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There is really no difference. You can frame the Kalman filter as a Bayesian posterior inference problem. For example, for a stationary linear Gaussian model,
by n3ur0n 8y ago
There is really no difference. You can frame the Kalman filter as a Bayesian posterior inference problem.
For example, for a stationary linear Gaussian model, you have a transition model of the form:
z_t = Az_{t-1} + Bu_t + e where e ~ Gaussian(0,Q)
and an observation model of the form:
x_t = Cz_{t} + Du_t + d, where, d ~ Gaussian (0,R)
Since, z_t and x_t are both multivariate gaussians in this model, you can compute the posterior distribution on z_t's, which will also be a Gaussian. That is basically the Kalman filter.
As the writeup mentions, you might choose a non-Gaussian noise model, in which case the posterior distribution is not a Gaussian and then you employ something like a unscented Kalman filter or extended Kalman filter.
- jvanderbot 8y agoIn the event of Gsussian noise, linear state transition and linear measurement, KF yields the exact posterior. Both unscented and extended refer to non-linear measurement or state transition equations. Neither are designed for non gaussian noise. Though the KF and it's variants are one of the simplest, well-performing estimation methods out there, so it wouldn't suprise me if it's used for everything, appropriate or not.