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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

Estimation of ARMA from state-space generated data

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

Alternative construction of ARMA(1,1) process

where the equivalence is proved.

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    $\begingroup$ "Any time series textbook" is not all that helpful. A concrete reference could be more helpful. $\endgroup$ Nov 13, 2019 at 19:28
  • $\begingroup$ Dear @Richard Hardy, I have added the reference. $\endgroup$
    – ABK
    Nov 13, 2019 at 21:47
  • $\begingroup$ Thank you very much. $\endgroup$ Nov 14, 2019 at 7:14

1 Answer 1

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There is a mistake in the ARMA(1,1) model. Look at Alternative construction of ARMA(1,1) process where the correct ARMA(1,1) was derived.

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