presentation of related material can be found here. $\endgroup$ – Richard Hardy Oct 28 '14 at 20:55
  • $\begingroup$ Yes, you are right. I fixed the indices and renamed $\mu_y$ as $\mu$ to make it clear that it is a parameter to be estimated. $\endgroup$ – javlacalle Oct 28 '14 at 20:57

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.