# Redundant sigma points in Unscented Kalman Filter?

According to the Unscented Transform equations in an Unscented Kalman Filter, sigma points are chosen via:

$$\chi^{[0]}=\mu$$

$$\chi^{[i]}=\mu+\left(\sqrt{(n+\lambda)\Sigma}\right)_i\;\;\;i=1,...,n$$

$$\chi^{[i]}=\mu-\left(\sqrt{(n+\lambda)\Sigma}\right)_{i-n}\;\;\;i=n+1,...,2n$$

where $$\mu$$ is a vector of variable means, $$\Sigma$$ is the covariance matrix of the variables, and $$n,\lambda$$ are tuning parameters.

In order to take the square root of $$\Sigma$$, the literature suggests performing a Cholesky LL' transformation and using the resulting L as the matrix "square root".

With L being a lower triangle matrix, the upper triangle is obviously all zeros. This ultimately populates the $$\chi$$ matrix (sigma points) with several redundant copies of the mean $$\mu$$.

These redundant sigma points would obviously be a poor choice, as:

• they wouldn't spread out to sample the nonlinear function well
• they waste repeated computation on the same values

A simple example:

Assume $$(n+\lambda)=1$$

Using Cholesky LL decomposition:

$$\Sigma=LL^*$$

$$\sqrt\Sigma=L=\begin{bmatrix}L_{0,0}&0&0\\L_{1,0}&L_{1,1}&0\\L_{2,0}&L_{2,1}&L_{2,2}\end{bmatrix}$$

Substituting this back in to the equations for $$\chi$$:

$$\chi= \begin{bmatrix} \mu_0&\mu_0+L_{0,0}&\mu_0&\mu_0&\mu_0-L_{0,0}&\mu_0&\mu_0\\ \mu_1&\mu_1+L_{1,0}&\mu_1+L_{1,1}&\mu_1&\mu_1-L_{1,0}&\mu_1-L_{1,1}&\mu_1\\ \mu_2&\mu_2+L_{2,0}&\mu_2+L_{2,1}&\mu_2+L_{2,2}&\mu_2-L_{2,0}&\mu_2-L_{2,1}&\mu_2-L_{2,2} \end{bmatrix}$$

As you can see, the top two variables (rows) have sigma points that are made up of redundant values of the variable mean.

Am I doing something wrong in my calculations, or does the use of Cholesky decomposition really result in a poor choice of sigma points?

• Looking at the equations and thinking about the "spirit" of taking the sqrt(covariance), it seems like the goal is to place sigma points at the mean and at +/- 1 standard deviation (with a scaling factor applied). I wonder if current literature using the cholesky decomp is taking the sqrt(covariance) too literally and missing the point? Commented Jul 22, 2020 at 4:43
• I had a colleague check over the math yesterday, and he was able to verify that the process is correct and the results are as listed. Commented Jul 23, 2020 at 16:39
• Just saw the post. I don't see the problem in that the 2 first rows are having redundant values, since you are taking 3D points, and as long as all the coordinates are not equal, those sigma points won't be the same. Also, I don't get why would you take $(n + \lambda) = 1$, isn't $\lambda \geq 1$ for the UKF? EDIT: just read the paper, and the latest of my statements is wrong I think. Commented Oct 4, 2022 at 10:26