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I would like to apply the Kalman filter in order to get a causal Hodrick-Prescott filter. The Hodrick-Prescott filter models a time series $(y_t)_{t=0}^T$ as $$ y_t = \tau_t + c_t $$ where $\tau_t$ is a trend component and $c_t$ is a cyclical component.

This reference defines a state space formulation of the form $$ y_t = \tau_t + c_t $$ as the measurement equation and $$ \tau_t = 2 \tau_{t−1} − \tau_{t−2} + \epsilon_t $$ for the unobservable trend.

I have three questions on this:

A) $c_t$ is assumed to be a random error here, right? Normally distributed with constant variance. This seems a difficult assumption to me.

B) What's the logic behind the equation for the trend?

C) Does anybody know a different state space formulation for this problem? Or a nother reference?

Thanks!

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    $\begingroup$ This material is covered quite well in the book Econometric Modelling with Time Series by Martin, Hurn, and Harris. Specifically, see Chapter 15 and pages 567-571. I may provide a full answer later. $\endgroup$
    – Graeme Walsh
    Feb 20, 2014 at 2:03

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A) Actually, your reference says "a stationary residual or cyclical component, $c_t$", while this calls $c_t$ a 'cyclical component'.

Maravall and del Rio (2001) say "a residual, $c_t$, to be called “cycle”."

B) This formulation follows from the form of $A$; the second difference of $\tau$ is the error term.

C) some other references you may find helpful:

http://www.reservebank.govt.nz/research/discusspapers/dp03_02.pdf

http://www.stat.itam.mx/seminarios/Resumens/p136.pdf (see p6 which is relevant to your B, but they appear to have a sign in the A-matrix wrong - c.f. your link, p13)

http://www3.istat.it/dati/pubbsci/contributi/Contributi/contr_2005/2005_07.pdf

http://www.terrapub.co.jp/journals/jjss/pdf/3801/38010041.pdf

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  • $\begingroup$ Thanks for your answer! Some remarks: ad A) No matter how we all it. $c_t$ in the classical formulation of HP an have any distribution as far as I know. In the Kalman filter setting it has to be normal. This is a strong assmption, isn't it? ad B) Now I see: this is the smoothness condition ... great! Thanks for C)! $\endgroup$
    – Richi W
    Jan 23, 2013 at 13:18

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