Writing process $\zeta_t =\epsilon_t + \eta_t - \phi\eta_{t-1}$ as a moving average, MA(1) Let  $X_t = \phi X_{t-1} + \epsilon_t$, $Y_t = X_t+\eta_t$
where $\{X_t\}$ is unobserved, $\{Y_t\}$ is observed and $\{\epsilon_t\}$ and $\{\eta_t\}$ are white noise sequences, then $X_t$ is an AR(1), we can write
$\zeta_t = Y_t - \phi Y_{t-1} = X_t+\eta_t - \phi (X_{t-1}+\eta_{t-1}) = \epsilon_t + \eta_t - \phi\eta_{t-1}$
It then says $\zeta_t$ is stationary and cov$(\zeta_t,\zeta_{t+k})=0$ for $k\geq 2$. Up to this point, I am following. It then says:
$$\zeta_t \text{ can be modelled as MA(1) process}$$
I do not follow this part the conclusion. Specifically, if $\epsilon_t$ is not present, that is a MA(1) but it appears to me that $\epsilon_t$ destroys this property. 
See the bottom of page 5 of http://www.statslab.cam.ac.uk/~rrw1/timeseries/t.pdf
(or page 9 of the pdf doc)
 A: 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.
A: You have a mistake, but you are correct in your last sentence. $\text{Cov}(\zeta_t,\zeta_{t+k})=0$ for $k>1$, not for $k \ge 0$. This quick cutoff is reminiscent of an MA model's autocovariance function, but you are right that they are not the same. 
In particular  
$$
\gamma(1) = \text{Cov}(\zeta_t,\zeta_{t+1})= -\phi \text{Var}(\eta_t)
$$
and
$$
\gamma(0) = \text{Var}(\zeta_t) = \text{Var}(\epsilon_t) + \text{Var}(\eta_t)(1+\phi^2) .
$$
If $\text{Var}(\epsilon_t) = 0$, then this autocovariance function would be the same as an MA(1).
Edit:
This is wrong. See above answer.
