The Kalman Filter Derived: The Power of Gaussians

[@honzasvoboda7987]

Really nice video! I would suggest putting every video of this kind into a playlist so that it is easier for you to get views on other videos, since all of them are really good and they deserve to be viewed more. :)

[@jameshan8]

Thank you and thanks for the suggestion! I'll do that now :)

[@SolvinSigurdson]

Wow, really useful - thanks for putting this together! Clear derivation and great animations!

[@santoshprajapat7709]

Great work It's really helpfull

[@jik4107]

are The variables μt bar and Σt​ bar are the same as the mean mt​ and the covariance Ψt (where here they are considered predicted to get a predicted x)?

[@jameshan8]

Great question! Nono it is not the same in this case. Notice how we're taking the derivatives with respect to x_{t-1}. When we complete the square for x_{t-1} we get m_t and Psi_t. The point of finding m_t and Psi_t is to complete the square for x_{t-1}. Once we have this form, we can exploit that fact that the integral of e^quadratic is simply a constant (it is a constant because the integral of any probability distribution is 1. For a Gaussian, if we match the quadratic form in the exponential, everything in front is a constant that normalizes the distribution). Does that make sense?