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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 $x_t$:


$$\hat{bel}(x_t) \hat{bel}(x_t) = \int{p(x_t|u_t, x_{t-1})bel(x_{t-1})dx_{t-1}}$$ x_{t-1})bel(x_{t-1})dx_{t-1}}$$$bel(x_t) bel(x_t) = \etap(z_t|x_t)\hat{bel}(x_t)$$ 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 $E(x_t)$?

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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 $E(x_t)$?

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