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Kalman filter: the bayesian approach derivation some clarifications

1

$begingroup$

I'm reading the book Methods and algorithms for signal processing from Moon Stirling at page 592 there is a derivation of Kalman filter using the Bayesian approach. I have some issues in understanding the propagate step which is the following:
$ p(textbfx_k|textbfz_1:k-1) = inttextbfx_k-1)p(textbfx_k-1$.
The book says under some assumption:
$ x_k-1 | z_1:k-1 sim mathcalN(m_k-1), P_k-1)$ and
$x_k|x_k-1 sim mathcalN(F_k x_k-1, Q_k)$
And inserting these in the prediction step and performing the integration (which involves expanding and completing the square) we find that $x_k$ is a Gaussian with mean $F_k m_k-1$ and variance $F_k P_k F_k ^T + Q_k$.
I don't understand when I insert that two gaussian in prediction formula how is possible to get that mean and that variance, there we have the product of two gaussian distribution but with different random variables maybe that's a gaussian? but then we have an integral of the resulting distribution. Can someone clarify me this?
Thank you

gaussian-integral kalman-filter

share|cite|improve this question

asked Mar 18 at 11:09

Francesco FisherFrancesco Fisher

61

$endgroup$

add a comment | 

1

$begingroup$

I'm reading the book Methods and algorithms for signal processing from Moon Stirling at page 592 there is a derivation of Kalman filter using the Bayesian approach. I have some issues in understanding the propagate step which is the following:
$ p(textbfx_k|textbfz_1:k-1) = inttextbfx_k-1)p(textbfx_k-1$.
The book says under some assumption:
$ x_k-1 | z_1:k-1 sim mathcalN(m_k-1), P_k-1)$ and
$x_k|x_k-1 sim mathcalN(F_k x_k-1, Q_k)$
And inserting these in the prediction step and performing the integration (which involves expanding and completing the square) we find that $x_k$ is a Gaussian with mean $F_k m_k-1$ and variance $F_k P_k F_k ^T + Q_k$.
I don't understand when I insert that two gaussian in prediction formula how is possible to get that mean and that variance, there we have the product of two gaussian distribution but with different random variables maybe that's a gaussian? but then we have an integral of the resulting distribution. Can someone clarify me this?
Thank you

gaussian-integral kalman-filter

share|cite|improve this question

asked Mar 18 at 11:09

Francesco FisherFrancesco Fisher

61

$endgroup$

add a comment | 

$begingroup$

I'm reading the book Methods and algorithms for signal processing from Moon Stirling at page 592 there is a derivation of Kalman filter using the Bayesian approach. I have some issues in understanding the propagate step which is the following:
$ p(textbfx_k|textbfz_1:k-1) = inttextbfx_k-1)p(textbfx_k-1$.
The book says under some assumption:
$ x_k-1 | z_1:k-1 sim mathcalN(m_k-1), P_k-1)$ and
$x_k|x_k-1 sim mathcalN(F_k x_k-1, Q_k)$
And inserting these in the prediction step and performing the integration (which involves expanding and completing the square) we find that $x_k$ is a Gaussian with mean $F_k m_k-1$ and variance $F_k P_k F_k ^T + Q_k$.
I don't understand when I insert that two gaussian in prediction formula how is possible to get that mean and that variance, there we have the product of two gaussian distribution but with different random variables maybe that's a gaussian? but then we have an integral of the resulting distribution. Can someone clarify me this?
Thank you

gaussian-integral kalman-filter

share|cite|improve this question

asked Mar 18 at 11:09

Francesco FisherFrancesco Fisher

61

$endgroup$

share|cite|improve this question

asked Mar 18 at 11:09

Francesco FisherFrancesco Fisher

61

share|cite|improve this question

asked Mar 18 at 11:09

Francesco FisherFrancesco Fisher

61

asked Mar 18 at 11:09

Francesco FisherFrancesco Fisher

61

asked Mar 18 at 11:09

Francesco FisherFrancesco Fisher

61