Exact Particle Filtering Derivation I am very confused with one of the derivations used by this paper http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.409.2389&rep=rep1&type=pdf in pg 1021 of this paper. It is a simple use of the applications of bayes theorem. Note $y^t=y_{1:t} $ and $x^t=x_{1:t}$. It uses this fact that $p(y_{t+1},x_{t+1}|x_t)=p(x_{t+1}|y_{t+1},x_t)p(y_{t+1}|x_t)$
$$p(x_{t+1}|y^{t+1}) \propto \int p(y_{t+1}|x_t)p(x_{t+1}|x_t, y_{t+1})p(x_t|y^t) dx_t$$
I am guessing that this uses a lot of hidden Markov assumptions to simplify the expression of the densities i.e. $x_{t+1}$ is only dependent on $x_{t}$ and $y_{t}$ is only dependent on $x_{t}$, but it does not seem to fall out in my case. What I tried is?
\begin{equation}
\begin{split}
p(x_{t+1}|y^{t+1}) & \propto \int p(x_{t+1},x_t|y^{t+1}) dx_t \\
& \propto \int p(x_{t+1}|x_t,y^{y+1})p(x_t|y^{t+1}) dx_t ~~(by~conditional)\\
& \propto \int p(y_{t+1}|x_{t+1},x_{t},y^{t})p(x_{t+1}|x_t,y^t)p(x_t|y^{t+1})dx_t~~(by~bayes)
\end{split}
\end{equation}
The last line can possibly simplify to $\propto \int p(y_{t+1}|x_{t+1})p(x_{t+1}|x_t)p(x_t|y^{t+1})dx_t$ by conditional dependence of hidden Markov model as I understand it, but as you can see the answer is different.So my question is What have I done wrong in the last step? Notice I did not make use of the fact they gave me. 
 A: I dislike the notation $y^{t+1}$ because it can easily be confused with an exponent. Thus, I  will use $y_{1:t+1}$. I also think in this case, this notation may have confused you. 
$$ 
\begin{array}{rll}
p(x_{t+1}|y_{1:t+1}) &
= \int p(x_{t+1},x_t|y_{1:t+1}) dx_t  \\
&\propto \int p(y_{t+1},x_{t+1}|x_t,y_{1:t})p(x_t|y_{1:t}) dx_t & \mbox{(by Bayes)} \\
&= \int p(y_{t+1},x_{t+1}|x_t)p(x_t|y_{1:t}) dx_t & \mbox{(cond. ind.)} \\
&= \int p(y_{t+1}|x_t)p(x_{t+1}|y_{t+1},x_t)p(x_t|y_{1:t}) dx_t & \mbox{(by fact provided)} \\
\end{array} $$
But I'm not sure how this is helpful since you generally don't know $p(y_{t+1}|x_t)$ or $p(x_{t+1}|y_{t+1},x_t)$. Typically the following derivation is more helpful.
$$ 
\begin{array}{rll}
p(x_{t+1}|y_{1:t+1}) &
= \int p(x_{t+1},x_t|y_{1:t+1}) dx_t &  \\
&\propto \int p(y_{t+1}|x_{t+1},x_t,y_{1:t})p(x_{t+1},x_t|y_{1:t}) dx_t & \mbox{(by Bayes)} \\
&= \int p(y_{t+1}|x_{t+1})p(x_{t+1},x_t|y_{1:t}) dx_t & \mbox{(cond. ind.)} \\
&= \int p(y_{t+1}|x_{t+1})p(x_{t+1}|x_t,y_{1:t})p(x_t|y_{1:t}) dx_t  \\
&= \int p(y_{t+1}|x_{t+1})p(x_{t+1}|x_t)p(x_t|y_{1:t}) dx_t & \mbox{(cond. ind.)}
\end{array} $$
Because $p(y_{t+1}|x_{t+1})$ and $p(x_{t+1}|x_t)$ are the observation and evolution equation respectively.
