How is $P(x_k | Z^k ) = P(z_k | x_k) \left[ P(z_k | Z^{k-1}) \right]^{-1} P(x_k |Z^{k-1} )$ an application of Bayes theorem? We are in the econometric context of a dynamic system given by 
$$x_{k+1}=\phi_k x_k +w_k$$
$$ z_k = H_k x_k + v_k$$
where $\phi_k \in R$ is the state transition, and $H_k$ is the observation matrix (for simplicity let's assume this is $1 \times 1$ as the state transition), $v_k, w_k$ are zero mean independent sequences which are also mutually independent. We also assume that the initial state $x_0$ is independent from the future disturbances $v_k, w_k$.
Given the vector of observations $(z_1, \dots ,z_k)$ we want to estimate the state  $x_k.$
As our optimal estimator we take the minimal variance estimator which is given by the conditional variance $\hat{x}_{k} = E [x_k | Z^k ]$ where $Z^k := (z_1, z_2, \dots, z_k) $
We also denote $P(z_k| Z^{k-1})$ as the density of $z_k$ conditioned on previous observations $(z_1, \dots ,z_{k-1})$ and similarly $P(x_k| Z^{k-1})$ then by Bayes law we have the following:
$$P(x_k | Z^k ) = P(z_k | x_k) \left[ P(z_k | Z^{k-1}) \right]^{-1} P(x_k |Z^{k-1} )$$
I don't see how this is Bayes law, there are three conditional expectations on the RHS, bayes law only has one. Could someone clarify this for me?
 A: Because $P(z_k|x_k)$ is akin to the likelihood, and $P(x_k|Z^{k-1})$ is akin to the prior. Multiplying these together gives you the joint $P(z_k,x_k|Z^{k-1})$. And if you take this joint and integrate out $x_k$ to get a normalizing constant, you get $p(z_k|Z^{k-1})$. The state variable $x_k$ takes the place of what might typically for you be the parameter.
With your Linear-Gaussian state space model, because the "prior" and "likelihood" are conjugate, the "posterior" filtering distribution is also Gaussian. With a Kalman filter, one only needs to keep track of the means and variances that parametrize these filtering "posterior" distributions. That's what the Kalman filter is: recursive formulas for the means and variances of these filtering/posterior distributions.
A: $P(x_k|Z^k)=P(z_k|x_k,Z^{k-1})\frac{P(x_k,Z^{k-1})}{P(Z^k)}.$
Notice that $P(z_k|x_k,Z^{k-1})=P(z_k|x_k)$, since $z_k$ depends only on $x_k$ and $v_k$ and $v_k$ is independent of everything. Now just expand the rest:
$$P(x_k,Z^{k-1})=P(x_k|Z^{k-1})P(Z^{k-1})$$,
$$P(Z^{k})=P(z_k|Z^{k-1})P(Z^{k-1}),$$
and the result will follow.
