1
$\begingroup$

Let's say

$$ x_t = a + bt + \varphi e_{t-1} + e_t $$

and

$$ y_t = c + y_{t-1} + \varphi e_{t-1} + e_t $$

where the $e_t$ term is white noise with zero mean.

Can I just sum everything as below:

$$ z_t = a + c + bt + y_{t-1} + \varphi e_{t-1} + e_t $$

to get an ARMA (1,1)?

$\endgroup$
1
$\begingroup$

By adding up the equations for $x_t$ and $y_t$ you will get

$$ \begin{align} (x_t + y_t) &= a + bt + \varepsilon_t + \varphi \varepsilon_{t-1} + c + y_{t-1} + \varepsilon_t + \varphi \varepsilon_{t-1} \\ &= (a + c) + bt + y_{t-1} + 2\varepsilon_t + 2\varphi \varepsilon_{t-1}. \end{align} $$

You may wish to get an ARMA(1,1) for a new variable $z_t := x_t + y_t$. For that you would need to be able to represent the process as

$$ z_t = \varphi_1 z_{t-1} + u_t + \theta_1 u_{t-1} $$

or equivalently

$$ (x_t + y_t) = \varphi_1 (x_{t-1} + y_{t-1}) + u_t + \theta_1 u_{t-1} $$

(possibly including a constant) where $u_t$ is an $i.i.d.$ sequence. However, in your case this is not possible. You cannot form a lagged $z_t$ on the right hand side since there is only $y_{t-1}$ there but no $x_{t-1}$ (note also that a unit coefficient in front of $y_{t-1}$ implies a unit root for the process $y_t$); hence, there will be no autoregressive part and no ARMA(1,1) (or actually no ARMA($p,q$) in general). I do not have a formal proof, though.


In general, summing up two ARMA(1,1) processes does not yield an ARMA(1,1) process. A sufficient condition for that seems to be that the autoregressive and the moving-average coefficients in the two original models be the same:

$$ x_t = \varphi_1 x_{t-1} + u_t + \theta_1 u_{t-1} $$ $$ + $$ $$ y_t = \varphi_1 y_{t-1} + v_t + \theta_1 v_{t-1} $$ $$ \downarrow $$ $$ (x_t + y_t) = \varphi_1 (x_{t-1} + y_{t-1}) + (u_t + v_t) + \theta_1 (u_{t-1} + v_{t-1}) $$

which is

$$ z_t = \varphi_1 z_{t-1} + \varepsilon_t + \theta_1 \varepsilon_{t-1} $$

for $z_t := x_t + y_t$ and $\varepsilon_t := u_t + v_t$, i.e. it can be represented as an ARMA(1,1) process.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks Richard that all makes sense. I am also struggling to see how the two processes could be turned into one ARMA process in term of z. $\endgroup$ – dcr Mar 3 '16 at 19:31
  • $\begingroup$ Is that a question or just a comment? $\endgroup$ – Richard Hardy Mar 3 '16 at 19:47
  • $\begingroup$ Sorry the whole comment was: Thanks Richard that all makes sense. Is it then not possible to get an ARMA process in terms of z then in this case? I am trying to spot it. Could I lag x by t-1, to give xt-1 = a+b(t-1)+φet−2+et-1 and then plug this into the et-1 from y? And then try stating the ARMA in terms of z? $\endgroup$ – dcr Mar 3 '16 at 20:23
  • $\begingroup$ Then you would lack contemporaneous $x_t$: you would have $y_t$ but not $x_t$, so no $z_t$ (which is $y_t+x_t$). $\endgroup$ – Richard Hardy Mar 3 '16 at 21:04
  • $\begingroup$ Ah ok. Is there no method of turning the two processes into one arma process in terms of z then? Y also appears to be a random walk since yt-1 has a coefficient of one, unless I am missing something. thanks. $\endgroup$ – dcr Mar 6 '16 at 18:36

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.