I am trying to prove a version of Bayes formula which is used in Beyond the Kalman Filter: Particle Filters for Tracking Applications, by Branko Ristic and Sanjeev Arulampalam, page 45-47. http://books.google.com provides free access to these pages. The formula is: $$p(\textbf x_k|\textbf x_{k-1},\textbf z_k)=\frac{p(\textbf z_k|\textbf x_k)p(\textbf x_k|\textbf x_{k-1})}{p(\textbf z_k|\textbf x_{k-1})}\tag{3.33}$$ which can be rearranged as:$$p(\textbf z_k|\textbf x_k)p(\textbf x_k|\textbf x_{k-1})=p(\textbf x_k|\textbf x_{k-1},\textbf z_k)p(\textbf z_k|\textbf x_{k-1})\tag{3.34}$$ $(3.34)$ is equivalent to $$\frac{p(\textbf x_k,\textbf z_k)}{p(\textbf x_k)}\frac{p(\textbf x_k,\textbf x_{k-1})}{p(\textbf x_{k-1})}=\frac{p(\textbf x_k,\textbf x_{k-1},\textbf z_k)}{p(\textbf x_{k-1},\textbf z_k)}\frac{p(\textbf x_{k-1},\textbf z_k)}{p(\textbf x_{k-1})}$$ which can be simplified to $$p(\textbf x_k,\textbf z_k)p(\textbf x_k,\textbf x_{k-1})=p(\textbf x_k,\textbf x_{k-1},\textbf z_k)p(\textbf x_k)$$ This is correct if these three random vectors are independent. But according to the context, they are not. Their relationship is $$\textbf x_k=f(\textbf x_{k-1})+\textbf v_{k-1}, \textbf z_k=\textbf H_k\textbf x_k+\textbf w_k$$ where $\textbf v_k$ and $\textbf w_k$ are independent $0$ mean normal vectors.
I deem the Bayes formula should be $$p(\textbf x_k|\textbf x_{k-1},\textbf z_k)=\frac{p(\textbf z_k|\textbf x_k,\textbf x_{k-1})p(\textbf x_k|\textbf x_{k-1})}{p(\textbf z_k|\textbf x_{k-1})}$$ I guess I may have failed to see through some consequences of the assumptions.
This is equivalent to the identity $$\mathrm{Law}(Hx+w\mid x,u)=\mathrm{Law}(Hx+w\mid x),$$ where $(u,v,w)$ are independent and $x=f(u)+v$. More generally, $$\mathbb E(h(x,w)\mid x,u)=\mathbb E(h(x,w)\mid x),$$ for every $h$ measurable, where $(u,v,w)$ are independent and $x=f(u,v)$. To prove the latter, note that $w$ is independent of $(x,u)$ hence, for every function $k$, $$\mathbb E(k(x,u,w)\mid x,u)=m(x,u),\qquad m(\cdot,\cdot)=\mathbb E(k(\cdot,\cdot,w)),$$ hence $$\mathbb E(h(x,w)\mid x,u)=\ell(x),\qquad\ell(\cdot)=\mathbb E(h(\cdot,w)).$$ In particular, $$\mathbb E(h(x,w)\mid x)=\mathbb E(\mathbb E(h(x,w)\mid x,u)\mid x)=\mathbb E(\ell(x)\mid x)=\ell(x)=\mathbb E(h(x,w)\mid x,u).$$