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Nonlinear Estimation

2 Literature review on quadrature based nonlinear estimation (Cont'd...)

2.3 Cubature based filtering

Cubature Kalman Filter (CKF) is a nonlinear filter which uses spherical radial cubature rule for numerically computing the multivariate moment integrals encountered in the nonlinear Bayesian filtering framework. This filter is considered as a benchmark in nonlinear filtering algorithms. It was first introduced by Arasaratnam et.al [Arasaratnam 2009]. In this technique, the intractable integrals encountered in nonlinear bayesian filtering are decomposed into spherical and radial integrals. This was achieved using a spherical-radial transformation. The spherical and radial integrals are then numerically computed by the spherical cubature rule and the Gaussian quadrature rule. Transformation has been done in such a way that the surface integral can be calculated over a unit hyper-sphere of dimension $n$.

Theorem: For an arbitrary function $f(x)$, $X \in R^n$, the integral \begin{equation*} I(f)=\dfrac{1}{\sqrt{|\Sigma|(2\pi)^n}}\int_{R^{n}}f(x)\exp^{\dfrac{-1}{2}(X-\mu)^T \Sigma^{-1} (X-\mu)}dX \end{equation*} can be expressed in spherical cordinate system as $$I(f)=\dfrac{1}{\sqrt{(2\pi)^n}}\int_{r=0}^{\infty}\int_{U_n}[f(CrZ+ \mu)ds(Z)]r^{n-1}e^{r^{2}/2} dr$$ where $X=CrZ+\mu$, $C$ is the cholesky decomposition of $\Sigma$, $\|Z\|=1$, $U_n$ is the surface of unit hyper-sphere.

Proof: Let us transform the integral $I(f)$ to a spherical coordinate system. Let $X=CY+\mu, Y \in R^n$, where $\Sigma=CC^T$ i.e. $C$ is the cholesky decomposition of $\Sigma$. Then $(X-\mu)^T \Sigma^{-1} (X-\mu)=Y^T C^TC^{-T}C^{-1}CY=Y^TY$ and $dX=|C|dY=\sqrt{|\Sigma|}dY$. So the desired integral, $$I(f)=\dfrac{1}{\sqrt{(2 \pi)^n}}\int_{R^n}f(CY+\mu)e^-\dfrac{1}{2}Y^TY dY$$ Now let $Y=rZ$, with $\|Z\|=\sqrt{Z^TZ}=1, Y^TY=Z^TrrZ=r^2$. The elementary volume of hyper-sphere at $n$ dimensional space is $dY=r^{n-1}drds(Z)$ where $ds(\bullet)$ is the element on $U_n$. $U_n$ is the surface of hyper-sphere defined by $U_n={Z \in R^n | ZZ^T=1}; r \in [0,\infty)$. Hence $$\begin{split} I(f)=\dfrac{1}{\sqrt{(2\pi)^n}} \int_{r=0}^{\infty} \int_{U_n}f(CrZ+ \mu)e^{-r^2/2} r^{n-1} dr ds(Z) \\ = \dfrac{1}{\sqrt{(2\pi)^n}} \int_{r=0}^{\infty} \int_{U_n}f(CrZ+ \mu)ds(Z)]r^{n-1}e^{-r^2/2}dr \end{split}$$