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I am puzzled by an equation, $$ y_t = \phi_1 y_{t-1} + \phi_2 y_{t-2} + u_t + \varepsilon_t - \varepsilon_{t-1}, $$ where $u_t$ and $\varepsilon_t$ are independent white-noise processes.

Is this an ARMA(2, 1) process? Or does the fact that there are two shocks at each timestep (but only one carried over) change this?

Secondly, can $\phi_1$ and $\phi_2$ be estimated consistently using 2SLS? I thought that $y_{t-2}$ would be a valid instrument for $y_{t-1}$, but then $y_{t-2}$ is also a regressor variable.

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    $\begingroup$ As far as I know, instrumental variables are not used for estimation of AR or ARMA models. $\endgroup$ Jul 20, 2021 at 17:16
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    $\begingroup$ There is an example on CV here. $\endgroup$ Jul 20, 2021 at 17:34
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    $\begingroup$ Cool! I learned something new today. $\endgroup$ Jul 20, 2021 at 18:30

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Your model $$ (1-\phi_1B-\phi_2B^2)y_t=u_t+(1-B)\epsilon_t \tag{1} $$ is ARMA(2,1) since the right hand side is clearly an MA(1) process since its autocovariance function cuts off for lags larger than 1. Hence, the right hand side can be represented by $$ (1-\theta_1 B)v_t \tag{2} $$
where $v_t$ is another white noise process. Equating the autocovariance of the right hand side of (1) to that of (2) at lag 0 and 1 yields a set of two non-linear equations \begin{align} \sigma_u^2+2 \sigma_\epsilon^2 &= \sigma_v^2(1+\theta_1^2) \\ -\sigma_\epsilon^2 &= -\theta_1\sigma_v^2 \end{align} which can be straightforwardly solved for $\sigma_v^2$ and $\theta_1$. There are two solutions. Only the one for which $|\theta_1|<1$ is invertible and relevant.

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I am puzzled by an equation, $$ y_t = \phi_1 y_{t-1} + \phi_2 y_{t-2} + u_t + \varepsilon_t - \varepsilon_{t-1}, $$ where $u_t$ and $\varepsilon_t$ are independent white-noise processes.

Is this an ARMA(2, 1) process? Or does the fact that there are two shocks at each timestep (but only one carried over) change this?

It is possible to show that in general

$ARMA(p_1,q_1) + ARMA(p_2,q_2) = ARMA(p_3,q_3)$

now, "noise" is a special case for ARMA. Then your ARMA plus noise admit an ARMA representation. Moreover, the fact that in your case the coefficient of MA component is imposed to $1$ is not a problem. You can obtain something like that:

$u_t + \varepsilon_t = r_t$

$$ y_t = \phi_1 y_{t-1} + \phi_2 y_{t-2} + r_t + \alpha r_{t-1}, $$

Regarding estimation problems I think that ML procedure is a better idea.

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    $\begingroup$ This is not an easy question. See Hamilton 1994 par 4.7 for more details: ru.ac.bd/stat/wp-content/uploads/sites/25/2019/03/… $\endgroup$
    – markowitz
    Jul 21, 2021 at 15:10
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    $\begingroup$ $u_t+\varepsilon_t=:r_t$ will not work to produce MA(1) in terms of $r_t$. Or at least I do not see how it could. $\endgroup$ Jul 21, 2021 at 18:20
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    $\begingroup$ The question is not about adding ARMA processes (e.g. ARMA + white noise) but instead about adding white noise to the moving average part of an ARMA model. This makes the whole process ARMA(2,1), see my answer below. $\endgroup$ Jul 21, 2021 at 21:07
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    $\begingroup$ The model given is ARMA(2,1) plus noise. We can consider it as AR(2) + MA(1) + noise = AR(2) + MA(1) = ARMA(2,1) where the 'new' MA(1) is different from the initial. Therefore my answer seems me correct. $\endgroup$
    – markowitz
    Jul 21, 2021 at 22:01
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    $\begingroup$ @Richard and Jarle, consider that we sum two process not two observations only. Moreover consider that AR, MA and noise are special cases of ARMA. Then we stay precisely in sum of ARMAs case. $\endgroup$
    – markowitz
    Jul 21, 2021 at 22:17
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To answer the second part of my question: using $y_{t-3}$ as an instrument, it is indeed possible to estimate $\phi_1$ and $\phi_2$ consistently by 2SLS. I verified this by computer simulation.

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    $\begingroup$ Do you perhaps mean 2SLS instead of OLS? In the context of instrumental variables, the term OLS is more typically used for estimating the original model as is (without use of IVs). $\endgroup$ Jul 21, 2021 at 6:58
  • $\begingroup$ Yes, I do mean 2SLS. In the first stage I regressed $y_{t-1}$ on $y_{t-2}, y_{t-3}$ to obtain fitted values $\hat{y}_{t-1}$. Then I regressed $y_t$ on $\hat{y}_{t-1}$ and $y_{t-2}$, giving me (apparently) consistent estimates of $\phi_1, \phi_2$. Does this seem correct to you? Unlike $y_{t-1}$, $\hat{y}_{t-1}$ is uncorrelated with the error term so the endogeneity problem should be fixed. $\endgroup$ Jul 21, 2021 at 10:32

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