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We consider the ARMA(1,1) model:

$$ y_t = \beta y_{t−1} + \varepsilon_t + \theta \varepsilon_{t−1}. $$

I would like to demonstrate that $y_{t-2}$ provides an instrumental variable estimator for $\beta$. How can I do that ?

Thank you!

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First of all, ARMA(1,1) process can be rewritten in this way:

$y_t$ = $\epsilon_t$ + ($\beta$+$\theta$)$\sum_{j=1}^{\infty}$$\beta^{j-1}$$\epsilon_{t-j}$

Therefore,
A = E[$y_{t-2}$$\epsilon_t$] = E[$\epsilon_{t-2}$$\epsilon_t$ + ($\beta$+$\theta$)$\sum_{j=3}^{\infty}$$\beta^{j-3}$$\epsilon_{t-j}$$\epsilon_t$] ---------- eq.(1)

Because $\epsilon_t$ is white noise, E[$\epsilon_{t-2}$$\epsilon_t$]=0, and similarly for other terms in eq.(1). Overall A=0.

Likewise, if you calculate
B = E[$y_{t-2}$$\epsilon_{t-1}$], you will get zero,

which means that $y_{t-2}$ is not correlated with error term $u_t$ = $\epsilon_t$ + $\theta$$\epsilon_{t-1}$. We also know that $y_{t-2}$ is correlated with $y_{t}$ through $y_{t-1}$. This all points to $y_{t-2}$ as a reliable instrument for $y_{t-1}$.

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  • $\begingroup$ nice answer. just two quick things: 1) In your second to last sentence, you meant not correlated with the error term $u_t$. 2) what I don't get about instrumental variables is the following: I understand that, by using $y_{t-2}$, you then meet the OLS assumption of uncorrelatedness with the error term so the regression is "legal". But doesn't the IV coefficient have totally different meaning since $y_{t-1} is gone ? Why is the IV helpful to the original model ? Thanks and apologies for the possibly stupid question. $\endgroup$
    – mlofton
    Commented May 1, 2023 at 6:25
  • $\begingroup$ Hi, thanks for the comment. Yup, I meant 'not correlated' and edited it accordingly. I am not sure what you mean by $y_{t-1}$ is gone. In this example, the estimate for $\alpha$ would be sum($y_{t-2}$$y_{t}$)/sum($y_{t-2}$$y_{t-1}$). There would still be $y_{t-1}$ in the formula for IV estimation. If I remember correctly, IV estimates generally tend to be larger than OLS estimates. But I have no idea if this continues to hold in time-series although I am inclined to say yes. $\endgroup$
    – yam3721
    Commented May 1, 2023 at 15:41
  • $\begingroup$ Thanks. I'm going to have review IV so don't worry about my question. Your answer to the OP was fine. I'm just vague on the IV specifics but I have a book that covers IV quite nicely so I'll dig it up. I'll also put the title of it in another comment once I dig up the book. $\endgroup$
    – mlofton
    Commented May 5, 2023 at 8:22

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