i am currently reading the Probabilistic robotics book where the filters are discussed.
Such filters as kalman filter or particle filters. 
Now I can understand one thing while reading about the Kalman filter.
First I want to say that I could successfully understand about Bayes filtering. 
I've read some of theory of random processes and I can understand it.
Let me give you some details about the problem, if you think it is not sufficient I will give more. Who knows about that book I write here the page: it is about Kalman filter at page 41.
Let me not to explaing the whole problem, I hope the reader could be able to understand it as it is related with Kalman filtering. However I will write more if it is needed.
1. The state transition probability $p(x\_t|u\_t, x\_{t-1})$ must be a linear function in its arguments with added Gaussian noise. This is expressed by the following equation:
$$x\_t = A\_t x\_{t-1} + B\_t u\_t + \epsilon\_t (1)$$
$x\_t$ and $x\_{t-1}$ are state vectors, and $u\_t$ is a control vector at time $t$.
$\epsilon\_t$ is a gaussian noise.


Also it is given the definition of multivariant normal distribution:
$$p(x) = det\left(2\pi\Sigma\right)^{-1/2}exp\left(-1/2(x-\mu)^T\Sigma^{-1}(x-\mu)\right) (2)$$
Where the $\mu$ and $\Sigma$ are the mean and covariance.
The $x\_t$ and $u\_t$ are of the form:
$$x\_t = \left(x\_{1,t}, x\_{2,t}, ..., x\_{n,t}\right)^T$$
$$u\_t = \left(u\_{1,t}, u\_{2,t}, ..., u\_{m,t}\right)^T$$
So it is said that in order to obatain the state transition probability $p(x\_t|u\_t, x\_{t-1})$ you need to plugg the equation $(1)$ to the multivariant normal distribution (2).


Ok, I got that, but it is also written there that (here is the problem):
The mean of the posterior state is given by: $A\_tx\_{t-1}+B\_tu\_t$ and the covariance by $R\_t$ 
What I can't understand is that how does the mean in the multivariant normal distribution equation $(2)$ is calculated to $A\_tx\_{t-1}+B\_tu\_t$?
The whole formula affter plugging $(1)$ to $(2)$ become:
$$p(x\_t|u\_t, x\_{t-1}) = det\left(2\pi R\_t\right)^{-1/2}exp\left(-1/2(x\_t-A\_tx\_{t-1}-B\_tu\_t)^TR\_t^{-1}(x\_t-A\_tx\_{t-1}-B\_tu\_t)\right) $$


I think that the mean of $x\_t$ should be calculated like this:
$$E{x\_t} = E(A\_tx\_{t-1}+B\_tu\_t+\epsilon\_t)$$
$$E{x\_t} = A\_tE(x\_{t-1})+E(B\_tu\_t)$$
And I can't understand anyway how does the 
$Ex\_t$ is equal to $A\_tx\_{t-1}+B\_tu\_t$
If you see any mistakes in my reasoning please tell me. If you need more details please comment this question for that. Thank you very much! Hope you help!


EDIT 1:


From book it is said that Kalman Filter (KF) is an implementation of Bayes Filter(BF).
I understood BF. Actually it calculates belief $bel(x\_t)$ at time $t$ from belief at time $t-1$. 
BF algorithm:



```
For all $x_t$:

```

$\hat{bel}(x\_t) = \int{p(x\_t|u\_t, x\_{t-1})bel(x\_{t-1})dx\_{t-1}}$


$bel(x\_t) = \etap(z\_t|x\_t)\hat{bel}(x\_t)$



```
end for
return $bel(x_t)$

```

So about KF:
$$\hat{bel}(x\_t) = \int{p(x\_t|x\_{t-1}, u\_t)bel(x\_{t-1})dx\_{t-1}}$$
where $bel(x\_{t-1})$ is represented by mean $\mu\_{t-1}$ and covariance $\Sigma\_{t-1}$
The state transition probability $p(x\_t|x\_{t-1}, u\_t)$ is given as a normal distribution over $x\_t$ with mean $A\_tx\_{t-1}+B\_tu\_t$ and covariance $R\_t$.


If $x\_t$ can't be observed directly, so then what is $E(x\_t)$?



kalman filter: understanding the mathematical part
Do you really mean $Ex\_t$? That's the unconditional mean without looking at any of the data, which is a constant. Normally, the Kalman filter tells you how to compute the conditional mean based on the data you have at a particular moment in time.


I'm not familiar with the book, but I assume that you mean what the [Wikipedia page](http://en.wikipedia.org/wiki/Kalman_filter) calls the predicted state estimate. 


You don't say anything about measurement error. Do you observe $x\_t$ exactly? By assumption, since you choose $u\_t$ you know it at time $t-1$. Then the formula is just telling you that
$$
E(x\_t | u\_t, x\_{t-1}) = A E(x\_{t-1} | u\_t, x\_{t-1}) + B E(u\_t | u\_t, x\_{t-1}) + E(\epsilon\_t | u\_t, x\_{t-1}).
$$
Since $E(x\_{t-1} | u\_t, x\_{t-1}) = x\_{t-1}$ and $E(u\_t | u\_t, x\_{t-1}) = u\_t$, and the noise is independent,
then 
$$
E(x\_t | u\_t, x\_{t-1}) = A x\_{t-1} + B u\_t.
$$


If you don't observe $x\_t$ exactly, then you observe it with some error, given by
$$
z\_t = H\_t x\_t + \nu\_t,
$$
where $z\_t$ is your observation at time $t$ and $\nu\_t$ is again Gaussian white noise, independent of everything else.


Now, at time $t-1$ you only know $z\_1, \ldots, z\_{t-1}$, so the best you can do is
$$
E(x\_t | u\_t, z\_1, \ldots, z\_{t-1}),
$$
which I'll write as 
$$
E\_{t-1} (x\_t),
$$
since it's the conditional mean given all information at time $t-1$ (including $u\_t$, which you choose at time $t-1$.


Using the definition, 
$$
E\_{t-1}(x\_t) = A E\_{t-1}(x\_{t-1}) + B E\_{t-1}(u\_t) + E\_{t-1}(\epsilon\_t),
$$
but the most this simplifies is to
$$
E\_{t-1}(x\_t) = A E\_{t-1}(x\_{t-1}) + B u\_t.
$$
With measurement error, the Kalman filter by design just works by calculating the two conditional means $E\_{t-1}(x\_t)$ and $E\_{t-1}(x\_{t-1})$, and no other conditional or unconditional means.


