Alternative construction of ARMA(1,1) process My question is related to the exercise 2.9, p. 79 in Brockwell & Davis, An Introduction to Time Series Analysis and Forecasting, 2nd edition, New-York, Springer, 2002 (It is also related to exercise 3.5, same reference).
Let {$Y_t$} be a process defined by
$$ Y_t = X_t + W_t,$$ where $\{W_t\}\sim \mbox{WN}(0, \sigma_w^2),$ and {$X_t$} is the following AR(1) process
$$ X_t - \phi X_{t-1}= Z_t,\quad \{Z_t\}\sim \mbox{WN}(0, \sigma_z^2),$$ and $E(W_s Z_t)=0$ for all $s$ and $t$.
The goal of this exercise is to show that $Y_t$ is in fact an ARMA(1,1) process. 
We define the process $\{U_t\}$ as 
$$U_t= Y_t - \phi Y_{t-1}$$
1) We compute the autocovariance function of $U_t$ at lag $h$ and we get 
$$\gamma_U(h) = 
\left\{
\begin{array}{ll}
\displaystyle \sigma^2_z + \sigma_w^2 (1+\phi^2) , & \text{ if }  h=0, \\
\displaystyle -\phi\ \sigma^2_w  ,& \text{ if } |h|=1, \\
\displaystyle 0, &  \text{ if } |h|>1.
\end{array}
\right.
$$
$\{U_t\}$ is 1-correlated and hence is a MA(1) process (by Proposition 2.1.1, B & D).
2) Thus, there exists a white noise sequence $\{\varepsilon_t\}$ with variance $\sigma_\varepsilon^2$ such that:
$$Y_t - \phi Y_{t-1} =  U_t = \varepsilon_t + \lambda \varepsilon_{t-1}.
$$
Then we want to express the parameters characterizing the MA(1) process $\{U_t\}$, namely $\lambda$ and $\sigma_\varepsilon^2$, in terms of the parameters characterizing $\{Y_t\}$ and $\{X_t\}$, namely, $\phi$, $\sigma_w^2$ and $\sigma^2_z$.
By equalizing the autocovariance function of the two representations, we obtain the following system:
$$
\left\{
\begin{array}{rcl}
\displaystyle  \sigma^2_\varepsilon (1+\lambda^2) &= & \sigma^2_z + \sigma_w^2 (1+\phi^2),  \\
\displaystyle  \lambda \sigma_\varepsilon^2 & = & -\phi\ \sigma^2_w.  \\
\end{array}
\right.
$$
If $\phi = 0$, we get $\lambda = 0 $ and the process $\{Y_t\}$ is a white noise with variance $\sigma_\varepsilon^2 = \sigma_z^2 + \sigma_w^2$. We now assume that $\phi \neq 0$ and $\lambda \neq 0$. Dividing the two equations of the system, we get:
$$ \frac{1+\lambda^2}{\lambda} = \frac{1}{-\phi} \frac{\sigma^2_z}{\sigma^2_w} -\frac{1+\phi^2}{\phi} \Leftrightarrow \frac{1+\lambda^2}{\lambda} = -\frac{k^2 + \phi^2 +1 }{\phi} . $$
where $k^2 = \frac{\sigma^2_z}{\sigma^2_w}$. We then get the following second order equation for $\lambda$:
$$\phi \lambda^2 + (k^2 + \phi^2 +1)\lambda + \phi.  $$
The latter equations admits two real (and positive) solutions, if I am not wrong.
Question: is there any issue with the non-identifiability of the MA(1) process defined by
$ \varepsilon_t + \lambda \varepsilon_{t-1}$? In other words, is that correct that I have, for the same process $\{Y_t\}$, two solutions for representing it in this way?
 A: Rather than working through the auto-covariance function, it is simpler to perform the analysis as an algebraic exercise working with the initial recursive equations for the two levels of the model.  Taking time back by one unit in the upper process gives the equation:
$$Y_{t-1} = X_{t-1} + W_{t-1}.$$
Multiplying both sides by $\phi$ and substituting the recursions from the lower and upper processses then gives:
$$\begin{aligned}
\phi Y_{t-1} 
&= \phi X_{t-1} + \phi W_{t-1} \\[6pt]
&= X_{t} - Z_t + \phi W_{t-1} \\[6pt]
&= Y_{t} - W_{t} - Z_t + \phi W_{t-1}. \\[6pt]
\end{aligned}$$
Re-arranging this equation gives the recursive form:
$$Y_{t} = \phi Y_{t-1} + W_{t} - \phi W_{t-1} + Z_{t}.$$
Now, to put this in the desired form, let $\varepsilon_t \equiv W_{t} + Z_{t} \sim \text{N}(0, \sigma^2)$ denote the new error term for the model, with variance $\sigma_\varepsilon^2 \equiv \sigma_w^2+\sigma_z^2$.  Equivalence to the ARMA(1,1) model is obtained by taking the MA coefficient:
$$\lambda = - \phi \cdot \sqrt{\frac{\sigma_w^2}{\sigma_w^2+\sigma_z^2}},$$
which gives the ARMA(1,1) model:
$$Y_{t} = \phi Y_{t-1} + \varepsilon_t + \lambda \varepsilon_{t-1}.$$
As to your main question, the answer is no, there is only one solution for the MA coefficient.  This solution is given directly by the requirement that $\lambda \sigma_\varepsilon^2 = - \phi \sigma_w^2$, which is one of the equations you have stated in your working.  You have divided the two equations and you are working only with this single equation, which is why you are getting two solutions instead of one.  Applying the second equation directly yields the unique solution.
