Trying to understand linear state-space notation I am trying to understand the state-space notation in this paper. Specifically, it has the observation model as

and the state/dynamic model as

$y$ are the observations, $x$ and $z$ are possible covariates, and $L$ the autoregression lag.
My question is why does the mean, $\mu$, not enter the state-space equation? As it is specified, does this not indicate that the value of the state-space variable will generally be $\mu$ lower (if $\mu$ is positive). That is, if we are interested in making inference on the state variable $\theta$, it will generally be $\mu$ distance away.
 A: SSMs aren't always identifiable without restrictions on the parameter space. There is a common distinction between centered and uncentered parameterizations.
What you have written would be called the uncentered parameterization of this particular model. Generally this is when the observation equation is loaded up with more parameters. In this case, the latent state will have a marginal mean of $0$.
An equivalent centered parameterization would have the state equation as
$$
\theta_{t}   \mid \theta_{t-L:t-1} \sim \mathcal{N}\left(\mu + \sum_{l=1}^L\Phi_l\left(\theta_{t-l} - \mu\right) + \Delta z_t, \Sigma_{\eta}  \right)
$$
and an observation equation
$$
y_t \mid \theta_t \sim \mathcal{N}\left(\Psi\theta_t + \Gamma x_t, \Sigma_{\epsilon} \right).
$$
The marginal mean of $\theta_t$ is now $\mu$, if you write it like this.
A: The intuition, I think, is that it wouldn't be identifiable.
A sketch of the argument: Add an intercept, call it $\mu_\theta$, to the equation for the expected value for $\theta_t$ in (1). Setting the first col/row/element of $\Psi$ to 1 for convenience, plug in this expected value to get the expectation of $Y_t$ conditional on $\theta_t$, and collect terms. $\mu$ and $\mu_\theta$ enter additively into the conditional expectation formula, but we don't have any way to distinguish them, so we can only really estimate the level shift $(\mu + \mu_\theta)$. Consequently, we can just set one of them to zero. Since $Y$ is observable, it will probably make more sense - although it doesn't really matter - to keep $\mu$ and interpret it like an regression intercept. However, you could keep $\mu_\theta$ in (1) instead, I guess.
