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I'm trying to work out the details of the proof of the following statement:

Suppose $\xi_t = \rho \xi_{t-1} + \epsilon_t$ is an AR(1) process.

Using Kalman filter, one can prove that $\mathbb{E}_t\{\xi_t\}$ follows an AR(2) dynamics with:

$$\mathbb{E}_t\{\xi_t\} = \left(1 - \frac{\lambda}{\rho}\right) \left(\frac{1}{1-\lambda L} \right) \xi_t$$ where $\lambda = \rho(1-g)$ and $g \in (0,1)$ is the Kalman gain and $L$ is the lag operator.

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    $\begingroup$ Are you familiar with the Kalman filter ? If so, then you need to write out the Kalman filter recursions by writing the AR(1) in state space form ( see Kohn and Ansley for how to transform from an ARIMA to state space form. I forget year and title ). If you're not familiar with the Kalman filter, then you need to understand that first before dealing with the question. There are so many books and papers introducing the Kalman filter that I don't know what to recommend. $\endgroup$
    – mlofton
    Apr 17, 2022 at 3:26
  • $\begingroup$ Yes I'm familiar, I was hoping though for a detailed proof of the statement $\endgroup$
    – Luca Gi
    Apr 18, 2022 at 13:20
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    $\begingroup$ I would try it but I forget how to write the AR(1) in state space form. Once you do that, it shouldn't be too bad. Maybe someone else can pipe in here because, if I could do it, I would but I forget the first step which is key. Google for "writing the AR(1) in state space form" and see what comes up. I'm too busy riight now but if this is still unanswered when I get free, I'll try to take a closer look. $\endgroup$
    – mlofton
    Apr 19, 2022 at 16:42
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    $\begingroup$ Hi: page 7 of this has it for AR(p) so let p = 1 and go from there. www-stat.wharton.upenn.edu/~stine/stat910/lectures/…. Like I said, if I get time later, I'll give it a shot. It's been a looooooooong time !!!! $\endgroup$
    – mlofton
    Apr 19, 2022 at 16:44
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    $\begingroup$ I didn't get very far and I'm already confused. Does equation(15) make any sense to you ? It doesn't to me because the same thing in the expectation is on the right hand side also ? I don't know what the greek letter is for it but it's possibly eta ? $\endgroup$
    – mlofton
    Sep 29, 2022 at 4:57

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