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)?
 A: 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.
