Short version:

How to calculate a likelihood of Unscented Kalman filter?

Long version:

Likelihood for linear Kalman filter (KF) is:

$$\mathcal{L} = \frac{1}{\sqrt{2\pi S}}\exp \left[-\frac{1}{2}\mathbf y^\mathsf T\mathbf S^{-1}\mathbf y\right]$$

Other words likelihood of KF depends on residual and uncertainty:

$$\mathcal{L} (\mathbf y, \mathbf S)$$

We can see that uncertainty depends on process noise $\mathbf Q$ which is function of a time $\mathbf t$:

$$\mathbf{S} (\mathbf P) \Rightarrow \mathbf{P} (\mathbf Q) \Rightarrow\mathbf{Q} (\mathbf t)$$

This allows conclude that likelihood of KF depends on time:

$$\mathcal{L} (\mathbf y, \mathbf t)$$

For Unscented Kalman filter (UKF) instead of uncertainty $\mathbf S$ is used covariance of sigma-points in measurement space $\mathbf P_z$, where:

$$\mathbf P_z = \sum w^c {\left(\mathcal Z - \mu_z\right)\left(\mathcal Z - \mu_z\right)^\mathsf{T} + \mathbf R}$$

In this case the likelihood can be found as:

$$\mathcal{L} = \frac{1}{\sqrt{2\pi P_z}}\exp \left[-\frac{1}{2}\mathbf y^\mathsf T\mathbf P_z^{-1}\mathbf y\right]$$

But, neither $\mathcal Z$ no $\mathbf R$ depends on process noise $\mathbf Q$, so likelihood of UKF doesn't depend on time.

In practice it means the following: if I try to find likelihood of KF consequentially increasing time which passed after last measurement then likelihood will be reduce, but for UKF likelihood will not be changed.

Where is my mistake?

  • $\begingroup$ You mean Unscented Kalman filter, right? not unscensed $\endgroup$
    – Ben
    Feb 7, 2018 at 18:11

1 Answer 1


Your mistake is that $\mathcal Z$ does depend on process noise. It's just a bit more obscure than the linear filter. $\mathcal Z$ is the projection of the sigma points through the observation function. The sigma points are a $\mathcal f(P)$. Therefore $\mathcal Z$ is a $\mathbf f(P)$, and therefore related to $\mathbf Q$.


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