# Correct form for State Space Equation for Kalman Filter for DNS

In this paper:

http://www.ssc.upenn.edu/~fdiebold/papers/paper55/DRAfinal.pdf

in eqns 3,5 the state eqn has the mean removed.

$(z_t-\mu)=A(z_{t-1}-\mu) + \epsilon_t$

$y_t=C z_t + \delta_t$

However I have looked at several implementations of Kalman Filters for state space models I haven't seen this "de-meaned" version any where?

Moreover if you look at implementations such as this by Kevin Murphy he doesn't de-mean the state process either:

https://github.com/kmatzen/FullBNT-1.0.1/blob/master/Kalman/kalman_update.m

Thus Murphy (and most other authors) take the state space model to be i.e the state equation is NOT demeaned:

$z_t=Az_{t-1} + \epsilon_t$

$y_t=Cz_t + \delta_t$

However this model also includes an initial observation x0 which is distributed as $N(\xi, \Lambda)$. (Note that in the paper the choice over what x0 to use is not covered.)

Q1.) From a theoretical point of view are the two representations equivalent? Clearly they are if $\mu=0$ but more generally for non zero values of $\mu$?Does the mean of $x_0=\xi$ offset the process so that the rest of the observations cab be treated as mean zero? If so can someone show me how this makes the processes equivalent algebraically!

When I test the model I use data generated using the non demeaned version of the state equation, with the appropriate x_0, and use the toolbox to estimate it (which also does not demean the state equation) and I get good estimates. This is to be expected as $\mu$ in this case is zero by design. However when I work with empirical data I can't be sure that the empirical state process will be mean zero.

Q.2) Therefore should I always de-mean the state process when working with empirical data.

Kind Regards

Baz