It may seem that the $\epsilon_t$ term makes it so that $\{\zeta_t\}$ can't be written as an MA(1), but that is not the case!
Your Prof is correct. You can write process $\{\zeta_t\}$ as an MA(1)
Let $b = -\phi$ and consider two representations of the process $\{\zeta_t\}$. Structurally, the process is:
\begin{align*}
\zeta_t &= \epsilon_t + \eta_t + b \eta_{t-1}\\
\end{align*}
The process $\{\zeta_t\}$ is stationary with auto-covariance function $\gamma(k) = 0$ for $k\geq 2$. Consequently, by the Wold Decomposition Theorem, the unique representation below also exists:
$$ \zeta_t = u_t + a u_{t-1} $$
where $\{u_t\}$ is a white noise process.
Note autocovariance function $\gamma(k)$ is given by:
\begin{align*}
\gamma(0) &= \sigma^2_\epsilon + (1 +b^2) \sigma^2_\eta = (1 + a^2) \sigma^2_u\\
\gamma(1) &= b \sigma^2_\eta = a \sigma^2_u \\
\gamma(k) &= 0 \quad \text{for }k \geq 2
\end{align*}
Computing the Wold Decomposition:
Let $L$ denote the lag operator. Using the lag operator to write $\zeta_t$ we have $\zeta_t = \left( 1 + a L \right) u_t$
If $|a| < 1$ then $\left( 1 + a L \right)^{-1}$ exists and equating the two representations of $\zeta_t$ we can write:
\begin{align*}
u_t &= \left( 1 + aL\right)^{-1} \left( \epsilon_t + \eta_t + b \eta_{t-1} \right)\\
&= \left(\sum_{j=0}^\infty(-aL)^j \right)\left( \epsilon_t + \eta_t + b \eta_{t-1}\right)\\
&= \left[ \epsilon_t + \eta_t + b \eta_{t-1}\right] -a\left[\epsilon_{t-1} + \eta_{t-1} + b \eta_{t-2} \right] + a^2\left[ \epsilon_{t-2} + \eta_{t-2} +b \eta_{t-3} \right] + \ldots
\end{align*}
Furthermore, you can show that $a$ is the solution to the quadratic equation in $a$:
$$ \frac{1}{a} + a = \frac{1}{b} \left( \frac{\sigma^2_{\epsilon}}{\sigma^2_\eta} + 1 \right) + b$$
One way to obtain the above equation is by equating the autocovariance function based upon the two representations of process $\{\zeta_t\}$ and solving for the root where $|a|< 1$.
This might be a bit imprecise, and there might be additional regularity conditions. (Also above I used that for $|a| < 1$ we have $(1 + a)^{-1} = 1 - a + a^2 - a^3 + a^4 - a^5 \ldots $.)
Some takeaways:
Process $\zeta_t = \epsilon_t + \eta_t - \phi \eta_{t-1}$ can be written as an MA(1) process $\zeta_t = u_t + a u_{t-1}$ where $a$ and $u_t$ are related to $\phi$, $\epsilon_t$, and $\eta_t$ by the above formulas.
The same time-series can be written in multiple ways. (Note the Wold representation is unique though.)
- For example, an AR(1) can be written as an MA($\infty$).
There may be a somewhat nuanced relation between structural shocks (eg. $\epsilon_t$ and $\eta_t$ here) and shocks that you might recover in reduced form estimation (in this case, $u_t$).
MATLAB simulation to illustrate principle:
T = 1000000;
s2e = 2;
s2eta = 5;
e = sqrt(s2e) * randn(T, 1);
eta = sqrt(s2eta) * randn(T, 1);
phi = -.4;
zeta = e + eta + phi * lagmatrix(eta, 1);
m = estimate(arima(0,0,1), zeta)
a_est = m.MA{1};
disp('both sides of the below vectors should be roughly equivalent')
[1 / a_est + a_est, (1/phi) * ( 1 + s2e / s2eta) + phi]
[m.Variance, (phi / a_est) * s2eta]
beta = 1/phi * ( s2e / s2eta + 1) + phi;
a = (beta + sqrt(beta^2 - 4)) / 2; %solve quadratic formula
u = zeros(T, 1);
for i=20:T
for j=0:18
u(i) = u(i) + (-a)^j * (zeta(i-j));
end
end
zeta2 = u + a * lagmatrix(u,1);
The two representations zeta and zeta2 are equivalent. And estimating an MA(1) model using zeta recovers a_est which is the same coefficient a as I calculate.
