# Student's t Kernel-Based Maximum Correntropy Criterion Extended Kalman Filter for GPS Navigation
## 3 Student's t Kernel-Based Maximum Correntropy Criterion Extended Kalman Filter  
According to [PERSON] et al. (2014), one can derive the STMCCEKF. From Equation (14), we obtain  
\[\mathbf{B}_{\mathbf{p}}^{-1}\mathbf{\overset{\star}{x}}_{k|k-1}=\mathbf{B}_{ \mathbf{p}}^{-1}\mathbf{x}_{k}+\mathbf{B}_{\mathbf{p}}^{-1}\mathbf{\delta}_{k, 1},\ \mathbf{B}_{\mathbf{R}}^{-1}\mathbf{z}_{k}=\mathbf{B}_{\mathbf{R}}^{-1}h( \mathbf{x}_{k})+\mathbf{B}_{\mathbf{R}}^{-1}\mathbf{\delta}_{k,2} \tag{31}\]  
where  
\[\mathbf{\delta}_{k,1}=(\mathbf{\overset{\star}{x}}_{k|k-1}-\mathbf{x}_{k})^{ \mathsf{T}},\ \mathbf{\delta}_{k,2}=\mathbf{v}_{k}^{\mathsf{T}}\]  
Furthermore, defining \(\mathbf{e}_{x}\) and \(\mathbf{e}_{z}\) as  
\[\mathbf{e}_{x}=\mathbf{B}_{\mathbf{p}}^{-1}\mathbf{\delta}_{k,1}=\mathbf{B}_{ \mathbf{p}}^{-1}\mathbf{\overset{\star}{x}}_{k|k-1}-\mathbf{B}_{\mathbf{p}}^ {-1}\mathbf{x}_{k};\ \mathbf{e}_{z}=\mathbf{B}_{k,\mathbf{R}}^{-1}\mathbf{\delta}_{k,2}=\mathbf{B}_ {k,\mathbf{R}}^{-1}\mathbf{z}_{k}-\mathbf{B}_{k,\mathbf{R}}^{-1}h(\mathbf{x} _{k})\]  
The above equations can be rewritten as  
\[\mathbf{e}_{x}=\mathbf{B}_{\mathbf{p}}^{-1}(\mathbf{x}_{k}-\mathbf{\overset{ \star}{x}}_{k|k-1});\ \mathbf{e}_{z}=\mathbf{B}_{\mathbf{R}}^{-1}(\mathbf{z}_{k}-h(\mathbf{x}_{k})) \tag{32}\]  
By considering Student's t, the cost function is given as  
\[J_{\text{MCC-ST}}=\frac{1}{L}\left[\sum_{j=1}^{n}S_{v,\sigma}(\mathbf{e}_{x,j} )+\sum_{j=1}^{m}S_{v,\sigma}(\mathbf{e}_{z,j})\right] \tag{33}\]  
where \(L=n+m\). Considering the Student's t kernel-based MCC, in order to obtain the optimal state estimation, we can solve the following equation:  
\[\frac{1}{L}\left[\frac{\partial}{\partial\mathbf{x}_{k}}\sum_{i=1}^{n}S_{v, \sigma}(\mathbf{e}_{x,i})+\frac{\partial}{\partial\mathbf{x}_{k}}\sum_{j=1}^{ m}S_{v,\sigma}(\mathbf{e}_{z,j})\right]=0 \tag{34}\]  
After derivation, we obtain  
\[\sum_{i=1}^{n}\,(\frac{w\sigma^{2}}{v\sigma^{2}+\mathbf{e}_{x,i}^{2}})S_{v, \sigma}(\mathbf{e}_{x,i})\mathbf{B}_{\mathbf{p},j}^{-1}\mathbf{B}_{\mathbf{p },j}^{-1}(\mathbf{x}_{k}-\mathbf{\overset{\star}{x}}_{k|k-1})-\sum_{j=1}^{m} \,(\frac{v\sigma^{2}}{v\sigma^{2}+\mathbf{e}_{z,j}^{2}})S_{v,\sigma}(\mathbf{ e}_{z,j})\mathbf{H}_{k}^{\mathsf{T}}\mathbf{B}_{\mathbf{R},j}^{-1}\mathbf{B}_{ \mathbf{R},j}^{-1}(\mathbf{z}_{k}-h(\mathbf{x}_{k}))=0 \tag{35}\]  
Replacing the summation form in Equation with its matrix form, we obtain  
\[\begin{array}{ccccc}\mathbf{\Lambda}_{k}^{(t-1)}&=diag\,(\ \ (\frac{w \sigma^{2}S_{v\sigma}(\mathbf{e}_{x,1})}{w\sigma^{2}+\mathbf{e}_{x,1}^{2}}),& \ldots&,(\frac{w\sigma^{2}S_{v\sigma}(\mathbf{e}_{x,n})}{w\sigma^{2}+\mathbf{ e}_{z,n}^{2}})\ )\\ \mathbf{\Lambda}_{\mathbf{z}}^{(t-1)}&=diag\,(\ \ (\frac{w\sigma^{2}S_{v\sigma}( \mathbf{e}_{z,1})}{w\sigma^{2}+\mathbf{e}_{z,1}^{2}}),&\ldots&,(\frac{w\sigma ^{2}S_{v\sigma}(\mathbf{e}_{z,m})}{w\sigma^{2}+\mathbf{e}_{z,m}^{2}})\ )\end{array} \tag{36}\]  
We can obtain the following equation:  
\[\mathbf{B}_{\mathbf{p}}^{-\mathsf{T}}\mathbf{\Lambda}_{\mathbf{x}}\mathbf{B }_{\mathbf{p}}^{-1}(\mathbf{x}_{k}-\mathbf{\overset{\star}{x}}_{k|k-1})- \mathbf{H}_{k}^{\mathsf{T}}\mathbf{B}_{\mathbf{R}}^{-1}\mathbf{\Lambda}_{ \mathbf{z}}\mathbf{B}_{\mathbf{R}}^{-1}(\mathbf{z}_{k}-h(\mathbf{x}_{k}))=0 \tag{37}\]  
By using \(\mathbf{\overset{\star}{P}}_{k|k-1}=\mathbf{B}_{\mathbf{P}}\mathbf{\Lambda}_ {x}^{-1}\mathbf{B}_{\mathbf{p}}^{\mathsf{T}}\) and \(\mathbf{\overset{\star}{R}}_{k}=\mathbf{B}_{\mathbf{R}}\mathbf{\Lambda}_{z} ^{-1}\mathbf{B}_{\mathbf{R}}^{\mathsf{T}}\) we have  
\[\mathbf{\overset{\star}{P}}_{k|k-1}(\mathbf{x}_{k}-\mathbf{\overset{\star}{x}}_ {k|k-1})-\mathbf{H}_{k}^{\mathsf{T}}\mathbf{\overset{\star}{R}}_{k}^{-1}( \mathbf{z}_{k}-h(\mathbf{x}_{k}))=0 \tag{38}\]  
\[\mathbf{\overset{\star}{P}}_{k|k-1}^{-1}(\mathbf{x}_{k}-\mathbf{\overset{ \star}{x}}_{k|k-1})-\mathbf{H}_{k}^{\mathsf{T}}\mathbf{\overset{\star}{R}}_{k}^ {-1}(\mathbf{z}_{k}-h(\mathbf{\overset{\star}{x}}_{k|k-1})-\mathbf{H}_{k}( \mathbf{x}_{k}-\mathbf{\overset{\star}{x}}_{k|k-1}))=0\]