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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?

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Here is my answer, feel free to comment and or correct. By assuming the invertibility of the process, I can discard one of the two solutions of the equation in $\lambda$.

We remind the second order equation for $\lambda$: $$\phi \lambda^2 + (k^2 + \phi^2 +1)\lambda + \phi. $$ The discriminant of this equation is simply given by: $$\Delta = (k^2 + \phi^2 +1)^2 - 4\phi^2 = (k^2 + \phi^2 +1 - 2\phi) (k^2 + \phi^2 +1 + 2\phi)$$ $$ 1+k^2 - 4(1+k^2) = -3(1+k^2)$$ % $$(1 + k^2 - 2 \phi + \phi^2) (1 + k^2 + 2 \phi + \phi^2).$$ Both factors of the discriminant are strictly positive. Indeed, the polynomial $$ \phi^2 - 2\phi + k^2 +1 $$ does not admit any root and is hence strictly positive since its own discriminant is given by $$ 4 - 4(1+k^2) = -4k^2 <0.$$ Hence, $\lambda$ finally admits two solutions: \begin{align*} \lambda_1 & = - \frac{1 + k^2 + \phi^2 - \sqrt{(k^2 + \phi^2 +1)^2 - 4\phi^2}}{2 \phi}, \\ \lambda_2 & = - \frac{1 + k^2 + \phi^2 + \sqrt{(1 + k^2 + \phi^2)^2-4 \phi^2 }}{2 \phi}. \end{align*} Note that we have $$(1 + k^2 + \phi^2)^2 > (1 + k^2 + \phi^2)^2 - 4\phi^2 \Leftrightarrow 1 + k^2 + \phi^2 > \sqrt{(1 + k^2 + \phi^2)^2 - 4\phi^2}, $$ and thus both solutions $\lambda_1$ and $\lambda_2$ are of opposite sign of $\phi$ since the numerator of the expression of $\lambda_1$ and $\lambda_2$ is positive, which is compliant with the fact that the variance $\sigma_\varepsilon$, given by, $\sigma_\varepsilon^2 = \frac{\phi\ \sigma_w^2}{\lambda}$, must be positive. We know that the process will be invertible if and only if $|\lambda|< 1$. We will show that $|\lambda_1| \leq1$ and $|\lambda_2|\geq 1$.

Consider now a right-angled triangle with hypotenuse $1+k^2+ \phi^2$, and a cathetus of length $2|\phi|$. Then, by the Pythagorean theorem, the last cathetus has length $\sqrt{(1 + k^2 + \phi^2)^2 - 4\phi^2}$. By the triangle inequality, we have: \begin{eqnarray*} & & 2|\phi| + \sqrt{(1 + k^2 + \phi^2)^2 - 4\phi^2} \geq 1 + k^2 + \phi^2\\ & \Leftrightarrow & 1+ k^2 + \phi^2 - \sqrt{(1 + k^2 + \phi^2)^2 - 4\phi^2} \leq 2|\phi|\\ &\Leftrightarrow &\frac{1 + k^2 + \phi^2 - \sqrt{(1 + k^2 + \phi^2)^2 - 4\phi^2}}{2|\phi|}\leq 1\\ &\Leftrightarrow & |\lambda_1| = \frac{1 + k^2 + \phi^2 - \sqrt{(1 + k^2 + \phi^2)^2 - 4\phi^2}}{2|\phi|}\leq 1. \end{eqnarray*} On the other hand, note that $$ \sqrt{(1 + k^2 + \phi^2)^2 - 4\phi^2} \geq 0 \Rightarrow \sqrt{(1 + k^2 + \phi^2)^2 - 4\phi^2} \geq -[(1-|\phi|)^2 + k^2]. $$ Then, by some simple computations, we get: \begin{eqnarray*} & & \sqrt{(1 + k^2 + \phi^2)^2 - 4\phi^2} \geq -[(1-|\phi|)^2 + k^2]\\ & \Leftrightarrow & \sqrt{(1 + k^2 + \phi^2)^2 - 4\phi^2} \geq -[1-2|\phi| + \phi^2 + k^2]\\ & \Leftrightarrow & 1 + k^2 + \phi^2 + \sqrt{(1 + k^2 + \phi^2)^2 - 4\phi^2} \geq 2|\phi| \\ &\Leftrightarrow & |\lambda_2| = \frac{1 + k^2 + \phi^2 + \sqrt{(1 + k^2 + \phi^2)^2 - 4\phi^2}}{2|\phi|}\geq 1. \end{eqnarray*} Hence, $\lambda_2$ does not yield a invertible solution to the process defined above and we retain $\lambda_1 := \lambda$ as the unique invertible solution.

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