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Suppose the stochastic process ${X_t}$ satisfies the equation

$$X_t=\phi X_{t-1} + Z_t \tag{A}$$

where $\phi>1$ and $Z_t$ is a white noise. Then iterating forward we get that the only stationary solution of the equation above can be written as:

$$X_t=-\sum_{j=1}^{\infty}Z_{t+j} {\phi}^{-j}$$

Therefore, we can write

$$X_t-\phi^{-1} X_{t-1} = -\sum_{j=1}^{\infty}Z_{t+j} {\phi}^{-j} +\sum_{j=1}^{\infty}Z_{t-1+j} {\phi}^{-j-1}=\phi^{-2}Z_t$$

And thus, defining $\tilde{Z}_t=\phi^{-2}Z_t$, we have that $X_t$ also satisfies the following AR equation:

$$X_t=\phi^{-1}X_{t-1}+\tilde{Z}_t\tag{B}$$

My question is simple. Suppose I have some data on $X_t$ and I run an OLS regression with one lag and no intercept on it. What solution will I estimate, (A) or (B)? Why?

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    $\begingroup$ Sorry but (A) and (B) are not compatible. The first mistake is when "iterating forward", which does not yield the first series written in the post. $\endgroup$
    – Did
    Sep 27, 2014 at 15:27

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If you regress $X_t$ against $X_{t-1}$ you will get item (A). This is because the defining equation here is $p(X_t | X_{t-1}) = N(X_{t-1}, \sigma^2)$ where $\sigma^2$ is the variance of the process.

Note that the Gaussian AR(1) process is time reversible.

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  • $\begingroup$ But the variance of the error is smaller in (B). So I thought that I would get (B) using OLS. But now comes the question: if that is true, what guarantees that (B) is the AR representation of $X_t$ with minimal variance of the white noise term? $\endgroup$
    – Pcw.
    Jul 28, 2014 at 23:36
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    $\begingroup$ I don't think that your equation is correct. You have an infinite sum written above. But with regression you have only finite data. There is some sensitivity to the length of your series. In any event, you can show numerically that your (B) series is clearly not the same as your (A) series. $\endgroup$
    – alpha137
    Jul 29, 2014 at 4:46

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