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Consider this exercise taken from Brockwell and Davis (1991):

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I'm a bit confused as to how that implies $Y_t$ is $ARMA(1,1)$. I've tried to show it, but I end up going back in circles, and I'm not sure how exactly you prove it's $ARMA(1,1)$.

I'm not sure if I did a) and b) wrong. I keep getting for b) that $Cov(U_t,U_{t+2})\neq 0$ But means it's not 1 correlated so I'm not sure how to fix that.

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  • $\begingroup$ could you please cite the source of your exercise? thanks! $\endgroup$
    – wonderer
    Mar 26, 2023 at 2:55
  • $\begingroup$ sorry about that, it's from Brockwell and Davis: home.iitj.ac.in/~parmod/document/… $\endgroup$
    – eddie
    Mar 26, 2023 at 3:12
  • $\begingroup$ Please type your question as text, do not just post a photograph or screenshot (see here). When you retype the question, add the self-study tag & read its wiki. $\endgroup$ Mar 26, 2023 at 5:15

2 Answers 2

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I'm confident that wonderer's use of the wold decomposition is correct but I wanted to provide an answer that doesn't use it, in case a reader is not familar with that theorem.

We have that 1) $Y_t = \phi Y_{t-1} + U_t$

We are also given that 2) $Y_t = X_t + W_t$.

First, we take 2) and use the lag operator $(1 - \phi L)$ on both sides. This results in:

$Y_{t} - \phi Y_{t-1} = X_{t} - \phi X_{t-1} + W_{t} - \phi W_{t-1}$.

But, if one calculates the autocorrelations of of $Y_{t} - \phi Y_{t-1}$, (i.e. the process above ) one finds that it is autocorrelated only at lag one and no other lag so this implies that, the process on the RHS is MA(1). So, introducing new variables called $\epsilon$ and $\theta$, we can rewrite the previous relation as :

$Y_{t} - \phi Y_{t-1} = U_{t} = \epsilon + \theta \epsilon_{t-1}$.

This relation is justified by the fact that we know that the RHS is an MA(1).

Finally moving the second term on the LHS to the RHS,

we obtain, 4) $Y_{t} = \phi Y_{t-1} + \epsilon + \theta \epsilon_{t-1}$

Notice the first term on the RHS of 4) is the AR(1) component and the last two terms on the RHS are the MA(1) component. So, it has been shown that the process $Y_{t} - \phi Y_{t-1}$ is ARMA(1,1).

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From part (b) you know that $Y_t-\phi Y_{t-1}$ is MA(1), thus you can immediately conclude that \begin{align*} Y_t = \phi Y_{t-1} + U_t, \end{align*} where $U_t = \psi_0\varepsilon_t + \psi_1\varepsilon_{t-1}$ for some $\psi_0$ and $\psi_1$ that you determined in part (b). Therefore, \begin{align*} Y_t = \phi Y_{t-1} +\psi_0\varepsilon_t + \psi_1\varepsilon_{t-1} \end{align*} for some non-zero $(\phi,\psi_0,\psi_1)$, i.e. $\{Y_t\}$ is $ARMA(1,1)$. Note that $\psi_0$ and $\psi_1$ are going to be functions of $(\phi,\sigma_w,\sigma_z)$.

PS: If you wonder why we can always write $U_t = \psi_0\varepsilon_t + \psi_1\varepsilon_{t-1}$ for some white noise process $\{\varepsilon_t\}$ and coefficients $\psi_0$ and $\psi_1$, it is because $U_t$ is stationary and we can apply the Wold's decomposition theorem.

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  • $\begingroup$ here's what i'm confused about: in part b), $U_t = Y_t-\phi Y_{t-1}$ So, $U_t$ is written in terms of $Y$, so I get $Y_t$ in terms of $Y$ terms again. $\endgroup$
    – eddie
    Mar 26, 2023 at 3:12
  • $\begingroup$ Part b is telling you that $U_t= Y_t-\phi Y_{t-1}$ is a moving average process. Since $U_t$ is an $MA(1)$ process, by wold's theorem we can also write $U_t=\psi_0\varepsilon_t+\psi_1\varepsilon_{t-1}$. The two expressions involving $U_t$, if equated, give you the ARMA(1,1). In part b is asking you to precisely determine the expression $U_t=\psi_0\varepsilon_t+\psi_1\varepsilon_{t-1}$ $\endgroup$
    – wonderer
    Mar 26, 2023 at 3:19
  • $\begingroup$ Oh, so Wold's Theorem will allow me to write it in terms of $\epsilon$s instead? $\endgroup$
    – eddie
    Mar 26, 2023 at 3:20
  • $\begingroup$ In general, yes. In this particular exercise you already know that $U_t$ is a linear combination of a white noise process. This is because it is a function of $W_t$ (white noise) and $X_t$ (which is a function of white noise) $\endgroup$
    – wonderer
    Mar 26, 2023 at 3:22
  • $\begingroup$ I think I messed up part b) also, could you please provide me with some guidance on it $\endgroup$
    – eddie
    Mar 26, 2023 at 15:49

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