*3.7. Li's MoN*

For a scalar random variable *x*, the un-normalized MoN proposed by Li [43,44] represents the square root of the minimum mean square distance between the nonlinear measurement function *h* and the set of all affine functions *L*,

$$J = \min \left( E\{ (L(\mathbf{x}) - h(\mathbf{x}))^2 \} \right)^{1/2},\tag{59}$$

where *L*(*x*) = *Ax* + *B*. The scalar parameters *A* and *B* are determined in the minimization process. For the current problem, where *x* is non-random, the un-normalized MoN *J* and normalized MoN *ν* ar given, respectively, by

$$J = \sigma\_h \sqrt{1 - \frac{c\_{hx}^2}{\sigma\_h^2 \sigma\_x^2}}.\tag{60}$$

$$\nu = \mathbf{J}/\sigma\_{\mathbf{h}} = \sqrt{1 - \frac{c\_{\mathbf{h}\mathbf{x}}^2}{\sigma\_{\mathbf{h}}^2 \sigma\_{\mathbf{x}}^2}}.\tag{61}$$

Given *x*ˆ and *σ<sup>x</sup>* (30), the unscented transformation (UT) [14,15], cubature transformation (CT) [16], or Monte Carlo method [8] can be used to compute *σ*<sup>2</sup> *<sup>h</sup>* and *chx*. We find that the UT gives good results in calculating the two MoNs. Next we dscribe computing *J* and *ν* using the UT. We use *κ*UT = 2 [14]. The three weights and sigma points are given, respectively, by

$$w\_0 = 2/3, \quad w\_1 = 1/6, \quad w\_2 = 1/6,\tag{62}$$

$$
\chi\_0 = \mathfrak{x}, \quad \chi\_1 = \mathfrak{x} + \sqrt{3}\sigma\_{\mathfrak{x}\prime} \quad \chi\_2 = \mathfrak{x} - \sqrt{3}\sigma\_{\mathfrak{x}\prime} \tag{63}
$$

The measurement transformed points are

$$h\_i = a \,\chi\_i^n, \quad i = 0, 1, 2. \tag{64}$$

Then the mean and variance of *h* a given by

$$\bar{h} = \sum\_{i=0}^{2} w\_i h\_{i\prime} \tag{65}$$

$$
\sigma\_h^2 = \sum\_{i=0}^2 w\_i (h\_i - \bar{h})^2,\tag{66}
$$

The cross-covariance *chx* is computed by

$$\boldsymbol{\omega}\_{\text{hx}} = \sum\_{i=0}^{2} w\_i (\boldsymbol{h}\_i - \boldsymbol{\bar{h}}) (\boldsymbol{\chi}\_i - \boldsymbol{\mathfrak{k}}).\tag{67}$$
