Mathematics Asked by pcdangio on November 18, 2021
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:
A simple example:
Assume $(n+lambda)=1$
Using Cholesky LL decomposition:
$Sigma=LL^*$
$sqrtSigma=L=begin{bmatrix}L_{0,0}&0&0\L_{1,0}&L_{1,1}&0\L_{2,0}&L_{2,1}&L_{2,2}end{bmatrix}$
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?
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