# Sum of two ARMA processes that have the same innovations

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)?

## 1 Answer

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.

• 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. – dcr Mar 3 '16 at 19:31
• Is that a question or just a comment? – Richard Hardy Mar 3 '16 at 19:47
• 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? – dcr Mar 3 '16 at 20:23
• Then you would lack contemporaneous $x_t$: you would have $y_t$ but not $x_t$, so no $z_t$ (which is $y_t+x_t$). – Richard Hardy Mar 3 '16 at 21:04
• 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. – dcr Mar 6 '16 at 18:36