Deriving the Bayes Filter Correction Equation The correction rule for Bayes filters is: $$p\left(x_{k}|D_{k}\right)=\dfrac{p\left(y_{k}|x_{k}\right)\cdot p\left(x_{k}|D_{k-1}\right)}{p\left(y_{k}|D_{k-1}\right)}
 $$
For: 


*

*State at time $k$ is $x_{k}$

*Observation at time $k$ is $y_{k}$,

*Past Observations at time $k$ and earlier are $D_{k}=(y_{k},y_{k-1},...)$


I am trying to derive it. In all derivations/explinations of Bayes filters i have found,
it just says for this step "Simply apply Bayes Rule.".
To my knowledge, bayes rule is: $p\left(x_{k}|y_{k}\right)=p\left(y_{k}\right)p\left(x_{k}\cap y_{k}\right)=p\left(x_{k}\right)p\left(x_{k}\cap y_{k}\right)
 $
So I apply it on LHS:
$$p\left(x_{k}|D_{k}\right)=\dfrac{p\left(x_{k}\cap D_{k}\right)}{p\left(D_{k}\right)}$$
Then I take a look at the right hand side:
$$RHS=\dfrac{p\left(y_{k}|x_{k}\right)\cdot p\left(x_{k}|D_{k-1}\right)}{p\left(y_{k}|D_{k-1}\right)}$$
$RHS=\dfrac{\dfrac{p\left(y_{k}\cap x_{k}\right)}{p\left(x_{k}\right)}\cdot\dfrac{p\left(x_{k}\cap D_{k-1}\right)}{p\left(D_{k-1}\right)}}{\dfrac{p\left(y_{k}\cap D_{k-1}\right)}{p\left(D_{k-1}\right)}}$
  Expand Conditionals
$RHS=\dfrac{p\left(y_{k}\cap x_{k}\right)\cdot p\left(x_{k}\cap D_{k-1}\right)}{p\left(x_{k}\right)\cdot p\left(y_{k}\cap D_{k-1}\right)}$
  Cancel terms
$RHS=\dfrac{p\left(y_{k}\cap x_{k}\right)\cdot p\left(x_{k}\cap D_{k-1}\right)}{p\left(x_{k}\right)\cdot p\left(D_{k}\right)}$
 substitute definition of
$p\left(D_{k}\right)=p\left(y_{k}\cap D_{k-1}\right)$
But then I get stuck. I don't see any where to go from here.
 A: The formula$$p\left(x_{k}|D_{k}\right)=\dfrac{p\left(y_{k}|x_{k}\right)\cdot p\left(x_{k}|D_{k-1}\right)}{p\left(y_{k}|D_{k-1}\right)}$$ is Bayes' rule (or Bayes' Theorem) conditional on $D_{k-1}$. Not what you wrote as your "knowledge of bayes rule", i.e., $p\left(x_{k}|y_{k}\right)=p\left(y_{k}\right)p\left(x_{k}\cap y_{k}\right)=p\left(x_{k}\right)p\left(x_{k}\cap y_{k}\right)$ which does not hold, the lhs is the joint distribution of $x_k$ and $y_k$. The proper way to define the conditional pdf $p\left(x_{k}|y_{k}\right)$ is
$$p\left(x_{k}|y_{k}\right) = \dfrac{p\left(x_{k},y_{k}\right)}{p\left(y_{k}\right)}$$ 
And the generic Bayes rule is the inversion of the above
$$p\left(x_{k}|y_{k}\right) = \dfrac{p\left(y_k|x_{k}\right)p\left(x_{k}\right)}{p\left(y_{k}\right)}$$
(Avoid using the symbol '$\cap$' in joint densities as it only applies for discrete variables.)
The key is in the above filter formula is that it relies on the assumption that $y_k$ is independent from $D_{k-1}$ given $x_k$. 
\begin{align*}
p(x_k|D_k) &= \frac{p(x_k,D_k)}{p(D_k)}\\
&=\frac{p(x_k,y_k,D_{k-1})}{p(D_k)}\\
&=\frac{p(y_k|x_k,D_{k-1})p(x_k,D_{k-1})}{p(D_k)}\qquad\text{conditional+marginal}\\
&=\frac{p(y_k|x_k)p(x_k,D_{k-1})}{p(D_k)}\ \qquad\text{independence assumption}\\
&=\frac{p(y_k|x_k)p(x_k|D_{k-1})p(D_{k-1})}{p(y_k,D_{k-1})}\\
&=\frac{p(y_k|x_k)p(x_k|D_{k-1})p(D_{k-1})}{p(y_k|D_{k-1})p(D_{k-1})}\\
&=\dfrac{p\left(y_{k}|x_{k}\right)p\left(x_{k}|D_{k-1}\right)}{p\left(y_{k}|D_{k-1}\right)}
\end{align*}
which gives the result. Once again, the independence assumption is crucial in this representation. Otherwise, $p(y_k|x_k)$ would have to be replaced with $p(y_k|x_k,D_{k-1})$.
