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


  • 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.


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 $p(y_k|x_k,D_{k-1})$.

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    $\begingroup$ I was thinking of $y_x$, and $x_k$ as being discrete. But now I realise they are not. Not at all. $\endgroup$ – Lyndon White Jan 2 '15 at 12:56
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    $\begingroup$ Ok, now that I have found time to check thought it, it clears things up. I think it would be abit clearer if you finished the proof with the final line being the result being proved. It would also have helped me it you included a definition that $p(a|b)=p(a,b)/p(b)$ But it is a pretty solid answer alround, thanks. $\endgroup$ – Lyndon White Jan 5 '15 at 0:41

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