# Variance of AR(1) plus noise and its “equivalent” ARMA(1,1)

Let us consider the following state-space model $$z_{t} = x_{t} + v_{t}\\ x_{t} = \phi x_{t-1} + w_{t}$$ where $$\phi< 1$$, the errors $$v_{t}\sim \mathcal{N}(0,V^{2})$$ and $$w_{t}\sim \mathcal{N}(0,W^{2})$$ are independent

The stationary variance of $$x_{t}$$ is given be $$Var(x_{t}) = \frac{W^2}{1-\phi^2}$$, therefore, the stationary variance of $$z_{t}$$ is given by $$Var(z_{t}) = V^{2} + \frac{W^2}{1-\phi^2}$$

The state space model above is equivalent to ARMA(1,1) process $$z_{t} = \phi z_{t-1} + \theta \varepsilon_{t-1} + \varepsilon_{t},$$ where $$\theta = - \phi \frac{V}{\sqrt{V^{2} + W^{2}}}$$ and $$\varepsilon_{t}\sim \mathcal{N}(0,V^{2} + W^{2})$$ are i.i.d. This is actually AR(1) plus noise, which is equivalent to ARMA(1,1). The prof can be found, for example, here http://www.stats.ox.ac.uk/~reinert/time/notesht10short.pdf

Next, let us consider "equivalent" ARMA(1,1) model. Its stationary variance is given by $$Var(z_{t}) = \frac{1+2\phi\theta +\theta^{2}}{1 - \phi^{2}}(V^2 + W^2),$$ where $$\theta = - \phi \frac{V}{\sqrt{V^{2} + W^{2}}}$$, see, for example, https://math.unice.fr/~frapetti/CorsoP/chapitre_23_IMEA_1.pdf

I can not see that the variance of equivalent model is equal to the state-space model.

This question is related to

where it was shown by simulations, that "equivalence" is not working and

Alternative construction of ARMA(1,1) process

where the equivalence is proved.

• "Any time series textbook" is not all that helpful. A concrete reference could be more helpful. – Richard Hardy Nov 13 '19 at 19:28
• Dear @Richard Hardy, I have added the reference. – ABK Nov 13 '19 at 21:47
• Thank you very much. – Richard Hardy Nov 14 '19 at 7:14